Digital Signal Processing (Solution Manual) - 3rd Edition by Mitra

SOLUTIONS MANUAL to accompany

Digital Signal Processing: A Computer-Based Approach Third Edition

Sanjit K. Mitra

Prepared by Chowdary Adsumilli, John Berger, Marco Carli, Hsin-Han Ho, Rajeev Gandhi, Chin Kaye Koh, Luca Lucchese, and Mylene Queiroz de Farias

Not for sale.

1

Chapter 2 2.1

(a) x1 1 = 22.85, x1 2 = 9.1396, x1 ∞ = 4.81, (b) x 2 1 = 18.68, x 2 2 = 7.1944, x 2 ∞ = 3.48.

2.2

1, n ≥ 0, 1, n < 0, µ[ n] = ⎧⎨ Hence, µ[ −n − 1] = ⎧⎨ Thus, x[ n] = µ[ n] + µ[ − n − 1]. ⎩0, n < 0. ⎩0, n ≥ 0.

2.3

(a) Consider the sequence defined by x[ n] = ∑ δ[ k ]. If n < 0, then k = 0 is not included

n

k = −∞

in the sum and hence, x[n] = 0 for n < 0. On the other hand, for n ≥ 0, k = 0 is included in the sum, and as a result, x[n] =1 for n ≥ 0. Therefore, n 1, n ≥ 0, x[ n] = ∑ δ[ k ] = ⎧⎨ = µ[ n]. 0 ⎩ , n < 0, k = −∞

2.4

2.5

1, n ≥ 0, 1, n ≥ 1, it follows that µ[ n − 1] = ⎧⎨ Hence, (b) Since µ[ n] = ⎧⎨ 0 , n 0 , < ⎩ ⎩0, n < 1. 1, n = 0, µ[ n] − µ[ n − 1] = ⎧⎨ = δ[ n]. ⎩0, n ≠ 0, Recall µ[ n] − µ[ n − 1] = δ[ n]. Hence, x[ n] = δ[ n] + 3δ[ n − 1] − 2δ[ n − 2] + 4δ[ n − 3] = (µ[ n] − µ[ n − 1]) + 3(µ[ n − 1] − µ[ n − 2]) − 2(µ[ n − 2] − µ[ n − 3]) + 4(µ[ n − 3] − µ[ n − 4]) = µ[ n] + 2µ[ n − 1] − 5µ[ n − 2] + 6µ[ n − 3] − 4µ[ n − 4]. (a) c[ n] = x[ − n + 2] = {2 0 − 3 − 2 1 5 − 4}, ↑

(b) d[ n] = y[ − n − 3] = {− 2 7 8 0 − 1 − 3 6 0 0}, ↑

(c) e[ n] = w[ − n] = {5 − 2 0 − 1 2 2 3 0 0}, ↑

(d) u[ n] = x[ n] + y[ n − 2] = {− 4 5 1 − 2 3 − 3 1 0 8 7 − 2}, ↑

(e) v[ n] = x[ n] ⋅ w[ n + 4] = {0 15 2 − 4 3 0 − 4 0}, ↑

(f) s[ n] = y[ n] − w[ n + 4] = {− 3 4 − 5 0 0 10 2 − 2}, ↑

(g) r[ n] = 3.5 y[ n] = {21 − 10.5 − 3.5 0 2.8 24.5 − 7}. ↑

2.6

(a) x[ n] = −4δ[ n + 3] + 5δ[ n + 2] + δ[ n + 1] − 2δ[ n] − 3δ[ n − 1] + 2δ[ n − 3], y[ n] = 6δ[ n + 1] − 3δ[ n] − δ[ n − 1] + 8δ[ n − 3] + 7δ[ n − 4] − 2δ[ n − 5], w[ n] = 3δ[ n − 2] + 2δ[ n − 3] + 2δ[ n − 4] − δ[ n − 5] − 2δ[ n − 7] + 5δ[ n − 8], (b) Recall δ[ n] = µ[ n] − µ[ n − 1]. Hence, x[ n] = −4(µ[ n + 3] − µ[ n + 2]) + 5(µ[ n + 2] − µ[ n + 1]) + (µ[ n + 1] − µ[ n]) − 2(µ[ n] − µ[ n − 1]) − 3(µ[ n − 1] − µ[ n − 2]) + 2(µ[ n − 3] − µ[ n − 4])

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2

= −4µ[ n + 3] + 9µ[ n + 2] − 4µ[ n + 1] − 3µ[ n] − µ[ n − 1] + 3µ[ n − 2] + 2µ[ n − 3] − 2µ[ n − 4],

2.7

(a) x[n-1]

_ z 1

x[n]

x[n-2]

_ z 1

h[2]

h[1]

h[0]

y[n]

+

+

From the above figure it follows that y[ n] = h[ 0] x[ n] + h[1] x[ n − 1] + h[2] x[ n − 2]. (b) x[n]

h[0]

w[n]

+ _1

+

y[n]

_1

z

z

β11

x[n _ 1]

+

_ z 1

+

_ z 1

β 21

x[n _ 2]

β12

w[n _ 1]

w[n _ 2]

β 22

From the above figure we get w[ n] = h[0]( x[ n] + β11 x[ n − 1] + β 21 x[ n − 2]) and y[ n] = w[ n] + β12 w[ n − 1] + β 22 w[ n − 2]. Making use of the first equation in the second we arrive at y[ n] = h[0]( x[ n] + β11 x[ n − 1] + β 21 x[ n − 2]) + β12 h[0]( x[ n − 1] + β11 x[ n − 2] + β 21 x[ n − 3]) + β 22 h[0]( x[ n − 2] + β11 x[ n − 3] + β 21 x[ n − 4]) = h[0]( x[ n] + (β11 + β12 ) x[ n − 1] + (β 21 + β12 β11 + β 22 ) x[ n − 2] + (β12 β 21 + β 22 β11 ) x[ n − 3] + β 22 β 21 x[ n − 4]). (c) Figure P2.1(c) is a cascade of a first-order section and a second-order section. The input-output relation remains unchanged if the ordering of the two sections is interchanged as shown below. x[n]

0.6

w[n]

+ _

+ _ z 1

0.8

u[n]

+

y[n+1] _ z 1

0.3

+

+

0.4

y[n]

w[n _ 1] _ z 1 _ 0.5

0.2

w[n _ 2]

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3

The second-order section can be redrawn as shown below without changing its inputoutput relation. x[n]

0.6

w[n]

+ _

+

+

+

y[n+1] _ z 1

_ _ z 1 z 1 0.8 w[n _ 1]

0.3

0.4

y[n]

+

_ z 1

_ z 1 _ 0.5

u[n]

0.2

w[n _2]

The second-order section can be seen to be cascade of two sections. Interchanging their ordering we finally arrive at the structure shown below: 0.6

+

x[n]

s[n]

_ z 1

_

0.3 x[n _ 1]

u[n]

+ 0.8

_ z 1

x[n _ 2]

y[n+1] _ z 1

_ z 1

+

+

+

0.4

u[n _ 1]

y[n]

_ z 1 _ 0.5

0.2

u[n _ 2]

Analyzing the above structure we arrive at s[ n] = 0.6 x[ n] + 0.3 x[ n − 1] + 0.2 x[ n − 2], u[ n] = s[ n] − 0.8u[ n − 1] − 0.5u[ n − 2], y[ n + 1] = u[ n] + 0.4 y[ n]. From u[ n] = y[ n + 1] − 0.4 y[ n]. Substituting this in the second equation we get after some algebra y[ n + 1] = s[ n] − 0.4 y[ n] − 0.18 y[ n − 1] + 0.8 y[ n − 2]. Making use of the first equation in this equation we finally arrive at the desired input-output relation y[ n] + 0.4 y[ n − 1] + 0.18 y[ n − 2] − 0.2 y[ n − 3] = 0.6 x[ n − 1] + 0.3 x[ n − 2] + 0.2 x[ n − 3]. (d) Figure P2.19(d) is a parallel connection of a first-order section and a second-order section. The second-order section can be redrawn as a cascade of two sections as indicated below: x[n]

w[n]

+

+

_ _ z 1 z 1 _ 0.8 w[n _ 1]

_ z 1 _ 0.5

+

y 2[n]

_ z 1

w[n _2]

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0.3

0.2

4

Interchanging the order of the two sections we arrive at an equivalent structure shown below: 0.3

x[n]

_1

+

z

q[n]

y 2[n]

+

_ z 1

_

0.2

_ z 1

0.8

+

y 2 [n _ 1] _ z 1

_ 0.5

y 2[n _ 2]

Analyzing the above structure we get q[ n] = 0.3 x[ n − 1] + 0.2 x[ n − 2], y2 [ n] = q[ n] − 0.8 y2 [ n − 1] − 0.5 y2 [ n − 2]. Substituting the first equation in the second we have y2 [ n] + 0.8 y2 [ n − 1] + 0.5 y2 [ n − 2] = 0.3 x[ n − 1] + 0.2 x[ n − 2].

(2-1)

Analyzing the first-order section of Figure P2.1(d) given below u[n] _ 1 u[n _ 1] 0.6 y 1 [n] z x[n] + 0.4

we get

u[ n] = x[ n] + 0.4u[ n − 1], y1[ n] = 0.6u[ n − 1]. Solving the above two equations we have y1[ n] − 0.4 y1[ n − 1] = 0.6 x[ n − 1]. (2-2) The output y[n] of the structure of Figure P2.19(d) is given by y[ n] = y1[ n] + y2 [ n]. (2-3) From Eq. (2-2) we get 0.8 y1[ n − 1] − 0.32 y1[ n − 2] = 0.48 x[ n − 2] and 0.5 y1[ n − 2] − 0.2 y1[ n − 3] = 0.3 x[ n − 3]. Adding the last two equations to Eq. (2-2) we arrive at y1[ n] + 0.4 y1[ n − 1] + 0.18 y1[ n − 2] − 0.2 y1[ n − 3] (2-4) = 0.6 x[ n − 1] + 0.48 x[ n − 2] + 0.3 x[ n − 3]. Similarly, from Eq. (2-1) we get − 0.4 y2 [ n − 1] − 0.32 y2 [ n − 2] − 0.2 y2 [ n − 3] = −0.12 x[ n − 2] − 0.08 x[ n − 3]. Adding this equation to Eq. (2-1) we arrive at y2 [ n] + 0.4 y2 [ n − 1] + 0.18 y2 [ n − 2] − 0.2 y2 [ n − 3] (2-5) = 0.3 x[ n − 1] + 0.08 x[ n − 2] − 0.08 x[ n − 3]. Adding Eqs. (2-4) and (2-5), and making use of Eq. (2-3) we finally arrive at the inputoutput relation of Figure P2.1(d) as: y[ n] + 0.4 y[ n − 1] + 0.18 y[ n − 2] − 0.2 y[ n − 3] = 0.9 x[ n − 1] + 0.56 x[ n − 2] + 0.22 x[ n − 3]. Not for sale.

5

2.8

(a) x1* [ n] = {1 − j 4 − 2 − j 5 3 + j 2 − 7 − j 3 − 1 − j}, ↑

x1* [ −n] = {− 1 − j − 7 − j 3 3 + j 2 − 2 − j 5 1 − j 4}. Therefore ↑

x1, cs [n] =

1 2

x1, ca [n] =

1 2

(x [n] + x [−n]) = { j1.5

− 4.5 + j 3 − 4.5 − j − j1.5},

* 1

1

↑

(x1[n] − x1*[−n]) = {1 + j2.5

(b) x 2 [ n] = e

jπn / 3

. Hence, x 2* [ n] = e

(

2.5 + j 4 − j 2 − 2.5 + j 4 − 1 + j 2.5}. ↑ − jπn / 3

)

and thus, x 2* [ −n] = e jπn / 3 = x 2 [ n].

Therefore, x2, cs [n] = x2 [n] + x2* [−n] = e j 2 πn / 3 = x2 [n], and 2 1

x2, ca [n] =

1 2

(x2 [n] − x2* [−n]) = 0.

(c) x 3 [ n] = j e − jπn / 5 . Hence, x 3* [ n] = − j e jπn / 5 and thus,

(

)

x 3* [ − n] = − j e − jπn / 5 = − x 3 [ n]. Therefore, x3, cs [n] = x3 [n] + x3*[− n] = 0, and 2 1

x3, ca [n] = 2.9

1 2

(x3[n] − x3*[−n]) = x3[n] = j e − jπn / 5 .

(a) x[ n] = {− 4 5 1 − 2 − 3 0 2}. Hence, x[ − n] = {2 0 − 3 − 2 1 5 − 4}. Therefore, x ev [ n] =

↑ 1 ( x[ n] + 2

↑

x[ − n]) =

1 {− 2 2

5 − 2 − 4 − 2 5 − 2} ↑

= {− 1 2.5 − 1 − 2 − 1 2.5 − 1}

and x od [ n] =

↑ 1 1 ( x[ n] − x[ −n]) = {− 6 2 2

5 4 0 − 4 − 5 6} ↑

= {− 3 2.5 2 0 − 2 − 2.5 3}. ↑

(b) y[ n] = {0 0 0 0 6 − 3 − 1 0 8 7 − 2}. Hence, ↑

y[ − n] = {− 2 7 8 0 − 1 − 3 6 0 0 0 0}. ↑

Therefore, yev [ n] = and y od [ n] =

1 ( y[ n] + 2

y[ − n]) = {− 1 3.5 4 0 2.5 − 3 2.5 0 4 3.5 − 1} ↑

1 ( y[ n] − y[ −n]) = {1 2

− 3.5 − 4 0 3.5 0 − 3.5 0 4 3.5 − 1}. ↑

(c) w[ n] = {0 0 0 0 0 0 0 0 0 0 3 2 2 − 1 0 − 2 5}. Hence, ↑

w[ − n] = {5 − 2 0 − 1 2 2 3 0 0 0 0 0 0 0 0 0 0}. Therefore ↑

wev [ n] =

1 (w[ n] + w[ − n]) 2

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6

= {2.5 − 1 0 − 0.5 1 1 1.5 0 0 0 1.5 1 1 − 0.5 0 − 1 2.5} and ↑

wod [ n] =

1 (w[ n] − w[ − n]) 2

= {− 2.5 1 0 0.5 − 1 − 1 − 1.5 0 0 0 1.5 1 1 − 0.5 0 − 1 2.5}. ↑

2.10

(a) x1[ n] = µ[ n + 2]. Hence, x1[ −n] = µ[ −n + 2]. Therefore, n ≥ 3, ⎧⎪1 / 2, 1 x1, ev [ n] = (µ[ n + 2] + µ[ − n + 2]) = ⎨ 1, − 2 ≤ n ≤ 2, and 2 ⎪⎩1 / 2, − 3 ≤ n, n ≥ 3, ⎧⎪ 1 / 2, 1 − 2 ≤ n ≤ 2, x1, od [ n] = (µ[ n + 2] − µ[ − n + 2]) = ⎨ 0, 2 ⎪⎩− 1 / 2, − 3 ≤ n. (b) x 2 [ n] = α n µ[ n − 3]. Hence, x 2 [ −n] = α − n µ[ −n − 3]. Therefore,

⎧ 1 αn, n ≥ 3, ⎪⎪ 2 n −n α µ[ n − 3] + α µ[ −n − 3] = ⎨ 0, − 2 ≤ n ≤ 2, and 1 − n ⎪ α , − 3 ≤ n, ⎪⎩ 2 ⎧ 1 αn, n ≥ 3, ⎪⎪ 2 n −n α µ[ n − 3] − α µ[ −n − 3] = ⎨ 0, − 2 ≤ n ≤ 2, 1 ⎪− α − n , − 3 ≤ n. ⎪⎩ 2

x 2, ev [ n] =

1 2

(

)

x 2, od [ n] =

1 2

(

)

(c) x 3 [ n] = n α n µ[ n]. Hence, x 3 [ − n] = − n α − n µ[ − n]. Therefore,

( (

) )

1 n 1 n α n µ[ n] + ( − n) α − n µ[ − n] = n α 2 2 1 n 1 x 3, od [ n] = n α n µ[ n] − ( − n) α − n µ[ − n] = n α . 2 2

x 3, ev [ n] =

n

(d) x 4 [ n] = α . Hence, x 4 [ −n] = α 1

1

1

1

−n

=α

n

and

= x 4 [ n]. Therefore, n

x 4, ev [ n] = ( x 4 [ n] + x 4 [ − n]) = ( x 4 [ n] + x 4 [ n]) = x 4 [ n] = α 2 2

and

x 4, od [ n] = ( x 4 [ n] − x 4 [ −n]) = ( x 4 [ n] − x 4 [ n]) = 0. 2 2 2.11 x ev [ n] = (x[ n] + x[ − n]). Thus, x ev [ −n] = (x[ − n] + x[ n]) = x ev [ n]. Hence, x ev [n] is 2 2 1

1

an even sequence. Likewise, x od [ n] = (x[ n] − x[ − n]). Thus, 2 1

x od [ − n] =

1 2

(x[ −n] − x[ n]) = − xod [ n].

Hence, x od [n] is an odd sequence.

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7

2.12 (a) g[ n] = x ev [ n]x ev [ n]. Thus, g[ −n] = x ev [ − n]x ev [ − n] = x ev [ n]x ev [ n] = g[ n]. Hence, g[n] is an even sequence. (b) u[ n] = x ev [ n]x od [ n]. Thus, u[ − n] = x ev [ − n]x od [ − n] = x ev [ n](− x od [ n]) = −u[ n]. Hence, u[n] is an odd sequence. (c) v[ n] = x od [ n]x od [ n]. Thus, v[ −n] = x od [ − n]x od [ − n] = (− x od [ n])(− x od [ n]) = x od [ n]x od [ n] = v[ n]. Hence, v[n] is an even sequence. 2.13 (a) Since x[n] is causal, x[ n] = 0, n < 0. Also, x[ − n] = 0, n > 0. Now, Hence, x ev [0] = (x[0] + x[0]) = x[0] and 2 ⎧2 x ev [ n], n > 0, ⎪ 1 x ev [ n] = x[ n], n > 0. Combining the two equations we get x[ n] = ⎨ x ev [ n], n = 0, 2 ⎪⎩ 0, n < 0.

x ev [ n] =

1 2

(x[ n] + x[ −n]).

Likewise, x od [ n] =

x od [ n] =

1 2

1 2

1

(x[ n] − x[−n]).

Hence, x od [0] =

1 2

(x[0] − x[0]) = 0

and

2 x [ n], n > 0, x[ n], n > 0. Combining the two equations we get x[ n] = ⎧⎨ ev n ≤ 0. ⎩ 0,

(b) Since y[n] is causal, y[ n] = 0, n < 0. Also, y[ − n] = 0, n > 0. Let y[ n] = yre [ n] + jyim [ n], where yre [n] and yim [n] are real causal sequences. 1

(

)

1

(

)

Now, yca [ n] = y[ n] − y ∗ [ − n] . Hence, yca [0] = y[0] − y ∗ [0] = jyim [0] and 2 2

yca [ n] =

1 y[ n], n 2

> 0. Since yre [0] is not known, y[n] cannot be fully recovered from

yca [n] . 1

(

)

1

(

)

Likewise, ycs [ n] = y[ n] + y ∗ [ − n] . Hence, ycs [0] = y[0] + y ∗ [0] = yre [0] and 2 2

ycs [ n] =

1 y[ n], n 2

> 0. Since yim [ 0] is not known, y[n] cannot be fully recovered from

ycs [n] . 2.14 Since x[n] is causal, x[ n] = 0, n < 0. From the solution of Problem 2.13 we have ⎧2 x ev [ n], n > 0, ⎧2 cos(ωo n), n > 0, ⎪ ⎪ 1 n = 0, = 2 cos(ωo n)µ[ n] − δ[ n]. x[ n] = ⎨ x ev [ n], n = 0, = ⎨ 0 , n < 0, ⎪⎩ 0, ⎪ n < 0, ⎩ 2.15 (a) {x[ n]} = {Aα n } where A and α are complex numbers with α < 1. Since for

n < 0, α

n

can become arbitrarily large, {x[ n]} is not a bounded sequence.

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8

⎧ n (b) y[ n] = Aα n µ[ n] = ⎨ Aα , n ≥ 0, where A and α are complex numbers with n < 0, ⎩ 0, n

α < 1. Here, α ≤ 1, n ≥ 0. Hence y[ n] ≤ A for all values of n. Hence, {y[ n]} is a bounded sequence. (c) {h[ n]} = Cβ n µ[ n] where C and β are complex numbers with β > 1. Since for

n > 0, β

n

can become arbitrarily large, {h[ n]} is not a bounded sequence.

(d) {g[ n]} = 4 cos(ωo n). Since g[ n] ≤ 4 for all values of n,{g[ n]} is a bounded sequence. 1 ⎧⎛ 1 1 ⎪⎜1 − 2 ⎞⎟, n ≥ 1, (e) v[ n] = ⎨⎝ n ⎠ Since 2 < 1 for n > 1 and 2 = 1 for n = 1, v[ n] < 1 for n n ⎪⎩ 0, n ≤ 0. all values of n. Thus {v[ n]} is a bounded sequence. ∞ ∞ ( −1) n +1 ∞ 1 (−1) n +1 µ[ n − 1]. Now ∑ x[ n] = ∑ = ∑ = ∞. Hence {x[ n]} is 2.16 x[ n] = n n n = −∞ n =1 n =1 n not absolutely summable.

∞

2.17 (a) x1[ n] = α n µ[ n − 1]. Now

∑

n = −∞

∞

∞

n =1

n =1

x 2 [ n] = ∑ α n = ∑ α

n

=

α 1− α

< ∞ , since

α < 1. Hence, {x1[ n]} is absolutely summable. ∞

(b) x 2 [ n] = α n µ[ n − 1]. Now

α

2

∑

n = −∞

∞

∞

n =1

n =1

x 2 [ n] = ∑ nα n = ∑ n α

n

α

=

(1 − α ) 2

< ∞, since

< 1. Hence, {x 2 [ n]} is absolutely summable.

(c) x 3 [ n] = n 2 α n µ[ n − 1]. Now 2

3

∞

∑

n = −∞

∞

∞

n =1

n =1

x 3 [ n] = ∑ n 2 α n = ∑ n 2 α

n

4

= α + 2 2 α + 32 α + 4 2 α + K 2

3

4

2

3

4

3

4

5

= ( α + α + α + α + K) + 3( α + α + α + K) + 5( α + α + α + K)

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9

4

5

6

+ 7( α + α + α + K) =

= =

α 1− α

⎛ ∞ 1 n⎞ ⎜⎜ ∑ (2n − 1) α ⎟⎟ = ⎠ 1− α ⎝ n =1

1 1− α

α (1 + α ) (1 − α ) 3

+

3α

2

1− α

+

5α

3

7α

+

1− α

4

1− α

+K

∞ ⎛ ∞ 1 n n⎞ ⎜⎜ 2 ∑ n α − ∑ α ⎟⎟ = n =1 ⎠ 1− α ⎝ n =1

⎛ 2α α ⎜ − ⎜ (1 − α ) 2 1 − α ⎝

⎞ ⎟ ⎟ ⎠

< ∞. Hence, {x 3 [ n]} is absolutely summable.

∞ ∞ 1 ∞ 1 1 1 2.18 (a) x a [ n] = = ∑ = = 2 < ∞. Hence, µ[ n]. Now ∑ x a [ n] = ∑ n n 1 n = −∞ n=0 2 n=0 2 2n 1− 2

{x a [ n]} is absolutely summable. ∞ ∞ 1 1 µ[ n]. Now ∑ x b [ n] = ∑ (n + 1)(n + 2) n = −∞ n = 0 (n + 1)(n + 2) ∞ ⎛ 1 1 ⎞ ⎛ 1⎞ ⎛1 1⎞ ⎛1 1⎞ ⎛1 1⎞ = ∑ ⎜ − ⎟ = ⎜1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + K = 1 < ∞. Hence, 2⎠ ⎝2 3⎠ ⎝3 4⎠ ⎝ 4 5⎠ n=0 ⎝ n + 1 n + 2 ⎠ ⎝ {x b [ n]} is absolutely summable.

(b) x b [ n] =

2.19 (a) A sequence x[n] is absolutely summable if

∞

∑

n = −∞

x[ n] < ∞. By Schwartz inequality

∞

⎞⎛ ∞ ⎞ 2 ⎛ ∞ x[ n] ≤ ⎜⎜ ∑ x[ n] ⎟⎟⎜⎜ ∑ x[ n] ⎟⎟ < ∞. Hence, an absolutely summable n = −∞ ⎝ n = −∞ ⎠⎝ n = −∞ ⎠ sequence is square summable and has thus finite energy.

we have

∑

1 n

Now consider the sequence x[ n] = µ[ n − 1]. The convergence of an infinite series can be shown via the integral test. Let a n = f (x ), where a continuous, positive and ∞

∞

decreasing function is for all x ≥ 1. Then the series ∑ a n and the integral ∫ f ( x )dx n =1

1

1

1 ∞

∞

1

both converge or both diverge. For a n = , f ( x ) = . But ∫ dx = (ln x ) 1 = ∞ − 0 = ∞ . n x x 1 ∞

Hence,

∑

n = −∞

∞ 1 n =1 n

x[ n] = ∑

1 n

does not converge. As a result, x[ n] = µ[ n − 1] is not

absolutely summable.

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10

(b) To show that {x[ n]} is square-summable, we observe that here a n = ∞

1 n2

, and thus,

∞

∞ 1 1 ⎛ 1⎞ f ( x ) = 2 . Now, ∫ 2 dx = ⎜ − ⎟ = − + 1 = 1. Hence, ∑ 2 converges, or in other ∞ x ⎝ x ⎠1 n =1 n 1x

1

1

1 n

words, x[ n] = µ[ n − 1] is square-summable. 2.20 See Problem 2.19, Part (a) solution. 2

∞ ∞ ⎛ cos ω n ⎞ ∞ cos ωc n 1 2 c µ[ n − 1]. Now, ∑ x 2 [ n] = ∑ ⎜ 2.21 x 2 [ n] = . Since, ⎟ ≤ ∑ 2 2 πn n = −∞ n =1 ⎝ πn ⎠ n =1 π n 2 π 2 ∞ ⎛ cos ωc n ⎞ 1 , ∑ ⎜ ⎟ ≤ . Therefore x2 [ n] is square-summable. ∑ 2 = 6 n =1 ⎝ πn ⎠ 6 n =1 n

∞

1

Using the integral test (See Problem 2.19, Part (a) solution) we now show that x2 [ n] is

cos ωc x ∞ cos ω x πx 1 c ⋅ cos int(ωc x ) dx = ⋅ not absolutely summable. Now, ∫ πx π cos ωc x 1

∞

where 1

∞ cos ω x c

cos int is the cosine integral function. Since ∫

πx diverges. Hence, x2 [ n] is not absolutely summable.

∞

cos ωc n also πn n =1

dx diverges, ∑

1

∞ K 1 1 2 2 ∑ x [ n] = lim ∑ x ev [ n] + x od [ n] K → ∞ 2 K +1 n = − K K → ∞ 2 K +1 K = −∞

(

2.22 Px = lim

(

)

)

K 1 2 2 ∑ x ev [ n] + x od [ n] + 2 x ev [ n]x od [ n] K → ∞ 2 K +1 n = − K

= lim

= Px ev + Px od +

K 1 1 lim ∑ x[ n] + 2 K → ∞ 2 K +1 n = − K

(

1

1⎛ K

x[ − n])(x[ n] − x[ −n]) ∞

⎞

⋅ ⎜⎜ ∑ x 2 [ n ] − ∑ x 2 [ − n ] ⎟⎟ = Px + Px = Px ev + Px od + lim ev od + K 2 1 2 ⎝ n=− K n = −∞ ⎠ K →∞

as

K

K

n = −K

n = −K

2 2 ∑ x [ n] = ∑ x [ − n]. Now for the given sequence, K K 1 6 1 1 2 ∑ x od [ n] = lim ∑ ⎛⎜ ⎞⎟ K → ∞ 2 K +1 n = − K K → ∞ 2 K +1 n = 0 ⎝ 3 ⎠

Px od = lim

6

K 1 1 = ⎛⎜ ⎞⎟ lim ∑1 ⎝ 3 ⎠ K → ∞ 2 K +1 n =0

6 K +1 1 1 6 1 ⎛ 1 ⎞6 1 = ⎛⎜ ⎞⎟ . Hence, Px = Px − Px = ⎛⎜ ⎞⎟ lim = − 10 ⎜ ⎟ . ev od 2 ⎝3⎠ ⎝ 3 ⎠ K → ∞ 2 K +1 2 ⎝ 3 ⎠

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N −1

2

N −1

2.23 x[ n] = sin( 2 πkn / N ), 0 ≤ n ≤ N − 1. Now E x = ∑ x[ n] = ∑ sin 2 (2πkn / N ) n=0

=

1 N −1 ∑ 1 − cos(4πkn / N ) 2 n=0

(

) = N2 − 12

n=0 N −1

N −1

∑ cos(4πkn / N ). Let C = ∑ cos(4 πkn / N ) and

n=0

N −1

N −1

n=0

n=0

S = ∑ sin( 4 πkn / N ). Then C + jS = ∑ e − j 4 πkn / N =

∞

α

2.24 (a) x[ n] = A µ[ n]. Then E x a = ∑

n = −∞

2.25

1 n2

1− e

1 − e − j 4 πk / N

= 0. This implies

N . 2

C = 0. Hence E x =

(b) x b [ n] =

n=0 − j 4 πkn

∞ 2 x a [ n] = A 2 ∑ α 2 n =

∞

µ[ n − 1]. Then E x b ∑

n = −∞

2

A2

n=0

1− α

∞

∞

x b [ n] = ∑

1

2 n =1 n

= ∑

2

1

4 n =1 n

. =

π4 . 90

(a) x1[ n] = (−1) n . Then average power K 1 ∑ K → ∞ 2 K +1 n = − K

Px1 = lim ∞

E x1 = ∑

1 (2 K + 1) = 1, and energy K →∞ 2K + 1

∞

2

n = −∞

2

x1[ n] = lim

x1[ n] = ∑ 1 = ∞. n = −∞

(b) x 2 [ n] = µ[ n]. Then average power K K K +1 1 1 1 2 Px 2 = lim = , and energy ∑ x 2 [ n] = lim ∑ 1 = lim K → ∞ 2K + 1 n = −K K →∞ 2K + 1 n = 0 K → ∞ 2K + 1 2 ∞

E x2 = ∑

n = −∞

∞

2

x 2 [ n] = ∑ 1 = ∞. n=0

(c) x 3 [ n] = nµ[ n]. Then average power K K K ( K + 1)(2 K + 1) 1 1 2 2 Px 3 = lim = ∞, ∑ x3 [ n] = lim ∑ n = lim 6 K → ∞ 2K + 1 n = −K K → ∞ 2 K + 1 n =1 K →∞ ∞

and energy E x 3 = ∑

n = −∞

2

∞

x3 [ n] = ∑ n 2 = ∞. n=0

K 1 2 ∑ x 4 [ n] K → ∞ 2K + 1 n = −K

(d) x 4 [ n] = A0 e jω 0 n . Then average power Px 4 = lim

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12

2 K K 1 1 1 jω n 2 = lim ⋅ A02 (2 K + 1) = A02 , ∑ A0 e 0 ∑ A0 = lim K → ∞ 2K + 1 n = −K K →∞ 2K + 1 n = −K K →∞ 2K + 1

= lim

∞

and energy E x 3 = ∑

n = −∞

2

∞

x 3 [ n] = ∑

n = −∞

A0 e jω 0 n

2

∞

= ∑ A02 = ∞. n = −∞

2 πn (e) x 5 [ n] = A cos⎛⎜ + φ ⎞⎟. Note x 5 [ n] is a periodic sequence. Then average power M ⎠ ⎝

Px 5 =

2 1 M −1 1 M −1 1 A 2 M −1⎛ ⎛ 4 πn 2 πn 2 ⎞ + 2 φ ⎞⎟ + 1⎟. ⋅ + φ ⎞⎟ = ∑ ⎜ cos⎜ ∑ A cos⎛⎜ ∑ x 5 [ n] = M M ⎝ ⎠ ⎠ ⎝ ⎠ M n=0 M n=0 M 2 n=0 ⎝

M −1 M −1 4πn 4πn Let C = ∑ cos⎛⎜ + 2φ ⎞⎟ and S = ∑ cos⎛⎜ + 2φ ⎞⎟. Then M M ⎠ ⎠ ⎝ ⎝ n=0 n=0

j ⎛⎜ C + jS = ∑ e ⎝ n=0 M −1

4 πn + 2 φ ⎞⎟ M ⎠

=e

j 2φ

M −1

∑ e

j 4 πn / M

n =0

=e

j 2φ

⋅

1 − e j 4π 1 − e j 4π / M

= 0.

1 A 2 M −1 A 2 ⋅ . ∑1 = M 2 n=0 2 Since x 5 [ n] is a periodic sequence, it has infinite energy.

Hence C = 0. Therefore Px 5 =

2.26 In each of the following parts, N denotes the fundamental period and r is a positive integer. 2π x [ n] = 4 cos(2πn / 5). Here N and r must satisfy the relation ⋅ N = 2 πr. (a) ~ 1

5

Among all positive solutions for N and r , the smallest values are N = 5 and r = 1. Hence the average power is given by Px1 =

2 1 4 1 N −1 ~ 2 2 πn ⎞ ⎟ = 8. ∑ x1[ n] = ∑ 4 cos⎛⎜ ⎝ 5 ⎠ N n=0 5 n=0

3π x 2 [ n] = 3 cos(3πn / 5). Here N and r must satisfy the relation (b) ~ ⋅ N = 2πr. 5

Among all positive solutions for N and r , the smallest values are N = 10 and r = 3. Hence the average power is given by 2 1 9 1 N −1 ~ 2 3πn ⎞ ⎛ Px 2 = ⎟ = 4 .5 . ∑ x 2 [ n] = ∑ 3 cos⎜ ⎝ 5 ⎠ N n=0 10 n = 0 3π (c) ~ x3 [ n] = 2 cos(3πn / 7). Here N and r must satisfy the relation ⋅ N = 2πr.

7

Among all positive solutions for N and r , the smallest values are N = 14 and r = 3. Hence the average power is given by 2 1 13 1 N −1 ~ 2 3πn ⎞ ⎛ Px 3 = ⎟ = 2. ∑ x 3 [ n] = ∑ 2 cos⎜ ⎝ 7 ⎠ N n=0 14 n = 0

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5π x 4 [ n] = 4 cos(5πn / 3). Here N and r must satisfy the relation (d) ~ ⋅ N = 2 πr. 3

Among all positive solutions for N and r , the smallest values are N = 6 and r = 5. Hence the average power is given by 2

1 5 1 N −1 ~ 2 5πn ⎞ ⎟ = 8. ∑ x 4 [ n] = ∑ 4 cos⎛⎜ ⎝ 3 ⎠ N n=0 6 n=0 (e) ~ x 5 [ n] = 4 cos(2πn / 5) + 3 cos(3πn / 5). We first determine the fundamental period

Px 4 =

2π

N1 of cos( 2 πn / 5). Here N1 and r must satisfy the relation ⋅ N1 = 2 πr. Among all 5 positive solutions for N1 and r , the smallest values are N1 = 5 and r = 1. We next determine the fundamental period N 2 of cos(3πn / 5). Here N 2 and r must satisfy the 3π ⋅ N2 5

= 2πr. Among all positive solutions for N 2 and r , the smallest values are N 2 = 10 and r = 3. The fundamental period of ~ x 5 [ n] is then given by relation

LCM ( N1 , N 2 ) = LCM (5,10) = 10. Hence the average power is given by 2 1 4 1 N −1 ~ 2 2 πn ⎞ 3πn ⎞ ⎛ ⎛ Px 5 = ⎟ + 3 cos⎜ ⎟ ∑ x 5 [ n] = ∑ 4 cos⎜ ⎝ 5 ⎠ ⎝ 5 ⎠ N n=0 10 n = 0 11 11 1 ⎛ 11 3πn ⎞ 2 πn ⎞ 3πn ⎞ ⎞ 2 πn ⎞ = ⎜⎜ ∑ 16 cos 2 ⎛⎜ ⎟ ⎟ ≅ 8 + 4.5 + 0 = 12.5. ⎟ cos⎛⎜ ⎟ + ∑ 24 cos⎛⎜ ⎟ + ∑ 9 cos 2 ⎛⎜ ⎝ 5 ⎠ ⎝ 5 ⎠ ⎟⎠ ⎝ 5 ⎠ n=0 ⎝ 5 ⎠ n=0 10 ⎝ n = 0

x6 [ n] = 4 cos(5πn / 3) + 3 cos(3πn / 5). We first determine the fundamental period (f) ~ 5π

N1 of cos( 5πn / 3). Here N1 and r must satisfy the relation ⋅ N1 = 2πr. Among all 3 positive solutions for N1 and r , the smallest values are N1 = 6 and r = 5. We next determine the fundamental period N 2 of cos(3πn / 5). Here N 2 and r must satisfy the 3π ⋅ N2 5

= 2πr. Among all positive solutions for N 2 and r , the smallest values are N 2 = 10 and r = 3. The fundamental period of ~ x6 [ n] is then given by relation

LCM ( N1 , N 2 ) = LCM (6,10) = 30. Hence the average power is given by 2 1 N −1 ~ 1 29 2 5πn ⎞ 3πn ⎞ ⎛ ⎛ Px 6 = ⎟ + 3 cos⎜ ⎟ ∑ x 6 [ n] = ∑ 4 cos⎜ ⎝ 3 ⎠ ⎝ 5 ⎠ N n=0 30 n = 0 29 30 ⎞ 1 ⎛ 29 ⎜⎜ ∑ 16 cos 2 ⎛⎜ 5πn ⎞⎟ + ∑ 9 cos 2 ⎛⎜ 3πn ⎞⎟ + ∑ 24 cos⎛⎜ 5πn ⎞⎟ cos⎛⎜ 3πn ⎞⎟ ⎟⎟ ≅ 8 + 4.5 + 0 = 12.5. = ⎝ 3 ⎠ ⎝ 5 ⎠⎠ ⎝ 5 ⎠ n =0 ⎝ 3 ⎠ n=0 30 ⎝ n = 0

2.27

∞ Now , from Eq. (2.38) we have ~ y[ n] = ∑ x[ n + kN ]. Therefore k = −∞

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14

∞ ~ y[ n + N ] = ∑ x[ n + kN + N ]. Substituting r = k + 1 we get k = −∞ ∞

~ y[ n + N ] = ∑ x[ n + rN ] = ~ y[ n]. Hence ~ y[ n] is a periodic sequence with a period N . r = −∞

2.28

∞ x p [ n] in the range 0 ≤ n ≤ 4 is (a) N = 5. Now ~ x p [ n] = ∑ x[ n + k 5]. The portion of ~ n = −∞

given by x[ n − 5] + x[ n] + x[ n + 5] = {0 0 − 4 5 1} + {− 2 − 3 0 2 0} + {0 0 0 0 0} = {− 2 − 3 − 4 7 1}, 0 ≤ n ≤ 4. x p [ n] is given by {− 2 − 3 − 4 7 1}, 0 ≤ n ≤ 4. Hence, one period of ~ ∞ y p [ n] in the range 0 ≤ n ≤ 4 is given by Now ~ y p [ n] = ∑ y[ n + k 5]. The portion of ~ n = −∞

y[ n − 5] + y[ n] + y[ n + 5] = {0 0 0 0 6} + {− 3 − 1 0 8 7} + {− 2 0 0 0 0} = {− 5 − 1 0 8 13}, 0 ≤ n ≤ 4. y p [ n] is given by {− 5 − 1 0 8 13}, 0 ≤ n ≤ 4. Hence, one period of ~ ∞ ~ [ n] in the range 0 ≤ n ≤ 4 is given by ~ [ n] = ∑ Now w w[ n + k 5]. The portion of w p p n = −∞

w[ n − 5] + w[ n] + w[ n + 5] = {0 0 0 0 0} + {0 0 3 2 2} + {− 1 0 − 2 5 0} = {− 1 0 1 7 2}, 0 ≤ n ≤ 4. ~ [ n] is given by {− 1 0 1 7 2}, 0 ≤ n ≤ 4. Hence, one period of w p ∞ x p [ n] in the range 0 ≤ n ≤ 6 is (b) N = 7. Now ~ x p [ n] = ∑ x[ n + k 7]. The portion of ~ n = −∞

given by x[ n − 7] + x[ n] + x[ n + 7] = {0 0 0 0 − 4 5 1} + {− 2 − 3 0 2 0 0 0} + {0 0 0 0 0 0 0} = {− 2 − 3 0 2 − 4 5 1}, 0 ≤ n ≤ 6. Hence, one period of ~ x p [ n] is given by

{− 2 − 3 0 2 − 4 5 1}, 0 ≤ n ≤ 6. ∞ y p [ n] in the range 0 ≤ n ≤ 6 is given by Now ~ y p [ n] = ∑ y[ n + k 7]. The portion of ~ n = −∞

x[ n − 7] + x[ n] + x[ n + 7] = {0 0 0 0 0 0 6} + {− 3 − 1 0 8 7 − 2 0} + {0 0 0 0 0 0 0} = {− 3 − 1 0 8 7 − 2 6}, 0 ≤ n ≤ 6. Hence, one period of ~ y p [ n] is given by {− 3 − 1 0 8 7 − 2 6}, 0 ≤ n ≤ 6. ∞ ~ [ n] in the range 0 ≤ n ≤ 6 is given by ~ [ n] = ∑ Now w w[ n + k 7]. The portion of w p p n = −∞

w[ n − 7] + w[ n] + w[ n + 7] = {0 0 0 0 0 0 0} + {0 0 3 2 2 − 1 0} + {− 2 5 0 0 0 0 0}

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2.29

~ [ n] is given by = {− 2 5 3 2 2 − 1 0}, 0 ≤ n ≤ 6. Hence, one period of w p {− 2 5 3 2 2 − 1 0}, 0 ≤ n ≤ 6. ~ x[ n] = A cos(ωo n + φ). x[ n] = {1 − 1 − 1 1 1 − 1 − 1 1}. Hence A = 2 , ω = π / 2, φ = π / 4. (a) ~ o

(b) ~ x[ n] = {0 − 3 0 3 0 − 3 0 3}. Hence A = 3 , ωo = π / 2, φ = π / 2. x[ n] = {1 − 0.366 − 1.366 − 1 0.366 1.366}. Hence A = 2 , ωo = π / 3, (c) ~ φ = π / 4. x[ n] = {2 0 − 2 0 2 0 − 2 0}. Hence A = 2, ωo = π / 2, φ = 0. (d) ~ 2.30

The fundamental period N of a periodic sequence with an angular frequency ω o satisfies Eq. (2.47a) with the smallest value of N and r . (a) ωo = 0.5π. Here Eq. (2.47a) reduces to 0.5πN = 2πr which is satisfied with N = 4, r = 1. (b) ωo = 0.8π. Here Eq. (2.47a) reduces to 0.8πN = 2πr which is satisfied with N = 5, r = 2. (c) We first determine the fundamental period N1 of Re{e jπn / 5} = cos(0.2πn). In this case, Eq. (2.47a) reduces to 0.2πN 1= 2πr1 which is satisfied with N1 = 10, r1 = 1. We next determine the fundamental period N 2 of Im{e jπn / 10 = j sin(0.1πn). In this case, Eq. (2.47a) reduces to 0.1πN 2 = 2πr2 which is satisfied with N 2 = 20, r2 = 1. Hence the fundamental period N of ~ x c [ n] is given by

LCM ( N1 , N 2 ) = LCM (10,20) = 20. (d) We first determine the fundamental period N1 of 3 cos(1.3πn). In this case, Eq. (2.47a) reduces to 1.3πN 1= 2πr1 which is satisfied with N1 = 20, r1 = 13. We next determine the fundamental period N 2 of 4 sin( 0.5πn + 0.5π). In this case, Eq. (2.47a) reduces to 0.5πN 2 = 2πr2 which is satisfied with N 2 = 4, r2 = 1. Hence the x 4 [ n] is given by LCM ( N1 , N 2 ) = LCM (20,4) = 20. fundamental period N of ~ (e) We first determine the fundamental period N1 of 5 cos(1.5πn + 0.75π). In this case, Eq. (2.47a) reduces to 1.5πN 1= 2πr1 which is satisfied with N1 = 4, r1 = 3. We next determine the fundamental period N 2 of 4 cos( 0.6 πn). In this case, Eq. (2.47a) reduces to 0.6πN 2 = 2πr2 which is satisfied with N 2 = 10, r2 = 3. We finally determine the fundamental period N 3 of sin( 0.5πn). In this case, Eq. (2.47a) reduces to 0.5πN 3= 2πr3 which is satisfied with N 3 = 4, r3 = 1. Hence the fundamental period x 5 [ n] is given by LCM ( N1 , N 2 , N 3 ) = LCM (4,10,4) = 20. N of ~ 2.31

The fundamental period N of a periodic sequence with an angular frequency ω o satisfies Eq. (2.47a) with the smallest value of N and r . Not for sale.

16

(a) ωo = 0.6π. Here Eq. (2.47a) reduces to 0.6πN = 2πr which is satisfied with N = 10, r = 3. (b) ωo = 0.28π. Here Eq. (2.47a) reduces to 0.28πN = 2πr which is satisfied with N = 50, r = 7. (c) ωo = 0.45π. Here Eq. (2.47a) reduces to 0.45πN = 2πr which is satisfied with N = 40, r = 9. (d) ωo = 0.55π. Here Eq. (2.47a) reduces to 0.55πN = 2πr which is satisfied with N = 40, r = 11. (e) ωo = 0.65π. Here Eq. (2.47a) reduces to 0.65πN = 2πr which is satisfied with N = 40, r = 13. 2.32

ωo = 0.08π. Here Eq. (2.47a) reduces to 0.08πN = 2πr which is satisfied with x 2 [ n] = sin(ω2 n) with a fundamental period of N = 25 , N = 25, r = 1. For a sequence ~ Eq. (2.47a) reduces to 25ω2 = 2πr. For example, for r = 2 we have ω2 = 4π / 25 = 0.16π. Another sequence with the same fundamental period is obtained by setting r = 3 which leads to ω3 = 6π / 25 = 0.24 π. The corresponding periodic x [ n] = sin(0.16πn) and ~ x [ n] = sin(0.24 πn). sequences are therefore ~ 2

2.33

3

The three parameters A, Ω o , and φ of the continuous-time signal x a (t ) can be determined from x[ n] = x a (nT ) = A cos(Ω o nT + φ) by setting 3 distinct values of n. For example x[ 0] = A cos φ = α, x[ −1] = A cos(−Ω o T + φ) = A cos(Ω o T ) cos φ + A sin(Ω o T ) sin φ = β, ,

x[1] = A cos(Ω o T + φ) = A cos(Ω o T ) cos φ − A sin(Ω o T ) sin φ = γ. Substituting the first equation into the last two equations and then adding them we get β+γ which can be solved to determine Ω o . Next, from the second cos(Ω o T ) = 2α equation we have A sin φ = β − A cos(Ω o T ) cos φ = β − α cos(Ω o T ). Dividing this β − α cos(Ω o T ) equation by the last equation on the previous page we arrive at tan φ = α sin(Ω o T which can be solved to determine φ. Finally, the parameter is determined from the first equation of the last page.

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17

2π = 2Ω o . In this case x[ n] = A cos( nπ + φ) = β and T x[ n + 1] = A cos((n + 1)π + φ) = A cos(nπ + φ) = β . Since all sample values are equal, the three parameters cannot be determined uniquely.

Now consider the case Ω T =

2π < 2Ω o . In this case x[ n] = A cos(Ω o nT + φ) T = A cos(ωo n + φ) implying ωo = Ω o T > π. As explained in Section 2.2.1, a digital

Finally consider the case Ω T =

sinusoidal sequence with an angular frequency ω o greater than π assumes the identity of a sinusoidal sequence with an angular frequency in the range 0 ≤ ω < π. . Hence, Ω o cannot be uniquely determined from x[ n] = A cos(Ω o nT + φ) . 2.34

x[ n] = cos(Ω o nT ). If x[n] is periodic with a period N , then x[ n + N ] = cos(Ω 0 nT + Ω 0 NT ) = x[ n] = cos(Ω 0 nT ). This implies Ω o NT = 2πr with r any nonzero positive integer. Hence the sampling rate must satisfy the relation π T = 2πr / Ω o N . If Ω o = 20, i.e., T = π / 8, then we must have 20 N ⋅ = 2 πr . The 8 smallest value of N and r satisfying this relation are N = 4 and r = 5. The fundamental period is thus N = 4 .

2.35

(a) For an input xi [ n], i = 1,2, the output is yi [ n] = b0 xi [ n] + b1 xi [ n − 1] + b2 xi [ n − 2] + a1 yi [ n − 1] + a 2 yi [ n − 2], i = 1,2. Then, for an input x[ n] = Ax1[ n] + Bx2 [ n], the output is y[ n] = b0 ( Ax1[ n] + Bx 2 [ n])

+ b1 ( Ax1[ n − 1] + Bx2 [ n − 1]) + b2 ( Ax1[ n − 2] + Bx2 [ n − 2]) + a1 ( Ay1[ n − 1] + By2 [ n − 1]) + a1 ( Ay1[ n − 2] + By2 [ n − 2]) = A(b0 x1[ n] + b1 x1[ n − 1] + b2 x1[ n − 2] + a1 y1[ n − 1] + a 2 y2 [ n − 1]) + B(b1 x 2 [ n] + b2 x 2 [ n − 1] + b3 x 2 [ n − 2] + a1 x 2 [ n − 1] + a 2 x 2 [ n − 2]) = Ay1[ n] + By2 [ n]. Hence, the system of Eq. (2.18) is linear. ⎧ x [ n / L ], n = 0, ± L, ± 2 L,L (b) For an input xi [ n], i = 1,2, the output is yi [ n] = ⎨ i otherwise. ⎩ 0, For an input x[ n] = Ax1[ n] + Bx2 [ n], the output for n = 0, ± L, ± 2 L,K is y[ n] = x[ n / L ] = Ax1[ n / L ] + Bx2 [ n / L ] = Ay1[ n] + By2 [ n] . For all other values of n, y[ n] = A ⋅ 0 + B ⋅ 0 = 0. Hence the system of Eq. (2.20) is linear.

(c) For an input xi [ n], i = 1,2, the output is yi [ n] = x i [ n / M ], i = 1,2. Then, for an input

x[ n] = Ax1[ n] + Bx2 [ n], the output is y[ n] = Ax1[ n / M ] + Bx2 [ n / M ] = Ay1[ n] + By 2 [ n]. Hence the system of Eq. (2.21) is linear.

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(d) For an input xi [ n], i = 1,2, the output is yi [ n] =

1 M −1 ∑ xi [ n − k ], i = 1,2. Then, for M k =0

an input x[ n] = Ax1[ n] + Bx2 [ n], the output is y[ n] =

1 M −1 ∑ ( Ax1[ n − k ] + Bx 2 [ n − k ]) M k =0

⎞ ⎛ 1 M −1 ⎞ ⎛ 1 M −1 = A⎜⎜ ∑ x 2 [ n − k ] ⎟⎟ = Ay1[ n] + By2 [ n]. Hence the system of ∑ x1[ n − k ] ⎟⎟ + B⎜⎜ ⎠ ⎝ M k =0 ⎠ ⎝ M k =0 Eq. (2.61) is linear.

(e) The first term on the RHS of Eq. (2.65) is the output of a factor-of-2 up-sampler. The second term on the RHS of Eq. (2.65) is simply the output of an unit delay followed by a factor-of-2 up-sampler, whereas, the third term is the output of an unit advance operator followed by a factor-of-2 up-sampler. We have shown in Part (b) that the up-sampler is a linear system. Moreover, the unit delay and the unit advance operator are linear systems. A cascade of two linear systems is linear and the linear combination of linear systems is also linear. Hence, the factor-of-2 interpolator of Eq. (2.65) is a linear system. (f) Following the arguments given in Part (e), we can similarly show that the factor-of3 interpolator of Eq. (2.66) is a linear system. 2.36

(a) y[ n] = n 3 x[ n]. For an input xi [ n], i = 1,2, the output is yi [ n] = n 3 x i [ n], i = 1,2. Then, for an input x[ n] = Ax1[ n] + Bx2 [ n], the output is y[ n] = n 3 ( Ax1[ n] + Bx 2 [ n]) = Ay1[ n] + By2 [ n]. Hence the system is linear. For an input x[ n] = δ[ n], the output is the impulse response h[ n] = n 3 δ[ n]. As h[ n] = 0 for n < 0, and the system is causal. Let x[ n] = 1 for all values of n. Then y[ n] = n 3 and y[n] → ∞ as n → ∞. Since a bounded input results in an unbounded output, the system is not BIBO stable. Finally, let y[n] and y1[ n] be the outputs for inputs x[n] and x1[ n], respectively. If

x1[ n] = x[ n − no ] then y1[ n] = n 3 x1[ n] = n 3 x[ n − no ]. However, y[ n − no ] = ( n − n o ) 3 x[ n − n o ]. Since y1[ n] ≠ y[ n − no ], the system is not time-invariant.

(b) y[ n] = ( x[ n]) 5 . For an input xi [ n], i = 1,2, the output is yi [ n] = ( x i [ n]) 5 , i = 1,2. Then, for an input x[ n] = Ax1[ n] + Bx2 [ n], the output is y[ n] = ( Ax1[ n] + Bx 2 [ n])5

≠ A( x1 [ n]) 5 + B( x 2 [ n]) 5 . Hence the system is nonlinear. For an input x[ n] = δ[ n], the output is the impulse response h[ n] = (δ[ n]) 5 . As h[ n] = 0 for n < 0, and the system is causal.

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19

For a bounded input x[ n] ≤ B < ∞, the magnitude of the output samples are 5

y[ n] = ( x[ n]) 5 = x[ n] ≤ B 5 < ∞. As the output is also a bounded sequence, the

system is BIBO stable. Finally, let y[n] and y1[ n] be the outputs for inputs x[n] and x1[ n], respectively. If x1[ n] = x[ n − no ] then y1[ n] = ( x 1 [ n]) 5 = (x[ n − no ])5 = y[ n − no ]. Hence, the system is time-invariant.

3

(c) y[ n] = β + ∑ x[ n − l] with β a nonzero constant. For an input xi [ n], i = 1,2, the l=0

3

output is yi [ n] = β + ∑ x i [ n − l], i = 1,2. Then, for an input x[ n] = Ax1[ n] + Bx2 [ n], l =0

3

3

3

l =0

l =0

l=0

the output is y[ n] = β + ∑ ( Ax1[ n − l] + Bx 2 [ n − l]) = β + ∑ Ax1[ n − l] + ∑ Bx 2 [ n − l]

≠ Ay1[ n] + By2 [ n]. Hence the system is nonlinear.

∞

For an input x[ n] = δ[ n], the output is the impulse response h[ n] = β + ∑ δ[ n − l]. As l=0

h[ n] ≠ 0 for n < 0, the system is noncausal. For a bounded input x[ n] ≤ B < ∞, the magnitude of the output samples are

y[ n] ≤ β + 4 B < ∞. As the output is also a bounded sequence, the system is BIBO stable. Finally, let y[n] and y1[ n] be the outputs for inputs x[n] and x1[ n], respectively. If 3

x1[ n] = x[ n − no ] then y1[ n] = β + ∑ x[ n −n o −l] = y[ n − no ]. Hence, the system is l=0

time-invariant. (d) y[ n] = ln(2 + x[ n] ). For an input xi [ n], i = 1,2, the output is yi [ n] = ln(2 + xi [ n] ), i = 1,2. Then, for an input x[ n] = Ax1[ n] + Bx2 [ n], the output is

y[ n] = ln (2 + Ax1[ n] + Bx2 [ n] ) ≠ Ay1[ n] + By2 [ n]. Hence the system is nonlinear.

For an input x[ n] = δ[ n], the output is the impulse response h[ n] + ln(2 + δ[ n] ) . For n < 0, h[ n] = ln(2) ≠ 0. Hence, the system is noncausal. For a bounded input x[ n] ≤ B < ∞, the magnitude of the output samples are

y[ n] ≤ ln(2 + B ) < ∞. As the output is also a bounded sequence, the system is BIBO stable. Finally, let y[n] and y1[ n] be the outputs for inputs x[n] and x1[ n], respectively. If

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20

x1[ n] = x[ n − no ] then y1[ n] = ln(2 + x[ n − no ] ) = y[ n −n o ]. Hence, the system is time-invariant. (e) y[ n] = αx[ − n + 2], with a nonzero constant. For an input xi [ n], i = 1,2, the output is yi [ n] = αxi [ −n + 2], i = 1,2. Then, for an input x[ n] = Ax1[ n] + Bx2 [ n], the output is

y[ n] = Aαx1[ −n + 2] + Bαx 2 [ −n + 2] = Ay1[ n] + By2 [ n]. Hence the system is linear. For an input x[ n] = δ[ n], the output is the impulse response h[ n] = αδ[ −n + 2]. For n < 0, h[ n] = 0. Hence, the system is causal. For a bounded input x[ n] ≤ B < ∞, the magnitude of the output samples are y[ n] = αB < ∞. As the output is also a bounded sequence, the system is BIBO stable. Finally, let y[n] and y1[ n] be the outputs for inputs x[n] and x1[ n], respectively. If x1[ n] = x[ n − no ] then y1[ n] = αx1[ − n + 2] = αx[ −(n − no ) + 2] = y[ n −n o ]. Hence, the system is time-invariant. (f) y[ n] = x[ n − 4]. For an input xi [ n], i = 1,2, the output is yi [ n] = xi [ n − 4], i = 1,2. Then, for an input x[ n] = Ax1[ n] + Bx2 [ n], the output is y[ n] = Ax1[ n − 4] + Bx2 [ n − 4] = Ay1 [ n] + By2 [ n]. Hence the system is linear. For an input x[ n] = δ[ n], the output is the impulse response h[ n] = δ[ n − 4]. For n < 0, h[ n] = 0. Hence, the system is causal. For a bounded input x[ n] ≤ B < ∞, the magnitude of the output samples are

y[ n] = B < ∞. As the output is also a bounded sequence, the system is BIBO stable. Finally, let y[n] and y1[ n] be the outputs for inputs x[n] and x1[ n], respectively. If x1[ n] = x[ n − no ] then y1[ n] = x[ n −n o −4] = y[ n − no ]. Hence, the system is timeinvariant. 2.37

Let y[n] and y1[ n] be the outputs of a median filter of length 2 K + 1 for inputs x[n] and x1[ n] , respectively. If x1 [ n] = x[ n − no ] , then y1[ n] = med{x1[ n − K ], K, x1[ n − 1], x1[ n], x1[ n + 1], K, x1[ n + K ]} = med{x[ n − no − K ], K, x[ n − no − 1], x[ n − no ], x[ n − no + 1], K, x[ n − no + K ]} = y[ n − no ]. Hence, the system is time-invariant.

2.38

y[ n] = x[ n + 1] − 2 x[ n] + x[ n − 1]. For an input x i [ n], i = 1,2, the output is yi [ n] = x i [ n + 1] − 2 xi [ n] + xi [ n − 1], i = 1,2. Then, for an input x[ n] = Ax1[ n] + Bx2 [ n], the output is y[ n] = Ax1[ n + 1] + Bx 2 [ n + 1] − 2 Ax1[ n] − 2 Bx 2 [ n] + Ax1[ n − 1] + Bx 2 [ n − 1] = Ay1[ n] + By2 [ n]. Hence the system is linear. If x1[ n] = x[ n − no ], then y1[ n] = x[ n − no + 1] − 2 x[ n − no ] + x[ −no n − 1] = y[ n − no ]. Hence, the system is time-invariant.

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21

The impulse response of the system is h[ n] = δ[ n + 1] − 2δ[ n] + δ[ n − 1] . Now h[ −1] = δ[0] = 1. Since h[ n] ≠ 0 for all values of n < 0, the system is noncausal. 2.39

y[ n] = x 2 [ n] − x[ n − 1]x[ n + 1]. For an input xi [ n], i = 1,2, the output is yi [ n] = x i2 [ n] − x i [ n − 1]x i [ n + 1], i = 1,2. Then, for an input x[ n] = Ax1[ n] + Bx2 [ n], the

output is y[ n] = ( Ax1[ n] + Bx 2 [ n])2 − ( Ax1[ n − 1] + Bx 2 [ n − 1])( Ax1[ n + 1] + Bx 2 [ n + 1]) ≠ Ay1[ n] + By2 [ n]. Hence the system is nonlinear. If x1[ n] = x[ n − no ], then y1[ n] = x12 [ n] − x1[ n − 1]x1[ n + 1] = x 2 [ n − no ] − x[ n − no − 1] x[ n − no + 1] = y[ n − no ]. Hence, the system is timeinvariant.

The impulse response of the system is h[ n] = δ 2 [ n] − δ[ n − 1]δ[ n + 1] = δ[ n]. Since h[ n] = 0 for all values of n < 0, the system is causal. 2.40

1⎛ x[ n] ⎞ ⎜⎜ y[ n − 1] + ⎟. Now for an input x[ n] = αµ[ n], the output y[n] 2⎝ y[ n − 1] ⎟⎠ converges to some constant K as n → ∞. The input-output relation of the system as α⎞ 1⎛ n → ∞ reduces to K = ⎜ K + ⎟ from which we get K 2 = α or in other words 2⎝ K⎠ K = α. x [ n] ⎞ 1⎛ ⎟, i = 1,2. Then, For an input xi [ n], i = 1,2, the output is y i [ n] = ⎜⎜ yi [ n − 1] + i 2⎝ yi [ n − 1] ⎟⎠ y[ n] =

for an input x[ n] = Ax1[ n] + Bx2 [ n], the output is y[ n] =

Ax [ n] + Bx 2 [ n] ⎞ 1⎛ ⎜⎜ y[ n − 1] + 1 ⎟⎟. y[ n − 1] 2⎝ ⎠

On the other hand,

Bx 2 [ n] ⎞ Ax1 [ n] ⎞ 1 ⎛ 1⎛ ⎟ ≠ y[ n]. ⎟⎟ + ⎜⎜ By 2 [ n − 1] + ⎜⎜ Ay1 [ n − 1] + y 2 [ n − 1] ⎟⎠ y1 [ n − 1] ⎠ 2 ⎝ 2⎝ Hence the system is nonlinear. x[ n − no ] ⎞ 1⎛ ⎟ = y[ n − no ]. Hence, the If x1[ n] = x[ n − no ], then y1[ n] = ⎜⎜ y1[ n − 1] + y1[ n − 1] ⎟⎠ 2⎝ system is time-invariant. Ay1[ n] + By2 [ n] =

2.41

y[ n] = x[ n] − y 2 [ n − 1] + y[ n − 1]. For an input xi [ n], i = 1,2, the output is yi [ n] = x i [ n] − yi2 [ n − 1] + yi [ n − 1], i = 1,2. Then, for an input x[ n] = Ax1[ n] + Bx2 [ n], the output is

y[ n] = Ax1[ n] + Bx 2 [ n] − y 2 [ n − 1] + y[ n − 1]. On the other hand, Ay1[ n] + By2 [ n]

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22

= Ax1[ n] − Ay12 [ n − 1] + Ay1[ n − 1] + Bx 2 [ n] − By22 [ n − 1] + By2 [ n − 1] ≠ y[ n]. Hence the system is nonlinear. Let y[n] be the output for an input x[n] , i.e., y[n] = x[n] − y 2 [n] + y[n − 1]. Then, for an input x[n − no ] the output is given by y[ n − n o ] = x[ n − n o ] − y 2 [ n − n o −1] + [ n − n o −1] , or in other words, the system is time-invariant. Now, for an input x[ n] = αµ[ n] , the output y[n] converges to some constant K as n → ∞ . The difference equation describing the system as n → ∞ reduces to K = α − K 2 + K , or K 2 = α, i.e., K = α . 2.42

The impulse response of the factor-of-3 interpolator of Eq. (2.66) is the output for an input x u [ n] = δ[ n] and is given by 2 3 2 , 1, 3

1 3

h[ n] = δ[ n] + (δ[ n − 1] + δ[ n + 1]) + (δ[ n − 2] + δ[ n + 2]) or equivalently by {h[ n]} =

2.43

{

1 , 3

2 , 3

}

1 , − 2 ≤ n ≤ 2. 3

The input-output relation of a factor-of- L interpolator is given by L −1 L − k (xu [ n − k ] + xu [ n + k ]). Its impulse response is the output for y[ n] = xu [ n] + ∑ k =1 L L −1 L − k

an input xu [ n] = δ[ n] and is thus given by h[ n] = δ[ n] + ∑

{,

k =1

L

K,

2 , L

(δ[ n − k ] + δ[ n + k ])

or equivalently by

{h[ n]} = 2.44

1 L

2 , L

K,

L −2 , L

L −1 , L

1,

L −1 , L

L −2 , L

1 L

}, − L + 1 ≤ n ≤ L − 1.

The impulse response h[n] of a causal discrete-time system satisfies the difference equation h[ n] − a h[ n − 1] = δ[ n]. Since the system is causal, we have h[ n] = 0 for n < 0. Evaluating the above difference equation for n = 0, we arrive at h[0] − ah[ −1] = 1 and thus h[0] = 1. Next, for n = 1 , we have h[1] − a h[ 0] = 0 and thus

h[1] = a. Continuing we get for n = 2, h[ 2] − ah[1] = 0 , i.e., h[2] = ah[1] = a 2 . Assume

h[ n − 1] = a n −1 with n > 0. From the difference equation we then have h[ n] − a h[ n − 1] = 0 , i.e., h[ n] = ah[ n − 1] = a n . Since the last equation holds for n = 0,1, 2, by induction, it holds for n ≥ 3.

2.45

As x[n] and h[n] are right-sided sequences, assume x[ n] = 0 for all n < N1 and h[ n] = 0 and n < N 2 . Hence, y[ n] = h[ n] O * x[ n] = 0 for all n < N1 + N 2 and thus

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23

∞

y[n] is also a right-sided sequence. Therefore,

=

∞

∞

∞

n = N1 + N 2 k = N 2 ∞

n = N1 + N 2

∞

* x[ n] ∑ h[ n] O

n = N1 + N 2 ∞

∑ h[ k ]x[ n − k ] = ∑ h[ k ]

k = N 2 n = N1 + N 2

∞

∞

∞

m = N1 + N 2 − k

k = N2

m = N1

k =N2

∞

∑ x[ n − k ]

n = N1 + N 2

∑ x[ m] = ∑ h[ k ] ∑ x[ m] as x[ m] = 0 for all m < N1 . Hence,

= ∑ h[ k ] k = N2

∞

∑ h[ k ]x[ n − k ] = ∑

∑

∑ y[ n] =

⎞ ⎞⎛ ⎛ ∑ y[ n] =⎜⎜ ∑ h[ n] ⎟⎟⎜⎜ ∑ x[ n] ⎟⎟. n ⎠ ⎠⎝ n ⎝n 2.46

2.47

∞ ∞ ⎧∑ n α k , n ≥ 0, k k (a) α n µ[n] O * µ[n] = ∑ α µ[k ]µ[n − k ] = ∑ α µ[n − k ] = ⎨ k = 0 0, n < 0, k = −∞ k =0 ⎩ ⎛ 1 − α n +1 ⎞ ⎟µ[n]. =⎜ ⎜ 1− α ⎟ ⎠ ⎝ ∞ ∞ ⎧∑ n kα k , n > 0, k k (b) nα n µ[ n] O * µ[ n] = ∑ kα µ[ k ]µ[ n − k ] = ∑ kα µ[ n − k ] = ⎨ k = 0 0, n ≤ 0. k = −∞ k =0 ⎩

Now from Eq. (2.72) an arbitrary input x[n] can be expressed as ∞

x[ n] = ∑ x[ k ]δ[ n − k ] which can be rewritten using Eq. (2.41b) as k = −∞ ∞

∞

∞

k = −∞

k = −∞

k = −∞

x[ n] = ∑ x[ k ](µ[ n − k ] − µ[ n − k − 1]) = ∑ x[ k ]µ[ n − k ] − ∑ x[ k ]µ[ n − k − 1]. Since s[n] is the response of an LTI system for an input µ[n], s[ n − k ] is the response for an input µ[ n − k ] and s[ n − k − 1] is the response for an input µ[ n − k − 1]. Hence, the output for an input ∞

∞

∞

k = −∞ ∞

k = −∞

∑ x[ k ]µ[ n − k ] − ∑ x[ k ]µ[ n − k − 1] is given by

* s[ n] − x[ n − 1] O * s[ n − 1]. y[ n] = ∑ x[ k ]s[ n − k ] − ∑ x[ k ]s[ n − k − 1] = x[ n] O k = −∞

2.48

y[ n] =

k = −∞

∞ x[ n − m]. Hence, ∑ h[ m]~

m = −∞

y[ n + kN ] =

∞ x[ n + kN − m] = ∑ h[ m ]~

m = −∞

∞

∑ h[ m]x[ n − m] = y[ n]. Thus, y[n] is also a

m = −∞

periodic sequence with a period N . 2.49

In this problem we make use of the identity δ[ n − m ] O * δ[ n − r ] = δ[ n − m − r ].

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24

(a) y1[ n] = x1[ n] O * h1[ n] = (3δ[ n − 2] − 2δ[ n + 1]) O * (− δ[ n + 2] + 4δ[ n] + 2δ[ n − 1]) = −3δ[ n − 2] O * δ[ n + 2] * δ[ n + 2] + 12δ[ n − 2] O * δ[ n] − 6δ[ n − 2] O * δ[ n − 1] + 2δ[ n + 1] O − 8δ[ n + 1] O * δ[ n] + 4δ[ n + 1] O * δ[ n − 1] . Hence

y1[ n] = −3δ[ n] + 12δ[ n − 2] − 6δ[ n − 3] + 2δ[ n + 3] − 8δ[ n + 1] + 4δ[ n] = 2δ[ n + 3] − 8δ[ n + 1] + δ[ n] + 12δ[ n − 2] − 6δ[ n − 3]. (b) y 2 [ n] = x 2 [ n] O * h2 [ n] = (5δ[ n − 3] + 2δ[ n + 1]) O * (3δ[ n − 4] + 1.5δ[ n − 2] − δ[ n + 1]) = 15δ[ n − 3] O * δ[ n − 4] + 7.5δ[ n − 3] O * δ[ n − 2] − 5δ[ n − 3] O * δ[ n + 1] + 6δ[ n + 1] O * δ[ n − 4] + 3δ[ n + 1] O * δ[ n − 2] − 2δ[ n + 1] O * δ[ n + 1] = 15δ[ n − 7] + 7.5δ[ n − 5] − 5δ[ n − 2]

+ 6δ[ n − 3] + 3δ[ n − 1] − 2δ[ n + 2]. (c) y3 [ n] = x1[ n] O * h2 [ n] = (− 3δ[ n − 2] − 2δ[ n + 1]) O * (3δ[ n − 4] + 1.5δ[ n − 2] − δ[ n + 1]) = 9δ[ n − 2] O * δ[ n − 4] + 4.5δ[ n − 2] O * δ[ n − 2] − 3δ[ n − 2] O * δ[ n + 1] − 6δ[ n + 1] O * δ[ n − 4] − 3δ[ n + 1] O * δ[ n − 2] + 2δ[ n + 1] O * δ[ n + 1] = 9δ[ n − 6] + 4.5δ[ n − 4] − 3δ[ n − 1]

− 6δ[ n − 3] − 3δ[ n − 1] − 3δ[ n − 1] + 2δ[ n + 2] = 2δ[ n + 2] − 6δ[ n − 1] − 6δ[ n − 3] + 4.5δ[ n − 4] + 9δ[ n − 6]. (d) y 4 [ n] = x 2 [ n] O * h1[ n] = (5δ[ n − 3] + 2δ[ n + 1]) O * (− δ[ n + 2] + 4δ[ n] − 2δ[ n − 1]) = −5δ[ n − 3] O * δ[ n + 2] + 20δ[ n − 3] O * δ[ n] − 10δ[ n − 3] O * δ[ n − 1] − 2δ[ n + 1] O * δ[ n + 2] + 8δ[ n + 1] O * δ[ n] − 4δ[ n + 1] O * δ[ n − 1] = −5δ[ n − 1] + 20δ[ n − 3] − 10δ[ n − 4] − 2δ[ n + 3]

+ 8δ[ n + 1] − 4δ[ n] = −2δ[ n + 3] + 8δ[ n + 1] − 4δ[ n] − 5δ[ n − 1] + 20δ[ n − 3] − 10δ[ n − 4]. 2.50

(a) u[ n] = x[ n] O * y[ n]

= {− 24, 42, − 5, − 20, − 45, 23, 66, − 25, − 42, − 17, 22, 14, − 4}, − 4 ≤ n ≤ 8. (b) v[ n] = x[ n] O * w[ n]

= {− 12, 7, 5, 10, − 16, − 3, − 28, 30, 13, − 6, − 15, − 4, 10}, − 1 ≤ n ≤ 11. (c) g[ n] = w[ n] O * y[ n]

= {18, 3, 3, − 14, 25, 26, 60, − 11, − 16, − 14, 26, 39, − 10}, 1 ≤ n ≤ 13.

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25

2.51

N2 y[ n] = ∑ m g[ m]h[ n − m]. Now, h[ n − m] is defined for M1 ≤ n − m ≤ M 2 . Thus, =N 1

for m = N1 , h[ n − m] is defined for M1 ≤ n − N1 ≤ M 2 , or equivalently, for M1 + N1 ≤ n ≤ M 2 + N1 . Likewise, for m = N 2 , h[ n − m] is defined for M1 ≤ n − N 2 ≤ M 2 , or equivalently, for M1 + N 2 ≤ n ≤ M 2 + N 2 . For the specified sequences N1 = −3, N 2 = 4, M1 = 2, M 2 = 6. (a) The length of y[n] is M 2 + N 2 − M1 − N1 + 1 = 6 + 4 − 2 − (−3) + 1 = 12 . (b) The range of n for y[ n] ≠ 0 is min(M1 + N1 , M 2 + N 2 ) ≤ n ≤ max(M1 + N1 , M 2 + N 2 ) , i.e., M1 + N1 ≤ n ≤ M 2 + N 2 . For the specified sequences the range of n is − 1 ≤ n ≤ 10. ∞

2.52

* x 2 [ n] = ∑ x 1 [ n − k ]x 2 [ k ]. Now, y[ n] = x1 [ n] O k = −∞

∞

* x 2 [ n − N 2 ] = ∑ x1 [ n − N1 − k ] x 2 [ k − N 2 ]. Let k − N 2 = m. v[ n] = x1 [ n − N1 ] O k = −∞

Then v[ n] = 2.53

∞

∑ x 1 [ n − N1 m = −∞

− N 2 − m ]x 2 [ m] = y[ n − N1 − N 2 ].

g[ n] = x1[ n] O * x 2 [ n] O * x 3 [ n] = y[ n] O * x 3 [ n] where y[ n] = x1[ n] O * x 2 [ n]. Now v[ n] = x1[ n − N1 ] O * x 2 [ n − N 2 ]. Define h[ n] = v[ n] O * x 3 [ n − N 3 ]. Then from the

results of Problem 2.52, v[ n] = y[ n − N1 − N 2 ]. Hence, h[ n] = y[ n − N1 − N 2 ] O * x 3 [ n − N 3 ]. Therefore, making use of the results of Problem 2.52 again we get h[ n] = y[ n − N1 − N 2 − N 3 ]. ∞

2.54

* h[ n] = ∑ x[ n − k ]h[ k ]. Substituting k by n − m in this expression, we y[ n] = x[ n] O k = −∞

get y[ n] =

∞

* x[ n]. Hence the convolution operation is ∑ x[ m]h[ n − m] = h[ n] O

m = −∞

commutative. ∞

* (h1[ n] + h2 [ n]) = ∑ x[ n − k ](h1[ k ] + h2 [ k ]) Let y[ n] = x[ n] O k = −∞

∞

∞

k = −∞

k = −∞

* h1[ n] + x[ n] O * h2 [ n]. Hence the = ∑ x[ n − k ]h1[ k ] + ∑ x[ n − k ]h2 [ k ] = x[ n] O

convolution operation is also distributive. 2.55

x 3 [ n] O * x 2 [ n] O * x1[ n] = x 3 [ n] O * ( x 2 [ n] O * x1[ n]). As x2 [ n] O * x1[ n] is an unbounded

sequence, the result of this convolution cannot be determined. But

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26

x 2 [ n] O * x 3 [ n] O * x1[ n] = x 2 [ n] O * ( x 3 [ n] O * x1[ n]). Now x 3 [ n] O * x1[ n] = 0 for all values

of n , and hence the overall result is zero. As a result, for the given sequences x 3 [ n] O * x 2 [ n] O * x1[ n] ≠ x 2 [ n] O * x 3 [ n] O * x1[ n] 2.56

w[ n] = x[ n] O * h[ n] = ∑ x[ k ]h[ n − k ] and * h[ n] O * g[ n]. Define y[ n] = x[ n] O k

f [ n] = h[ n] O * g[ n] = ∑ g[ k ]h[ n − k ]. Consider w1 [ n] = ( x[ n] O * h[ n]) O * g[ n] k

= y[ n] O * g[ n] = ∑ g[ m ]∑ x[ k ]h[ n − m − k ]. Now consider w 2 [ n] = x[ n] O * (h[ n] O * g[ n]) m

k

= x[ n] O * f [ n] = ∑ x[ k ]∑ g[ m ]h[ n − k − m ]. The difference between the k

m

expressions for w1[ n] and w2 [ n] is that the order of the summations is changed. A) Assumptions: h[n] and g[n] are causal sequences, and x[ n] = 0 for n < 0. This n 0, for m < 0, ⎧ implies y[ m] = ⎨ m Thus, w[ n] = ∑ g[ m]y[ n − m] ⎩∑ k = 0 x[k]h[m - k], for m ≥ 0. m=0 n

n−m

m=0

k =0

= ∑ g[ m] ∑ x[ k ]h[ n − m − k ]. All sums have only a finite number of terms. Hence, the interchange of the order of the summations is justified and will give correct results. B) Assumptions: h[n] and g[n] are stable sequences, and x[n] is a bounded sequence

⎛ k2 ⎞ with x[ n] ≤ B < ∞ . Here, y[ m] = ∑ ∞ k = −∞ h[ k ] x[ m − k ] = ⎜⎝ ∑ k = k1 h[ k ] x[ m − k ] ⎟⎠

+ε k1 , k 2 [ m] with ε k1 ,k 2 [ m ] ≤ ε n B. In this case, all sums have effectively only a finite number of terms and the error ε k1 , k 2 [ m] can be reduced by choosing k1 and

k2 sufficiently large. As a result, in this case the problem is again effectively reduced to that of the one-sided sequences. Thus, the interchange of the order of the summations is again justified and will give correct results. Hence, for the convolution to be associative, it is sufficient that the sequences be stable and single-sided. 2.57

y[ n] = ∑ ∞ k = −∞ x[ n − k ]h[ k ]. Since h[k ] is of length M and defined for 0 ≤ k ≤ M − 1, −1) the convolution sum reduces to y[ n] = ∑ (kM = 0 x[ n − k ]h[ k ]. y[n] will be nonzero for

all those values of n and for k which n − k satisfies 0 ≤ n − k ≤ N − 1. Minimum value of n − k = 0 and occurs for lowest n at n = 0 and k = 0. Maximum value of n − k = N − 1 and occurs for maximum value of k at M − 1. Thus n − k = M − 1 ⇒ n = N + M − 2. Hence the total number of nonzero samples = N + M − 1.

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2.58

y[ n] = ∑ kN=−01x[ n − k ] x[ k ]. The maximum value of y[n] occurs at n = N − 1 when all

product terms are present. The maximum value is given by y[ N − 1] = ∑ kN=−01a N −1− k a k . 2.59

y[ n] = ∑ kN=−01x[ n − k ]h[ k ]. The maximum value of y[n] occurs at n = N − 1 when all

product terms are present. The maximum value is given by y[ N − 1] = ∑ kN=−01a N −1− k bk . 2.60

* hev [ n] = ∑ ∞ (a) y[ n] = g ev [ n] O k = −∞ hev [ n − k ]g ev [ k ]. Now,

y[ −n] = ∑ ∞ k = −∞ hev [ − n − k ]g ev [ k ]. Replace k by − k. Then the summation on the left ∞ becomes y[ − n] = ∑ ∞ k = −∞ hev [ − n + k ]g ev [ − k ] = ∑ k = −∞ hev [ −( n − k )]g ev [ − k ] * hev [ n] is an even sequence. = y[n]. Hence g ev [ n] O * hod [ n] = ∑ ∞ (b) y[ n] = g ev [ n] O k = −∞ hod [ n − k ]g ev [ k ]. Now,

∞ y[ −n] = ∑ ∞ k = −∞ hod [ − n − k ]g ev [ k ] = ∑ k = −∞ hod [ − n + k ]g ev [ − k ] ∞ = ∑∞ k = −∞ hod [ −( n − k )]g ev [ − k ] = − ∑ k = −∞ hod [ n − k ]g ev [ k ] = − y[ n]. * hod [ n] is an odd sequence. Hence g ev [ n] O * hod [ n] = ∑ ∞ (c) y[ n] = g od [ n] O k = −∞ hod [ n − k ]g od [ k ]. Now,

∞ y[ −n] = ∑ ∞ k = −∞ hod [ − n − k ]g od [ k ] = ∑ k = −∞ hod [ − n + k ]g od [ − k ] ∞ = ∑∞ k = −∞ hod [ −( n − k )]g od [ − k ] = ∑ k = −∞ hod [ n − k ]g od [ k ] = y[ n]. * hod [ n] is an even sequence. Hence g od [ n] O

2.61

The impulse response of the cascade is given by h[ n] = h1[ n] O * h2 [ n] where

(

)

h1[ n] = α n µ[ n] and h2 [ n] = β n µ[ n]. Hence, h[ n] = ∑ nk = 0 α k β n − k µ[ n]. 2.62

k ∞ Now h[ n] = α n µ[ n]. Therefore y[ n] = ∑ ∞ k = −∞ h[ k ] x[ n − k ] = ∑ k = 0 α x[ n − k ]

k k ∞ = x[ n] + ∑ ∞ k =1 α x[ n − k ] = x[ n ] + α ∑ k = 0 α x[ n − 1 − k ] = x[ n] + α y[ n − 1]. Hence, x[ n] = y[ n] − αy[ n − 1]. Thus the inverse system is given by y[ n] = x[ n] − αx[ n − 1]. The impulse response of the inverse system is given by h[ n] = {1, α}, 0 ≤ n ≤ 1.

2.63

(

)

From the results of Problem 2.62 we have h[ n] = ∑ nk = 0 α k β n − k µ[ n]. Now,

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∞ ⎛ k ∞ ⎛ k ∞ ⎞ ⎞ y[ n] = ∑ x[ n − k ]h[ k ] = ∑ ⎜⎜ ∑ α m β k − m ⎟⎟µ[ m]x[ n − k ] = ∑ ⎜⎜ ∑ α m β k − m ⎟⎟x[ n − k ] k = −∞ k = −∞ ⎝ m = 0 k =0 ⎝ m =0 ⎠ ⎠ ∞ ⎛ k ⎞ = x[ n] + ∑ ⎜⎜ ∑ α m β k − m ⎟⎟x[ n − k ]. Substituting r = k − 1 in the last expression we get k =1 ⎝ m = 0 ⎠ ∞ ⎛ r ∞ ⎛ r +1 ⎞ ⎞ y[ n] = x[ n] + ∑ ⎜⎜ ∑ α m β r +1− m ⎟⎟x[ n − r − 1] = x[ n] + ∑ ⎜⎜ ∑ α m β r +1− m + α r +1 ⎟⎟x[ n − r − 1] r =0 ⎝ m =0 r =0 ⎝ m =0 ⎠ ⎠ ∞ ∞ ⎛ r ⎞ = x[ n] + β ∑ ⎜⎜ ∑ α m β r − m ⎟⎟x[ n − r − 1] + ∑ α r +1 x[ n − r − 1] r =0 ⎝ m =0 r =0 ⎠

= x[ n] + βy[ n − 1] + αx[ n − 1] + α 2 x[ n − 2] + α 3 x[ n − 3] + K. The inverse system is therefore given by x[ n] = y[ n] − (α + β) y[ n − 1] + αβ y[ n − 2]. 2.64

(a) h[ n] = h1[ n] O * h2 [ n] O * h3 [ n] O * h3 [ n] + h1[ n] O * h2 [ n] + h3 [ n] O * h4 [ n]. (b) h[ n] = h4 [ n] +

2.65

h1[ n] O * h2 [ n] O * h3 [ n] 1 − h1[ n] O * h2 [ n] O * h5 [ n]

.

h[ n] = h1[ n] O * h2 [ n] + h3 [ n]. Now h1[ n] O * h2 [ n] = (2δ[ n − 2] − 3δ[ n + 1]) O * (δ[ n − 1] + 2δ[ n + 2])

= 2δ[ n − 2] O * δ[ n − 1] − 3δ[ n + 1] O * δ[ n − 1] + 2δ[ n − 2] O * 2δ[ n + 2] − 3δ[ n + 1] O * 2δ[ n + 2]

= 2δ[ n − 3] − 3δ[ n] + 4δ[ n] − 6δ[ n + 3] and h3 [n] = 5δ[n − 5] + 7δ[n − 3] + 2δ[n − 1] − δ[n] + 3δ[n + 1]. Therefore, h[ n] = 2δ[ n − 3] − 3δ[n] + 4δ[n] − 6δ[ n + 3] + 5δ[ n − 5] + 7δ[n − 3] + 2δ[ n − 1] − δ[ n] + 3δ[ n + 1] = 5δ[n − 5] + 9δ[n − 3] + 2δ[n − 1] + 3δ[n + 1] − 6δ[n + 3].

2.66

3 ⎫ 1 ⎧ ⎨ y[ n] − ∑ h[ k ]x[ n − k ]⎬ we h[0] ⎩ k =0 ⎭

(a) The length of x[n] is 8 − 4 + 1 = 5. Using x[ n] = arrive at {x[ n]} = {3, − 2, 0, 1, 2}, 0 ≤ n ≤ 4. (b) The length of x[n] is 7 − 4 + 1 = 4. Using x[ n] =

3 ⎫ 1 ⎧ ⎨ y[ n] − ∑ h[ k ]x[ n − k ]⎬ we h[0] ⎩ k =0 ⎭

arrive at {x[ n]} = {1, 2, 3, 4}, 0 ≤ n ≤ 3. (c) The length of x[n] is 8 − 5 + 1 = 4. Using x[ n] =

4 ⎫ 1 ⎧ ⎨ y[ n] − ∑ h[ k ]x[ n − k ]⎬ we h[0] ⎩ k =0 ⎭

arrive at {x[ n]} = {1, − 2, 3, − 1}, 0 ≤ n ≤ 3.

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2.67

y[ n] = ay[ n − 1] + bx[ n]. Hence, y[0] = ay[ −1] + bx[ 0]. Next, y[1] = ay[0] + bx[1]

= a(ay[ −1] + bx[0]) + bx[1] = a 2 y[ −1] + abx[0] + bx[1]. Continuing further in a similar way we obtained y[ n] = a n +1 y[ −1] + ∑ nk = 0 a n − k bx[ k ]. (a) Let y1[ n] be the output due to an input x1[ n]. Then

y1[ n] = a n +1 y[ −1] + ∑ nk = 0 a n − k bx1[ k ]. If x1[ n] = x[ n − no ], then y1[ n] = a n +1 y[ −1] + ∑ nk = 0 a n − k bx[ k − no ] = a n +1 y[ −1] + ∑ rn=−0n o a n − n o − r bx[r ]. However, y[ n − no ] = a n +1 y[ −1] + ∑ nk = 0 a n − k bx[ k − no ] = a n − n o +1 y[ −1] + ∑ rn=−0n o a n − n o − r bx[r ]. Hence y1[ n] ≠ y[ n − no ] if y[ −1] ≠ 0, i.e., the system is time-variant. The system is time-invariant if and only if y[ −1] = 0, as then y1[ n] = y[ n − no ] . (b) Let y1[ n] and y1[ n] be the outputs due to inputs x1[ n] and x1[ n] , respectively. Let y[n] be the output due to an input αx1[ n] + βx 2 [ n]. However, αy1[ n] + βy2 [ n] =

α a n +1 y[ −1] + β a n +1 y[ −1] + α ∑ nk = 0 a n − k bx1[ k ] + β∑ nk = 0 a n − k bx 2 [ k ], whereas, y[ n] = a n +1 y[ −1] + α ∑ nk = 0 a n − k bx1[ k ] + β∑ nk = 0 a n − k bx 2 [ k ]. Hence, the system is

nonlinear if y[ −1] ≠ 0 and is linear if and only if y[ −1] = 0.

(c) Generalizing the above result it can be shown that an N –th order causal discretetime system is linear and time-invariant if and only if y[ −r ] = 0,1 ≤ r ≤ N . 2.68

y[ n] = p0 x[ n] + p1 x[ n − 1] − d1 y[ n − 1] leads to x[ n] =

1 p0

y[ n] +

d1 p0

y[ n − 1] −

p1 p0

x[ n − 1],

which is the difference equation characterizing the inverse system. In other words, simply solve the equation for x[n] in terms of present and past values of y[n] and x[n]. 2.69

s[ n] = ∑ nk = 0 h[ k ]µ[ n − k ] = ∑ nk = 0 h[ k ], n ≥ 0, and s[ n] = 0, n < 0. Since h[k ] is

nonnegative, s[n] is a monotonically increasing function of n for n ≥ 0 , and is not oscillatory. Hence, there is no overshoot. 2.70

(a) f [ n] = f [ n − 1] + f [ n − 2]. Let f [ n] = αr n , then the difference equation reduces to αr n − αr n −1 − αr n − 2 = 0 which reduces further to r 2 − r − 1 = 0 resulting in

r=

n n 1± 5 ⎛ 1+ 5 ⎞ ⎛ 1− 5 ⎞ + α1 ⎜ . . Thus, f [ n] = α1 ⎜ ⎟ ⎟ 2 ⎝ 2 ⎠ ⎝ 2 ⎠

As f [ 0] = 0, hence α1 + α 2 = 0. Also f [1] = 1, , and hence

⎛⎜ α 1 + α 2 ⎞⎟ + 5 ⎛⎜ α 1 − α 2 ⎞⎟ = 1. Solving for α and α , we get α = −α = 1 . Hence, 1 2 1 2 5 ⎝ 2 ⎠ ⎝ 2 ⎠

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n

n

1 ⎛ 1− 5 ⎞ ⎛ 1+ 5 ⎞ f [ n] = ⎜ ⎟ − ⎜ ⎟ . 5⎝ 2 ⎠ 5⎝ 2 ⎠ 1

(b) y[ n] = y[ n − 1] + y[ n − 2] + x[ n − 1]. As the system is LTI, the initial conditions are equal to zero. Let x[ n] = δ[ n]. Then y[ n] = y[ n − 1] + y[ n − 2] + δ[ n − 1]. Hence, y[0] = y[ −1] + y[ −2] = 0 and y[1] = y[ 0] + y[ −2] + δ[ 0] = 1. For n > 1, the corresponding difference equation is y[ n] = y[ n − 1] + y[ n − 2] with initial conditions y[ 0] = 0 and y[1] = 1, which are the same as those for the solution of the Fibonacci’s n

n

1 ⎛ 1− 5 ⎞ ⎛ 1+ 5 ⎞ sequence. Hence y[ n] = ⎜ ⎟ − ⎜ ⎟ . Thus denotes the impulse 5⎝ 2 ⎠ 5⎝ 2 ⎠ response of a causal LTI system described by the difference equation y[ n] = y[ n − 1] + y[ n − 2] + x[ n − 1]. 1

2.71 y[ n] = αy[ n − 1] + x[ n]. Denoting y[ n] = yre [ n] + j yim [ n], and α = a + j b, we get yre [ n] + j yim [ n] = (a + j b)( yre [ n − 1] + j yim [ n − 1]) + x[ n]. Equating the real and the imaginary parts, and noting that x[n] is real, we get yre [ n] = ayre [ n − 1] − byim [ n − 1] + x[ n], yim [ n] = byre [ n − 1] + ayim [ n − 1]. From the second equation we have yim [ n − 1] =

1 b yim [ n] − yre [ n − 1]. a a

Substituting this

equation in the top left equation we arrive at b a

yre [ n] = ayre [ n − 1] − yim [ n] +

b2 a

yre [ n − 1] + x[ n], from which we get

byim [ n − 1] = − ayre [ n − 1] + ( a 2 + b 2 ) yre [ n − 2] + ax[ n − 1]. Substituting this equation

in the equation yre [ n] = ayre [ n − 1] − byim [ n − 1] + x[ n] we arrive at yre [ n] = 2 ayre [ n − 1] − ( a 2 + b 2 ) yre [ n − 2] + x[ n] − ax[ n − 1] which is a second-order

difference equation representing yre [n] in terms of x[n]. 2.72

The first-order causal LTI system is characterized by the difference equation y[ n] = p0 x[ n] + p1 x[ n − 1] − d1 y[ n − 1]. Letting x[ n] = δ[ n] we obtain the difference equation representation of its impulse response h[ n] = p0 δ[ n] + p1δ[ n − 1] − d1h[ n − 1]. Solving it for n = 0,1, 2, we get h[0] = p0 , h[1] = p1 − d1h[0] = p1 − d1 p0 , and h[2] = −d1h[1] = − d1 p0 ( p1 − d1 p0 ). Solving these equations we get p0 = h[0], d1 = −

2.73

h[ 2 ] , h[1]

and p1 = h[1] −

h[ 2 ]h[ 0 ] . h[1]

M

N

M

N

k =0

k =0 N d h[r ∑k =0 k

k =0

k =0

∑ pk x[ n − k ] = ∑ d k y[ n − k ]. Let x[ n] = δ[ n]. Then ∑ p k δ[ n − k ] = ∑ d k h[ n − k ].

Thus, pr =

− k ]. Since the system is assumed to be causal, h[r − k ] = 0

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31

for all k > r. Hence, pr = ∑ kN= 0 d k h[r − k ] = ∑ kN= 0 h[ k ]d r − k . 2.74

For a filter with a complex-valued impulse response, the first part of the proof is the same as that for a filter with a real-valued impulse response. From ∞ ∞ y[ n] = ∑ ∞ k = −∞ h[ k ]x[ n − k ] we get y[ n] = ∑ k = −∞ h[ k ]x[ n − k ] ≤ ∑ k = −∞ h[ k ] x[ n − k ] .

Since the input is bounded x[ n] ≤ B x . Therefore y[ n] ≤ B x ∑ ∞ k = −∞ h[ k ] . So if ∑∞ k = −∞ h[ k ] = S < ∞, then y[ n] ≤ B x S indicating that y[n] is also bounded.

To prove the converse we need to show that if a bounded input is produced by a bounded input then S < ∞. Consider the following bounded input defined by ∞ h * [ − k ]h[ k ] ∞ h * [ − n] . Then y[ n] = ∑ x[ n] = = ∑ h[ k ] = S. Now since the output h[ −n] h[ k ] k = −∞ k = −∞ is bounded, S < ∞. Thus for a filter with a complex impulse response is BIBO stable if and only if ∑ ∞ k = −∞ h[ k ] = S < ∞. 2.75

The impulse response of the cascade is g[ k ] = ∑ ∞ k = −∞ h1[ k − r ]h 2 [r ]. Thus ∞

∞

∞ ⎞ ⎞⎛ ∞ ⎛ ∞ ∑ h1[ k − r ] h2 [r ] ≤ ⎜⎜ ∑ h1[ k ] ⎟⎟ ⎜⎜ ∑ h2 [r ] ⎟⎟. Since k = −∞ k = −∞ r = −∞ ⎠ ⎠ ⎝ r = −∞ ⎝ k = −∞ h1[ n] and h2 [ n] are stable, ∑ h1[ k ] < ∞ and ∑ h2 [ k ] < ∞. Hence ∑ g[ k ] < ∞ and as a

∑ g[ k ] = ∑

k

k

k

result, a cascade of two stable LTI systems is also stable. 2.76

The impulse response of the parallel structure is g[ n] = h1[ n] + h2 [ n]. Now, ∞

∞

∞

∞

k = −∞

k = −∞

k = −∞

k = −∞

∑ g[ k ] = ∑ h1[ k ] + h2 [ k ] ≤ ∑ h1[ k ] + ∑ h2 [ k ] . Since h1[ n] and

h2 [ n] are stable, ∑ h1[ k ] < ∞ and ∑ h2 [ k ] < ∞. Hence ∑ g[ k ] < ∞ and as a result, a k

k

k

parallel connection of two stable LTI systems is also stable. 2.77

Consider a cascade connection of two passive LTI systems with an input x[n] and an output y[n]. Let y1[ n] and y2 [ n] be the outputs of the two systems for the input ∞ ∞ ∞ x[n]. Now ∑ ∞ n = −∞ y1[ n] ≤ ∑ n = −∞ x[ n] and ∑ n = −∞ y2 [ n] ≤ ∑ n = −∞ x[ n] . Let y1[ n] = y2 [ n] = x[ n] satisfying the above inequalities. Then y[ n] = y1[ n] + y2 [ n] 2

2

2

2

2

2

2

∞ ∞ = 2 x[ n] and as a result, ∑ ∞ n = −∞ y[ n] = 4∑ n = −∞ x[ n] > ∑ n = −∞ x[ n] . Hence, the parallel connection of two passive LTI systems may not be passive.

2.78

Consider a parallel connection of two passive LTI systems with an input x[n] and an output y[n]. Let y1[ n] and y2 [ n] be the outputs of the two systems for the input Not for sale.

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∞ ∞ ∞ x[n]. Now ∑ ∞ n = −∞ y1[ n] ≤ ∑ n = −∞ x[ n] and ∑ n = −∞ y2 [ n] ≤ ∑ n = −∞ x[ n] . Let y1[ n] = y2 [ n] = x[ n] satisfying the above inequalities. Then y[ n] = y1[ n] + y2 [ n] 2

2

2

2

2

2

2

∞ ∞ = 2 x[ n] and as a result, ∑ ∞ n = −∞ y[ n] = 4∑ n = −∞ x[ n] > ∑ n = −∞ x[ n] . Hence, the parallel connection of two passive LTI systems may not be passive.

2.79

Let the difference equation ∑ kM= 0 p k x[ n − k ] = y[ n] + ∑ kN=1 d k y[ n − k ] represents the causal IIR digital filter. For an input x[ n] = δ[ n], the corresponding output is then y[ n] = h[ n], the impulse response of the filter. As the number of coefficients { p k } is M + 1 and the number of coefficients {d k } is N , there are a total of N + M + 1 unknowns. To determine these coefficients from the impulse response samples, we compute only the first N + M + 1 impulse response samples. To illustrate the method, without any loss of generality, we assume N = M = 3. Then, from the difference equation we arrive at the following N + M + 1 = 7 equations: h[0] = p0 ,

h[1] + h[0]d1 = p1 , h[2] + h[1]d1 + h[0]d 2 = p2 , h[3] + h[2]d1 + h[1]d 2 + h[0]d 2 = p2 , h[ 4] + h[3]d1 + h[2]d 2 + h[1]d 2 = 0, h[5] + h[ 4]d1 + h[3]d 2 + h[2]d 2 = 0, h[6] + h[5]d1 + h[ 4]d 2 + h[3]d 2 = 0. Writing the last three equations in matrix form we arrive at −1 ⎡ d1 ⎤ ⎡ h[3] h[2] h[1] ⎤ ⎡h[ 4]⎤ ⎡h[ 4]⎤ ⎡ h[3] h[2] h[1] ⎤ ⎡ d1 ⎤ ⎡0⎤ ⎢ h[5]⎥ = ⎢h[ 4] h[3] h[2]⎥ ⎢d 2 ⎥ = ⎢0⎥, and hence, ⎢d 2 ⎥ = − ⎢h[ 4] h[3] h[2]⎥ ⎢ h[5]⎥. ⎢d ⎥ ⎢⎣ h[5] h[ 4] h[3]⎥⎦ ⎢⎣h[6]⎥⎦ ⎢⎣h[6]⎥⎦ ⎢⎣ h[5] h[ 4] h[3]⎥⎦ ⎢ d ⎥ ⎢⎣0⎥⎦ ⎣ 3⎦ ⎣ 3⎦ Substituting these values in the first four equations written in matrix form we get ⎡ p0 ⎤ ⎡h[0] 0 0 0 ⎤⎡ 1 ⎤ ⎢ p1 ⎥ ⎢ h[1] h[0] 0 0 ⎥ ⎢ d1 ⎥ ⎢ p ⎥ = ⎢h[2] h[1] h[0] 0 ⎥ ⎢d ⎥. 2 ⎢ 2⎥ ⎢ ⎢d ⎥ ⎥ [ 3 ] [ 2 ] [ 1 ] [ 0 ] h h h h ⎦⎣ 3 ⎦ ⎣ p3 ⎦ ⎣

2.80

y[ n] = y[ −1] + ∑ ln= 0 x[l] =y[ −1] + ∑ ln= 0 lµ[l] =y[ −1] + ∑ ln= 0 l =y[ −1] + (a) For y[ −1] = 0, y[ n] =

n( n +1) . 2

n( n +1) . 2

(b) For y[ −1] = −2, y[ n] = −2 +

n( n +1) 2

=

n2 +n−4 . 2

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2.81

y(nT ) = y((n − 1)T ) + ∫(nT n −1)T x ( τ)dτ =y(( n − 1)T ) + T ⋅ x (( n − 1)T ). Therefore, the

difference equation representation is given by y[ n] = y[ n − 1] + T ⋅ x[ n − 1] where y[ n] = y( nT ) and x[ n] = x ( nT ). 2.82

y[ n] = ∑ ln=1 x[l] = ∑ ln=−11 x[l] + x[ n], n ≥ 1. Now y[ n − 1] = 1 n

1 n

1 n

1 ∑ n −1 x[l], n ≥ 1, n −1 l =1

i.e,

∑ ln=−11 x[l] = (n − 1) y[ n − 1]. Thus, the difference equation representation is given by 1 n −1 ⎞ y[ n] = ⎛⎜ ⎟ y[ n − 1] + x[ n]. n ⎝ n ⎠

2.83

y[ n] − 0.35 y[ n − 1] = 2.4µ[ n] with y[ −1] = 3. The total solution is given by y[ n] = yc [ n] + y p [ n], where yc [n] is the complementary solution and y p [n] is the

particular solution.

yc [n] is obtained by solving yc [ n] − 0.35 yc [ n − 1] = 0. To this end we set y c [ n] = λn , which yields λn − 0.35λn −1 = 0 resulting in the solution λ = 0.35. Hence y c [ n] = α(0.35) n .

For the particular solution we choose y p [ n] = β. Substituting this solution in the difference equation representing the system we get β − 0.35β = 2.4µ[ n]. For n = 0 we get β − 0.35β = 2.4 , i.e., (1 − 0.35)β = 2.4 and hence β = 2.4 / 0.65 = 48 / 13. Therefore y[ n] = yc [ n] + y p [ n] = α(0.35) n +

y[ −1] = 3 = α(0.35) −1 + y[ n] = 2.84

48 implying 13 48 −0.2423(0.35) n + , n ≥ 0. 13

48 , 13

n ≥ 0. For n = −1, we thus have

α = −0.2423. The total solution is thus given by

y[ n] − 0.3 y[ n − 1] − 0.04 y[ n − 2] = 3 n µ[ n] with y[ −1] = 2 and y[ −2] = 1. The total solution is given by y[ n] = yc [ n] + y p [ n], where yc [n] is the complementary solution and y p [n] is the particular solution.

yc [n] is obtained by solving yc [ n] − 0.3 yc [ n − 1] − 0.04 yc [ n − 2] = 0. To this end we set y c [ n] = λn , which yields λn − 0.3λn −1 − 0.04 λn − 2 = 0 resulting in the solutions

λ = 0.4 or λ = −0.1. Hence y c [ n] = α1 (0.4) n + α 2 ( −0.1) n . For the particular solution we choose y p [ n] = β(3) n . Substituting this solution in the difference equation representing the system we get

β(3) n − 0.3β(3) n −1 − 0.04β(3) n − 2 = 3 n µ[ n]. For n = 0 we have β − 0.3β(3) −1 − 0.04β(3) −2 = 1 which yields β = 1.1166 . Therefore

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34

y[ n] = yc [ n] + y p [ n] = α1 (0.4) n + α 2 (−0.1) n + 1.1166(3) n , n ≥ 0. For n = −1 and

n = −2 we thus have y[ −1] = α1 (0.4) −1 + α 2 (−0.1) −1 + 1.1166(3) −1 = 2 and y[ −2] = α1 (0.4) −1 2 + α 2 (−0.1) −2 + 1.1166(3) −2 = 1. Solving these two equations we get α1 = 0.5489 and α 2 = −0.0255. Hence, y[ n] = 0.5489(0.4) n − 0.0255(−0.1) n + 1.1166(3) n , n ≥ 0. 2.85

y[ n] − 0.3 y[ n − 1] − 0.04 y[ n − 2] = x[ n] + 2 x[ n − 1] with x[ n] = 3 n µ[ n], y[ −1] = 2 and y[ −2] = 1. The total solution is given by y[ n] = yc [ n] + y p [ n], where yc [n] is the

complementary solution and y p [n] is the particular solution. From the solution of Problem 2.84, the complementary solution is of the form y c [ n] = α1 (0.4) n + α 2 ( −0.1) n . To determine y p [n] we observe that the it is given by the sum of the particular solution

y p1[ n] of the difference equation y1[ n] − 0.3 y1[ n − 1] − 0.04 y1[ n − 2] = x[ n] = 3 n µ[ n] and the particular solution y p2 [ n] of the difference equation

y2 [ n] − 0.3 y2 [ n − 1] − 0.04 y2 [ n − 2] = 2 x[ n − 1] = 2 ⋅ 3 n −1 µ[ n − 1]. From the solution of Problem 2.84, we have y p1[ n] = β(3) n . Hence, y p2 [ n] = 2 y p1[ n − 1] = 2β(3) n −1 . Therefore, y p [ n] = y p1[ n] + 2 y p2 [ n] = β(3) n + 2β(3) n −1 = 3 n µ[ n] + 2 ⋅ 3 n −1 µ[ n − 1]. For n = 1 the above equation reduces to 3β + 2β = 3 + 2. Thus, β = 1. Therefore, the total solution is given by

y[ n] = yc [ n] + y p [ n] = α1 (0.4) n + α 2 (−0.1) n + (3) n + 2(3) n −1 , n ≥ 0. For n = −1 and

n = −2 we thus have y[ −1] = α1 (0.4) −1 + α 2 (−0.1) −1 + (3) −1 + 2(3) −2 = 2 and y[ −2] = α1 (0.4) −2 + α 2 (−0.1) −2 + (3) −2 + 2(3) −3 = 1. Solving these two equations we get α1 = 0.4883 and α 2 = −0.0224. Hence, y[ n] = 0.4883(0.4) n − 0.0224(−0.1) n + (3) n + 2(3) n −1 , n ≥ 0. 2.86

h[ n] − 0.35h[ n − 1] = δ[ n]. The solution is given by h[ n] = hc [ n] + h p [ n], where hc [n] is the complementary solution and h p [n] is the particular solution. If h[n] is the impulse response, then h p [ n] = 0. From Problem 2.83 we note that hc [ n] = α(0.35) n . Thus, h[0] − 0.35h[ −1] = h[0] = 1. This implies α = 1. Hence, h[ n] = (0.35) n , n ≥ 0.

2.87

The overall system can be regarded as the cascade of two causal LTI systems: S1: y[n] − 0.3 y[n − 1] − 0.04 y[n − 2] = x1[n] and S2: x1[n] = x[n] + 2 x[n − 1] .

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The impulse response h1[ n] of the system S1 can be found by solving the complementary solution of h1[ n] − 0.3h1[ n − 1] − 0.04h1[ n − 2] = δ[ n] . Let the complementary solution be h1c [ n] = λn , we have λn − 0.3λn −1 − 0.04λn − 2 = 0 hence λ = {0.4, − 0.1} . Therefore, the impulse response h1[ n] is given by h1[ n] = h1c [ n] = A(0.4) n + B( −0.1) n , n ≥ 0 . Solving constants A, B, we get A = 0.8

and B = 0.2. Hence h1[ n] = 0.8(0.4) n + 0.2(−0.1) n , n ≥ 0. . The impulse response h2 [ n] of the system S2 is given by h2 [ n] = δ[ n] + 2δ[ n − 1]. . The impulse response h[n] of the overall system is

(

)

(

)

h[ n] = h1[ n] * h2 [ n] = 0.8(0.4) n + 0.2(−0.1) n µ[ n] + 2 0.8(0.4) n −1 + 0.2(−0.1) n −1 µ[ n − 1] = δ[ n] + 1.92(0.4) 2.88

n −1

µ[ n − 1] + 0.38(−0.1)

n −1

µ[ n − 1].

h[ n] = (−α) n µ[ n], 0 < α < 1. Step response is then given by s[ n] = h[ n] O * µ[ n] ∞ ⎧∑ n (−α) k , n ≥ 0, k = (−α) n µ[ n] O * µ[ n] = ∑ (−α) µ[ k ]µ[ n − k ] = ⎨ k = 0 0, n<0 k = −∞ ⎩ ⎧⎪1− ( − α ) n +1 = ⎨ 1+ α , n ≥ 0, ⎪⎩ 0, n < 0.

2.89

K

K

n +1 = 1. Since there λ i . Now lim n→∞ n 1 + λi A < 1. Hence exists a positive integer N o such that for all n > N o , 0 < n +1 < 2 An A n +1 Let An = n (λ i ) . Then n +1 = An n K

n

∑∞ n =0 An converges. 2.90

{x[ n]} = {− 4, 5, 1, − 2, − 3, 0, 2}, − 3 ≤ n ≤ 3,

{y[ n]} = {6, − 3, − 1, 0, 8, 7, − 2}, − 1 ≤ n ≤ 5, {w[ n]} = {3, 2, 2, − 1, 0, − 2, 5}, 2 ≤ n ≤ 8. (a) rxx [l] = ∑ 3n = −3 x[ n]x[ n − l], − 6 ≤ l ≤ 6. {rxx [l]} = {− 8, 10, 14, − 11, − 23, − 11, 59, − 11, − 23, − 11, 14, 10, − 8}, − 6 ≤ l ≤ 6,

ryy [l] = ∑ 5n = −5 y[ n]y[ n − l], − 6 ≤ l ≤ 6. {ryy [l]} = {− 12, 48, 29, − 31, − 30, 27, 163, 27, − 30, − 31, 29, 48, − 12}, − 6 ≤ l ≤ 6, rww [ l] = ∑ 6n = −6 w[ n]w[ n − l], − 6 ≤ l ≤ 6.

{rww [l]} = {15, 4, 6, − 12, 6, − 2, 47, − 2, 6, − 12, 6, 4, 15}, − 6 ≤ l ≤ 6, Not for sale.

36

(b) rxy [l] = ∑ 3n = −5 x[ n]y[ n − l], − 8 ≤ l ≤ 4.

{rxy [l]} = {8, − 38, 1, 51, 4, − 30, − 68, 43, 31, − 3, − 20, − 6, 12}, − 8 ≤ l ≤ 4, rxw [l] = ∑ 3n = −8 x[ n]y[ n − l], − 11 ≤ l ≤ 1. {rxw [l]} = {− 20, 33, − 5, − 8, − 24, 7, 12, 12, − 7, − 14, − 5, 4, 6}, − 11 ≤ l ≤ 1, 2.91

n n−l ∞ (a) x1[ n] = α n µ[ n]. rx 1 x1 [l] = ∑ ∞ µ[ n − l] n = −∞ x1[ n]x1[ n − l] = ∑ n = −∞ α µ[ n] α

⎧ α −l , l < 0, 2n − l ∞ ⎧ , l < 0, ⎪⎪1− α 2 ⎪∑ n = 0 α 2n − l ∞ = ∑n=0 α µ[ n − l] = ⎨ =⎨ l 2n − l ⎪⎩∑ ∞ , l ≥ 0, ⎪ α , l ≥ 0. n=l α ⎪⎩1− α 2

Note for l ≥ 0, rx1 x1 [l] =

αl 1− α 2

, and for l < 0, rx1 x1 [l] =

in the second expression we get rx1 x1 [ −l] =

α −( −l ) 1− α 2

=

α −l 1− α 2

αl 1− α 2

. Replacing

l with − l

= rx x [l]. Hence, rx x [l] 1 1 1 1

is an even function of l. Maximum value of rx1 x1 [l] occurs at l = 0 since α l is a decaying function for increasing when α < 1.

1, 0 ≤ n ≤ N − 1, Now rx 2 x 2 [l] = ∑ nN=−01x 2 [ n − l], where (b) x 2 [ n] = ⎧⎨ 0 , otherwise. ⎩ for l < −( N − 1), ⎧ 0, ⎪ N + l, for − ( N − 1) ≤ l ≤ 0, ⎪ 1, l ≤ n ≤ N − 1 + l, x 2 [ n − l] = ⎧⎨ Therefore, rx 2 x 2 [l] = ⎨ N , for l = 0, 0 , otherwise. ⎩ ⎪ N − l, for 0 < N − l ≤ N − 1, ⎪⎩ 0, for l > N − 1. It follows from the above that rx 2 x2 [l] is a triangular function of l , and hence is an even function with a maximum value of N at l = 0. 2.92

πn (a) x1[ n] = cos⎛⎜ ⎞⎟ where M is a positive integer. Period of x1[ n] is 2 M , and ⎝M⎠ 1 2 M −1 1 2 M −1 ⎛ πn ⎞ ⎛ π(n + l) ⎞ hence rx1 x1 [l] = ⎟ ∑ x 2 [ n] x 2 [ n + l ] = ∑ cos⎜ ⎟ cos⎜ 2M n =0 2M n =0 ⎝M⎠ ⎝ M ⎠

=

1 2 M −1 ⎛ πn ⎞⎧ ⎛ πn ⎞ ⎛ πl ⎞ ⎛ πn ⎞ ⎛ πl ⎞⎫ ∑ cos⎜ ⎟⎨cos⎜ ⎟ cos⎜ ⎟ − sin⎜ ⎟ sin⎜ ⎟⎬ 2M n =0 ⎝ M ⎠ ⎝ M ⎠⎭ ⎝ M ⎠⎩ ⎝ M ⎠ ⎝ M ⎠

=

1 ⎛ πn ⎞ ⎛ πl ⎞2 M −1 cos⎜ ⎟ ∑ cos 2 ⎜ ⎟. 2M ⎝M⎠ ⎝ M ⎠ n=0

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37

Now 2 M −1

N −1 N −1 ⎧ ⎛ 4 πn ⎞⎫ N 1 N −1 ⎛ 4 πn ⎞ 2 ⎛ πn ⎞ 2 ⎛ 4 πn ⎞ 1 ∑ cos ⎜ ⎟ = ∑ cos ⎜ ⎟ = ∑ ⎨1 + cos⎜ ⎟⎬ = + ∑ cos⎜ ⎟. ⎝ M ⎠ n=0 ⎝ N ⎠ 2 n=0 ⎩ ⎝ N ⎠⎭ 2 2 n = 0 ⎝ N ⎠ n=0 N −1 N −1 ⎛ 4 πn ⎞ N −1 ⎛ 4πn ⎞ j ( 4 πn / N ) Let C = ∑ cos⎜ ⎟ and S = ∑ sin⎜ ⎟. Then C + jS = ∑ e ⎝ N ⎠ ⎝ N ⎠ n=0 n=0 n=0

e j 4π − 1

= e

j 4π / N

−1

= 0. This implies C = 0. Thus

Hence, rx1 x1 [l] =

2 M −1

2 ⎛ πn ⎞ N ∑ cos ⎜ ⎟ = = M . ⎝M⎠ 2 n=0

M ⎛ πl ⎞ ⎛ πl ⎞ 1 cos⎜ ⎟ = cos⎜ ⎟. 2M ⎝M⎠ ⎝M⎠ 2

(b) {x 2 [ n]} = 〈 n〉 6 = {0, 1, 2, 3, 4, 5}, 0 ≤ n ≤ 5. It is a periodic sequence with a 1

period 6. Thus, rx 2 x 2 [l] = ∑ 5n = 0 x 2 [ n]〈 x 2 [ n + l]〉 6 , 0 ≤ l ≤ 5. rx 2 x2 [l] is also a 6 periodic sequence with a period 6.

(

)

1 55 x [0]x 2 [0] + x 2 [1]x 2 [1] + x 2 [2]x 2 [2] + x 2 [3]x 2 [3] + x 2 [ 4]x 2 [ 4] + x 2 [5]x 2 [5] = , 6 2 6 40 1 rx2 x2 [1] = x 2 [0]x 2 [1] + x 2 [1]x 2 [2] + x 2 [2]x 2 [3] + x 2 [3]x 2 [ 4] + x 2 [ 4]x 2 [5] + x 2 [5]x 2 [0] = , 6 6 1 32 rx2 x2 [2] = x 2 [0]x 2 [2] + x 2 [1]x 2 [3] + x 2 [2]x 2 [ 4] + x 2 [3]x 2 [5] + x 2 [ 4]x 2 [0] + x 2 [5]x 2 [1] = , 6 6 28 1 rx2 x2 [3] = x 2 [0]x 2 [3] + x 2 [1]x 2 [ 4] + x 2 [2]x 2 [5] + x 2 [3]x 2 [0] + x 2 [ 4]x 2 [1] + x 2 [5]x 2 [2] = , 6 6 1 31 rx2 x2 [ 4] = x 2 [0]x 2 [ 4] + x 2 [1]x 2 [5] + x 2 [2]x 2 [0] + x 2 [3]x 2 [1] + x 2 [ 4]x 2 [2] + x 2 [3]x 2 [ 4] = , 6 6 40 1 rx2 x2 [5] = x 2 [0]x 2 [5] + x 2 [1]x 2 [0] + x 2 [2]x 2 [1] + x 2 [3]x 2 [2] + x 2 [ 4]x 2 [3] + x 2 [5]x 2 [ 4] = . 6 6

rx 2 x 2 [0] =

(

)

(

)

(

)

(

)

(

)

(c) x 3 [ n] = ( −1) n is a periodic sequence with a period 2. Thus, 1 2 1 2

rx 3 x 3 [l] = ∑1n = 0 x3 [ n]x3 [ n + l], 0 ≤ l ≤ 1. Hence, rx 3 x 3 [0] =

(x3 [0]x3 [0] + x3 [1]x3 [1]) = 1,

rx 3 x 3 [1] =

1 2

(x3 [0]x3 [1] + x3 [1]x3 [0]) = −1.

rx3 x3 [l] is also a periodic sequence with a period 2. M2.1 (a) The input data entered during the execution of Program 2_2.m are: Type in real exponent = -1/12 Type in imaginary exponent = pi/6 Type in gain constant = 1 Type in length of sequence = 41 (b) The input data entered during the execution of Program 2_2.m are: Type in real exponent = -1/12 Type in imaginary exponent = pi/6 Type in gain constant = 1 Type in length of sequence = 41 Not for sale.

38

Real part

Imaginary part

0.5

0 Amplitude

Amplitude

0 -0.5 -1

-0.5 -1 -1.5

-1.5 0

10

20 Time index n

30

40

0

5

10 15 20 Time index n

25

30

M2.2 (a) ~ x a [ n] = e − j 0.4 πn . The plots generated using Program 2_2.m are shown below: Imaginary part

Real part

1

1

0.5 Amplitude

Amplitude

0.5 0

0 -0.5

-0.5

-1

-1 0

10

20 Time index n

30

0

40

10

20 Time index n

30

40

(b) The code fragment used to generate ~ x b [ n] = sin(0.8πn + 0.8π) is as follows: x = sin(0.8*pi*n + 0.8*pi); The plot of the periodic sequence is given below: 1

Amplitude

0.5 0 -0.5 -1

0

10

20 n

30

(

40

) (

)

(c) The code fragment used to generate ~ x c [ n] = Re e jπn / 5 + Im e jπn / 10 is as follows: x = real(exp(i*pi*n/5)+ imag(exp(i*pi*n/10); The plot of the periodic sequence is given below:

Not for sale.

39

2

Amplitude

1 0 -1 -2

0

10

20 n

30

40

(d) The code fragment used to generate ~ x d [ n] = 3 cos(1.3πn) − 4 sin(0.5πn + 0.5π) is as follows: x = 3*cos(1.3*pi*n)-4*sin(0.5*pi*n+0.5*pi); The plot of the periodic sequence is given below: 10

Amplitude

5 0 -5 -10

0

10

20 n

30

40

(e) The code fragment used to generate ~ x e [ n] = 5 cos(1.5πn + 0.75π) + 4 cos(0.6πn) − sin(0.5πn) is as follows: x = 5*cos(1.5*pi*n+0.75*pi)+4*cos(0.6*pi*n)-sin(0.5*pi*n); The plot of the periodic sequence is given below: 10

Amplitude

5 0 -5 -10

0

10

20 n

30

40

M2.3 (a) L = input('Desired length = '); A = input('Amplitude = '); omega = input('Angular frequency = '); phi = input('Phase = '); n = 0:L-1;

Not for sale.

40

x = A*cos(omega*n + phi); stem(n,x); xlabel('Time Index'); ylabel('Amplitude'); title(['\omega_{o} = ',num2str(omega/pi),'\pi']); (b)

ωo = 0.28π 1.5

1

1

0.5

0.5

Amplitude

Amplitude

ωo = 0.6π 1.5

0 -0.5

0 -0.5

-1 0

10

20 Time Index ωo = 0.45π

30

-1.5

40

1.5

1

1

0.5

0.5

0 -0.5 -1 -1.5

0

10

20 Time Index

30

40

ωo = 0.55π

1.5

Amplitude

Amplitude

-1.5

-1

0 -0.5 -1

0

10

20 Time Index

30

-1.5

40

0

10

20 Time Index

30

40

ωo = 0.65π 1.5

Amplitude

1 0.5 0 -0.5 -1 -1.5

0

10

20 Time Index

30

40

M2.4 t = 0:0.001:1; fo = input('Frequency of sinusoid in Hz = '); FT = input('Sampling frequency in Hz = '); g1 = cos(2*pi*fo*t); plot(t,g1,'-'); xlabel('time'); ylabel('Amplitude'); hold n = 0:1:FT; gs = cos(2*pi*fo*n/FT); plot(n/FT,gs,'o'); hold off M2.5 t = 0:0.001:0.85; Not for sale.

41

g1 = cos(6*pi*t); g2 = cos(14*pi*t); g3 = cos(26*pi*t); plot(t/0.85,g1,'-', t/0.85, g2, '--', t/0.85, g3,':'); xlabel('time'); ylabel('Amplitude'); hold n = 0:1:8; gs = cos(0.6*pi*n); plot(n/8.5,gs,'o'); hold off M2.6 As the length of the moving average filter is increased, the output of the filter gets more smoother. However, the delay between the input and the output sequences also increases (This can be seen from the plots generated by Program 2_4.m for various values of the filter length.) M2.7 alpha = input('Alpha = '); y0 = 1; y1 = 0.5*(y0 + (alpha/y0)); while abs(y1-y0)>0.00001 y2 = 0.5*(y1+(alpha/y1)); y0 = y1; y1 = y2; end disp('Squre root of alpha is'); disp(y1); M2.8 format long alpha = input('Alpha = '); y0 = 0.3; y = zeros(1,61); L = length(y) - 1; y(1) = alpha - y0*y0 + y0; n = 2; while abs(y(n-1) - y0) > 0.00001 y2 = alpha - y(n-1)*y(n-1) + y(n-1); y0 = y(n-1); y(n) = y2; n = n+1; end disp('Square root of alpha is');disp(y(n-1)); m = 0:n-2; err = y(1:n-1) - sqrt(alpha); stem(m,err); axis([0 n-2 min(err) max(err)]); xlabel('Time index n'); ylabel('Error'); title(['\alpha = ',num2str(alpha)]); The displayed out is Square root of alpha is 0.84178104293115

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42

α = 0.7086 0.06

Error

0.04 0.02 0 -0.02 -0.04 0

5

10 15 Time index n

20

25

M2.9 {x[ n]} = {− 4, 5, 1, − 2, − 3, 0, 2}, − 3 ≤ n ≤ 3, {y[ n]} = {6, − 3, − 1, 0, 8, 7, − 2}, − 1 ≤ n ≤ 5, {w[ n]} = {3, 2, 2, − 1, 0, − 2, 5}, 2 ≤ n ≤ 8. {rxx [ n]} = {− 8, 10, 14, −11, − 23, −11, 59, −11, − 23, −11, 14, 10, − 8}, − 6 ≤ n ≤ 6. {ryy [ n ]} = {−12, 48, 29, − 31, − 30, − 27, 163, − 27, − 30, − 31, 29, 48, −12}, − 6 ≤ n ≤ 6.

{rww [ n]} = {15, 4, 6, −12, 6, − 2, 47, − 2, 6, −12, 6, 10, 15}, − 6 ≤ n ≤ 6. ryy [n]

rxx [n]

200

60

150 Amplitude

Amplitude

40 20 0

50 0

-20 -40 -6

100

-4

-2

0 Lag index

2

4

-50 -6

6

-4

rww[n]

-2

0 Lag index

2

4

6

rxy [n]

60

60 40 20 Amplitude

Amplitude

40 20

0 -20 -40

0

-60 -20 -6

-4

-2

0 Lag index

2

4

6

Not for sale.

-80

-5

0 Lag index

5

43

rxw[n]

ryw[n]

40

60 40 Amplitude

Amplitude

20 0 -20 -40 -6

20 0 -20

-4

-2

0 Lag index

2

4

-40 -6

6

-4

-2

0 Lag index

2

4

6

M2.10 N = input('Length of sequence = '); n = 0:N-1; x = exp(-0.8*n); y = rand(1,N)-0.5+x; n1 = length(x)-1; r = conv(y,fliplr(y)); k = (-n1):n1; stem(k,r); xlabel('Lag_index'); ylabel('Amplitude'); 3

Amplitude

2 1 0 -1 -30

-20

-10

0 10 Lag index

Not for sale.

20

30

44

Chapter 3 3.1

∞

∞

−∞ ∞

−∞

X a ( jΩ) = ∫ x a (t ) e − jΩt dt. Now, ∫ x a (t ) dt < ∞. Hence, ∞

∞

X a ( jΩ) = ∫ x a (t ) e − jΩt dt ≤ ∫ x a (t ) e − jΩt dt −∞

−∞

∞

3.2

≤ ∫ x a (t ) dt < ∞.

1 −∞

−∞

(

)

(a) Ya ( jΩ) = ∫ cos(Ω o t ) e − jΩt dt = ∫ e jΩot + e − jΩot e − jΩt dt 2 −∞ −∞ =

1 2

(δ(Ω − Ωo ) + δ(Ω + Ωo )) . ∞ −α t − jΩt

(b) U a ( jΩ) = ∫ e

=

1 α − jΩ

−∞

[

e

∞

0

dt = ∫ e αt e − jΩt dt + ∫ e − αt e − jΩt dt −∞

]

0 ⋅ e ( α − jΩ ) t − ∞ −

1 α + jΩ

0 0 ⋅ e − ( α + jΩ ) t − ∞ =

[

∞

∞

−∞

−∞

]

1 1 2α + = . 2 α − j Ω α + jΩ α + Ω 2

(c) Va ( jΩ) = ∫ e jΩ o t e − jΩt dt = ∫ e − j (Ω − Ω o )t dt = δ(Ω − Ω o ). ∞ ⎛ ∞ ∞ ⎛ ∞ ⎞ ⎞ (d) Pa ( jΩ) = ∫ ⎜⎜ ∑ δ(t − lT ) ⎟⎟ e − jΩt dt = ∑ ⎜⎜ ∫ δ(t − lT )e − jΩt dt ⎟⎟ using the l = −∞ ⎝ − ∞ ⎠ − ∞ ⎝ l = −∞ ⎠ linearity property of the CTFT. Next, using the shifting property of the CTFT we get

∞

Pa ( jΩ) = ∑ e − jΩlT which can be alternately expressed in the form l = −∞

Pa ( jΩ) =

2π ∞ 2 πl ) making use of the results of Problem 3.2(c). ∑ δ(Ω − T l = −∞ T ∞

3.3

(a) Va ( jΩ) = ∫ e − jΩt dt = δ(Ω). (b)

µ ( jΩ ) =

−∞ ∞

∞ ⎛ t ⎞ − jΩt 1 − j Ωt µ = ( ) + πδ(Ω) . t e dt dt = ∫ ∫ ⎜⎜ ∫ δ(τ) dτ ⎟⎟ e Ω j −∞ −∞ ⎝ −∞ ⎠ ∞

(c) The function x a (t ) is also denoted by rect (t ). Thus, X a ( jΩ) = ∫ x a (t ) e − jΩt dt

=

1/ 2

− jΩt dt = − ∫ e

−1 / 2

(

)

−∞

1 − jΩ / 2 2 sin(Ω / 2) sin(Ω / 2) − e jΩ / 2 = = e = sinc(Ω / 2). Ω jΩ Ω/2

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⎧1 − 2 t , t < 0.5, (d) y a (t ) = ⎨ A more convenient way to determine the CTFT of y a (t ) ⎩ 0, t ≥ 0.5. is to differentiate it twice with respect to t , determine the CTFT of

d 2 y a (t ) dt 2

and then

make use of the time-differentiation property given in Problem 3.6(e) and time-shifting dy (t ) ⎧ 2, t < 0.5, dy ( t ) property given in Problen 3.6(a). Now, a = ⎨ As a has jump dt dt t − 2 , ≥ 0 . 5 . ⎩

discontinuities with a positive jump of value 2 at t = ±0.5, a negative jump of value − 4 at t = 0, and zero everywhere else,

d 2 y a (t ) dt 2

and an impulse of strength − 4 at t = 0, i.e.,

has only impulses of strength 2 at t = ±0.5, d 2 y a (t ) dt 2

= 2δ(t + 0.5) − 4δ(t ) + 2δ(t − 0.5). If

Ya ( jΩ) denotes the CTFT of y a (t ) , then using of the time-differentiation property we d 2 y a ( t ) CTFT

have

dt

↔ ( jΩ) 2 Ya ( jΩ). Using the time-shifting property, we arrive at the

2

d 2 y a (t )

CTFT of

dt

2

given by 2 e jΩ / 2 − 4 + 2 e − jΩ / 2 . Therefore ( jΩ) 2 Ya ( jΩ) =

− Ω 2Ya ( jΩ) = 2 e jΩ / 2 − 4 + 2 e − jΩ / 2 = 4(cos(Ω / 2) − 1) , i.e,

Ya ( jΩ) = −

3.4

h(t ) =

4 Ω

1 σ 2π

2

(cos(Ω / 2) − 1) =

1 Ω Ω sin 2 ⎛⎜ ⎞⎟ = sinc 2 ⎛⎜ ⎞⎟. ⎝4⎠ 2 ⎝4⎠ Ω 8

2

∞ 2 2 ⋅ e − (t − µ) / 2σ . Thus, H ( jΩ) = ∫

1

− ∞ σ 2π

Making a change of variable t − µ = τ we get H ( jΩ) = =

1 σ 2π

=e e

3.5

−(

∞

1

−∞

σ 2π

2 2 e − jΩµ ∫ e − τ / 2 σ e − jΩτ dτ =

2

2 2 ⋅ e − (t − µ ) / 2σ e − jΩt dt.

1

∞

∫e

− τ 2 / 2 σ 2 − jΩ ( τ + µ )

σ 2π − ∞

e

2 2 ⋅ e − j Ωµ ⋅ σ 2 π ⋅ e − σ Ω / 2

2

σ Ω + jΩµ ) 2 .

For a zero mean impulse response, we then have the CTFT pair

− t / 2 σ CTFT 2

x a (t ) =

2

−σ2Ω2 / 2 . ↔ σ 2 πe

(

)

1 1 jΩt 1 sin( t ) 1 ∞ jΩt . e jΩ − e − jΩ = ∫ X ( jΩ ) e dΩ = ∫ e dΩ = 2 π −1 πt 2π − ∞ j 2 πt

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dτ

∞

3.6

(a)

∫ x a (t − t o ) e

− j Ωt

−∞

∞

dt = ∫ x a ( τ) e − jΩ( τ + t o ) dτ obtained using a change of variable −∞

∞

∞

−∞

−∞

t − t o = τ. Therefore ∫ x a (t − t o ) e − jΩt dt = e − jΩt o ∫ x a (τ) e − jΩτ dτ = e − jΩt o X a ( jΩ). ∞

(b)

∫ x a (t ) e

j Ω o t − j Ωt

e

−∞

(c) x a (t ) =

∞

dt = ∫ x a (t ) e − j (Ω − Ω o )t dt = X ( j (Ω − Ω o ) ). −∞

∞

∞ 1 j Ωt − j Ωt dΩ. ∫ X a ( jΩ) e dΩ. Therefore 2 π x a ( −t ) = ∫ X a ( jΩ) e 2π − ∞ −∞ ∞

Interchanging t and Ω we get 2 π x a (−Ω) = ∫ X a ( jt ) e − jΩt dt. −∞

(d) For a positive real constant a the CTFT of x a (at ) is given by ∞

− jΩ t dt = ∫ x a ( at ) e

−∞

1 ∞ − j (Ω / a )τ dτ ∫ x a ( τ) e a −∞

=

1 a

Xa ( j

Ω ). a

In a similar manner we Ω

1

can show that for a negative constant a the CTFT of x a (at ) is given by − X a ( j ). a a CTFT 1 Ω X a ⎛⎜ j ⎞⎟. Therefore x a (at ) ↔ a ⎝ a⎠

(e) Differentiating both sides of x a (t ) =

1 ∞ j Ωt ∫ X a ( jΩ) e dΩ get 2 π −∞

dx a (t ) CTFT dx a (t ) 1 ∞ j Ωt = ↔ jΩ X a ( jΩ). ∫ jΩ X a ( jΩ) e dΩ. Therefore dt dt 2π − ∞ ∞

3.7

X a ( jΩ) = ∫ x a (t ) e − jΩt dt = X a ( jΩ) e jθ a (Ω) , where θ a (Ω) = arg{X a ( jΩ)}. Thus, −∞ ∞

X a ( − jΩ) = ∫ x a (t ) e jΩt dt. If x a (t ) is a real function of then it follows from the −∞

definition of X a ( jΩ) and the expression for X a (− jΩ) that X a ( jΩ) and X a (− jΩ) are complex conjugates. Therefore X a (− jΩ) = X a ( jΩ) and θ a (−Ω) = −θ a (Ω). Or in other words, for a real , the magnitude spectrum X a ( jΩ) is an even function of Ω and the phase spectrum θa (Ω) is an odd function of Ω. 3.8

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∞

3.9

xˆ (t ) = ∫ hHT (t − τ)x ( τ)dτ. where hHT (t ) is the impulse response of the Hilbert −∞

transformer. Taking the CTFT of both sides we get Xˆ ( jΩ) = H HT ( jΩ) X ( jΩ) where Xˆ ( jΩ ) and H ( jΩ) denote the CTFTs of xˆ (t ) and h (t ), respectively. Rewriting HT

(

HT

)

Xˆ ( jΩ) = H HT ( jΩ) X p ( jΩ) + X n ( jΩ) = − j X p ( jΩ) + j X n ( jΩ). As the magnitude and

phase of Xˆ ( jΩ ) are an even and odd function, xˆ (t ) is seen to be real signal. Consider the complex signal y(t ) = x (t ) + j xˆ (t ) . Its CTFT is then given by Y ( jΩ) = X ( jΩ) + j Xˆ ( jΩ) = 2 X ( jΩ). p

3.10

[

]

∞ ∞ 2 ∞ 1 1 The total energy ε x = ∫ e − αt dt = ∫ e − 2αt dt = e − 2 αt 0 = = 1. − 2α 2 α α =1 / 2 −∞ −∞ The total energy can also be computed using using the Parsevals’ theorem

εx=

1 ∞ 1 dΩ. ∫ 2 2 2π −∞ α +Ω

Ωc

1

1

dΩ Therefore, the 80% bandwidth Ω c can be found by evaluating ∫ 2 2 2π α + Ω −Ωc Ω 1 ⎡1 −1 ⎛ Ω ⎞ ⎤ c = ⋅ tan ⎜ ⎟ 2 π ⎢⎣ α ⎝ α ⎠⎥⎦ − Ω

=

2 tan −1 (2Ω c ) = 0.8 . π

= c

1 ⎡ −1 ⎛ Ω c ⎞⎟ − tan −1 ⎛⎜ − Ω c ⎢ tan ⎜⎜ ⎟ ⎜ α 2 πα ⎣ ⎝ ⎝ α ⎠

⎞⎤ 1 −1 ⎛ Ω c ⎟⎟ ⎥ = πα ⋅ tan ⎜⎜ α ⎝ ⎠⎦

⎞ ⎟⎟ ⎠ α =1 / 2

1 0.8 π ⎞ Therefore, Ω c = ⋅ tan⎛⎜ ⎟ = 1.5388. 2 ⎝ 2 ⎠ 1

1

1

3.11 y[ n] = µ[ n] = y ev [ n] + y od [ n], where yev [ n] = ( y[ n] + y[ − n]) = + δ[ n] and 2 2 2 1 2

1 2

1 2

1 2

yod [ n] = ( y[ n] − y[ −n]) = (µ[ n] − µ[ − n]) = µ[ n] − − δ[ n]. Now,

Yev (e jω ) =

∞ ∞ ⎤ 1 1⎡ 1 2 ( 2 k ) + = π δ(ω + 2πk ) + . Since π δ ω + π ∑ ∑ ⎥ ⎢ 2 ⎣ k = −∞ 2 k = −∞ ⎦ 2 1

1

1

1

y od [ n] = µ[ n] − − δ[ n], yod [ n − 1] = µ[ n − 1] − − δ[ n − 1]. As a result, 2 2 2 2 1 2

1 2

yod [ n] − y od [ n − 1] = µ[ n] − µ[ n − 1] − δ[ n − 1] + δ[ n] =

1 2

(δ[ n] + δ[ n − 1]).

Taking the DTFT of both sides of the above equation, we get 1 ⎛ 1 + e − jω ⎞ 1 1 1 Yod (e jω ) − e − jωYod (e jω ) = 1 + e − jω or Yod (e jω ) = ⎜ − . ⎟= − j ω − j ω 2 ⎝ 1− e 2 2 ⎠ 1− e

(

)

Hence, Y (e jω ) = Yev (e jω ) + Yod (e jω ) =

1 1 − e − jω

∞

+ π ∑ δ(ω + 2 πk ). k = −∞

∞

3.12 The inverse DTFT of X (e jω ) = ∑ 2 πδ(ω + 2 πk ) is given by k = −∞

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x[ n] =

1 π 2π jωn = 1. ∫ 2πδ(ω) e dω = 2π − π 2π ∞

3.13 Y (e jω ) = ∑ α e − jωn with α < 1. Rewriting we get n

n = −∞ −1

∞

n = −∞

n=0

∞

∞

n =1

n=0

Y (e jω ) = ∑ α − n e − jωn + ∑ α n e − jωn = ∑ (α e jω ) n + ∑ (α e − jω ) n =

αe

jω

1 − α e jω

+

1 1 − α e − jω

=

1− α

2

1 − 2α cos ω + α 2

.

∞ ⎛ sin(ωc n) ⎞ − jωn 3.14 G(e jω ) = ∑ ⎜ δ[ n] − = 1 − H LP (e jω ). ⎟e n π ⎠ n = −∞ ⎝

G(e jω ) 1 _π

3.15 x[ n] =

_ω

c

0

ωc

π

ω

1 π 1 π j ω j ωn jω − jωn dω. ∫ X (e ) e dω. Hence, x * [ n] = ∫ X * (e ) e 2π − π 2π − π

(a) Since x[n] is real and even, we have X (e jω ) = X * (e jω ). Thus

x[ −n] = x[ n] =

1 π j ω − j ωn dω. Therefore, ∫ X (e ) e 2π − π

1 1 π jω jω − jω ( x[ n] + x[ −n]) = ). ∫ X (e ) cos(ωn)dω. As x[n] is even, X (e ) = X (e 2 2π − π

As a result, the term X (e jω ) cos n(ωn) inside the above integral is even, and hence

x[ n] =

1π jω ∫ X (e ) cos(ωn)dω. π0

(b) Since x[n] is real and odd, we have x[ n] = − x[ −n] and X (e jω ) = − X (e − jω ). Thus,

x[ n] =

j π 1 jω jω ( x[ n] − x[ −n]) = ∫ X (e ) sin(ωn)dω. As a result, the term X (e ) sin(ωn) 2 2π − π

inside the above integral is even, and hence x[ n] =

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j π jω ∫ X (e ) sin(ωn)dω. 2π 0

49

⎛ jω 0 n e jφ − e − jω 0 n e − jφ ⎞ ⎟µ[ n] 3.16 x[ n] = Aα sin(ω0 n + φ)µ[ n] = Aα ⎜ ⎟ 2 j ⎝ ⎠ n⎜ e

n

)

(

)

(

n n A A jφ e α e jω 0 µ[ n] − e − jφ α e − jω 0 µ[ n]. Therefore, the DTFT of x[n] is given 2j 2j 1 1 A A jφ by X (e jω ) = − e − jφ . e j ω j j ω j2 j2 1− αe e 0 1 − α e ω e − jω 0

=

3.17 Let x[ n] = α n µ[ n1] with α < 1. Its DTFT was computed in Example 3.6 and is given by X ( e jω ) =

1 1 − α e − jω

. ∞

(a) x1[ n] = α n µ[ n − 1] with α < 1. Its DTFT is given by X1 (e jω ) = ∑ α n e − jωn n =1

∞

∞

= ∑ ( α e − jω ) n = ∑ ( α e − jω ) n − 1 = n =1

n=0

1 1− αe

− jω

−1 =

αe

− jω

1 − α e − jω

.

(b) x 2 [ n] = nα n µ[ n] with α < 1. Note x 2 [ n] = n x[ n] . Therefore, using the differentiation-in-frequency property in Table 3.4 we get ⎞ dX (e jω ) d ⎛⎜ α e − jω 1 ⎟= X 2 ( e jω ) = j = j . dω dω ⎜⎝ 1 − α e − jω ⎟⎠ (1 − α e − jω ) 2 (c) x 3 [ n] = α n µ[ n + 1] with α < 1. Its DTFT is given by X 3 (e

jω

∞

n − jωn

) = ∑α e n = −1

−1 jω

=α e

∞

n − jωn

+ ∑α e n=0

−1 jω

=α e

1 ⎛⎜ e jω − α + = α ⎜⎝ 1 − α e − jω 1 − α e − jω 1

(d) x 4 [ n] = nα n µ[ n + 2] with α < 1. Its DTFT is given by ∞

∞

X 4 (e jω ) = ∑ nα ne − jωn = ∑ nα ne − jωn − 2α − 2e j 2ω − α −1e jω. From the results of Part n = −2

n =0

(b) we observe that X 4 ( e jω ) =

∞

α e − jω

n=0

(1 − α e − jω ) 2

n − jω n = ∑ nα e

α e − jω (1 − α e

− jω 2

. Hence,

− 2 α − 2 e j 2 ω − α − 1e jω .

)

(e) x 5 [ n] = nα n µ[ − n − 1] with α > 1. Its DTFT is given by

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⎞ ⎟. ⎟ ⎠

−1

∞

∞

1

n = −∞

m =1

m =0

1 − α − 1e j ω

X 5 (e jω ) = ∑ α n e − jωn = ∑ α − m e jωm = ∑ α − m e jωm − 1 =

−1 =

e jω α − e jω

.

⎧⎪ n (f) x 6 [ n] = ⎨α , n ≤ M , Its DTFT is given by ⎪⎩ 0, otherwise. −1

M

X 6 ( e j ω ) = ∑ α n e − j ωn + n=0

∑α

− n − jω n

e

n =−M

=

1 − α M +1e − jω( M +1) 1 − α e − jω

+ α M e j ωM ⋅

1 − α M +1e − jω( M +1) 1 − α e − jω

3.18 (a) x a [ n] = µ[ n] − µ[ n − 5]. Let denote µ (e jω ) the DTFT of µ[n]. Using the timeshifting property of the DTFT given in Table 3.4, the DTFT of x a [n] is thus given by X a (e jω ) = (1 − e − j 5ω ) µ (e jω ) . From Table 3.3, we have

µ ( e jω ) =

∞ 1 − e − j 5ω . + ∑ πδ(ω + 2πk ). Therefore, X a (e jω ) = 1 − e jω k = −∞ 1 − e jω

1

(b) x b [ n] = α n (µ[ n] − µ[ n − 8]). Let x[ n] = α n µ[ n1] with α < 1. Its DTFT was 1 computed in Example 3.6 and is given by X (e jω ) = . Now 1 − α e − jω x b [ n] = x[ n] − x[ n − 8]. Using the time-shifting property of the DTFT given in Table 3.4, the DTFT of x a [n] is thus given by X b (e jω ) = (1 − e − j 8ω ) X (e jω ) =

1 − e − j 8ω 1 − α e − jω

.

(c) x c [ n] = ( n + 1)α n µ[ n] = n α n µ[ n] + α n µ[ n] with α > 1. We can rewrite it as x c [ n] = x 2 [ n] + x[ n]. The DTFT of x2 [ n] was computed in Problem 3.17(b) and is given by X 2 (e jω ) =

α e − jω (1 − α e

and is given by X (e jω ) = =

3.19

αe

− jω

(1 − α e

− jω 2

+

)

− jω 2

1− αe

)

1 1− αe

1 − jω

and the DTFT of x[n] was computed in Example 3.6

=

− jω

. Therefore X c (e jω ) = X 2 (e jω ) + X (e jω ) 1

(1 − α e − jω ) 2

.

⎛ 1 − e − jω(2 N +1) ⎞ N 1, − N ≤ n ≤ N , ⎟ Then Y1 (e jω ) = ∑ e − jωn = e − jωN ⎜ (a) y1[ n] = ⎧⎨ j − ω 0 , otherwise. ⎜ ⎟ ⎩ n = −N ⎝ 1− e ⎠ 1 sin⎛⎜ ω[ N + ] ⎞⎟ 2 ⎠ ⎝ = . sin(ω / 2)

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.

− jω( N +1)

N 1− e 1, 0 ≤ n ≤ N , (b) y2 [ n] = ⎧⎨ Then Y2 (e jω ) = ∑ e − jωn = ⎩0, otherwise. n =0 1 − e − jω

⎛ sin(ω[ N + 1] / 2) ⎞ = e − jωN / 2 ⎜⎜ ⎟⎟ . ⎝ sin(ω / 2) ⎠ n ⎧⎪ (c) y3 [ n] = ⎨1 − N , − N ≤ n ≤ N , Assume N ⎪⎩ 0, otherwise. ⎧⎪1, − 1 y 3 [ n] = y 0 [ n] O * y 0 [ n] where y 0 [ n] = ⎨ N ⎪⎩0,

Y3 (e jω ) =

1 1 Y0 (e jω )Y0 (e jω ) = Y02 (e jω ). N N

to be odd. Then we can express N −1 2

≤n≤

N −1 , 2

Therefore,

otherwise.

Now, from the results of Part (a), we have

sin(ωN / 2 ) 1 sin 2 (ωN / 2 ) . Hence, Y3 (e jω ) = ⋅ . N sin 2 (ω / 2) sin(ω / 2) Note: The above result also holds for N even. Y0 (e jω ) =

⎧N + 1 − n , (d) y 4 [ n] = ⎨ 0, ⎩ considered in Part (a) and

− N ≤ n ≤ N, = y1[ n] + Ny3 [ n], where y1[ n] is the sequence otherwise, y3 [ n] is the sequence considered in Part (c). Hence,

1 sin⎛⎜ ω[ N + ]⎞⎟ 2 2 ⎠ sin (ωN / 2 ) ⎝ + . Y4 (e jω ) = Y1 (e jω ) + N ⋅ Y3 (e jω ) = sin(ω / 2) sin 2 (ω / 2)

cos(πn / 2 N ), − N ≤ n ≤ N , Then (e) y5 [ n] = ⎧⎨ 0, otherwise. ⎩ 1 N − j ( πn / 2 N ) − jωn 1 N j ( πn / 2 N ) − jωn + Y5 (e jω ) = e e ∑e ∑e 2 n=−N 2 n = −N ⎛

π

⎞

⎛

π

⎞

(

π

1

)

(

π

1

)

1 N − j ⎜⎝ ω − 2 N ⎟⎠ n 1 N − j ⎜⎝ ω + 2 N ⎟⎠ n 1 sin (ω − 2 N )( N + 2 ) 1 sin (ω + 2 N )( N + 2 ) = + = ⋅ + ⋅ . ∑e ∑e 2 n = −N 2 n = −N 2 sin (ω − π ) / 2 2 sin (ω + π ) / 2 2N 2N 3.20

(

)

(

)

(n + m − 1)! n α µ[ n] with α < 1. We shall prove by induction that the n! (m − 1)! 1 DTFT of x m [n] is given by X m (e jω ) = . From Table 3.3, it follows that (1 − α e − jω ) m it holds for m = 1. Let m = 2. Then (n + 1)! n x 2 [ n] = α µ[ n] = (n + 1) x1[ n] = nx1[ n] + x1[ n]. Therefore, n! Denote x m [ n] =

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X 2 ( e jω ) =

α e − jω (1 − α e − jω ) 2

+ 1 − α e − jω =

1 (1 − α e − jω ) 2

, and it also holds for m = 2.

Now, assume that it holds for m. Consider next x m +1[ n] =

(n + m)! n α µ[ n] n! (m)!

1 ⎛n+m⎞ ⎛ n + m ⎞ (n + m − 1)! n α µ[ n] = ⎜ =⎜ ⎟ x m [ n] = ⋅ n ⋅ x m [ n] + x m [ n]. Hence, ⎟ m ⎝ m ⎠ ⎝ m ⎠ n! (m − 1)! ⎞ α e − jω 1 d ⎛⎜ 1 1 1 ⎟+ = + X m + 1 ( e jω ) = j m dω ⎜⎝ (1 − α e − jω ) m ⎟⎠ (1 − α e − jω ) m (1 − α e − jω ) m +1 (1 − α e − jω ) m 1 . = (1 − α e − jω ) m +1 ∞

3.21 (a) X a (e jω ) = ∑ δ(ω + 2 πk ). Hence, x a [ n] = k = −∞

(b) X b (e jω ) =

(

e jω 1 − e jωN 1 − e jω

) = e jω N∑−1e jωn . n =0

1 π jωn ∫ δ(ω) e dω = 1. 2π − π − N +1

Let m = −n. X b (e jω ) = e jω ∑ e − jωm . m =0

− N +1

Consider the DTFT X (e jω ) = ∑ e − jωm . Its inverse is given by m =0

1, − ( N − 1) ≤ n ≤ 0, x[ n] = ⎧⎨ Therefore, by the time-shifting property of the DTFT, the otherwise. ⎩0, 1, − N ≤ n ≤ −1, inverse DTFT of X b (e jω ) = e jω X (e jω ) is given by x b [ n] = x[ n + 1] = ⎧⎨ ⎩0, otherwise. ⎧3, n = 0, N N ⎪ (c) X c (e jω ) = 1 + 2 ∑ cos(ωl) = 2 + ∑ e − jωl . Hence, x c [ n] = ⎨1, 0 < n ≤ N , l=0 l = −N ⎪⎩0, otherwise. (d) X d (e jω ) =

X d ( e jω ) =

− α e − jω (1 − α e − j

jω 2

)

with α < 1. We can rewrite X d (e jω ) as

dX o (e jω ) 1 where X o (e jω ) = . From Table 3.3, the inverse dω 1 − α e − jω

DTFT of X d (e jω ) is given by x o [ n] = α n µ[ n]. From Table 3.4, using the differentiation-in-frequency property the inverse DTFT of X d (e jω ) is thus given by x d [ n] = n x o [ n] = n α n µ[ n].

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e j 4ω − e − j 4ω 2j

3.22 (a) H a (e jω ) = sin(4ω) =

1 j 4ω e 2j

=

ha [ n] = {− j 0.5, 0, 0, 0, 0, 0, 0, 0, (b) H b (e jω ) = cos(4ω) =

e j 4ω + e − j 4ω 2

e j 5ω − e − j 5 ω 2j

1 − j 4ω e . Therefore, 2j

j 0.5}, − 4 ≤ n ≤ 4.

= 0.5e j 4ω + 0.5e − j 4ω . Therefore,

hb [ n] = { 0.5, 0, 0, 0, 0, 0, 0, 0, (c) H c (e jω ) = sin(5ω) =

−

=

0.5}, − 4 ≤ n ≤ 4.

1 j 5ω e 2j

−

1 − j 5ω e . Therefore, 2j

hc [ n] = {− j 0.5, 0, 0, 0, 0, 0, 0, 0, 0, 0, (d) H d (e jω ) = cos(5ω) =

e j 5 ω + e − j 5ω 2

j 0.5}, − 5 ≤ n ≤ 5.

= 0.5e j 5ω + 0.5e − j 5ω . Therefore,

hd [ n] = { 0.5, 0, 0, 0, 0, 0, 0, 0, 0, 0,

0.5}, − 5 ≤ n ≤ 5.

⎛ e jω + e − jω ⎞ ⎛ e j 2 ω + e − j 2 ω ⎞ 3.23 (a) H1 (e jω ) = 1 + 2 cos(ω) + 3 cos(2ω) = 1 + 2⎜ ⎟ + 3⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠ = 1 + e jω + e − jω + 1.5e j 2ω + 1.5e − j 2 ω . Therefore, {h1[ n]} = {1.5, 1, 1, 1, 1.5}, − 2 ≤ n ≤ 2. ω (b) H 2 (e jω ) = (3 + 2 cos(ω) + 4 cos(2ω)) cos⎛⎜ ⎞⎟ e − jω / 2 ⎝2⎠

⎡ ⎛ e jω + e − jω ⎞ ⎛ e j 2 ω + e − j 2 ω ⎞⎤ ⎛ e jω / 2 + e − jω / 2 ⎞ − jω / 2 = ⎢3 + 2⎜ ⎟e ⎟ + 4⎜ ⎟⎥ ⎜ 2 2 2 ⎠ ⎝ ⎠ ⎝ ⎠⎦ ⎝ ⎣ 1 = 3 + e jω + e − jω + 2 e j 2 ω + 2 e − j 2 ω 1 + e − jω 2

(

)(

)

= 2 + 1.5 e jω + 2 e − jω + e j 2ω + 1.5 e − j 2ω + e − j 3ω . Hence, {h2 [ n]} = {1, 1.5, 2, 2, 1.5, 1}, − 2 ≤ n ≤ 3. (c) H 3 (e jω ) = j [3 + 4 cos( ω) + 2 cos( 2ω)]sin( ω)

⎡ ⎛ e jω + e − jω ⎞ ⎛ e j 2 ω + e − j 2 ω ⎞⎤ ⎛ e jω − e − jω ⎞ = j ⎢3 + 4⎜ ⎟ ⎟ + 2⎜ ⎟⎥ ⎜ 2 2 2j ⎠ ⎝ ⎠ ⎝ ⎠⎦ ⎝ ⎣ =

1 2

(3 + 2 e jω + 2 e − jω + e j2ω + e − j2ω )(e jω − e − jω )

= 3 + 2 e jω + 2 e − jω + 0.5 e j 2ω + 0.5 e − j 2ω . Hence, {hc [ n]} = {0.5, 2, 3, 2, 0.5}, − 2 ≤ n ≤ 2.

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54

(d) H 4 (e jω ) = j[4 + 2 cos(ω) + 3 cos(2ω)]sin(ω / 2) e jω / 2

⎡ ⎛ e jω + e − jω ⎞ ⎛ e j 2 ω + e − j 2 ω ⎞⎤ ⎛ e jω / 2 − e − jω / 2 ⎞ − jω / 2 = j ⎢4 + 2⎜ ⎟e ⎟ + 3⎜ ⎟⎥ ⎜ 2 2 2j ⎠ ⎝ ⎠ ⎝ ⎠⎦ ⎝ ⎣ =

1 2

(4 + e jω + e − jω + 1.5 e j2ω + 1.5 e − j2ω )(1 − e − jω )

= 1.5 − 0.25 e jω − 1.5 e − jω + 1.5 e j 2ω − 0.5 e − j 2ω − 0.75 e − j 3ω . Hence, {h4 [ n]} = {− 1.5, − 0.5, − 3, 3, − 0.5, − 3}, − 3 ≤ n ≤ 2. 3.24 Let H (e jω ) and G(e jω ) denote the DTFTs of the sequences h[n] and g[n] , respectively. (a) Linearity Theorem: F {α h[ n ] + β g [ n ]} = ∞

∞

n = −∞

n = −∞

∞

∑ ( α h[ n ] + β g [ n ]) e

− jω n

n = −∞

= α ∑ h[ n] e − jωn + β ∑ g[ n] e − jωn = α H (e jω ) + β G(e jω ). ∞

(b) Time-reversal Theorem:

∑ h[ −n] e

n = −∞ ∞

(c) Time-shifting Theorem:

= e − jωn o

∞

∑ h[ m] e

m = −∞

− jωm

∑ h[ n − no ] e

∞

∑ h[ m] e

jωm

m = −∞

=

∞

∑ h[ m] e

= H (e − jω ).

− jω( m + n o )

m = −∞

= e − jωn o H (e jω ). ∞

= H (e

=

− jωn

n = −∞

(d) Frequency-shifting Theorem: j ( ω − ωo )

− jωn

(

)

∞

jω n − jωn = ∑ h[ n] e − j (ω − ω o ) n ∑ e o h[ n] e

n = −∞

n = −∞

).

3.25 Let H 1(e jω ) = F {h1[ n]}and H 2 (e jω ) = F {h2 [ n]}. From Example 3.8 we have ∞ ⎛ sin(ω n) ⎞ 2 H 2 ( e jω ) = ∑ ⎜ ⎟ πn ⎠ n = −∞ ⎝

⎧1, 0 ≤ ω ≤ ω2 , =⎨ From the result of Problem 3.14 we get ⎩0, ω2 < ω ≤ π. ∞ ⎛ sin(ω1n) ⎞ − jωn ⎧0, 0 ≤ ω ≤ ω1 , H1 (e jω ) = ∑ ⎜ δ[ n] − =⎨ ⎟e πn ⎠ n = −∞ ⎝ ⎩1, ω1 < ω ≤ π. As the impulse response of the cascade is given by h[ n] = h1[ n] O * h2 [ n] , using the convolution theorem we obtain the DTFT of the cascade: H (e jω ) = H 1(e jω )H 2 (e jω ) Not for sale

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⎧0, 0 ≤ ω < ω1 , ⎪ = ⎨1, ω1 ≤ ω ≤ ω2 , ⎪0, ω < ω ≤ π. 2 ⎩ H(e jω ) 1 _ π _ ω _ ω1 2

0

ω1

ω2 π ω

(

∞ ) n = −∞ ∞ ∞ ∞ Y (e jω ) = ∑ y[ n] e − jωn = X ((e jω ) 4 ) = ∑ x[ n] (e − jωn ) 4 = ∑ x[ m / 4] e − jωm . n = −∞ n = −∞ m = −∞

3.26 Y (e jω ) = X (e j 4ω ) = X (e jω )4 . Now, X (e jω ) = ∑ x[ n] e − jωn . Hence,

x[ n], n = 0, ± 4, ± 8, ± 16,K Therefore y[ n] = ⎧⎨ otherwise. ⎩ 0, ∞

∞

3.27 X (e jω ) = ∑ x[ n] e − jωn . Therefore, X (e jω / 2 ) = ∑ x[ n] e − j (ω / 2)n and n = −∞

n = −∞

∞

X (−e jω / 2 ) = ∑ x[ n] (−1) n e − j (ω / 2)n . Thus, we can write n = −∞

Y (e jω ) =

∞

∑ y[ n] e − jωn =

n = −∞

1 2

{X (e jω / 2 ) + X (−e jω / 2 )}= 12 ∑∞ (x[n] + x[n](−1)n )e − j(ω / 2)n . n = −∞

∞

⎧ ⎪ x[ n] e − j (ω / 2)n , for n even, x[2n], for n even, Thus, y[ n] = ⎨ ∑ Hence, y[ n] = ⎧⎨ n = −∞ for n odd. ⎩ 0, ⎪ 0, for n odd. ⎩ 3.28 F {x * [ −n]} = 3.29 X ca (e jω ) =

1 2

∞

∑ x * [ − n] e

n = −∞

[X (e

jω

− jωn

∞

∞

n = −∞

n = −∞

= ∑ x * [ n] e jωn = ( ∑ x[ n] e − jωn )* = X *(e jω ) .

]

) − X * ( e − jω ) =

∞ ∞ ∞ ⎞ 1⎛ ∞ = ⎜⎜ ∑ x re [ n] e − jωn + j ∑ x im [ n] e − jωn − ∑ x re [ n] e − jωn − j ∑ x im [ n] e − jωn ⎟⎟ 2 n = −∞ n = −∞ n = −∞ n = −∞ ⎠ ⎝ ∞

ω, = j ∑ x im [ n] e − jωn = F {j xim [n]}. n = −∞

3.30

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|X(e jω )|

0

0.6π

1.4π

2π

3.31 From Table 3.2 we observe that an even real-valued sequence has a real-valued DTFT and an odd real-valued sequence has an imaginary-valued DTFT. (a) Since x1[ n] is an odd sequence, it has an imaginary-valued DTFT. (b) Since x 2 [ n] is an even sequence, it has a real-valued DTFT.

sin(−ωc n) − sin(ωc n) sin(ωc n) = = = x3 [ n]. Since, x3 [ n] is an even − πn − πn πn sequence, it has a real-valued DTFT.

(c) x3 [ −n] =

(d) Since x 4 [ n] is an odd sequence, it has an imaginary-valued DTFT. (e) Since x 5 [ n] is an odd sequence, it has an imaginary-valued DTFT. 3.32 From Table 3.2 we observe that an even real-valued sequence has a real-valued DTFT and an odd real-valued sequence has an imaginary-valued DTFT. (a) Since Y1 (e jω ) is a real-valued function of ω, its inverse is an even sequence. (b) Since Y2 (e jω ) is an imaginary-valued function of ω, its inverse is an odd sequence. (c) Since Y3 (e jω ) is an imaginary-valued function of ω, its inverse is an odd sequence. 3.33 (a) Since H1 (e jω ) is a real-valued function of ω, its inverse is an even sequence. (b) Since H 2 (e jω ) is a real-valued function of ω, its inverse is an even sequence. 3.34 Let u[ n] = x * [ −n], and let X (e jω ) and U (e jω ) denote the DTFTs of x[n] and u[n], respectively. From the convolution property of the DTFT given in Table 3.4, the DTFT jω jω jω jω of y[ n] = x[ n] O * u[ n] is given by Y (e ) = X (e )U (e ). From Table 3.1, U (e )

= X * (e jω ). Therefore, Y (e jω ) = X (e jω ) X * (e jω ) = X (e jω )

2

which is a real-valued

function of ω. 3.35 From the frequency-shifting property of the DTFT given in Table 3.4,

F {x[ n] e − jπn / 3} = X (e j (ω + π / 3) ). A sketch of this DTFT is shown below Not for sale

57

X(e j(ω+π/3) )

_π

n

3.36 {−α µ[ − n − 1]} = X (e

jω

_ 2π _ π 3 3 −1

=

jω

∞

2

Now, 4 ∫ X (e jω ) dω = 2 ∫ X (e jω ) dω = 4π ∑ n

(b) X (e jω )

2

n = −∞

−π n

0 ∞

∞

⎛ e jω ⎞ ⎟ . ∑ ⎜ n=0 ⎝ α ⎠

2

1 1− α e

= −α e

n

∞

.

⇒ X (e jω ) =

1 2

1+ α − 2α cos(ω)

.

Hence α = −2. Therefore, x[ n] = (−2) n µ[ −n − 1].

π

2

/ α)

=

− 1 jω

2 ∞ 1 π 2 jω ∫ X (e ) dω = ∑ x[ n] . 2π − π n = −∞

1 . 5 + 4 cos(ω)

π

1− (e

jω

e

n =1

1

ω

− n jω n

= − ∑α

n = −∞

From Parseval’s relation, 2

∞

n − j ωn

) = ∑−α e

For α > 1, X (e jω ) = −α −1e jω

(a) X (e jω )

π

0

2

−1

x[ n] = 4π ∑ (−2) n

2

n = −∞

4π 1 1 = 4 π ∑ ⎛⎜ ⎞⎟ = π ∑ ⎛⎜ ⎞⎟ = . 3 n =1 ⎝ 4 ⎠ n=0 ⎝ 4 ⎠

π

=

1 . 3.25 − 3 cos(ω)

2

Now, ∫ X (e jω ) dω = 0 ∞

2 2 ∞ −1 1 π 2 jω n ∫ X (e ) dω = π ∑ x[ n] = π ∑ (1.5) 2 −π n = −∞ n = −∞

∞

n

Hence α = 1.5. Therefore, x[ n] = −(1.5) n µ[ −n − 1].

n

4π 9 4π 4π 4 4 = π ∑ ⎛⎜ ⎞⎟ = ⋅ = . ∑ ⎛⎜ ⎞⎟ = 9 5 5 9 9 9 n =1 ⎝ ⎠ n=0 ⎝ ⎠

(c) Using the differentiation-in-frequency property of the DTFT as given in Table 3.4, − jω ⎞ d ⎛ 1 ⎟ = αe ⎜ the inverse DTFT of X (e jω ) = j is x[ n] = −nα n µ[ −n − 1]. ⎟ ⎜ − j ω 2 − j ω dω ⎝ 1 − α e ) ⎠ (1− α e Hence, the inverse DTFT of 2

Y ( e jω ) = π

1 ( 5 − 4 e − jω ) 2

2

.

1 (1− α e − jω ) 2

is − (n + 1)α n µ[ −n − 1].

Hence α = 2. Therefore, y[ n] = −(n + 1)2 n µ[ −n − 1] . π

2

∞

Now, 4 ∫ X (e jω ) dω = 2 ∫ X (e jω ) dω = 4π ∑ 0

n = −∞

−π

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2

−1

x[ n] = 4π ∑ (n + 1) 2 ⋅ 2 2 n n = −∞

58

∞

n

9/4 1 = π ∑ ⎛⎜ ⎞⎟ ⋅ n 2 = π = 4 π. 9 / 16 n=0 ⎝ 4 ⎠ 2 ∞ 6 2 2 dX (e jω ) dω = 2π ∑ n x[ n] = 2π ∑ n x[ n] = 152π. (Using Parseval’s relation with ∫ dω −∞ −3 −π

π

3.37

differentiation-in-frequency property) ∞

6

n = −∞ ∞

n = −2

3.38 (a) X (e j 0 ) = ∑ x[ n] = ∑ x[ n] = 10. 6

(b) X (e jπ ) = ∑ x[ n] e jπn = ∑ x[ n] e jπn = −6. n = −∞

π

∫ X (e

(c)

jω

n = −2

)dω = 2πx[0] = −2π.

−π π

∞

2

jω ∫ X (e ) dω = 2 π ∑

(d)

n = −∞

−π

6

2

2

x[ n] = 2π ∑ x[ n] = 120π. n = −2

π dX (e jω ) 2 6 ∞ 2 2 dω = 2 π ∑ n x[n] = 2π ∑ n x[n] = 1724π. (e) ∫ dω −∞ −2 −π ∞

2

n = −∞ ∞

n = −6

3.39 (a) X (e j 0 ) = ∑ x[ n] = ∑ x[ n] = 12. 2

(b) X (e jπ ) = ∑ x[ n] e jπn = ∑ x[ n] e jπn = −12. n = −∞

π

∫ X (e

(c)

jω

n = −6

)dω = 2πx[0] = −4π.

−π π

∞

2

jω ∫ X (e ) dω = 2 π ∑

(d)

n = −∞

−π π

(e)

∫

−π

dX (e jω ) dω

2

∞

2

2

2

x[ n] = 2π ∑ x[ n] = 160π. n = −6

2

2

2

dω = 2π ∑ n x[ n] = 2π ∑ n x[ n] = 136π. −∞

−6

3.40 From the differentiation-in-frequency property of the DTFT given in Table 3.4 we ∞

have ∞

∑ n x[ n] e

n = −∞

∑ n x[ n] = j

n = −∞

− jωn

dX (e jω ) dω

=j

dX (e jω ) where X (e jω ) = F {x[ n]}. Therefore, dω ∞

. From the definition of the DTFT X (e jω ) = ∑ x[ n] e − jωn n = −∞

ω= 0

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j

∞

it follows that

j0 ∑ x[ n] = X (e ). Therefore, C g =

n = −∞

From Table 3.3, X (e jω ) = F {α n µ[ n]} = j

dX (e jω ) dω

= ω= 0

α e − jω (1− α e − jω ) 2

= ω= 0

1 1 − α e − jω

α (1 − α ) 2

dX (e jω ) dω

X (e

ω= 0

j0

.

)

, α < 1. As a result,

and X (e j 0 ) =

1 . 1− α

α

Hence, C g = . 1− α

3.41 Let G1 (e jω ) = F {g1[ n]}. (b) Note g 2 [ n] = g1 [ n] + g1 [ n − 4]. Hence, F {g 2 [ n]} = G2 (e jω ) = G1 (e jω ) + e − j 4ωG1 (e jω ) = 1 + e − j 4ω G1 (e jω ).

(

)

(c) Note g 3 [ n] = g1 [ −(n − 3)] + g1 [ n − 4]. Now, F {g1[ −n]} = G1 (e − jω ). Hence,

F {g3 [ n]} = G3 (e jω ) = e − j 3ωG1 (e − jω ) + e − j 4ωG1 (e jω ). (d) Note g 4 [ n] = g1 [ n] + g1 [ −(n − 7)]. Hence, F {g 4 [ n]} = G 4 (e jω )

= G1 (e jω ) + e − j 7ωG1 (e − jω ). ∞ ⎞ ⎞⎛ ∞ ⎛ ∞ 3.42 Y (e jω ) = X1 (e jω ) X 2 (e jω ), i.e., ∑ y[ n] e − jωn = ⎜⎜ ∑ x1[ n] e − jωn ⎟⎟⎜⎜ ∑ x 2 [ n] e − jωn ⎟⎟. n = −∞ ⎠ ⎠⎝ n = −∞ ⎝ n = −∞ ∞ ∞ ∞ ⎞ ⎞⎛ ⎛ (a) Setting ω = 0 in the above we get ∑ y[ n] = ⎜⎜ ∑ x1[ n] ⎟⎟⎜⎜ ∑ x 2 [ n] ⎟⎟. n = −∞ ⎠ ⎠⎝ n = −∞ ⎝ n = −∞ ∞ ⎛ ∞ ⎞⎛ ∞ ⎞ (b) Setting ω = π we get ∑ (−1) n y[ n] = ⎜⎜ ∑ (−1) n x1[ n] ⎟⎟⎜⎜ ∑ (−1) n x2 [ n] ⎟⎟. n = −∞ ⎝ n = −∞ ⎠⎝ n = −∞ ⎠

3.43 x[ n] = α n µ[ n], α < 1. From Table 3.3, F {x[ n]} = X (e jω ) = of x[n] is E x =

1 π 1 ∫ − jω 2π − π 1− α e

80% bandwidth of the signal, we

2

∞

dω = ∑ ( α 2 ) n = n=0

1 set E x,80 = 2π

ωc

∫

1 1 − α e jω

. The total energy

4 . To determine the 1 − α 2 α =1 / 2 3 1

=

2 1

− jω − ω c 1− α e

dω = 0.8 E x = 0.8 ⋅

4 3

⎡ 1 ωc ⎤ 3.2 1 dω⎥ = and solve for ω c , i.e., set E x,80 = ⎢ . ∫ ⎢⎣ 2 π − ω c (1− α cos ω) 2 + α 2 sin 2 ω ⎥⎦ α =1 / 2 3 A numerical solution of the above equation yields ωc = 0.5081π.

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1 2

3.44 Recall x[ n] = x ev [ n] + x od [ n], where x ev [ n] = ( x[ n] + x[ −n]) and As x[n] is causal,

x[ n] = 0 for n < 0 and x[ −n] = 0 for n > 0. Hence, there is no overlap between the nonzero portions of x[n] and x[ −n] except at n = 0, and we have x[ n] = 2 x ev [ n]µ[ n] − x ev [0]δ[ n] and x[ n] = 2 x od [ n]µ[ n] + x od [0]δ[ n]. Moreover, since

x[n] is real, it follows from Table 3.2 that F {x ev [ n]} = X re (e jω ) and

F {x ev [ n]} = j X im (e jω ). Taking the DTFT of x[ n] = 2 x ev [ n]µ[ n] − x ev [0]δ[ n] we arrive at

X ( e jω ) =

1 π jν j (ω − ν ) )dν − x[0]. ∫ X re (e ) µ (e π −π

From Table 3.3 we have ∞ ∞ ω ⎞ 1⎛ ( 2 k ) 1 j cot( ) + πδ(ω + 2πk ). + πδ ω + π = − ∑ ∑ ⎟ ⎜ 2⎝ 2 ⎠ k = −∞ 1− e − jω k = −∞ Substituting the above in the equation preceding it we get

µ (e jω ) = F {µ[ n]} = X (e jω ) =

1

1 π ⎡1 X re (e jν )⎢ ⎛⎜1 − ∫ π ⎣2 ⎝ −π ∞

ω − ν ⎞ ⎞⎤ j cot⎛⎜ ⎟ ⎟ dν ⎝ 2 ⎠ ⎠⎥⎦ π

+ ∑

∫ X re (e

jν

k = −∞ − π

=

) δ((ω − ν) + 2 πk ) dν − x[0]

j π 1 π ω− ν ⎞ jν jν ⎟dν + X re (e jω ) − x[0]. ∫ X re (e ) dν − ∫ X re (e ) cot⎛⎜ ⎝ 2 ⎠ 2π − π 2π − π

Comparing the last equation with X (e jω ) = X re (e jω ) + j X im (e jω ), we arrive at 1 π ω− ν ⎞ jν X im (e jω ) = − ⎟dν. ∫ X re (e ) cot⎛⎜ ⎝ 2 ⎠ 2π − π Likewise, taking the DTFT of x[ n] = 2 x od [ n]µ[ n] + x od [0]δ[ n] we get

X ( e jω ) =

j π jν j (ω − ν ) )dν + x[0]. ∫ X im (e ) µ (e π −π jω

Substituting the expression for µ (e

X ( e jω ) =

j π ⎡1 X im (e jν )⎢ ⎛⎜1 − ∫ π ⎣2 ⎝ −π ∞

+j ∑

) given earlier in the above equation we get

ω − ν ⎞ ⎞⎤ j cot⎛⎜ ⎟ ⎟ dν ⎝ 2 ⎠ ⎠⎥⎦ π

∫ X im (e

jν

k = −∞ − π 1 π

) δ((ω − ν) + 2πk ) dν + x[0]

j π ω− ν ⎞ jν jν ⎟dν + jX im (e jω ) + x[0]. ∫ X im (e ) dν + 2 π ∫ X im (e ) cot⎛⎜ 2π 2 ⎠ ⎝ −π −π Comparing the last equation with X (e jω ) = X re (e jω ) + j X im (e jω ), we arrive at

=

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X re (e jω ) =

1 π ω− ν ⎞ jν ⎟dν + x[0]. ∫ X im (e ) cot⎛⎜ ⎝ 2 ⎠ 2π − π

3.45 If u[ n] = z n is the input to the LTI discrete-time system, then its output is given by ∞

∞

∞

k = −∞ ∞

k = −∞

k = −∞

y[ n] = ∑ h[ k ] u[ n − k ] = ∑ h[ k ] z n − k = z n ∑ h[ k ] z − k = z n H ( z). where H ( z) = ∑ h[ k ] z − k . Hence is an eigenfunction of the system. k = −∞

n

If v[ n] = z µ[ n] is the input to the LTI discrete-time system, then its output is given by ∞

∞

n

k = −∞

k = −∞

k = −∞

y[ n] = ∑ h[ k ] v[ n − k ] = z n ∑ h[ k ] µ[ n − k ] z n − k = z n ∑ h[ k ] z − k . Since in this case the summation depends upon is not an eigenfunction of the system. 3.46 F {h1[ n]} = H1 (e jω ) = F {δ[ n] + δ[ n − 1]} = 1 + 0.5e − jω , 2 1

F {h2 [ n]} = H 2 (e jω ) = F {1 δ[ n] − 1 δ[ n − 1]} = 0.5 − 0.25e − jω , 2

4

jω

F {h3 [ n]} = H 3 (e ) = F {2δ[ n]} = 2, n

F {h4 [ n]} = H 4 (e jω ) = F {−2⎛⎜ 1 ⎞⎟ µ[ n]} =

−2

. ⎝2⎠ 1 − 0.5e − jω The overall frequency response of the structure of Figure 2.35 is given by H ( e jω ) = H 1 ( e jω ) + H 2 ( e jω ) H 3 ( e jω ) + H 2 ( e jω ) H 4 ( e jω )

= 1 + 0.5e − jω + 2(0.5 − 0.25e − jω ) −

2(0.5 − 0.25e − jω ) 1 − 0.5e − jω

= 1.

3.46 Denote H i (e jω ) = F {h1[ n]}, 1 ≤ i ≤ 5. (a) The overall frequency of Figure P2.2(a) is then given by H i (e jω ) = H1 (e jω ) H 2 (e jω ) + H 3 (e jω ) H 4 (e jω ) + H1 (e jω )H 2 (e jω ) H 3 (e jω ) H 5 (e jω ). (b) The structure of Figure P2.2(b) can be redrawn as shown below

h 4[n] ho[n]

h3[n]

+

where the block with an impulse response represents the part of Figure P2.2(b) with a feedback loop as shown below

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w[n]

u[n]

+

h1[n]

h2[n]

v[n]

h5[n] Let U (e jω ) = F {u[ n]}, V (e jω ) = F {v[ n]}, and W (e jω ) = F {w[ n]}. Then we have W (e jω ) = U (e jω ) + H 3 (e jω )V (e jω ) and V (e jω ) = H 1(e jω )H 2 (e jω )W (e jω ).

Eliminating W (e jω ) from these two equations we get

(1 −H 1(e jω )H 2(e jω )H 5(e jω ))V (e jω ) =H 1(e jω )H 2(e jω )U (e jω )

which leads to the frequency response of the feedback structure given by

H o ( e jω ) =

V ( e jω )

=

H 1(e jω )H 2 (e jω )

U (e jω ) 1 − H 1(e jω )H 2 (e jω )H 5(e jω ) The overall frequency of Figure P2.2(a) is thus given by H ( e jω ) = H 4 ( e jω ) + H o ( e jω ) H 3 ( e jω ) = H 4 ( e jω ) +

H 1(e jω )H 2 (e jω )

1 − H 1(e jω )H 2 (e jω )H 5(e jω )

3.48 F {h1[ n]} = H1 (e jω ) = F {2δ[ n − 2] − 3δ[ n + 1]} = 2e − j 2ω − 3e jω , F {h2 [ n]} = H 2 (e jω ) = F {δ[ n − 1] + 2δ[ n + 2]} = e − jω + 2e j 2ω ,

F {h3 [ n]} = H 3 (e jω ) = F {5δ[ n − 5] + 7δ[ n − 3] + 2δ[ n − 1] − δ[ n] + 3δ[ n + 1]} = 5e − j 5ω + 7e − j 3ω + 2e − jω − 1 + 3e jω . The overall frequency of Figure P2.3 is given by H (e jω ) = H 3(e jω ) + H 1(e jω ) H 2 (e jω ) = 5e − j 5ω + 7e − j 3ω + 2e − jω − 1 + 3e jω

(

)(

)

+ 2e − j 2ω − 3e jω e − jω + 2e j 2ω = 5e − j 5ω + 9e − j 3ω + 2e − jω + 3e jω − 6e j 3ω . 3.49 Now, hev [n] is the inverse DTFT of H re (e jω ). Rewriting we get ⎛ e jω + e − jω ⎞ ⎛ e j 2ω + e − j 2ω ⎞ ⎛ e j 3ω + e − j 3ω ⎞ ⎟ + 3⎜ ⎟ + 4⎜ ⎟ H re (e jω ) = 1 + 2⎜⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 2 2 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = 1 + e jω + e − jω + 1.5e j 2 ω + 1.5e − j 2ω + 2e j 3ω + 2e − j 3ω . Its inverse DTFT is hev [ n] = δ[ n] + δ[ n + 1] + δ[ n − 1] + 1.5δ[ n + 2] + 1.5δ[ n − 2] + 2δ[ n + 3] + 2δ[ n − 3].

Since h[n] is real and causal, and its DTFT H (e jω ) exists, it is also absolutely summable. Hence, we can reconstruct h[n] from hev [n] as

h[ n] = 2hev [ n]µ[ n] − hev [0]δ[ n] = 2(δ[ n] + δ[ n − 1] + 1.5δ[ n − 2] + 2δ[ n − 3]) − δ[ n] = δ[ n] + 2δ[ n − 1] + 3δ[ n − 2] + 4δ[ n − 3].

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63

.

π π 1 1 πn 3.50 (a) y a [ n] = sin⎛⎜ ⎞⎟ cos(ωo n) = sin⎛⎜ (ωo + )n ⎞⎟ − sin⎛⎜ (ωo − )n ⎞⎟. Hence, the angular 2 3 ⎠ 2 3 ⎠ ⎝ 3⎠ ⎝ ⎝ π

frequencies present in the output are (ωo ± )n. 3 1 1 (b) yb [ n] = cos3 (ωo n) = cos 2 (ωo n) cos(ωo n) = ⎡ + cos(2ωo n)⎤ cos(ωo n) ⎢⎣ 2 2 ⎥⎦

= cos(ωo n) + cos(2ωo n) cos(ωo n) = cos(ωo n) + [cos(3ωo n) + cos(ωo n)] 2 2 2 4 1

1

3

1

1

1

= cos(ωo n) + cos(3ωo n). Hence, the angular frequencies present in the output are 4 4 3ωo and ωo . (c) yc [ n] = cos(3ωo n). Hence, the angular frequency present in the output is 3ωo . 3.51 F {δ[ n] − αδ[ n − R]} = H (e jω ) = 1 − α e − jωR . Let α = α e jφ . Then the maximum value of H (e jω ) is 1 + α and the minimum value is 1 − α . There are R peaks of H (e jω ) located at ω = 2 πk / R, 0 ≤ k ≤ R − 1, and R dips located at ω = (2 k + 1)π / R, 0 ≤ k ≤ R − 1 in the frequency range 0 ≤ ω < 2π. 2 Phase, in radians

2

Magnitude

1.5 1 0.5 0

0

0.5

M −1

1 /

1.5

3.52 G(e jω ) = ∑ α n e − jωn = n=0 j0

to have G(e

K=

1− α 1− α M

0 -1 -2

2

1 − α M e − jωM 1 − αe

1

− jω

0

0.5

1 /

1.5

2

. Note G(e jω ) = H (e jω ) for α = 1. In order

) = 1, the impulse response should be multiplied by a factor K , where

.

3.53 H (e jω ) = [a1 cos(2ω) + ( a 2 + a3 ) cos(ω) + a 3 ] + j [− a1 sin(2ω) + ( a 2 − a 3 ) sin(ω)]. The frequency response will have zero-phase for a 2 = a3 and a1 = 0 . 3.54 H (e jω ) = a1 + a 2 e − jω + a 3 e − j 2ω + a 4 e − j 3ω + a 5 e − j 4 ω

(

)

(

)

= a1e j 2 ω + a 5 e − j 2ω e − j 2 ω + a 2 e jω + a 4 e − jω e − j 2ω + a 3 e − j 2ω .

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64

If a1 = a 5 and a 2 = a 4 , then we can rewrite the above equation as

H (e jω ) = [2 a1 cos(2ω) + 2 a 2 cos(ω) + a3 ]e − j 2 ω which is seen to have a linear phase.

3.55 H (e jω ) = a1e jkω + a 2 e j ( k −1)ω + a 2 e − j ( k − 2)ω + a1e − j ( k − 3)ω

= e jkω (a1 + a 2 e − jω + a 2 e − j 2ω + a1e − j 3ω ) 3 2

j ( k − )ω

[

]

=e a1 (e j 3ω / 2 + e − j 3ω / 2 ) + a 2 (e jω / 2 + e − jω / 2 ) . Hence, H (e jω ) will be a real function of ω if k = 3 / 2, in which case we have

H (e jω ) = a1 (e j 3ω / 2 + e − j 3ω / 2 ) + a 2 (e jω / 2 + e − jω / 2 ). 3.56 F {aδ[ n] + b δ[ n − 1] + δ[ n − 2]} = H1 (e jω ) = a + b e − jω + e − j 2ω , 1 1 F {c n µ[ n]} = H 2 (e jω ) = , and F {d n µ[ n]} = H 3 (e jω ) = . − jω 1− ce 1 − d e − jω The overall frequency response is then H (e jω ) = H1 (e jω ) H 2 (e jω ) H 3 (e jω ) =

a + b e − jω + e − j 2 ω (1 − c e − jω )(1 − d e − jω )

H ( e jω ) = =

2

=

. Therefore,

a + b e − jω + e − j 2 ω

a + b e jω + e j 2 ω

(1 − c e − jω )(1 − d e − jω ) (1 − c e jω )(1 − d e jω )

(a 2 + b 2 + 1) + 2b(a + 1) cos(ω) + 2 a cos(2ω)

(1 − 2c cos(ω) + c 2 )(1 − 2d cos(ω) + d 2 )

(a 2 + b 2 + 1) + 2b(a + 1) cos(ω) + 2 a cos(2ω) (1 + c 2 + d 2 + c 2 d 2 + 2cd ) − 2(c + d )(1 + cd ) cos(ω) + 2cd cos(2ω)

. Hence,

2 H (e jω ) = 1, if a 2 + b 2 + 1 = 1 + c 2 + d 2 + c 2 d 2 + 2cd,

b( a + 1) cos( ω) = −(c + d )(1 + cd ), and a = cd. Substituting a = cd in the equation on the left we get b = −(c + d ). α

3.57 Y (e jω ) = X (e jω ) e j arg X (e

jω )

. Therefore, H (e jω ) =

Y ( e jω ) X ( e jω )

= X ( e jω )

α −1

.

Since H (e jω ) is real function of ω, it has zero-phase. 3.58 y[ n] = x[ n] − α y[ n − R ]. Taking the DTFT of both sides we get

Y (e jω ) = X (e jω ) − α e − jωR Y (e jω ). Hence, H (e jω ) =

Not for sale

Y ( e jω ) X ( e jω )

=

1 1+ α e − jωR

.

65

The maximum value of H (e jω ) is

1 1− α

and the minimum value is

1 . 1+ α

There are

peaks and dips in the range 0 ≤ ω < 2π. The locations of the peaks and the dips are given α by 1 − α e − jωR = 1 ± α or e − jωR = ± . The locations of the peaks are given by α

ω = ωk =

2 πk R

and the locations of the dips are given by ω = ω k =

(2 π +1)k , 0 ≤ k ≤ R − 1. R

Plots of the magnitude and the phase responses of H (e jω ) for α = 0.8 and R = 5 are shown below: 5

1 Phase, in radians

Magnitude

4 3 2

0.5 0 -0.5

1 0

0.5

1 /

1.5

In this case the maximum value is

3.59 A(e

=

jω

)=

b0 + b1e − jω 1 + a1e − jω

-1

2

1 1 − 0 .8

0

0.5

1 /

= 5 and the minimum value is

1.5

2

1 1+ 0.8

= 0.5556.

2

. Thus, we set A(e jω ) = A(e jω ) A * (e jω ) = A(e jω ) A(e − jω )

(b0 + b1e − jω )(b0 + b1e jω ) b02 + b12 + 2b0 b1 cos(ω) = = 1. (1 + a1e − jω )(1 + a1e jω ) 1 + a12 + 2a1 cos(ω)

Solution #1: b0 = ±1 and b1 = sgn(b0 )a1 . In which case A(e

jω

)=±

1 + a1e − jω

= 1, a

1 + a1e − jω

trivial solution. Solution #2: b1 = ±1 and b0 = sgn(b1 )a1 . In which case A(e

3.60 A(e

jω

)=

b0 + b1e − jω 1 + a1e − jω

=

B0 e jφ 0 + B1e jφ1 e − jω 1 + A1e jθ e − jω

jω

)=±

a1 + e − jω

1 + a1e − jω

.

. Thus, we set

2 B e jφ 0 + B1e jφ1 e − jω B0 e − jφ 0 + B1e − jφ1 e jω A(e jω ) = A(e jω ) A * (e jω ) = 0 ⋅ 1 + A1e jθ e − jω 1 + A1e − jθ e jω

=

B02 + B12 + 2 B0 B1 cos(ω − φ1 + φ 0 ) 1 + A12 + 2 A1 cos(ω − θ)

= 1.

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66

Solution #1: B0 = ±1, B1 = sgn( B0 ) A1 , φ1 − φ 0 = θ. In which case

A(e jω ) =

⎛ 1 + A e jθ e − jω ⎞ ± e jφ 0 ± A1 e j (φ 0 + θ) e − jω 1 ⎟ = ±e jφ 0 implying = ±e jφ 0 ⎜ jθ − jω jθ − jω ⎟ ⎜ 1 + A1 e e ⎝ 1 + A1 e e ⎠

φ 0 = ± π. A trivial solution. Solution #2: B1 = ±1, B0 = sgn( B1 ) A1 , φ1 − φ 0 = θ. In which case

± A1e j (φ1 − θ) ± e − jω

⎛ A e − jθ + e − jω ⎞ ⎟ = ±e jφ1 implying φ = ± π. = ± e j φ1 ⎜ 1 1 jθ − jω jθ − jω ⎟ ⎜ 1 + A1 e e ⎝ 1 + A1 e e ⎠ ⎛ a * + e − jω ⎞ ⎟. Hence, a non-trivia solution is A(e jω ) = e jπ ⎜ 1 ⎜ 1 + a e − jω ⎟ 1 ⎝ ⎠ A(e jω ) =

cos(ω / 2) cos(ω) ω 3.61 (a) H a (e jω ) = cosec(ω) = cot⎛⎜ ⎞⎟ − cot(ω) = − ⎝2⎠ sin(ω / 2) sin(ω) =

j ( e jω / 2 + e − jω / 2 ) ( e jω / 2 − e − jω / 2 )

=j

−

j (e jω + e − jω ) ( e jω − e − jω )

(e jω / 2 + e − jω / 2 )(e jω − e − jω ) − (e jω / 2 − e − jω / 2 )(e jω + e − jω )

(e jω / 2 − e − jω / 2 )(e jω − e − jω ) ⎞ ⎛ e − jω − e − j 2 ω ⎟. Therefore, the input-output relation is given by = j 2⎜ ⎜ 1 − e − jω − e − j 4 ω + e − j 3ω ⎟ ⎠ ⎝ y[ n] − y[ n − 1] − y[ n − 2] + y[ n − 3] = j 2 x[ n − 1] − j 2 x[ n − 2].

1 2 2 e − jω . Therefore, the input-output = = cos(ω) e jω + e − jω 1 + e − j 2ω relation is given by y[ n] + y[ n − 2] = 2 x[ n − 1]. (b) H b (e jω ) = sec(ω) =

(c) H c (e jω ) = cot(ω) = =

j (1+ e − j 2 ω ) 1− e − j 2 ω

.

⎛ e jω + e − jω ⎞ j ( e jω + e − jω ) cos(ω) j e − j ω ( e j ω + e − jω ) ⎟⎟ = = = j ⎜⎜ sin(ω) e jω − e − jω 1− e − j 2 ω ⎝ e jω − e − jω ⎠

Therefore, the input-output relation is given by

y[ n] − y[ n − 2] = j x[ n] + j x[ n − 2].

sin(ω / 2) − j + j e − jω 1 − e − jω ω e j ω / 2 − e − jω / 2 (d) H d (e jω ) = tan⎛⎜ ⎞⎟ = = = . == ⎝ 2 ⎠ cos(ω / 2) j (e jω / 2 + e − jω / 2 ) j (1+ e − jω ) 1 + e − jω Therefore, the input-output relation is given by y[ n] + y[ n − 1] = − j x[ n] + j x[ n − 1].

3.62 From Eq. (2.20), the input-output relation of a factor-of- L up-sampler is given by x[ n / L ], n = 0, ± L, ± 2 L, ± 3L,K y[ n] = ⎧⎨ The DTFT of y[n] is thus given by otherwise. ⎩ 0,

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67

∞

∞

n = −∞

n = −∞ n = mL

Y (e jω ) = ∑ y[ n] e − jωn = ∑ x[ n / L ] e − jωn =

∞

∑ x[ m ] e

− jωmL

m = −∞

= X (e jLω ), where

X (e jω ) = F {x[ n]}. 3.63 G(e jω ) =

X ( e jω ) =

1 1− αe 1

− jLω − jω

, α < 1. Thus, we can write G(e jω ) = X (e jLω ), where

. From Table 3.3, the inverse DTFT of X (e jω ) is x[ n] = α n µ[ n].

1− αe Hence, from the results of Problem 3.62, it follows that x[ n / L ], n = 0, ± L, ± 2 L, ± 3L,K g[ n] = ⎧⎨ otherwise. ⎩ 0, 3.64 From Table 3.3, H (e jω ) =

1

. Thus, H (e jω ) =

1

and 1.25 − cos(ω) 1 − 0.5 e − jω ⎛ − 0.5 sin(ω) ⎞ ⎟⎟. H (e ± jπ / 5 ) = 1.3504 m j 0.6664. Therefore arg{H (e jω )} = θ(ω) = tan −1 ⎜⎜ 1 0 . 5 cos( ) − ω ⎝ ⎠ H (e ± jπ / 5 ) = 1.5059 and θ( ± π / 5) = m 0.4585 radians.

Now, for an input x[ n] = sin(ωo n)µ[ n], the steady-state output is given by y[ n] = H (e jω o ) sin (ωo n + θ(ωo ) ). For ωo = π / 5, the steady-state output is therefore π π π given by y[ n] = H (e jπ / 5 ) sin⎛⎜ n + θ( ) ⎞⎟ = 1.5059 sin sin⎛⎜ n − 0.4585 ⎞⎟. 5 5 ⎠ ⎝5 ⎠ ⎝

3.65 H (e jω ) = h[0] + h[1] e − jω + h[0] e − j 2ω = h[0](1 + e − j 2ω ) + h[1] e − jω

= e − jω (2h[0] cos(ω) + h[1]). We require H (e j 0.3 ) = 2h[ 0] cos(0.3) + h[1] = 0 and H (e j 0.6 ) = 2h[ 0] cos(0.6) + h[1] = 1. Solving these two equations we get

h[0] = 3.8461 and h[1] = −6.3487. 3.66 H (e jω ) = h[0](1 + e − j 2ω ) + h[1] e − jω = e − jω (2h[0] cos(ω) + h[1]). We require H (e j 0.3 ) = 2h[ 0] cos(0.3) + h[1] = 1 and H (e j 0.6 ) = 2h[ 0] cos(0.6) + h[1] = 0. Solving

these two equations we get h[0] = −3.8461 and h[1] = 7.3487. 3.67 3.68 H (e jω ) = h[0](1 + e − j 4ω ) + h[1] (e − jω + e − j 3ω ) + h[2] e − j 2ω

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68

= e − j 2ω (2h[0] cos(2ω) + 2h[1] cos(ω) + h[2]). We require H (e j 0.2 ) = 2 cos(0.4)h[ 0] + 2 cos(0.2)h[1] + h[ 2] = 0, H (e j 0.5 ) = 2 cos(1.0)h[ 0] + 2 cos(0.5)h[1] + h[ 2] = 1, H (e j 0.8 ) = 2 cos(1.6)h[ 0] + 2 cos(0.8)h[1] + h[ 2] = 0. Solving these three equations we

get h[ 0] = −13.4866, h[1] = 45.228, h[ 2] = −63.8089, i.e., {h[ n]} = {− 13.4866, 45.228, − 63.8089}, 0 ≤ n ≤ 2. 3.69 H (e jω ) = h[0] + h[1] e − jω + h[2] e − j 2ω + h[3] e − j 3ω . Therefore,

H (e j 0 ) = h[0] + h[1] + h[2] + h[3] = 2, H (e jπ / 2 ) = h[0] + h[1] e − jπ / 2 + h[2] e − jπ + h[3] e − j 3π / 2 = h[ 0] − j h[1] − h[ 2] + j h[3] = 7 − j 3, H (e jπ ) = h[0] − h[1] + h[2] − h[3] = 0. Since the impulse response is real, the value of H (e jω ) at ω = 3π / 2 is the conjugate of its value at ω = π / 2, i.e., H (e j 3π / 2 ) = H * (e jπ / 2 ) = h[ 0] + j h[1] − h[ 2] − j h[3] = 7 + j 3. Writing the four equations in matrix form we get −1 ⎡h[0]⎤ ⎡1 1 1 1 ⎤⎡ 2 ⎤ ⎡ 4 ⎤ 1 1⎤ ⎡ 2 ⎤ ⎡1 1 ⎢ h[1] ⎥ ⎢1 − j − 1 j ⎥ ⎢7 − j 3⎥ 1 ⎢1 j − 1 − j ⎥ ⎢7 − j 3⎥ ⎢ 2 ⎥ ⎢h[2]⎥ = ⎢1 − 1 1 − 1⎥ ⎢ 0 ⎥ = 4 ⎢1 − 1 1 − 1⎥ ⎢ 0 ⎥ = ⎢− 3⎥, and ⎢ h[3]⎥ ⎢⎣1 j − 1 − j ⎥⎦ ⎢⎣7 + j 3⎥⎦ ⎢⎣1 − j − 1 j ⎥⎦ ⎢⎣7 + j 3⎥⎦ ⎢⎣ − 1⎥⎦ ⎣ ⎦ hence {h[ n] = {4, 2, − 3, − 1}, 0 ≤ n ≤ 3.

3.70 (a) H (e jω ) = h[0] + h[1] e − jω + h[2] e − j 2ω + h[3] e − j 3ω

= h[0] + h[1] e − jω − h[1] e − j 2ω − h[0] e − j 3ω . Therefore, H (e jπ / 2 ) = h[0] + h[1] e − jπ / 2 − h[1] e − jπ − h[0] e − j 3π / 2 = h[ 0] − j h[1] + h[1] − j h[ 0] = −2 + j 2, H (e jπ ) = h[0] − h[1] − h[1] + h[0] = 8. Solving these two equations we get h[0] = 1 and h[1] = −3. Hence, {h[ n]} = {1, − 3, 3, − 1}, 0 ≤ n p 3. 3.71 (a) H (e jω ) = h[0] + h[1] e − jω + h[2] e − j 2ω = h[0] + h[1] e − jω + h[0] e − j 2ω . The two conditions to be satisfied by the filter are:

H (e j 0.4 π ) = h[0] + h[1] e − j 0.4 π + h[0] e − j 0.8π = 0 and H (e j 0 ) = h[0] + h[1] + h[0] = 1. Solving these two equations we get h[0] = 0.7236 and h[1] = −0.4472. (b) H (e jω ) = 0.7236 − 0.4472 e − jω + 0.7236 e − j 2ω .

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2

2

Phase, radians

Magnitude

1.5 1 0.5 0

0

0.2

0.4

0.6

0.8

1 0 -1 -2

1

0

0.2

0.4

ω/pi

0.6

0.8

1

ω/pi


= h[0] + h[1] e − jω + h[1] e − j 2ω + h[0] e − j 3ω = e − j 3ω / 2 (2h[0] cos(3ω / 2) + 2h[1] cos(ω / 2)). The two conditions to be satisfied by the filter are: H (e j 0.2 π ) = 2h[ 0] cos(0.3π) + 2h[1] cos(0.1π) = 0.8, H (e j 0.5π ) = 2h[ 0] cos(0.75π) + 2h[1] cos(0.25π) = 0.5. Solving these two equations we

get h[0] = 0.0414 and h[1] = 0.395. (b) H (e jω ) = 0.0414 + 0.395 e − jω + 0.395 e − j 2ω + 0.0414 e − j 3ω . 1

4

Phase, radians

Magnitude

0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

2 0 -2 -4

0

0.2

ω/pi

0.4

0.6

0.8

1

ω/pi


= h[0] + h[1] e − jω − h[1] e − j 2ω − h[0] e − j 3ω = j e − j 3ω / 2 (2h[0] sin s(3ω / 2) + 2h[1] sin(ω / 2)). The two conditions to be satisfied by the filter are: H (e j 0.5π ) = 2h[ 0] sin( 0.75π) + 2h[1] sin s(0.25π) = 0.2, H (e j 0.8π ) = 2h[ 0] sin(1.2 π) + 2h[1] cos(0.4 π) = 0.7. Solving these two equations we get

h[0] = −0.14 and h[1] = 0.2815. (b) H (e jω ) = −0.14 + 0.2815 e − jω − 0.2815 e − j 2ω + 0.14 e − j 3ω .

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1

0.8

0 Phase, radians

Magnitude

1

0.6 0.4 0.2 0

-1 -2 -3

0

0.2

0.4

0.6

0.8

-4

1

0

0.2

0.4

ω/π

0.6

0.8

1

ω/π

3.74 (a) H (e jω ) = h[0] + h[1] e − jω + h[2] e − j 2ω + h[3] e − j 3ω + h[ 4] e − j 4ω = h[0] + h[1] e − jω − h[1] e − j 3ω − h[0] e − j 4 ω = j e − j 2ω (2h[0] sin s(2ω) + 2h[1] sin( ω)). The two conditions to be satisfied by the filter are:

H (e j 0.5π ) = 2h[ 0] sin( 0.8π) + 2h[1] sin s(0.4 π) = 0.8, H (e j 0.8 π ) = 2h[ 0] sin(1.6 π) + 2h[1] cos(0.8π) = 0.2. Solving these two equations we get

h[0] = 0.112 and h[1] = 0.3514.

1

4

0.8

2

Phase, radians

Magnitude

(b) H (e jω ) = 0.112 + 0.3514e − jω − 0.3514 e − j 3ω − 0.112 e − j 4ω .

0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

0 -2 -4

0

0.2

0.4

0.6

0.8

1

ω/ π

ω/π

3.75 (a) H A (e jω ) = 0.3 − e − jω + 0.3 e − j 2ω , H B (e jω ) = 0.3 + e − jω + 0.3 e − j 2ω .

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71

|HA(e ω )|

|HB(e ω )| j

2

2

1.5

1.5 Magnitude

Magnitude

j

1

1

0.5 0

0.5

0

0.2

0.4

0.6

0.8

0

1

0

0.2

0.4

ω/π

0.6

0.8

1

ω/π

It can be seen from the above plots that H A (e jω ) is a highpass filter, whereas H B (e jω ) is a lowpass filter. (b) H C (e jω ) = H B (e jω ) = H A (e j (ω + π) ). 3.76 y[ n] = y[ n − 1] + 0.5(x[ n] + x[ n − 1]). Taking the DTFT of both sides we get

(

)

Y (e jω ) = e − jω Y (e jω ) + 0.5 X (e jω ) + e − jω X (e jω ) . Hence, the frequency response is given by H trap (e jω ) =

3.77 y[ n] = y[ n − 2] +

1 3

Y ( e jω ) X (e jω )

=

1 ⎛⎜ 1 + e − jω ⎞⎟ . 2 ⎜⎝ 1 − e − jω ⎟⎠

(x[n] + 4 x[ n − 1] + x[ n − 2]).

− jω + e − j 2ω 1 ⎛1+ 4e H simpson (e jω ) = ⎜ 3⎜ 1 − e − j 2ω ⎝

Hence,

⎞ ⎟. ⎟ ⎠

2 Trapezoidal Simpson

Magnitude

1.5 1 0.5 0

0

0.2

0.4

0.6

0.8

1

ω/π

Note: To compare the performances of the Trapezoidal numerical integration formula with that of the Simpson’s formula, we first observe that if the input is x a (t ) = e jωt , then the result of integration is y a (t ) = e jωt . Thus, the desired ideal frequency response is jω 1

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72

H (e jω ) =

1 . Hence, we take the ratio of the frequency responses of the approximation jω

to the ideal, and plot the two curves as indicated on the previous page. From this plot, it is evident that the Simpson’s formula amplifies high frequencies, whereas, the trapezoidal formula attenuates them. In the very low frequency range, both formulae yield results close to the ideal. However, Simpson’s formula is reasonably accurate for frequencies close to the midband range. 3.78 G (e jω ) = g 0 + g1e − jω + g 2 e − j 2 ω + g 3 e − j 3ω

(

= e − j 3ω / 2 g 0 e j 3ω / 2 + g1e jω / 2 + g 2 e − jω / 2 + g 0 e − j 3ω / 2

)

= e − j 3ω / 2 [ g 0 cos(3ω / 2) + jg 0 sin(3ω / 2) + g 3 cos( −3ω / 2) + jg 3 sin( −3ω / 2) + g1 cos(ω / 2) + jg1 sin(ω / 2) + g2 cos(−ω / 2) + jg2 sin(−ω / 2)] ] = e − j 3ω / 2 [( g 0 + g3 ) cos(3ω / 2) + ( g1 + g 2 ) cos(ω / 2) + j ( g 0 − g3 ) sin(3ω / 2) + j ( g1 − g 2 ) cos(ω / 2)].

Thus, if g 0 = g3 and g1 = g 2 , then G (e jω ) = e − j 3ω / 2 [( g 0 + g3 ) cos(3ω / 2) + ( g1 + g 2 ) cos(ω / 2)] which has a linear phase

θ(ω) = −

3ω , and 2

hence, a constant group delay.

Alternately, if g 0 = − g3 and g1 = − g 2 , then G (e jω ) = j e − j 3ω / 2 [( g 0 + g 3 ) sin(3ω / 2) + ( g1 + g 2 ) sin(ω / 2)] which has a linear phase

θ(ω) = −

3ω π + , and 2 2

hence, a constant group delay.

⎛ − b sin ω ⎞ ⎟⎟ . 3.79 (a) H a (e jω ) = a + b e − jω = a + b cos ω − j b sin ω. Thus, θ H a (ω) = tan −1 ⎜⎜ ⎝ a + b cos ω ⎠ d θ H a (ω) d ⎛ − b sin ω ⎞ 1 Hence, τ H a (ω) = − =− ⋅ ⎜ ⎟ 2 dω ⎝ a + b cos ω ⎠ dω b sin − ω ⎛ ⎞ 1+ ⎜ ⎟ ⎝ a + b cos ω ⎠ =− =−

(a + b cos ω) 2 2

(a + b cos ω) +(b sin ω)

2

⋅

− (a + b cos ω)b cos ω − (−b sin ω)(−b sin ω)

− ab cos ω − b 2 cos 2 ω − b 2 sin 2 ω a 2 + b 2 cos 2 ω + 2 ab cos ω + b 2 sin 2 ω

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(a + b cos ω) 2 =

b 2 + ab cos ω a 2 + b 2 + 2 ab cos ω

.

73

(b) Let G b (e jω ) = 1 + c e − jω = 1 + c cos ω − j c sin ω. From the results of Part (a) we have τ G b (ω) =

c 2 + c cos ω 2

1 + c + 2c cos ω

. Since H b (e jω ) =

1 G b (e

θ H b (ω) = −θGb (ω), we have τ H b (ω) = − τ G b (ω) = − a + b e − jω

(c) H c (e jω ) =

1+ ce

− jω

jω

= )

1 1 + c e − jω

c 2 + c cos ω 1 + c 2 + 2c cos ω

,

.

= H a (e jω )H b (e jω ), where H a (e jω ) is the frequency

response of Part (a) and H b (e jω ) is the frequency response of Part (b). Thus, θ H c (ω) = θ H a (ω) + θ H b (ω), and therefore, τ H c (ω) = τ H a (ω) + τ H b (ω) =

b 2 + ab cos ω

−

c 2 + c cos ω

. a 2 + b 2 + 2 ab cos ω 1 + c 2 + 2c cos ω 1 1 (d) H d (e jω ) = ⋅ = H b (e jω )H e (e jω ), where H b (e jω ) is the − jω − jω 1+ ce 1+ d e

frequency response of Part (b) and H e (e jω ) is similar in form to H b (e jω ) . Thus, θ H d (ω) = θ H d (ω) + θ H e (ω), and therefore, τ H d (ω) = τ H b (ω) + τ H e (ω) =

c 2 + c cos ω 1 + c 2 + 2c cos ω

+

d 2 + d cos ω 1 + d 2 + 2 d cos ω

.

3.80

The group delay of a causal LTI discrete-time system with a frequency response dθ(ω) H (e jω ) = H (e jω ) e j θ(ω) is given by τ g (ω) = − . Now, dω

d H ( e jω ) dH (e jω ) dθ(ω) . Hence, = e jθ(ω) + j H (e jω ) e jθ(ω) dω dω dω − j H (e

jω

)e

jθ(ω)

d H (e jω ) dH (e jω ) dθ(ω) jθ(ω) , or, equivalently, =e − dω dω dω

d H ( e jω ) dθ(ω) e jθ(ω) dH (e jω ) 1 − = ⋅ − ⋅ dω dω dω j H (e jω ) e jθ(ω) j H (e jω ) e jθ(ω) =

1 j H ( e jω )

⋅

d H ( e jω ) dω

+j

dH (e jω ) . The first term on the rightjω ω d H (e ) 1

⋅

hand side of the above equation is purely imaginary. Hence,

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74

⎧ dH (e jω ) ⎫ ⎪ j dω ⎪ dθ(ω) . τ g (ω) = − = Re ⎨ jω ⎬ dω ⎪ H (e ) ⎪ ⎭ ⎩

dH (e jω ) . Rewriting Eq. dω * ⎧ ⎛ dH (e jω ) ⎞ ⎫ ⎧ *⎫ j ω j ω j ω ⎞ ⎛ ⎪ dH (e ) ⎜ j ⎟ ⎪ ⎪ dH (e ) ⎜ j dH (e ) ⎟ ⎪ ⎜ dω ⎟ ⎪ 1 ⎪ j dω 1 ⎪ j dω ⎪ ⎠ ⎝ d ω ⎟ ⎬= ⎨ (3.127) we get τ g (ω) = ⎨ +⎜ + ⎬ j ω j ω j ω * j ω 2 ⎪ H (e ) ⎜ H ( e ) ⎟ ⎪ 2 ⎪ H ( e ) H ( e ) ⎪ ⎟ ⎜ ⎠ ⎪⎭ ⎝ ⎪ ⎪⎩ ⎪ ⎭ ⎩ * ⎫ ⎧ jω jω ⎞ ⎛ ⎪ j dH (e ) H * (e jω ) + ⎜ j dH (e ) ⎟ H (e jω ) ⎪ ⎜ dω dω ⎟ ⎪ 1⎪ ⎠ ⎝ = ⎨ ⎬ 2 2⎪ jω ⎪ H (e ) ⎪ ⎪ ⎭ ⎩ * 1 = Gre (e jω ) + j Gim (e jω ) H re (e jω ) + j H im (e jω ) 2 2 H ( e jω ) Since G(e jω ) is the Fourier transform of ng[n], G(e jω ) = j

3.81

[(

1

=

2 H (e =

jω

)

2

)(

[2G

)

(

)(

+ Gre (e jω ) + j Gim (e jω ) H re (e jω ) + j H im (e jω ) re (e

jω

) H re (e jω ) + 2Gim (e jω ) H im (e jω )

Gre (e jω ) H re (e jω ) + Gim (e jω ) H im (e jω ) H (e

jω

)

2

]

)* ]

.

3.82 (a) H a (e jω ) = 1 + 0.4 e − jω = 1 + 0.4 cos ω − j 0.4 sin ω and thus,

{

}

{

}

Re H a (e jω ) = 1 + 0.4 cos ω and Im H a (e jω ) = −0.4 sin ω. .

G a ( e jω ) = j

{

dH a (e jω ) = j (− j 0.4 e − jω ) = 0.4 cos ω − j 0.4 sin ω. Thus, dω

}

{

}

Re G a (e jω ) = 0.4 cos ω and Im G a (e jω ) = −0.4 sin ω. . Therefore, using Eq. (3.128) we

get τ H a (ω) = =

(1 + 0.4 cos ω)(0.4 cos ω) + (−0.4 sin ω)2 (1 + 0.4 cos ω)2 + (−0.4 sin ω)2

0.16 cos 2 ω + 0.16 sin 2 ω + 0.4 cos ω 1 + 0.16 cos 2 ω + 0.16 sin 2 ω + 0.8 cos ω

=

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0.16 + 0.4 cos ω . 1.16 + 0.8 cos ω

75

(b) Let Gb (e jω ) =

1 H b (e

jω

)

= 1 + 0.6 e − jω . Then τ G b (ω) = −τ H b (ω). Then using the

same procedure as in Part (a) we get τ G b (ω) = τ H b (ω) = −

0.36 + 0.6 cos ω . 1.36 + 1.2 cos ω

0.36 + 0.6 cos ω . Therefore, 1.36 + 1.2 cos ω

⎛ ⎞ 1 ⎟ = G a (e jω )Gb (e jω ), where (c) Let H c (e jω ) = (1 − 0.5 e − jω )⎜ ⎜ 1 + 0.3 e − jω ⎟ ⎝ ⎠ 1 . Therefore, G a (e jω ) = 1 − 0.5 e − jω and Gb (e jω ) = 1 + 0.3 e − jω τ H c (ω) = τ G a (ω) + τ G b (ω). Then using the same procedure as in Part (a) we get

0.25 − 0.5 cos ω and using the same procedure as in Part (b) we get 1.25 − cos ω 0.09 + 0.3 cos ω 0.25 − 0.5 cos ω 0.09 + 0.3 cos ω τ G b (ω) = − . Hence, τ H c (ω) = . − 1.09 + 0.6 cos ω 1.25 − cos ω 1.09 + 0.6 cos ω τ G a (ω) =

⎛ ⎞⎛ ⎞ 1 1 ⎟⎜ ⎟ = G a (e jω )Gb (e jω ), where (d) Let H d (e jω ) = ⎜ ⎜ 1 − 0.3 e − jω ⎟⎜ 1 + 0.5 e − jω ⎟ ⎝ ⎠⎝ ⎠ 1 1 G a ( e jω ) = and Gb (e jω ) = . Therefore, − jω 1 − 0.3 e 1 + 0.5 e − jω τ H d (ω) = τ G a (ω) + τ G b (ω). Then using the same procedure as in Part (b) we get τ G a (ω) =

− 0.09 + 0.3 cos ω − 0.25 − 0.5 cos ω and τ G a (ω) = . Hence, 1.09 − 0.6 cos ω 1.25 + cos ω

⎛ 0.09 − 0.3 cos ω 0.25 + 0.5 cos ω ⎞ ⎟. + τ G a (ω) = −⎜⎜ 1.25 + cos ω ⎟⎠ ⎝ 1.09 − 0.6 cos ω 1

3.83 From Table 3.3, H (e jω ) = H ( e jω ) =

1 + 0.5 e − jω 1 2

=

(1 + 0.5 cos ω) + (0.5 sin ω)

2

1 . Thus, 1 + 0.5 cos ω − j 0.5 sin ω =

1 1.25 + cos ω

and

⎛ 0.5 sin ω ⎞ ⎛ − 0.5 sin ω ⎞ ⎟⎟. Now ⎟⎟ = tan −1 ⎜⎜ arg{H (e jω )} = θ(ω) = − tan −1 ⎜⎜ ⎝ 1 + 0.5 cos ω ⎠ ⎝ 1 + 0.5 cos ω ⎠ 1 H ( e jπ / 5 ) = = 0.6821 + j 0.1427. Therefore, 1 + 0.5 cos(π / 5) − j 0.5 sin(π / 5)

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76

⎛ 0.1427 ⎞ H (e jπ / 5 ) = 0.6969 and arg{H (e jπ / 5 )} = θ(π / 5) = tan −1 ⎜ ⎟ = 0.2063 radians. ⎝ 0.6821 ⎠

Since for a frequency response with real coefficient impulse response, H (e jω ) is an even function of ω and θ(ω) is an odd function of ω, we have H (e − jπ / 5 ) = 0.6969 and θ( − π / 5) = −0.2063. Now, for an input x[ n] = sin(ωo n) µ[ n], the steady-state output is given by

y[ n] = H (e jω o ) sin (ωo n + θ(ωo ) ). Thus, for ωo = π / 5, the steady-state output is given π π by y[ n] = H (e jπ / 5 ) sin⎛⎜ n + θ(π / 5) ⎞⎟ = 0.6969 sin⎛⎜ n + 0.2063 ⎞⎟. ⎝5 ⎠ ⎝5 ⎠

(

)

3.84 H (e jω ) = h[0] 1 + e − j 2ω + h[1] e − jω = e − jω (2h[0] cos ω + h[1]). We require 2h[ 0] cos( 0.3) + h[1] = 1 and 2h[ 0] cos( 0.7) + h[1] = 0. Solving these two equations we get h[0] = 2.6248 and h[1] = − 4.015. 2.5

Magnitude

2 1.5 1 0.5 0

0

0.2

0.4

0.6

0.8

1

1.2

ω

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77

M3.1 r = 0.9, θ = 0.75 Imaginary part 2

6

0 Amplitude

Amplitude

Real part 8

4 2 0 -2

0

0.5 ω/π Magnitude Spectrum

-8

1

0

0.5

1

ω/π Phase Spectrum

Phase, radians

0.5

6 Magnitude

-4 -6

8

4 2 0

-2

0

0.5 ω/π

0 -0.5 -1 -1.5 -2

1

0

0.5

1

ω/π

r = 0.7, θ = 0.5 Real part

Imaginary part 0

3

Amplitude

Amplitude

4

2 1 0

0


0

0.5

1

ω/ π Phase Spectrum 0 Phase, radians

4 Magnitude

-2

-3

1

5

3 2 1 0

-1

0

0.5 ω/π

1

-0.5

-1

-1.5

0

0.5

1

ω/ π

M3.2 It should be noted that Program 3_1.m uses the function freqz to determine the samples of a DTFT that is rational function in e − jω , i.e., a ratio of polynomials in e − jω . Their inverse DTFTs are two-sided sequences. However, all sequences of

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78

Problem 3.19 except that in Part (b) are two-sided finite-length sequences of length

2 N + 1 , and their DTFTs have both positive and negative powers of e jω . As a result, the frequency sample computed using freqz should be multiplied by the vector e jω n N evaluated at the frequency points ω n used in freqz. In Parts (a), (c) and (d), the phase spectra are the plots of the unwrapped phase obtained using the function unwrap.

Moreover, the DTFTs of the sequences in Parts (a), (c) and (d) are real functions of ω and thus have zero phase. More accurate plots of the DTFTs are obtained using the function zerophase.

sin(21ω / 2 ) 1, − 10 ≤ n ≤ 10, Y1 (e jω ) = . The plots obtained using (a) y1[ n] = ⎧⎨ 0 , otherwise, sin(ω / 2) ⎩ Program 3_1.m are shown below: -14

20

1

10 0 -10

Magnitude

Amplitude

2

0


0.5

1

Phase Spectrum

20

0

15 10 5 0.5 ω/π

0

ω/π 2

0

Imaginary part

-1

25

0

x 10

0

-2

1

Phase, radians

Amplitude

Real part 30

-2 -4 -6 -8

1

0

0.5

1

ω/π

The plot obtained using the function zerophase is shown below: 25

Amplitude

20 15 10 5 0 -5

0

0.2

0.4

0.6

0.8

1

ω/π

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79

⎛ sin(ω[ N + 1] / 2) ⎞ 1, 0 ≤ n ≤ N , (b) y2 [ n] = ⎧⎨ Then Y2 (e jω ) = e − jωN / 2 ⎜⎜ ⎟⎟ . The plots 0 , otherwise. ⎩ ⎝ sin(ω / 2) ⎠ obtained using Program 3_1.m are shown below: Real part

Imaginary part

15

8 6 Amplitude

Amplitude

10 5 0 -5

0


-2

1

0

0.5

1

ω/π Phase Spectrum 3 Phase, radians

Magnitude

2 0

15

10

5

0

4

0

0.5 ω/π

2 1 0 -1 -2

1

0

0.5

1

ω/π

n ⎧⎪ 1 sin 2 (ωN / 2 ) . The plots obtained (c) y3 [ n] = ⎨1 − N , − N ≤ n ≤ N , Y3 (e jω ) = ⋅ 2 N ω sin ( / 2 ) ⎪⎩ 0, otherwise. using Program 3_1.m are shown below: Imaginary part 1.5

5

1

Amplitude

Amplitude

Real part 10

0

-5

0


0.5

0

1

1

Phase Spectrum 4 Phase, radians

8 Magnitude

0.5

ω/π

10

6 4 2 0

0

0

0.5 ω/π

1

3 2 1 0

0

0.5

1

ω/π

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80


Amplitude

20 15 10 5 0 -5

0

0.2

0.4

0.6

0.8

1

ω/π

1 sin⎛⎜ ω[ N + ] ⎞⎟ n 2 ⎧⎪ 2 ⎠ 1 sin (ωN / 2 ) + ⋅ . (d) y 4 [ n] = ⎨ N + 1 − N , − N ≤ n ≤ N , Y4 (e jω ) = N ⋅ ⎝ sin(ω / 2) N sin 2 (ω / 2) ⎪⎩ 0, otherwise. -14

Real part 40

2

20 10 0 -10

0

0.5 ω/ π Magnitude Spectrum

-1

0

0.5

1

ω/ π Phase Spectrum

Phase, radians

5

30 Magnitude

0

-2

1

40

20 10 0

Imaginary part

1 Amplitude

Amplitude

30

x 10

0

0.5 ω/ π

0

-5

-10

1

0

0.5

1

ω/ π


Amplitude

20 15 10 5 0 -5

0

0.2

0.4

0.6

0.8

1

ω/π

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81

cos(πn / 2 N ), − N ≤ n ≤ N , (e) y5 [ n] = ⎧⎨ 0, otherwise. ⎩ Y5 (e

jω

(

π

1

)

(

π

1

)

1 sin (ω − 2 N )( N + 2 ) 1 sin (ω + 2 N )( N + 2 ) + ⋅ . The plots obtained using )= ⋅ 2 sin (ω + π ) / 2 2 sin (ω − π ) / 2

(

)

2N

(

)

2N

Program 3_1.m are shown below: -14

Real part

x 10

15 0.5 Amplitude

Amplitude

10 5 0 -5

Imaginary part

0

0.5

0 -0.5

1

0

0.5

ω/π Magnitude Spectrum


Magnitude

15

10

5

0

1

ω/ π

0

0.5 ω/π

10 5 0 -5

1

0

0.5

1

ω/ π


Amplitude

20 15 10 5 0 -5

0

0.2

0.4

0.6

0.8

1

ω/π

M3.3 (a) X (e jω ) =

0.2418(1 + 0.139 e − jω − 0.3519 e − j 2 ω + 0.139 e − j 3ω + e − j 4ω ) 1 + 0.2386 e − jω + 0.8258 e − j 2ω + 0.1393 e − j 3ω + 0.4153 e − j 4 ω

The plots obtained using Program 3_1.m are shown below:

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82

Real part

Imaginary part

1

1 0.5 Amplitude

Amplitude

0.5

0

-0.5

0


0 -0.5 -1

1

1


0.8 Magnitude

0.5

ω/π

1

0.6 0.4 0.2 0

0

0

(b) X (e jω ) =

0.5 ω/ π

2 0 -2 -4

1

0

0.5

1

ω/π

0.1397(1 − 0.0911 e − jω + 0.0911 e − j 2ω − e − j 3ω ) 1 + 1.1454 e − jω + 0.7275 e − j 2 ω + 0.1205 e − j 3ω

.

The plots obtained using Program 3_1.m are shown below: Real part

Imaginary part

0.8

0.6 0.4 Amplitude

Amplitude

0.6 0.4 0.2

0.2 0

0 -0.2

0


-0.2

1

1


Magnitude

0.5

ω/ π

0.8 0.6 0.4 0.2 0

0

0

0.5 ω/ π

1

2 1 0 -1

0

0.5

1

ω/ π

M3.4 % Property 1

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83

N = 8; % Number of samples in sequence gamma = 0.5; k = 0:N-1; x = exp(-j*gamma*k); y = exp(-j*gamma*fliplr(k)); % r = x[-n] then y = r[n-(N-1)] % so if X1(exp(jw)) is DTFT of x[-n], then % X1(exp(jw)) = R(exp(jw)) = exp(jw(N-1))Y(exp(jw)) [Y,w] = freqz(y,1,512); X1 = exp(j*w*(N-1)).*Y; m = 0:511; w = -pi*m/512; X = freqz(x,1,w); % Verify X = X1 % Property 2 k = 0:N-1; y = exp(j*gamma*fliplr(k)); [Y,w] = freqz(y,1,512); X1 = exp(j*w*(N-1)).*Y; [X,w] = freqz(x,1,512); % Verify X1 = conj(X) % Property 3 y = real(x); [Y3,w] = freqz(y,1,512); m = 0:511; w0 = -pi*m/512; X1 = freqz(x,1,w0); [X,w] = freqz(x,1,512); % Verify Y3 = 0.5*(X+conj(X1)) % Property 4 y = j*imag(x); [Y4,w] = freqz(y,1,512); % Verify Y4 = 0.5*(X-conj(X1)) % Property 5 k = 0:N-1; y = exp(-j*gamma*fliplr(k)); xcs = 0.5*[zeros(1,N-1) x] + 0.5*[conj(y) zeros(1,N-1)]; xacs = 0.5*[zeros(1,N-1) x] - 0.5*[conj(y) zeros(1,N-1)]; [Y5,w] = freqz(xcs,1,512); [Y6,w] = freqz(xacs,1,512); Y5 = Y5.*exp(j*w*(N-1)); Y6 = Y6.*exp(j*w*(N-1)); % Verify Y5 = real(X) and Y6 = j*imag(X) M3.5 N = 8; % Number of samples in sequence gamma = 0.5; k = 0:N-1; x = exp(gamma*k); y = exp(gamma*fliplr(k)); xev = 0.5*([zeros(1,N-1) x] + [y zeros(1,N-1)]); xod = 0.5*([zeros(1,N-1) x] - [y zeros(1,N-1)]); [X,w] = freqz(x,1,512); [Xev,w] = freqz(xev,1,512); [Xod,w] = freqz(xod,1,512); Xev = exp(j*w*(N-1)).*Xev;

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Xod = exp(j*w*(N-1)).*Xod; % Verify real(X) = Xev, and j*imag(X) = Xod M3.5 N = input('The length of the seqeunce = '); k = 0:N-1; gamma = -0.5; g = exp(gamma*k); % g is an exponential seqeunce h = sin(2*pi*k/(N/2)); % h is a sinusoidal sequence with period = N/2 [G,w] = freqz(g,1,512); [H,w] = freqz(h,1,512); % Property 1 alpha = 0.5; beta = 0.25; y = alpha*g+beta*h; [Y,w] = freqz(y,1,512); % Plot Y and alpha*G+beta*H to verify that they are equal % Property 2 n0 = 12; % Sequence shifted by 12 samples y2 = [zeros(1,n0) g]; [Y2,w] = freqz(y2,1,512); G0 = exp(-j*w*n0).*G; % Plot G0 and Y2 to verify they are equal % Property 3 w0 = pi/2; % the value of omega0 = pi/2 r = 256; % the value of omega0 in terms of number of samples k = 0:N-1; y3 = g.*exp(j*w0*k); [Y3,w] = freqz(y3,1,512); k = 0:511; w = -w0+pi*k/512; % creating G(exp(w-w0)) G1 = freqz(g,1,w); % Compare G1 and Y3 % Property 4 k = 0:N-1; y4 = k.*g; [Y4,w] = freqz(y4,1,512); % To compute derivative we need sample at pi y0 = ((-1).^k).*g; G2 = [G(2:512)' sum(y0)]'; delG = (G2-G)*512/pi; % Compare Y4, delG % Property 5 y5 = conv(g,h); [Y5,w] = freqz(y5,1,512); % Compare Y5 and G.*H % Property 6

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y6 = g.*h; [Y6,w] = freqz(y6,1,512,'whole'); [G0,w] = freqz(g,1,512,'whole'); [H0,w] = freqz(h,1,512,'whole'); % Evaluate the sample value at w = pi/2 % and verify with Y6 at pi/2 H1 = [fliplr(H0(1:129)') fliplr(H0(130:512)')]'; val = 1/(512)*sum(G0.*H1); % Compare val with Y6(129) i.e., sample at pi/2 % Can extend this to other points similarly % Parsevals theorem val1 m = sum(g.*conj(h)); val2 = sum(G0.*conj(H0))/512; % Compare val1 with val2 M3.7 The DTFT of nh[n] is j

dH ( e jω ) . dω

Hence, the group delay τ g (ω) can be computed at a

set of N discrete frequency points ω k = 2πk / N , 0 ≤ k ≤ N − 1, as follows: DFT {n h[ n]} ⎞ τ g (ωk ) = Re⎛⎜ ⎟, ⎝ DFT {h[ n]} ⎠ where all DFTs are N –points in length with N greater than or equal to the length of {h[ n]}. M3.8 h = [3.8461 -6.3487 3.8461]; [H,w] = freqz(h,1,512); plot(w/pi,abs(H)); grid xlabel('\omega/\pi'); ylabel('Magnitude');

Magnitude

15

10

5

0

0

0.2

0.4

0.6

0.8

1

ω/π

M3.9 h = [-13.4866 45.228 -63.8089 45.228 -13.4866]; [H,w] = freqz(h,1,512); plot(w/pi,abs(H)); grid xlabel('\omega/\pi'); ylabel(Magnitude);

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200

Magnitude

150 100 50 0

0

0.2

0.4

0.6

0.8

1

ω/π

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Chapter 4 4.1

Let φ(t ) be an arbitrary continuous-time function with a CTFT Φ ( jΩ ), where ∞ ∞ ~ Φ( jΩ) = ∫ φ(t ) e − jΩt dt. Let φT (t ) = ∑ φ(t + nT ) denote the periodic continuous-

n = −∞

−∞

time function with a period T obtained by a periodic extension of φ(t ). Note that ~ φT (t ) is also given by the convolution of φ(t ) with the periodic impulse train ∞ ∞ ~ p(t ) = ∑ δ(t + nT ), i.e., φT (t ) = ∫ φ(τ)p(t − τ)dτ.

n = −∞

−∞

~ ~ Tthe CTFT of φT (t ) is then given by F{ φT (t )} = Φ( jΩ) ⋅ F { p(t ) }

= Φ ( jΩ ) ⋅

2π ∞ ∑δ T n = −∞

where Ω T =

( j(Ω − nΩT )) =

∞

2π Φ( jnΩ T )δ n = −∞ T

∑

( j(Ω − nΩT )),

(4-1)

2π . T

∞ ~ Now, the Fourier series expansion of φT (t ) = ∑ φ(t + nT ) is given by n = −∞

∞ ~ φT (t ) = ∑ a n e jnΩ T t . A CTFT of both sides of this equation is then n = −∞

∞ ~ F {φT (t )} = ∑ a n ⋅ 2πδ( j (Ω − nΩ T )).

(4-2)

n = −∞

1

Comparing Eqs. (4-1) and (4-2) we arrive at a n = Φ( jnΩ T ). Substituting this T ~ expression in the Fourier expansion of φT (t ) we therefore arrive at the Poisson’s sum ∞ ~ 1 ∞ jnΩ T t formula φT (t ) = ∑ φ(t + nT ) = . ∑ Φ( jnΩ T ) e T n = −∞

n = −∞

4.2

Consider the continuous-time signal g a (t ) = sin(Ω m t ) which is bandlimited to Ω m . If we sample ga (t ) at a rate Ω T = 2Ω m starting at t = 0 , then all its samples are zero. Hence, ga (t ) cannot be recovered from its samples obtained by sampling it at the Nyquist rate Ω T = 2Ω m . As a result, g a (t ) = sin(Ω m t ) must be sampled at a rate Ω T > 2Ω m to recover it fully from its samples.

4.3

(a) Now, the CTFT of y1 (t ) is given by Y1 ( jΩ) =

1 G ( jΩ ) O * G a ( jΩ ) 2π a

where

Ga ( jΩ) denotes the CTFT of g a (t ) and O * denotes the frequency-domain convolution. The highest frequency present in y1 (t ) is therefore twice that of g a (t ) and hence, the Nyquist frequency of y1 (t ) is 2Ω m .

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∞

(b) The CTFT of y2 (t ) is given by Y2 ( jΩ) = ∫ g a ( ) e − jΩt dt 3 −∞ t

∞

= 3 ∫ g a (τ) e − j 3Ωτ dτ = 3 G a ( j 3Ω). The highest frequency present in y2 (t ) is −∞

therefore one-third of that of g a (t ) and hence, the Nyquist frequency of y2 (t ) is Ω m / 3. ∞

(c) The CTFT of y3 (t ) is given by Y3 ( jΩ) = ∫ g a (3t ) e − jΩt dt −∞

=

1 ∞ − jΩτ / 3 dτ ∫ g a ( τ) e 3 −∞

1 3

Ω 3

= G a ( j ). The highest frequency present in y3 (t ) is

therefore three times of that of g a (t ) and hence, the Nyquist frequency of y3 (t ) is 3Ω m . (d) The CTFT of y 4 (t ) is given by ∞ ⎡∞ ∞ ⎤ ⎡∞ ⎤ Y4 ( jΩ) = ∫ ⎢ ∫ g a (t − τ)g a (τ)dτ⎥ e − jΩt dt = ∫ g a (τ) ⎢ ∫ g a (t − τ) e − jΩt dt ⎥ dτ − ∞ ⎣− ∞ −∞ ⎦ ⎣− ∞ ⎦ ∞

∞

−∞

−∞

= ∫ g a (τ) e − jΩτ G a ( jΩ) dτ = G a ( jΩ) ∫ g a (τ) e − jΩτ dτ = G a ( jΩ)G a ( jΩ). The

highest frequency present in y 4 (t ) is therefore the same as that of g a (t ) and hence, the Nyquist frequency of y 4 (t ) is Ω m . 1 ∞

(e) Now g a (t ) = G a ( jΩ) e jΩt dΩ. Differentiating both sides of this equation ∫ 2π −∞ we get y 5 (t ) =

dg a (t ) dt dg a ( t ) dt

=

1 ∞ jΩt ∫ jΩ G a ( jΩ) e dΩ. 2π −∞

Hence, it follows that the CTFT of

is simply jΩ G a ( jΩ). The highest frequency present in y5 (t ) is

therefore the same as that of g a (t ) and hence, the Nyquist frequency of y5 (t ) is Ωm . 4.4

By Parseval’s relation, the total energy of g a (t ) is given by ∞

2

1 ∞

1

2

Ωm

2

E g a (t ) = ∫ g a (t ) dt = ∫ G a ( jΩ) dΩ = 2 π ∫ Ga ( jΩ) dΩ. Likewise, the 2π −Ωm −∞ −∞ ∞

1 π

2

jω total energy of g[n] is given by E g[ n] = ∑ g[ n] = ∫ G ( e ) dω 2 π −∞ −π

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87

4.5

=

Ωm 2 1 π/T 1 1 π/T 1 2 2 G ( j Ω ) d ( Ω T ) = G ( j Ω ) d Ω = ∫ T a ∫ ∫ G a ( jΩ) dΩ a 2 πT 2 πT 2π −π / T −π / T −Ωm

=

1 T

E g a (t ) .

Sampling period T = FT =

1 T

2 .5 = sec. Hence, the sampling frequency is 5000

= 2000 Hz. Therefore, the highest frequency component that could be

present in the continuous-time signal has a frequency

4.6

20000 = 1000 Hz. 2

Since the continuous-time signal x a (t ) is being sampled at 2 kHz rate, the sampled version of its i -th sinusoidal component with a frequency Fi will generate discretetime sinusoidal signals with frequencies F i ±2000n, − ∞ < n < ∞. Hence, the frequencies Fim generated in the sampled version associated with the sinusoidal components present in are as follows:

F1 = 300 Hz ⇒ F1m = 300, 1700, 2300,K Hz F2 = 500 Hz ⇒ F2 m = 500, 1500, 2500,K Hz F3 = 1200 Hz ⇒ F3m = 1200, 800, 3200,K Hz F4 = 2150 Hz ⇒ F4 m = 2150, 150, 4150,K Hz F5 = 3500 Hz ⇒ F5m = 3500, 1500, 5500, 500, 7500,K Hz After filtering by a lowpass filter with a cutoff at 900 Hz, the frequencies of the sinusoidal components in y a (t ) are 150, 300, 500, 800 Hz. 4.7

One possible set of values for the frequencies present in y a (t ) are: F1 = 350 Hz,

F2 = 575 Hz, F3 = 815 Hz, and F4 = 9650 Hz. Another possible set of values for the frequencies present in y a (t ) are: F1 = 350 Hz, F2 = 575 Hz, F3 = 815 Hz, and F4 = 10575 Hz. Hence, the solution is not unique. 4.8

t = nT =

n . 50

Therefore,

20 πn ⎞ 24 πn ⎞ 120 πn ⎞ 176 πn ⎞ x[ n] = 4 sin⎛⎜ ⎟ − 5 cos⎛⎜ ⎟ + 3 sin⎛⎜ ⎟ + 2 cos⎛⎜ ⎟ ⎝ 50 ⎠ ⎝ 50 ⎠ ⎝ 50 ⎠ ⎝ 50 ⎠ 2 πn ⎞ 12 πn ⎞ (10 + 2)πn ⎞ (100 −12)πn ⎞ = 4 sin⎛⎜ ⎟ − 5 cos⎛⎜ ⎟ + 3 sin⎛⎜ ⎟ + 2 cos⎛⎜ ⎟ 5 25 ⎝ 5 ⎠ ⎝ 25 ⎠ ⎝ ⎠ ⎝ ⎠ 2 πn ⎞ 12 πn ⎞ 2 πn ⎞ (12 πn ⎞ = 4 sin⎛⎜ ⎟ − 5 cos⎛⎜ ⎟ + 3 sin⎛⎜ ⎟ + 2 cos⎛⎜ ⎟. ⎝ 5 ⎠ ⎝ 25 ⎠ ⎝ 5 ⎠ ⎝ 25 ⎠

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4.9

Both channels are being sampled at 45 kHz. Therefore, there are a total of 2 × 45000 = 90000 samples/sec. Each sample is quantized using 12 bits. Hence, the total bit rate of the two channels after sampling and digitization is 108 kpbs.

4.10 hr (t ) =

sin(Ω c t ) sin(Ω c nT ) . Therefore, hr (nT ) = . Since T = 2π / ΩT , we have ΩT t / 2 Ω T nT / 2

⎛ 2 πΩ c n ⎞ sin⎜ ⎟ ΩT ⎠ sin( πn) ⎝ . For Ω c = Ω T / 2, we thus have hr (nT ) = hr (nT ) = = δ[ n]. πn πn 4.11 The spectrum of the sampled signal is as shown below: Xp(jΩ) 1/T 1/2T _2Ω

Now, T =

2π 2Ω m

=

m

π Ωm

_

Ω m __ 3

0

Ωm

. As a result, ωc =

Ωmπ 3Ω m

Ωm

Ω

2 Ωm

π 3

= . Hence after lowpass filtering

the spectrum of the output continuous-time signal y a (t ) will be as shown below: Ya (jΩ)

m _Ω __

0

3

Ω m __ 3

Ω

4.12 (a) Ω1 = 100π, Ω1 = 150π. Thus, ∆Ω = Ω 2 − Ω1 = 50π. Note ∆Ω is an integer multiple of Ω 2 . Hence, we choose the sampling angular frequency as Ω T = 2∆Ω = 100π =

2 ×150 π , M

which is satisfied for M = 3. The sampling

frequency is therefore 50 Hz. The CTFT X p ( jΩ) of the sampled sequence and the frequency response H r ( jΩ) of the desired reconstruction filter are shown below. X p(jΩ) M=3 1/T

_ 200π _ 150π _ 100π

_ 50π

0

50π

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100π

150π

200π

Ω

89

H r (jΩ) T

_

150π

_ 100π

0

100π

Ω

150π

(b) Ω1 = 160π, Ω1 = 250π. Thus, ∆Ω = Ω 2 − Ω1 = 90π. Note ∆Ω is not an integer multiple of Ω 2 . Hence, we extend the bandwidth to the left by assuming the lowest frequency to be Ω 0 and choose the sampling angular frequency as Ω T = 2∆Ω = 2(Ω 2 −Ω 0 ) =

2 × 250 π , M

which is satisfied for Ω 0 = 125π and M = 2.

The sampling frequency is therefore 125 Hz. The CTFT X p ( jΩ) of the sampled sequence and the frequency response H r ( jΩ) of the desired reconstruction filter are shown below. X p (jΩ)

1/T

_

340π

_

250π _ 160π

_

90π

0

90π

160π

250π

Ω

340π

H r (jΩ) T

_

250π

_ 160π

0

160π

Ω

250π

(c) Ω1 = 110π, Ω1 = 180π. Thus, ∆Ω = Ω 2 − Ω1 = 70π. Note ∆Ω is not an integer multiple of Ω 2 . Hence, we extend the bandwidth to the left by assuming the lowest frequency to be Ω 0 and choose the sampling angular frequency as Ω T = 2∆Ω = 2(Ω 2 −Ω 0 ) =

2 ×180 π , M

which is satisfied for Ω 0 = 90π and M = 2.

The sampling frequency is therefore 90 Hz. The CTFT X p ( jΩ) of the sampled sequence and the frequency response H r ( jΩ) of the desired reconstruction filter are shown below. X p (jΩ)

1/T

_

250π

_

180π

_ 110π

_

70π

0

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70π

110π

180π

250π

Ω

90

H r (jΩ) T

_

4.13

180π

_ 110π

0

110π

Ω

180π

α p = −20 log10 (1 − δ p ) dB and α s = −20 log10 δ s dB. Therefore, −α p / 20

and δ s = 10 −α s / 20 . (a) α p = 0.21 dB and α s = 52 dB. Hence, δ p = 0.0239 and δ s = 0.025. (b) α p = 0.03 dB and α s = 69 dB. Hence, δ p = 0.0034 and δ s = 0.00355.

δ p = 1 − 10

(c) α p = 0.33 dB and α s = 57 dB. Hence, δ p = 0.0373 and δ s = 0.0014. 4.14

H a (s) =

a a , and hence, . Thus, H a ( jΩ) = jΩ + a s+a

H a ( jΩ )

2

= H a ( jΩ ) H a ( − jΩ ) =

a a a2 ⋅ = 2 2. j Ω + a − jΩ + a Ω +a

As Ω increases from 2

0 to ∞ , it can be seen that the square-magnitude function H a ( jΩ) and hence, the magnitude function H a ( jΩ) =

a Ω2 + a2

decreases monotonically from

H a ( j 0) = 1 to H a ( j∞) = 0. Let Ω c denote the 3 –dB cutoff frequency. Then 2

H a ( jΩ c ) = 4.15

G a (s ) = G a ( jΩ )

a2 Ω 2c + a 2

1

= , which implies Ω c = a. 2

jΩ s , and hence, . Thus, G a ( jΩ) = jΩ + a s+a 2

= G a ( jΩ)G a (− jΩ) =

− jΩ jΩ Ω2 ⋅ = 2 2. jΩ + a − jΩ + a Ω +a

As Ω increases from

0 to ∞ , it can be seen that the square-magnitude function G a ( jΩ) the magnitude function Ga ( jΩ) =

Ω Ω2 + a2

2

and hence,

increases monotonically from

Ga ( j 0) = 0 to G a ( j∞) = 1. Let Ω c denote the 3 –dB cutoff frequency. Then 2

G a ( jΩ c ) = 4.16

Ω c2 Ω c2 + a 2

1

= , which implies Ω c = a. 2

s a 1⎛ s−a⎞ 1 = ⎜1 − ⎟ = ( A1 ( s ) − A2 ( s )) and G a (s ) = s+a 2⎝ s+a⎠ 2 s+a 1⎛ s−a⎞ 1 s−a = ⎜1 + . Now, ⎟ = ( A1 (s ) + A2 (s )), where A1 (s) = 1 and A2 (s ) = 2⎝ s+a⎠ 2 s+a H a (s) =

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j Ω − a − jΩ − a jΩ − a jΩ + a ⋅ = ⋅ = 1 for all j Ω + a − jΩ + a jΩ + a jΩ − a values of Ω, A1 (s) and A2 (s) are allpass functions. 2

A1 ( jΩ) = 1 and A 2 ( jΩ) =

4.17

H a (s) =

bs

jbΩ

. Thus, H a ( jΩ) = 2

and hence, jbΩ + Ω o2 − Ω 2 − jbΩ jbΩ 2 ⋅ H a ( j Ω ) = H a ( jΩ ) H a ( − j Ω ) = 2 2 jbΩ + Ω o − Ω − jbΩ + Ω o2 − Ω 2

=

s 2 + bs + Ω o

b2Ω2 (Ω 2o − Ω 2 ) 2 + b 2 Ω 2

. At Ω = 0, H a ( j 0) = 0, at Ω = ∞, H a ( j∞) = 0, and at

Ω = Ω o , H a ( jΩ) has the maximum value of 1. Now, d H a ( jΩ )

2

dΩ

=

2b 2 Ω(Ω o2 − Ω 2 )(Ω o2 + Ω 2 )

(

(Ω o2

2 2

−Ω ) +b Ω

frequency range 0 ≤ Ω < Ω o , Ω o < Ω < ∞,

2

d H a ( jΩ)

)

2 2

d H a ( jΩ)

. It therefore follows that in the

2

dΩ

> 0, and in the frequency range

2

< 0. Hence, in the frequency range 0 ≤ Ω < Ω o ,

dΩ

2

H a ( jΩ) is a monotonically increasing function of Ω and in the frequency range 2

Ω o < Ω < ∞, H a ( jΩ) is a monotonically decreasing function of Ω. Or in other words, H a (s) has a bandpass magnitude response. The 3 –dB cutoff frequencies are given by the solution of 2

2 2

2

2

2

2

(Ω o − Ω c ) + b Ω c = 2 b Ω c ,

b 2 Ω 2c (Ω 2o − Ω 2c ) 2 + b 2 Ω 2c

=

1 , 2

or,

i.e., Ω c4 − (b 2 + 2Ω o2 )Ω c2 + Ω o4 = 0. Substituting

x = Ω 2c in the last equation we get x 2 − (b 2 + 2Ω 2o ) x + Ω o4 = 0. Let x1 = Ω12 and

x 2 = Ω 22 be the two roots of this quadratic equation. Then, x1 x 2 = Ω12 Ω 22 = Ω o4 and x1 + x 2 = Ω12 + Ω 22 = b 2 + 2Ω 2o . Therefore, Ω1Ω 2 = Ω 2o . From the last two equations we get Ω12 + Ω 22 − 2Ω1Ω 2 = (Ω 2 − Ω1 ) 2 = b 2 + 2Ω 2o − 2Ω 2o = b 2 . Hence, Ω 2 − Ω1 = b.

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4.18

G a (s) =

s 2 + Ω 2o 2

Ω o2 − Ω

2

(Ω o2 − Ω 2 ) 2 (Ω o2

Now,

2 2

2

−Ω ) +b Ω d G a ( jΩ ) dΩ

2

2

=

(

Ω o < Ω < ∞, 2

and hence,

Ω o2 − Ω 2

jbΩ + Ω o2 − Ω 2 − jbΩ + Ω o2 − Ω 2

2b 2 Ω 3 (Ω 2 − Ω o2 )

(Ω o2

d G a ( jΩ) dΩ

⋅

−Ω

2

. Note G a ( j 0) = G a ( j∞) = 1 and G a ( jΩ o ) = 0.

2 2

2

−Ω ) +b Ω

frequency range 0 ≤ Ω < Ω o ,

G a ( jΩ )

jbΩ + Ω o2 2

s + bs + Ω o

G a ( jΩ) = G a ( jΩ)G a (− jΩ) = =

Ω o2 − Ω 2

. Thus, G a ( jΩ) = 2

d G a ( jΩ)

)

2 2

. It therefore follows that in the

2

dΩ

< 0, and in the frequency range

2

> 0. Hence, in the frequency range 0 ≤ Ω < Ω o ,

is a monotonically decreasing function of Ω and in the frequency 2

range Ω o < Ω < ∞, G a ( jΩ) is a monotonically increasing function of Ω. Or in other words, G a (s ) has a bandstop magnitude response. The 3 –dB cutoff frequencies are given by the solution of

(Ω o2 − Ω c2 ) 2

=

1

2 (Ω o2 − Ω c2 ) 2 + b 2 Ω 2c 4 2 2 2 4 Ω c − (b + 2 Ω o )Ω c + Ω o = 0. This

,

or, 2(Ω o2 − Ω c2 ) 2 = (Ω o2 − Ω 2c ) 2 + b 2 Ω 2c , i.e., last equation is exactly the same as in solution of Problem 4.18 from which we get

4.19

Ω1Ω 2 = Ω 2o and Ω 2 − Ω1 = b. bs 1 ⎛ s 2 − bs + Ω 2o ⎞⎟ 1 = ( A1 (s ) − A2 (s)) and H a (s ) = = ⎜1 − s 2 + bs + Ω 2o 2 ⎜⎝ s 2 + bs + Ω 2o ⎟⎠ 2 s 2 + Ω o2 1 ⎛ s 2 − bs + Ω o2 ⎞⎟ 1 = ( A1 (s) + A2 (s)), where G a (s) = = ⎜1 + s 2 + bs + Ω o2 2 ⎜⎝ s 2 + bs + Ω o2 ⎟⎠ 2

A1 (s) = 1 and A2 (s ) =

s 2 − bs + Ω o2

. Now, A1 ( jΩ) = 1 and A 2 ( jΩ) 2

2

s 2 + bs + Ω o

− Ω 2 − jbΩ + Ω o2 − Ω 2 + jbΩ + Ω o2 = A2 ( jΩ) A2 (− jΩ) = ⋅ = 1, for all values of − Ω 2 + jbΩ + Ω o2 − Ω 2 − jbΩ + Ω 2o Ω, A1 (s) and A2 (s) are allpass functions.

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4.20

(a) Let Ai (s ) =

s + λ*i . Since the pole of Ai (s ) is strictly in the left-half s –plane s − λi

and hence, Ai (s ) is causal and stable. Now Ai ( jΩ) =

2

= Ai ( jΩ) Ai* ( jΩ)

jΩ + λ*i − jΩ + λ i jΩ + λ*i jΩ − λ i ⋅ = ⋅ = 1. Hence, Ai (s ) is an allpass jΩ − λ i − jΩ − λ*i jΩ − λ i jΩ + λ*i N

function. Since, A(s ) = ∏ A i (s ), it is a product of causal, stable allpass functions, i =1

and as a result, is also a causal, stable allpass function.

(b) Ai (s )

2

2

=

Ai (s ) Ai* (s ) =

2

s + λ i + 2 Re{sλ i } s + λ*i s * +λ i ⋅ . Let = 2 2 s − λ i s * −λ*i s + λ i − 2 Re{s * λ i }

s = σ + jΩ and λ i = ai + jbi . Then Ai (s )

2

=

2

2

− 2Ωbi ) + 2σai

2

2

− 2Ωbi ) − 2σai

( s + λi ( s + λi 2

Since ai < 0, it follows from the above that Ai (s ) < 1 for σ > 0, Ai (s ) for σ = 0, and Ai (s ) 4.21

H a ( jΩ )

2

=

follows then

1 + (Ω / Ω c )

d k D(Ω ) dΩ k

2N

. Define D(Ω) =

1 H a ( jΩ )

= 2 N (2 N − 1)L(2 N − k + 1)

Ω=0

= 1 + (Ω / Ω c ) 2 N . It

2

Ω2N − k Ω 2c N

= 0 for k = 1, 2,K , N − 1, or, equivalently,

for k = 1, 2,K , N − 1. 4.22

=1

> 1 for σ < 0.

1

d k D(Ω ) dΩ k

2

2

.

. Therefore,

d k H a ( jΩ ) dΩ k

=0 Ω=0

⎛ 1 ⎞ ⎟ = −0.25. Therefore, ε 2 = 10 0.025 − 1 = 0.0593. Next, from 10 log10 ⎜⎜ 2⎟ ⎝1+ ε ⎠ ⎛ 1 ⎞ 1 Ω 6 ⎟ = −25, we get A 2 = 10 2.5 = 316.2278. Now, = s = 10 log10 ⎜⎜ =4 2⎟ k Ω p 1 .5 ⎝A ⎠ log10 (1 / k1 ) A2 − 1 1 315.2278 = 3.0943. = = 72.9381. Hence, N = = log10 (1 / k ) k1 0.0593 ε2 We choose N = 4 as the filter order. To verify using MATLAB, we use the code fragment [N,Wn]=buttord(2*pi*1500,2*pi*6000,0.25,25,'s');

and

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which yields N = 4 and Wn = 18365.51286. 4.23 The poles are given by p l = e j π ( 5 + 2 l ) / , 1 ≤ l ≤ 6 . Hence,

p1 = e j (7π / 12) = −0.2588 + j 0.9659, p2 = e j (9π / 12) = −0.7071 + j 0.7071, p3 = e j (9π / 12 ) = −0.9659 + j 0.2588, p 4 = e j (11π / 12 ) = p3* = −0.9659 − j 0.2588,

p5 = e j (13π / 12 ) = p2* = −0.7071 − j 0.7071, p6 = e j (15π / 12 ) = p1* = −0.2588 − j 0.9659. The poles can also be determined in MATLAB using the statement [z,p,k]=buttap(6) which yields

p = -0.2588 -0.2588 -0.7071 -0.7071 -0.9659 -0.9659

+ + + -

0.9659i 0.9659i 0.7071i 0.7071i 0.2588i 0.2588i

4.24 From Eq. (4.41) of text, TN (Ω) = 2ΩTN −1 (Ω) − TN − 2 (Ω), where TN (Ω) is defined in Eq. (4.40). Case 1: Ω ≤ 1. Making use of Eq. (4.40) in Eq. (4.41) we get

(

)

( ) = 2Ω cos(N cos −1 Ω − cos −1 Ω ) − cos(N cos −1 Ω − 2 cos −1 Ω ) = 2Ω [cos( N cos −1 Ω) cos(cos −1 Ω) + sin( N cos −1 Ω) sin(cos −1 Ω)] − [cos( N cos −1 Ω) cos(2 cos −1 Ω) + sin( N cos −1 Ω) sin(2 cos −1 Ω)]

T N (Ω) = 2Ω cos ( N − 1) ⋅ cos −1 Ω − cos ( N − 2) ⋅ cos −1 Ω

= 2Ω cos( N cos −1 Ω) cos(cos −1 Ω) − cos( N cos −1 Ω) cos(2 cos −1 Ω)

[

]

= 2Ω 2 cos( N cos −1 Ω) − cos( N cos −1 Ω) 2 cos 2 (cos −1 Ω) − 1

[

]

= cos( N cos −1 Ω) 2Ω 2 − 2Ω 2 + 1 = cos( N cos −1 Ω). Case 2: Ω > 1. Making use of Eq. (4.40) in Eq. (4.41) we get

(

)

(

)

T N (Ω) = 2Ω cosh ( N − 1) ⋅ cosh −1 Ω − cosh ( N − 2) ⋅ cosh −1 Ω . Using the trigonometric identities cosh( A − B) = cosh( A) cosh( B) − sinh( A) sinh( B), sinh( 2 A) = 2 sinh( A) cosh( A), and

cosh(2 A) = 2 cosh 2 ( A) − 1, and following a similar algebra as in Case 1, we can show T N (Ω) = cosh( N cosh −1 Ω).

4.25

From the solution of Problem 4.22, we have

Not for sale

1 1 = 4 and = 72.9381. Hence, k k1

95

cosh −1 (1 / k1 )

= 2.4151. We choose the filter order as N = 3. cosh −1 (1 / k ) The filter order obtained using the MATLAB statement [N,Wn]=cheb1ord(2*pi*1500,2*pi*6000,0.25, 25, 's') results in N=3. N=

⎛ 1 4.26 10 log10 ⎜⎜ ⎝ 1 + ε2

⎞ ⎛ 1 ⎞ ⎟⎟ = −0.25, which yields ε = 0.2434. 10 log10 ⎜⎜ ⎟⎟ = −25, which ⎝ A2 ⎠ ⎠ Ω p 1500 0.2434 ε yields A 2 = 316.2278. Now, k = = = 0.25 and k1 = = = 2 Ω s 6000 315 . 2278 A −1

= 0.0137. Substituting the value of k in Eq. (4..55a) we get k ' = 0.9682. Then from Eq. (4.55b) we get ρ 0 = 0.004. Substituting the value ρ0 in Eq. (4.55c) we get ρ = 0.004. Finally, from Eq. (4.54) we arrive at N = 2.0591. We choose the next higher integer as the filter order N = 3. The filter order obtained using the MATLAB statement [N,Wn]=ellipord(2*pi*1500,2*pi*6000,0.25, 25, 's') results in N=3. 4.27 B N ( s ) = (2 N − 1) B N −1 ( s ) + s 2 B N − 2 (s ), where B1 (s) = s + 1 and B2 (s ) = s 2 + 3s + 3. (a) Thus, B3 ( s ) = 5 B2 ( s ) + s 2 B1 ( s ) = 5( s 2 + 3s + 3) + s 2 ( s + 1) = s 3 + 6 s 2 + 15s + 15, B4 ( s ) = 7 B3 ( s ) + s 2 B2 ( s ) = 7( s 3 + 6 s 2 + 15s + 15) + s 2 ( s 2 + 3s + 3) = s 4 + 10 s 3 + 45s 2 + 105s + 105.

(b) B5 ( s ) = 9 B4 ( s ) + s 2 B3 ( s ) = 9( s 4 + 10 s 3 + 45s 2 + 105s + 105) + s 2 ( s 3 + 6 s 2 + 15s + 15) = s 5 + 15s 4 + 105s 3 + 420 s 2 + 945s + 945. ˆ = 2 π × 3. The mapping is thus s = 4.28 Ω p = 2π × 0.24 and Ω p

ˆ Ω pΩ p sˆ

=

4 π 2 × 0.72 . sˆ

Denote K = 4 π 2 × 0.72 = 28.4245. Hence, the desired highpass transfer function is given 10 by H HP (sˆ ) = H LP (s ) s = K / sˆ = 3 2 ⎛⎜ K ⎞⎟ + 4.309⎛⎜ K ⎞⎟ + 9.2835⎛⎜ K ⎞⎟ + 10 ⎝ sˆ ⎠ ⎝ sˆ ⎠ ⎝ sˆ ⎠ 10 sˆ 3 10 sˆ 3 = = K 3 + 4.309 K 2 sˆ + 9.2835Ksˆ 2 + 10 sˆ 3 10 sˆ 3 + 263.8785 sˆ 2 + 3481.5sˆ + 22966 =

sˆ 3 sˆ 3 + 26.38785 sˆ 2 + 348.15sˆ + 2296.6

.

ˆ = 2 π × 3. The mapping is thus s = 4.29 Ω p = 2π × 0.9 and Ω p

ˆ Ω pΩ p sˆ

=

4π 2 × 2.7 . Denote sˆ

K = 4 π 2 × 2.7 = 106.5917. Hence, the desired lowpass transfer function is given by

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H LP (s ) = H LP (s) s = K / sˆ =

=

=

⎛⎜ K ⎞⎟ ⎝ sˆ ⎠ 3

3

2

⎛⎜ K ⎞⎟ + 9.238⎛⎜ K ⎞⎟ + 40.087⎛⎜ K ⎞⎟ + 100 ⎝ sˆ ⎠ ⎝ sˆ ⎠ ⎝ sˆ ⎠

K3 K 3 + 9.238 K 2 sˆ + 40.087 Ksˆ 2 + 100 sˆ 3 12110.735

. sˆ 3 + 42.729 sˆ 2 + 1049.602 sˆ + 12110.735 ˆ = 2 π × 3 = 6 π, Ω ˆ ˆ 4.30 Ω p = 2 π × 0.25 = 0.5π, Ω o p 2 − Ω p1 = 2 π(0.5) = π. The mapping is thus ˆ2 ⎛ sˆ 2 + 36π 2 ⎞ sˆ 2 + 36 π 2 sˆ 2 + Ω o ⎟= = 0.5π⎜ s = Ωp . ˆ ˆ ) ⎟ ⎜ ˆ ˆ π s s 2 − Ω sˆ(Ω p2 p1 ⎠ ⎝ 2 ⎡⎛ ˆ 2 ⎤ s + 36 π 2 ⎞ ⎟⎟ + 367.93⎥ 0.01⎢⎜⎜ ⎢⎝ 2 sˆ ⎥ ⎠ ⎣ ⎦ H BP (sˆ ) = H LP (s ) s = (sˆ 2 + 36π 2 ) / 2 sˆ = 2 ⎛ sˆ 2 + 36 π 2 ⎞ ⎛ sˆ 2 + 36 π 2 ⎞ ⎟⎟ + 3.895 ⎟⎟ + 2.269⎜⎜ ⎜⎜ ⎠ ⎝ 2 sˆ ⎠ ⎝ 2 sˆ 4 2 0.01(sˆ + 2182.33 sˆ + 126242.18) = . sˆ 4 + 4.538 sˆ 3 + 726.19 sˆ 2 + 1612.38 sˆ + 126242.18 ˆ = 2π × 6.5 × 10 3 and Ω ˆ = 2 π × 1.5 × 10 3. 4.31 Ω p

s

⎛ 1 ⎞ 10 log10 ⎜ ⎟ = −0.5, and hence, ε 2 = 10 0.05 − 1 = 0.122. 2 ⎝ 1+ ε ⎠ 1 A2 − 1 ⎛ 1 ⎞ = = 286.2632. 10 log10 ⎜ ⎟ = −40, and hence, A 2 = 10 4. Therefore, k1 ⎝ A2 ⎠ ε2 ˆ Ω 6.5 13 1 Ω 13 p Set Ω p = 1. Then Ω s = = = . Thus, = s = . The order of the ˆ k Ωp 1 .5 3 3 Ω s prototype lowpass filter is thus given by N =

log10 (1 / k1 ) = 3.8579. As a result, we log10 (1 / k )

choose the filter order as N = 4. The order of the prototype lowpass filter obtained using the MATLAB statement [N,Wn]=buttord(1,13/3,0.5, 40, 's') results in N=4. The order of the desired highpass filter is also 4. 4.32

Fˆ p1 = 20 × 10 3 , Fˆ p2 = 45 × 10 3 , Fˆs1 = 10 × 10 3 , and Fˆs 2 = 50 × 10 3. Thus,

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Fˆ p1 Fˆ p2 = 9 × 10 8 and Fˆs1 Fˆs 2 = 7.5 × 10 8. Since Fˆ p1 Fˆ p 2 > Fˆs1 Fˆs 2 , we can either increase left stopband edge Fˆs1 or decrease the left passband edge Fˆ p1 to make Fˆ p1 Fˆ p 2 = Fˆs1 Fˆs 2 . We choose to increase Fˆs1 to a new value given by Fˆs1 = 18 × 10 3 , in which case Fˆ p1 Fˆ p2 = Fˆs1 Fˆs 2 = Fˆo2 = 9 × 10 8. The center angular frequency of the ˆ = 2 π × 30 × 10 3. The passband width is bandpass filter is therefore Ω o

3 ˆ −Ω ˆ Bw = Ω p1 p 2 = 2 π × 25 × 10 .

To determine the bandedges of the prototype lowpass filter we set Ω p = 1 and thus, ˆ 2 −Ω ˆ2 Ω 30 2 − 18 2 o s1 Ωs = Ω p = = 1.28. ˆ B 18 × 25 Ω s1 w Ωp 1 Now, k = = = 0.78125. Hence, k ' = 1 − k 2 = 0.62421826 . Ω s 1.28

⎛ 1 Next, 10 log10 ⎜⎜ ⎝ 1 + ε2

⎞ ⎟⎟ = −0.25 or equivalently, log10 (1 + ε 2 ) = 0.025 which yields ⎠ ⎛ 1 ⎞ ⎟⎟ = −50 ε 2 = 10 0.025 − 1 = 0.05925372 5 or ε = 0.243421. Likewise, 10 log10 ⎜⎜ ⎝ A2 ⎠

or, equivalently, log10 ( A 2 ) = 5 which yields A 2 = 10 5 = 100000. Therefore, ε 1 − k' k1 = = 7.69768 × 10 − 4 , ρ 0 = = 0.058635856. As a result, 2(1 + k ' ) A2 − 1 ρ = ρ 0 + 2(ρ 0 ) 5 + 15(ρ 0 ) 9 + 150(ρ 0 )13 = 0.058637246 . Hence,

2 log10 (4 / k1 ) = 6.0328. We choose N = 7 as the order of the prototype lowpass log10 (1 / ρ) filter. Note that the order can also estimated using the specifications of the bandpass filter. To this end, the statement to use is [N,Wn]=ellipord([20 45],[15 50],0.25,50,'s') which also yields N=7 as the order of the prototype lowpass filter. The order of the desired bandpass filter is therefore 7 × 2 = 14. N=

4.33 Fˆ p1 = 10 × 10 6 , Fˆ p2 = 70 × 10 6 , Fˆ Fˆ = 70 × 1013 and Fˆ Fˆ

Fˆs1 = 20 × 10 6 , and Fˆs 2 = 45 × 10 6. Thus, 13 ˆ ˆ ˆ ˆ p1 p2 s1 s 2 = 90 × 10 . Since F p1 F p 2 < Fs1 Fs 2 , we can either increase left passband edge Fˆ or decrease the left stopband edge Fˆ to make p1

s1

Fˆ p1 Fˆ p 2 = Fˆs1 Fˆs 2 . We choose to increase Fˆ p1 to a new value given by

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Fˆ Fˆ Fˆ p1 Fˆ p2 = s1 s 2 = 12.8571 × 10 6 , in which case Fˆ p1 Fˆ p2 = Fˆs1 Fˆs 2 = Fo2 = 700 × 1012. Fˆ p2 ˆ −Ω ˆ = 2 π × 25 × 10 6 and the center angular The width of the stopband is Bw = Ω s2 s1

frequency of the stopband is Ω o2 = 4 π 2 × 700 × 1012. To determine the bandedges of the prototype lowpass filter we set Ω s = 1 resulting in its ˆ B Ω p1 w = 0.4375. passband edge Ω p = Ω p ˆ 2 −Ω ˆ2 Ω o p1

⎛ 1 Now, 10 log10 ⎜⎜ ⎝ 1 + ε2

⎞ ⎟⎟ = −0.5 or equivalently, log10 (1 + ε 2 ) = 0.05 which yields ⎠ ⎛ 1 ⎞ ⎟⎟ = −30 ε 2 = 10 0.05 − 1 = 0.1220184543 or ε = 0.349114. Likewise, 10 log10 ⎜⎜ ⎝ A2 ⎠

or, equivalently, log10 ( A 2 ) = 3 which yields A 2 = 10 3 = 1000. Therefore,

1 A2 − 1 999 1 Ωs 1 = = = 90.4836236. = = = 2.2857 and k1 ε 0.349114 k Ω p 0.4375 1 1 Substituting the values of and in Eq. (4.43) we get k k1 cosh −1 (90.4836236 )

= 3.5408. We therefore choose N = 4 as the order of the cosh −1 (2.2857) prototype lowpass filter. The order of the desired bandstop filter is thus 8. Using the statement [N,Wn]=cheb1ord(0.4375,1,0.5,30,'s') we get N=4. Note that the order can also estimated using the specifications of the bandstop filter. To This end, the statement to use is [N,Wn]=cheb1ord([10 70],[20 45],0.5,30,'s') which also yields N=4 as the order of the prototype lowpass filter. N=

4.34 From Eq. (4.71), the difference in dB in the attenuation levels at Ω p and Ω s is given by 20 N log10 (Ω p / Ω s ). Hence, for Ω o = 2Ω p , the attenuation difference in dB is equal to

20 N log10 2 = 6.0206 N . Likewise, for Ω o = 3Ω p , the attenuation difference in dB is equal to 20 N log10 3 = 9.5424 N . Finally, for Ω o = 4Ω p , the attenuation difference in dB is equal to 20 N log10 4 = 12.0412 N . 4.35 The equivalent representation of the D/A converter of Figure 4.48 reduces to the circuit shown below if j -th bit is ON and the remaining bits are OFF, i.e., a j = 1 and

a k = 0, k ≠ j.

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Vo,j _ 2N 1

_ 2j 1

2j

a j VR

_ 2j 2

20

GL

+ Yin

In the above circuit, Yin is the total conductance seen by the load conductance GL which is N −1

given by Yin = ∑ 2 i = 2 N − 1. The above circuit can be redrawn as indicated below: i =0

Vo,j _ Yin _ 2 j 1

_ 2j 1

a j VR

GL

+

2 j −1 ⋅ a j VR . Using the Using the voltage-divider relation we then get Vo, j = Yin + G L superposition theorem, the general expression for the output voltage Vo is thus given by ⎛ N RL 2 j −1 ⋅ a j V R = ∑ 2 j −1 a j ⎜ ⎜ 1 + (2 N − 1) R j =1 j =1 Yin + G L L ⎝ N

Vo = ∑

⎞ ⎟V . ⎟ R ⎠

4.36 The equivalent representation of the D/A converter of Figure 4.49 reduces to the circuit shown below on the left if N -th bit is ON and the remaining bits are OFF, i.e., a N = 1 and

a k = 0, k ≠ N , Vo,N G __ 2

G __ 2

aNVR

Vo,N G __ 2

GL

+

aNVR

G GL + __ 2

+

which simplifies to the circuit shown above on the right. Using the voltage-divider relation we then get Vo, N =

G 2 G G +GL + 2 2

⋅ a N VR =

RL ⋅ a N VR . 2( R + R L )

The equivalent representation of the D/A converter of Figure 4.49 reduces to the circuit shown below on the left if ( N − 1) -th bit is ON and the remaining bits are OFF, i.e., a N −1 = 1 and a k = 0, k ≠ N − 1, Not for sale

100

G

G

Vo,N _1

Vo,N _1

G __ 2

G __ 2

G __ 2 aN _ 1VR

G __ 2

G __ 2

GL

+

aN _ 1VR

__ GL + G 2

+

which simplifies to the circuit shown above on the right. Its Thevenin equivalent circuit is indicated below: Vo,N _1 G __ 2

G GL + __ 2

+ a___ N _1 VR 2 from which we readily obtain G 2

a a RL ⋅ N −1 V R = ⋅ N −1 V R . G + GL 2 2( RL + R) 2 Following the same procedure we can show that if the l –th bit is ON and the remaining bits are OFF, i.e., a l = 1, and a k = 0, k ≠ l, then a RL Vo, l = ⋅ l VR . 2( RL + R) 2 N − l Hence, in general we have N a RL ⋅ l VR . Vo = ∑ N −l l =1 2( R L + R ) 2 Vo, N −1 =

4.37 From the input-output relation of the first-order hold, we get the expression for the impulse δ(nT ) − δ(nT − T ) response as h f (t ) = δ(nT ) + (t − nT ), nT ≤ t < (n + 1)T . In the range T t δ(0) − δ(−T ) 0 ≤ t < T , the impulse response is given by h f (t ) = δ(0) + t =1+ . T T Likewise, in the range T ≤ t < 2T , the impulse response is given by δ(T ) − δ(0) t h f (t ) = δ(T ) + (t − T ) = 1 − . Outside these two ranges, h f (t ) = 0. Hence we T T have

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h f (t) 2

t , 0 < t < T, T t h f (t ) = 1 _ , T < t < 2T , T 0, otherwise.

{

1+

1

0

T

2T

t

_1

Using the step function we can write t⎞ t⎞ ⎛ ⎛ h f (t ) = ⎜1 + ⎟ [µ(t ) − µ(t − T )] + ⎜1 − ⎟ [µ(t − T ) − µ(t − 2T )] ⎝ T⎠ ⎝ T⎠ t 2( t − T ) (t − 2T ) = µ ( t ) + µ( t ) − µ(t − T ) − 2µ(t − T ) − µ(t − 2T ) + µ(t − 2T ) + 2µ(t − 2T ). T T T Taking the Laplace transform of the above equation we arrive at the transfer function 1 1 2 e − sT e − sT e − 2 sT 1 e − 2 sT e − 2 sT ⎛ 1 + sT H f (s) = + − ⋅ −2 − + ⋅ +2 =⎜ s Ts 2 T s 2 s s T s ⎝ T s2 Hence, the frequency response is given by − jΩT ⎛ 1 + jΩT ⎞⎛⎜ 1 − e H f ( jΩ ) = ⎜ ⎟ jΩ ⎝ T ⎠⎜⎝

− sT

⎞ ⎛⎜ 1 − e ⎟ s ⎠ ⎜⎝

2

2 ⎞ −1 ⎟ = T 1 + Ω 2 T 2 ⎛⎜ 2 sin(ΩT / 2) ⎞⎟ e − jΩT e j tan ΩT . A ⎟ ⎝ ΩT / 2 ⎠ ⎠ plot of the magnitude responses of the zero-order hold and the first-order hold is shown below:

Amplitude

1.5

← first-order hold

1

← zero-order hold

0.5

0

0

0.5

1

1.5

2

2.5

3

Ω

4.37

From the input-output relation of thelinear interpolator, we get the expression for the δ(nT ) − δ(nT − T ) impulse response as h f (t ) = δ(nT − T ) + (t − nT ), nT ≤ t < (n + 1)T . In the T δ(0) − δ(−T ) range 0 ≤ t < T , the impulse response is given by h f (t ) = δ(−T ) + t. Likewise, T

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102

2

⎞ ⎟ . ⎟ ⎠

in the range T ≤ t < 2T , the impulse response is given by h f (t ) = δ(0) +

δ(T ) − δ(0) (t − T ). T

⎧ t 0 ≤ t < T, ⎪ T, ⎪⎪ t Outside these two ranges, h f (t ) = 0. Hence we have h f (t ) = ⎨2 − , T ≤ t < 2T , ⎪ T otherwise. ⎪ 0, ⎪⎩ h f (t) 1

0

t

2T

T

Using the step function we can write t⎞ t ⎛ h f (t ) = [µ(t ) − µ(t − T )] + ⎜ 2 − ⎟ [µ(t − T ) − µ(t − 2T )] T⎠ T ⎝ t 2( t − T ) (t − 2T ) = µ(t ) − µ(t − T ) + µ(t − 2T ). T T T Taking the Laplace transform of the above equation we arrive at the transfer function ⎛ 1 − e − sT − + = T⎜ H f (s) = ⎜ sT s 2T s 2T s 2T ⎝ 1

2 e − sT

e − 2 sT

2

⎞ ⎟ . Hence, the frequency response is given ⎟ ⎠

2

2 ⎛ 1 − e − jΩT ⎞ ⎛ sin(ΩT / 2) ⎞ − jΩT ⎟ ⎜ by H f ( jΩ) = T . A plot of the magnitude = T⎜ ⎟ e ⎜ j ΩT ⎟ ⎝ ΩT / 2 ⎠ ⎠ ⎝ responses of the ideal filter, zero-order hold and the first-order hold is shown below:

Amplitude

1 0.8

← zero-order hold

0.6

← linear interpolator

0.4 0.2 0

0

0.5

1

1.5

2

2.5

3

Ω

M4.1 We use N = 4 and Wn = 18365.512865 computed in Problem 4.22 and use omega = 0:2*pi:2*pi*10000; to evaluate the frequency points. The gain plot obtained using Program 4_2 is shown below.

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Analog Lowpass Filter

Passband Details 0.5

-10

Gain, dB

Gain, dB

0

-20

0

-0.5

-30 -40

0

2000

4000 6000 Frequency, Hz

8000

-1

10000

0

500


2000

2500

M4.2 We use N = 3 computed in Problem 4.23 and Fp = 2*pi*1500 and Rp = 0.25 and use omega = 0:2*pi:2*pi*10000; to evaluate the frequency points. The gain plot obtained using Program 4_3 is shown below. Analog Lowpass Filter


0

Gain, dB

Gain, dB

-10 -20 -30

0

-0.5

-40 -50

0

2000


8000

-1

0

500


2000

M4.3 We replace the statement Fp = input('Passband edge frequency in Hz = '); with Fs = input('Stopband edge frequency in Hz = '); replace Rp = input('Passband ripple in dB = '); with Rs = input('Minimum stopband attenuation in dB = '); and replace [num,den] = cheby1(N,Rp,Fp,'s'); with [num,den] = cheby2(N,Rs,Fs,'s'); to modify Program 4_3. Next, we run the modified program using N = 3 and Rs = 25, and Fs = 2*pi*6000. The gain response plot generated by the modified program is shown below.

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Passband Details


0.5

0

Gain, dB

Gain, dB

-10 -20 -30

0

-0.5

-40 -50

0

2000


-1

8000

0

500


2000

2500

The numerator and the denominator coefficients of the 3rd order Type 2 Chebyshev lowpass filter can be obtained by typing num and den in the command window: H LP (s ) =

10138.1864 s 2 + 4.8663294 × 1010 s 3 + 7030.25552 5 s 2 + 2.41983322 54 × 10 7 s + 4.8663294 × 1010

.

M4.4 We use N = 3 and Wn = 9424.777960769379 computed in Problem 4.26 in Program 4_4 and use omega = [0: 200: 12000*pi]; to evaluate the frequency points. The gain plot generated by running this program is shown below: Analog Lowpass Filter


0

-20

Gain, dB

Gain, dB

-10

-30 -40

0

-0.5

-50 -60

0

1000

2000 3000 4000 Frequency, Hz

5000

6000

-1

0

500


2000

M4.5 The MATLAB program used is as given below: [N,Wn]=buttord(1,13/3,0.5, 40, 's'); [B,A] = butter(N,Wn,'s'); [num,den]=lp2hp(B,A,2*pi*6500); figure(1) [h,w]=freqs(B,A);gain = 20*log10(abs(h)); plot(w,gain);grid xlabel('\Omega');ylabel('Gain, dB'); title('Analog Lowpass Filter'); figure(2) [h,w]=freqs(num,den);gain = 20*log10(abs(h)); plot(w/(2*pi),gain);grid xlabel('Frequency, Hz');ylabel('Gain, dB'); title('Analog Highpass Filter'); Not for sale

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H LP (s ) = H HP (s ) =

3.5262257 4

3

s + 3.58086432 s + 6.41129464 s 2 + 6.72423556 s + 3.5262257

,

s4 s 4 + 7.7880 s 3 + 3.0326485 × 10 9 s 2 + 6.91763168 × 1013 s + 7.8897418 × 1017

Prototype Lowpass Filter


0

Gian, dB

Gian, dB

-10 -20 -30

0

-0.5

-40 -50

0

1

2

3

4

-1

5

0

0.5

Ω

1

1.5

Ω

Analog Highpass Filter


0

Gain, dB

Gain, dB

-10 -20 -30

0

-0.5

-40 -50

0

0.5

1 Frequency, Hz

1.5

2 4

x 10

-1 0.5

1 1.5 Frequency, Hz

2 4

x 10

M4.6 The MATLAB program used is given below: [N,Wn] = ellipord(1,1.28,0.25,50,'s'); [B,A] = ellip(N,0.25,50,Wn,'s'); [num,den] = lp2bp(B,A,2*pi*30e3,2*pi*25e3); figure(1) omega = [0:0.01:10]; h = freqs(B,A,omega); gain = 20*log10(abs(h)); plot(omega,gain); grid; axis([0 5 -80 5]); xlabel('\Omega'); ylabel('Gain, dB'); title('Analog Lowpass Filter'); figure(2) omega = [0:200:100e3*2*pi]; h = freqs(num,den,omega); gain = 20*log10(abs(h)); Not for sale

106

.

plot(omega/(2*pi),gain); grid; axis([0 60e3 -80 5]); xlabel('Frequency in Hz'); ylabel('Gain, dB'); title('Analog Bandpass Filter'); Analog Bandpass Filter

0

0

-20

-20 Gain, dB

Gain, dB


-40

-60

-60 -80

-40

-80

0

1

2

3

4

5

0

1

2 3 4 Frequency in Hz

Ω

H LP (s ) =

6

4

5

6 4

x 10

2

0.0185 s + 0.1364 s + 0.2887 s + 0.186835 7

6

s + 1.36426 s + 2.9795 s 5 + 2.7545 s 4 + 2.70025 s 3 + 1.56215 s 2 + 0.7275 s + 0.1868

The numerator and denominator coefficients of can be obtained by typing num and den in the Command Window. M4.7 The MATLAB program used is given below: [N,Wn] = cheb1ord(0.3157894, 1, 0.5, 30,'s'); [B,A] = cheby1(N,0.5, Wn,'s'); [num,den] = lp2bs(B,A,2*pi*sqrt(700)*10^6, 2*pi*15e6); figure(1) omega = [0:0.01:10]; h = freqs(B,A,omega); gain = 20*log10(abs(h)); plot(omega, gain); grid; axis([0 4 -70 5]); xlabel('\Omega'); ylabel('Gain, dB'); title('Analog Lowpass Filter'); figure(2) omega = [0:10000:160e6*pi]; h = freqs(num,den,omega); gain = 20*log10(abs(h)); plot(omega/(2*pi), gain); grid; axis([0 80e6 -70 5]); xlabel('Frequency in Hz'); ylabel('Gain, dB'); title('Analog Bandstop Filter');

H LP (s ) =

0.02253823 s 3 + 0.3956566 s 2 + 0.1530643 s + 0.02253823

Not for sale

.

107

.



-20

Gain, dB

Gain, dB

0

-40

0

-0.5

-60 0

1

2

3

-1

4

0

0.1

0.2

Ω

0.3

0.4

0.5

Ω

Stopband Details

Analog Bandstop Filter

0.5

-20

Gain, dB

Gain, dB

0

-40

0

-0.5

-60

-1 0

2

4 Frequency in Hz

6

8 7

x 10

0

2

4 Frequency in Hz

6

8 7

x 10

M4.8 The MATLAB program to generate the plots of Figure 4.56 is given below: % Droop Compensation w = 0:pi/100:pi; h1 = freqz([-1/16 9/8 -1/16],1,w); h2 = freqz(9, [8 1], w); w1 = 0; for n = 1:101; h3(n) = sin(w1/2)/(w1/2); w1 = w1 + pi/100; end m1 = 20*log10(abs(h1)); m2 = 20*log10(abs(h2)); m3 = 20*log10(abs(h3)); plot(w/pi,m3,’-’,w/pi,m1+m2,’--’,w/pi,m2+m3,’-.’);grid xlabel(‘Normalized frequency’);ylabel(Gain, dB’);

Not for sale

108

Chapter 5 5.1

N −1 N −1 ~ ~ ~ Let ~ y[ n] = ∑ ~ x[r ]h [ n − r ]. Then ~ y[ n + kN ] = ∑ ~ x[r ]h [ n + kN − r ]. Since h [ n] is r =0

r =0

~ ~ periodic in n with a period N , h [ n + kN − r ] = h [ n − r ]. Therefore ~ y[ n + kN ] N −1 ~ = ∑ ~ x[r ]h [ n − r ] = ~ y[ n], hence ~ y[ n] is also periodic in n with a period N . r =0

5.2

4 ~ ~ ~ ~ ~ ~ (a) ~ y[ 0] = ∑ ~ x[r ]h [ −r ] = ~ x[0]h [0] + ~ x[1]h [ 4] + ~ x[2]h [3] + ~ x[3]h [2] + ~ x[ 4]h [1] = −13, r =0

4

~ ~ ~ ~ ~ ~ ~ y[1] = ∑ ~ x[r ]h [1 − r ] = ~ x[0]h [1] + ~ x[1]h [0] + ~ x[2]h [ 4] + ~ x[3]h [3] + ~ x[ 4]h [2] = −13, r =0 4

~ ~ ~ ~ ~ ~ ~ y[ 2] = ∑ ~ x[r ]h [2 − r ] = ~ x[0]h [2] + ~ x[1]h [1] + ~ x[2]h [0] + ~ x[3]h [ 4] + ~ x[ 4]h [3] = −13, r =0 4

~ ~ ~ ~ ~ ~ ~ y[3] = ∑ ~ x[r ]h [3 − r ] = ~ x[0]h [3] + ~ x[1]h [2] + ~ x[2]h [1] + ~ x[3]h [0] + ~ x[ 4]h [ 4] = −13, r =0 4

~ ~ ~ ~ ~ ~ ~ y[ 4] = ∑ ~ x[r ]h [ 4 − r ] = ~ x[0]h [ 4] + ~ x[1]h [3] + ~ x[2]h [2] + ~ x[3]h [1] + ~ x[ 4]h [0] = −13. r =0

y[ n] = {− 13, − 13, − 13, − 13, − 13,}, 0 ≤ n ≤ 4. Therefore, ~ y[ n] = {1, 1, 1, 1, 1,}, 0 ≤ n ≤ 4. (b) ~ 5.3

~ [ n + rN ] = ψ ~ [ n], hence all the terms which are not in the range 0,1,K, N − 1, Since ψ k k ~ [ n], where 0 ≤ k ≤ N − 1. Hence, in this case the Fourier can be accumulated to ψ k series representation involves only N complex exponential sequences. Let 1 N −1 ~ j 2 πkn / N ~ x [ n] = , then ∑ X[ k ] e N k =0 N −1

1 N −1 ~ ⎛ N −1 j 2 π( k − r )n / N ⎞ 1 N − 1 N −1 ~ j 2 π( k − r )n / N ⎟⎟. = x[ n] e − j 2 πrn / N = ∑ ~ ∑ ∑ X[ k ] e ∑ X[ k ] ⎜⎜ ∑ e N n=0 k =0 N k =0 n =0 ⎝ n=0 ⎠

Now, from Eq. (5.11), the inner summation is equal to N if k = r , otherwise it is equal N −1 ~ ~ x[ n] e − j 2 πrn / N = X [r ]. Next, we observe X[ k + lN ] to 0. Thus, ∑ ~ n=0

N −1

N −1 N −1 ~ = ∑ ~ x[ n] e − j 2 π( k + lN )n / N = ∑ ~ x[ n] e − j 2 πkn / N e − j 2 πln = ∑ ~ x[ n] e − j 2 πkn / N = X[ k ]. n=0

n=0

Not for sale.

n=0

109

5.4

{

}

1 πn ~ x1[n] is N = 8. x1[ n] = cos⎛⎜ ⎞⎟ = e jπn / 4 + e − jπn / 4 . The period of ~ ⎝ 4⎠ 2 7 ⎫ 1⎧ 7 ~ X1[ k ] = ⎨ ∑ e j 2 πn / 8 e − j 2 πkn / 8 + ∑ e − j 2 πn / 8 e − j 2 πkn / 8 ⎬ 2 ⎩n = 0 n =0 ⎭

(a)

=

7 ⎫ 1 ⎧ 7 − j 2 πn( k −1) / 8 + e e − j 2 πn( k +1) / 8 ⎬. Now, from Eqn. (5.11) we observe ∑ ∑ ⎨ 2 ⎩n = 0 n =0 ⎭

7

∑e

− j 2 πn( k −1) / 8

n=0

7 8, for k = 1, 8, for k = 7, and ∑ e − j 2 πn( k +1) / 8 = ⎧⎨ = ⎧⎨ 0 , otherwise , ⎩0, otherwise. ⎩ n=0

~ 4, for k = 1,7, Hence, X1[ k ] = ⎧⎨ 0 ⎩ , otherwise. 1 3 πn πn (b) ~ x 2 [ n] = sin⎛⎜ ⎞⎟ + 3 cos⎛⎜ ⎞⎟ = {e jπn / 3 − e − jπn / 3} + {e jπn / 4 + e − jπn / 4 }. The 2 2 j ⎝ 3⎠ ⎝ 4⎠ πn πn x [ n] is the is 6 and the period of cos is 8. Hence, the period of ~ period of sin

( )

( )

3

4

2

GCM of (6,8) and is 24.

1 ⎧ 23 j8πn / 24 − j 2 πkn / 24 23 − j8πn / 24 − j 2 πkn / 24 ⎫ ~ − ∑e X 2 [k ] = e e ⎨ ∑e ⎬ 2 j ⎩n = 0 n =0 ⎭ 23 ⎫ 3 ⎧ 23 + ⎨ ∑ e j 6 πn / 24 e − j 2 πkn / 24 − ∑ e − j 6 πn / 24 e − j 2 πkn / 24 ⎬ 2 ⎩n = 0 n=0 ⎭

=

1 ⎧ 23 − j 2 πn( k − 3) / 24 23 − j 2 πn( k + 3) / 24 ⎫ − ∑e ⎨ ∑e ⎬ 2 j ⎩n = 0 n=0 ⎭

k = 3, ⎧− j12, ⎪ j12, 3 ⎧ 23 − j 2 πn( k − 4) / 24 23 − j 2 πn( k + 4) / 24 ⎫ ~ k = 21, + ⎨ ∑e − ∑e ⎬. Hence X 2 [ k ] = ⎨ 36, = 4, 20, k 2 ⎩n = 0 n =0 ⎭ ⎪ 0 , otherwise. ⎩ 5.5

~ Let P[ k ] denote the coefficients of the Fourier series representation of ~ p[ n]. Since ~ p[ n] is periodic with a period N , then from Eq. (5.185b), we have N −1 ~ P[ k ] = ∑ ~ p[ n] e − j 2 πkn / N = 1. Hence, from Eq. (5.185a) we get

~ p[ n] =

5.6

n=0 1 N −1 ~

j 2 πln / N = ∑ P[l] e

N l =0

~ X [ k ] = X ( e jω )

1 N −1 j 2 πln / N . ∑ e N l=0 ∞

ω = 2 πk / N

= X (e j 2 πk / N ) = ∑ x[ n] e − j 2 πk / N , − ∞ < k < ∞. n = −∞

~ ~ Now, X[ k + lN ] = X (e j 2 π( k + lN ) / N ) = X (e j 2 πk / N e jπl ) = X (e j 2 πk / N ) = X[ k ].

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110

1 N −1 ~ 1 N −1⎛ ∞ j 2 πkn / N − j 2 πln / N ⎞⎟ j 2 πkn / N x [ n] = = Likewise, ~ ∑ ⎜⎜ ∑ x[l] e ∑ X[ k ] e ⎟e N k =0 N k = 0 ⎝ l = −∞ ⎠ 1 N −1 ∞ j 2 π( k − l )n / N = . Let l = n + rN. Then ∑ ∑ x[l] e N k =0 l = −∞ N −1 ⎛ N −1 − j 2 πkr ⎞ 1 ∞ ~ ⎟⎟. But ∑ e − j 2 πkr = N . Hence, x [ n] = ∑ x[ n + rN ]⎜⎜ ∑ e N r = −∞ k =0 ⎠ ⎝ k =0

∞ ~ x[ n] = ∑ x[ n + rN ]. r = −∞

5.7

N −1 N −1 ~ (a) G[ k ] = ∑ g~[ n] e − j 2 πkn / N = ∑ ~ x [ n ]~ y[ n] e − j 2 πkn / N . Now, n=0 N −1 ~

n=0

1 − j 2 πrn / N ~ x [ n] = . Therefore, ∑ X [r ] e N r =0 1 N −1 N −1 ~ t 1 N −1 ~ N −1t ~ − j 2 π( k − r )n / N − j 2 π( k − r ) n / N = G[ k ] = ∑ ∑ X[r ] y[ n] e ∑ X[r ] ∑ y[ n] e N n=0 r =0 N r =0 n=0 1 N −1 ~ ~ = ∑ X[r ]Y [ k − r ]. N r =0 ~ 1 N −1 ~ ~ 1 N −1 N −1 ~ ~ j 2 πkn / N j 2 πk ( n − r ) / N = (b) h [ n] = ∑ X[ k ]Y [ k ] e ∑ ∑ x[r ]Y [ k ] e N k =0 N k =0 r =0 N −1 N −1 ⎛ 1 N −1 ~ j 2 πk ( n − r ) / N ⎞⎟ x[r ] ⎜⎜ x [r ] ~ y[ n − r ] . = = ∑ ~ ∑~ ∑ Y [k] e ⎟ N r =0 k r = 0 = 0 ⎠ ⎝ 5.8

(a) x a [ n] = sin(2πn / N ) = (e j 2 πn / N − e − j 2 πn / N ). Therefore, 2j 1

X a [k ] =

1 N −1 j 2 πn / N − j 2 πkn / N 1 N −1 − j 2 πn / N − j 2 πkn / N − e e ∑ e ∑ e 2 j n=0 2 j n=0

1 N −1 − j 2 π( k −1)n / N 1 N −1 − j 2 π( k +1)n / N − . From Eq. (5.11), the first sum is ∑ e ∑ e 2 j n=0 2 j n=0 equal to N when k = 1 and 0 othereise. Likewise, from Eq. (5.11), the second sum is equal to N when k = N − 1 and 0 otherwise. Therefore, k = 1, ⎧ N / 2 j, ⎪ X a [ k ] = ⎨− N / 2 j, k = N − 1, ⎪⎩ 0, otherwise. =

N 2 πn ⎞ 1 1 4 πn ⎞ 1 for (b) x b [ n] = cos 2 ⎛⎜ ⎟ = + cos⎛⎜ ⎟. Now the N –point DFT of is 2 2 ⎝ N ⎠ 2 2 ⎝ N ⎠

( ) is

k = 0 and 0 otherwise. From Example 5.2, the N –point DFT of cos k = 2 and k = N − 2 and 0 otherwise. Therefore,

Not for sale.

4 πn N

N 2

for

111

k = 0, ⎧⎪ N / 2, X b [ k ] = ⎨ N / 4, k = 2, N − 2, ⎪⎩ 0, otherwise. 2 πn ⎞ 1 6 πn ⎞ (c) x c [ n] = cos 3 ⎛⎜ ⎟ = cos⎛⎜ ⎟+ ⎝ N ⎠ 4 ⎝ N ⎠ N 6 πn ⎞ DFT of cos⎛⎜ for k = 3 and ⎟ is 2 ⎝ N ⎠

3 2 πn ⎞ cos⎛⎜ ⎟. 4 ⎝ N ⎠

From Example 5.2, the N –point

k = N − 3 and 0 otherwise. Likewise, from

N 2 πn ⎞ Example 5.2, the N –point DFT of cos⎛⎜ for k = 1 and k = N − 1 and 0 ⎟ is 2 ⎝ N ⎠ ⎧ N / 8, k = 3, N − 3, ⎪ otherwise. Therefore, X c [ k ] = ⎨3N / 8, k = 1, N − 1, ⎪⎩ 0, otherwise.

5.9

N −1

N −1

1 − αWNkN

n=0

n=0

1 − αWNk

(a) Ya [ k ] = ∑ α nWNkn = ∑ (αWNk ) n =

=

1− α 1 − αWNk

.

(b) Yb [ k ] = 2 ∑ WNkn − 3 ∑ WNkn . Assume first N is even, i.e., N = 2L. Then k even

k odd

(

)

⎛ 1 − W kL L −1 L −1 L −1 L −1 L Yb [ k ] = 2 ∑ W2kL2r − 3 ∑ W2kL(2r +1) = 2 ∑ WLkr − 3W2kL ∑ WLkr = 2 − 3WNk ⎜ k ⎜ r =0 r =0 r =0 r =0 ⎝ 1 − WL L

L −1

r =0

r =0

⎞ ⎟ = 0. ⎟ ⎠

Next, assume N is odd, i.e., N = 2 L + 1. Then Yb [ k ] = 2 ∑ W2kL2r − 3 ∑ W2kL(2r +1) =

L

2 ∑ WLkr r =0

− 3W2kL

L −1

∑

r =0

kL ⎛ 1 − W k ( L +1) ⎞ ⎛ k ⎜ 1 − WL L ⎜ ⎟ =2 − 3W2 L ⎜ 1−W k ⎟ ⎜ 1−W k L L ⎝ ⎝ ⎠

WLkr

k ⎛ ⎞ ⎟ = 2⎜ 1 − W L ⎜1− W k ⎟ L ⎝ ⎠

⎞ ⎟ = 2. ⎟ ⎠

5.10 x[ n] = cos(ωo n) = (e jω o n + e − jω o n ), 0 ≤ n ≤ N − 1. Therefore, 1 2

X[ k ] =

1 N −1 jω o n − j 2 πkn / N 1 N −1 + ∑ e − j jω o n e − j 2 πkn / N e ∑ e 2 n=0 2 n=0

1 N −1 − j ( = ∑ e 2 n=0 =

2 πk − ω o )n N

2 πk − j ⎛⎜ −ωo ⎞⎟ N N ⎝ ⎠ 1 1− e

2

⋅

2 πk − j ⎛⎜ −ωo ⎞⎟ N ⎝ ⎠ 1− e

1 N −1 − j ( + ∑ e 2 n=0 +

2 πk + ω o )n N

2 πk − j ⎛⎜ +ωo ⎞⎟ N N ⎝ ⎠ 1 1− e

2

⋅

2 πk − j ⎛⎜ +ωo ⎞⎟ N ⎝ ⎠ 1− e

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112

⎛ 2 πk −ω ⎞⎛ N −1 ⎞ ⎟ o ⎟⎜ N ⎠⎝ 2 ⎠

1 − j⎜ = e ⎝ 2

( ⋅ sin (

N −1

5.11 X [ k ] = ∑ x[ n]WNnk = n=0 ( N / 2 ) −1

)+ 1 e ) 2

ωo N 2 πk ωo − 2 N

sin πk −

( N / 2 ) −1

∑

2 πk N −1 ⎞ + ωo ⎞⎟⎛⎜ − j ⎛⎜ ⎟ ⎝ N ⎠⎝ 2 ⎠

sin

x[2r ]WN2rk +

( N / 2 ) −1

∑

r =0 r =0 ( N / 2 ) −1 = ∑ x[2r ]WNrk/ 2 + WNk ∑ x[2r + 1]WNrk/ 2 r =0 r =0 ( N / 2 ) −1 ( N / 2) −1 = ∑ x0 [r ]WNrk/ 2 + WNk ∑ x1[r ]WNrk/ 2 r =0 r =0 k = X 0 [〈 k 〉 N / 2 ] + WN X1[ 〈 k 〉 N / 2 ], 0 ≤ k ≤ N − 1. N −1

5.12 X [ k ] = ∑ x[ n]WNnk = =

n=0 ( N / 2 ) −1

( N / 2 ) −1

∑

n=0

x[ n]WNnk + WN( N / 2 )k

∑

x[ n]WNnk +

( N / 2 ) −1

n=0 ( N / 2 ) −1 ⎛

∑

n=0

x[

N 2

⋅

(

ωo N 2 πk ωo + N 2

sin πk +

(

)

).

x[2r + 1]WN(2r +1)k

N −1

nk ∑ x[ n]WN

n= N / 2

+ n]WNnk

⎞ nk k N ⎜ x[ n] + (−1) x[ + n] ⎟WN . For k = 2l, we get 2 ⎠ n=0 ⎝ ( N / 2 ) −1 ⎛ ( N / 2 ) −1 ⎛ N N ⎞ ⎞ X [2l] = ∑ ⎜ x[ n] + x[ + n] ⎟WN2 nl = ∑ ⎜ x[ n] + x[ + n] ⎟WNnl/ 2 = X 0 [l] 2 2 ⎠ ⎠ n=0 ⎝ n=0 ⎝ ( N / 2 ) −1 ⎛ N ⎞ and for k = 2l + 1 we get X [2l + 1] = ∑ ⎜ x[ n] − x[ + n] ⎟WN(2 l +1)n 2 ⎠ n=0 ⎝ =

=

∑

( N / 2 ) −1 ⎛

N N ⎞ n nl ⎜ x[ n] + x[ + n] ⎟WN ⋅ WN / 2 = X1[l] where 0 ≤ l ≤ − 1. 2 2 ⎠ n=0 ⎝ ∑

1 2

1 2

5.13 g[ n] = ( x[2n] + x[2 n + 1]), h[ n] = ( x[2n] − x[2n + 1]), 0 ≤ n ≤

N − 1. 2

Solving for

x[2n] and x[2n + 1], we get x[2 n] = g[ n] + h[ n] and x[2 n + 1] = g[ n] − h[ n]. Therefore, N −1

X [ k ] = ∑ x[ n]WNnk = n=0 ( N / 2 ) −1

( N / 2 ) −1

∑

x[2r ]WN2rk +

( N / 2 ) −1

∑

x[2r + 1]WN(2r +1)k

r =0 r =0 ( N / 2 ) −1 = ∑ x[2r ]WNrk/ 2 + WNk ∑ x[2r + 1]WNrk/ 2 r =0 r =0 ( N / 2 ) −1 ( N / 2 ) −1 = ∑ ( g[ n] + h[ n]) WNrk/ 2 + WNk ∑ ( g[ n] − h[ n]) WNrk/ 2 r =0 r =0 ( N / 2 ) −1 ( N / 2 ) −1 = (1 + WNk ) ∑ g[ n]WNrk/ 2 + (1 − WNk ) ∑ h[ n]WNrk/ 2 r =0 r =0

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113

= (1 + WNk )G[ 〈 k 〉 N / 2 ] + (1 − WNk )H [ 〈 k 〉 N / 2 ], 0 ≤ k ≤ N − 1.

5.14 g[ n] = a1 x[2n] + a 2 x[2n + 1], h[ n] = a3 x[2 n] − a 4 x[2 n + 1], 0 ≤ n ≤

N − 1, 2

with

a g[ n] − a 2 h[ n] a1a 4 ≠ a 2 a3 . Solving for x[2n] and x[2n + 1], we get x[2n] = 4 and a1a 4 − a 2 a3 − a3 g[ n] + a1h[ n] x[2n + 1] = . Therefore, a1a 4 − a 2 a3 N −1

X [ k ] = ∑ x[ n]WNnk = =

n=0 ( N / 2 ) −1

∑

( N / 2 ) −1

∑

x[2r ]WN2rk +

r =0 ( N / 2 ) −1 x[2r ]WNrk/ 2 + WNk ∑ x[2r r =0

r =0 ( N / 2 ) −1 ⎛ a

( N / 2 ) −1

∑

r =0

x[2r + 1]WN(2r +1)k

+ 1]WNrk/ 2

( N / 2 ) −1 ⎛ − a g[ n] + a h[ n] ⎞ g[r ] − a 2 h[r ] ⎞ rk 1 ⎜⎜ 4 ⎟⎟ WN / 2 + WNk ∑ ⎜⎜ 3 ⎟⎟ WNrk/ 2 r = 0 ⎝ a1a 4 − a 2 a3 ⎠ r = 0 ⎝ a1a 4 − a 2 a3 ⎠ 1 = (a 4 − a3WNk )G[〈 k 〉 N / 2 ] + (− a 2 + a1WNk )H[〈 k 〉 N / 2 ] , 0 ≤ n ≤ N − 1. a1a 4 − a 2 a3

=

∑

(

)

N −1

N −1

n=0

n=0

2 ln 5.15 (a) G[ k ] = ∑ x[ n]W2nk N . For k even, i.e., k = 2l, G[ 2l] = ∑ x[ n]W2 N N −1

= ∑ x[ n]2lnN = X[l], 0 ≤ l ≤ N − 1. n=0

2 N −1

(b) H[ k ] = ∑ x[ n − N ]W2nk N . Let m = n − N or n = m + N . Then n=N

N −1

N −1

m + N )k = (−1) k ∑ x[ m]W2mk H[ k ] = ∑ x[ m]W2(N N . For k even, i.e., k = 2l, m=0 N −1

N −1

m=0

n=0

n=0

2 N −1

2 N −1

n=0

n=0

H[2l] = ∑ x[ n]W22Nln = ∑ x[ n]WNln = X[l], 0 ≤ l ≤ N − 1.

nk 5.16 Y [ k ] = ∑ y[ n]W2nk N = ∑ ( g[ n] + h[ n]) W2 N = G[ k ] + H[ k ]. For k even, i.e.,

k = 2l, Y [ 2l] = G[ 2l] + H [ 2l] = 2 X [ l], 0 ≤ l ≤ N − 1. N −1

N −1 n=0

N −1

n=0 N −1

n=0

n=0

2l +1)n = ∑ x[ n]W2nN WNln and For k odd, i.e., k = 2l + 1, G[2l + 1] = ∑ x[ n]W2(N 2l +1)n H[2l + 1] = − ∑ x[ n]W2(N = − ∑ x[ n]W2nN WNln = −G[2l + 1], 0 ≤ l ≤ N − 1.

Hence, for k = 2l + 1, Y [ 2l + 1] = G[ 2l + 1] − G[ 2l + 1] = 0, 0 ≤ l ≤ N − 1.

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114

5.17 Y [ k ] =

MN −1

N −1

nk nk ∑ y[ n]WMN = ∑ x[ n]WMN . Thus,

n=0 N −1

n=0 N −1

n=0

n=0

nkM Y [ kM ] = ∑ x[ n]WMN = ∑ x[ n]WNnk = X[ k ].

5.18 N −1

N −1

5.19 (a) Now, X[ N / 2] = ∑ x[ n]WNnN / 2 = ∑ (−1) n x[ n]. Hence if x[ n] = x[ N − 1 − n] n=0

n=0

N −1

and N is even, then X[ N / 2] = ∑ (−1) n x[ n] = 0. n=0

N −1

(b) X[0] = ∑ x[ n]. Hence if x[ n] = − x[ N − 1 − n], then X [0] = 0. n=0

N −1

( N / 2 ) −1

n =0

n=0

(c) X[2l] = ∑ x[ n]WN2 nl = =

( N / 2 ) −1

∑

n=0

x[ n]WN2 nl +

∑

( N / 2 ) −1

∑

n=0

x[ n +

x[ n]WN2 nl +

N −1

2 nl ∑ x[ n]WN

n = ( N / 2 ) −1

M −1

nl = ∑ (x[ n] + x[ n + M ])W22M . Hence if

N ]WN2 nl 2

n=0

x[ n] = − x[ n + M ], then X [2l] = 0 for 0 ≤ l ≤ M − 1. N −1

5.20 X [2m] = ∑ x[n]W N2mn = =

n=0 ( N / 2 ) −1

∑

n=0

0≤m≤

( N / 2 ) −1

∑

n=0 N / 2 −1

x[n]W N2mn +

N −1

2 mn ∑ x[n]W N

n= N / 2 ( N / 2 ) −1 N x[n]W N2 mn + ∑ x[n + ]W N2mnW NmN = ∑ ⎛⎜ x[n] + 2 n=0 ⎝ n=0 N N − 1. This implies x[n] + x[ n + ] = 0. 2 2

x[n +

N 2

2 mn ⎟⎞ W N = 0, ⎠

5.21 (a) Using the circular convolution property of the DFT given in Table 5.3, we get km1

DFT{x[〈 n − m1 〉 N } = W N

km2

X [k ] and DFT{x[〈 n − m2 〉 N } = W N km1

Hence, W [k ] = DFT {w[n]} = W N (b) g[n] =

1 2

km2

X [k ] + W N

km1

X [k ] = (W N

(x[n] + (−1) n x[n]) = 12 (x[n] + WN− ( N / 2)n x[n]).

X [k ]. km1

+ WN

) X [k ].

Using the circular

convolution property of the DFT given in Table 5.3, we get G[k ] = DFT{g[n]} =

1 2

{X [k ] + X [〈k −

N 〉 2 N

}.

(c) Using the circular convolution property of the DFT given in Table 5.3, we get Y [k ] = DFT { y[n]} = X [k ] ⋅ X [k ] = X 2 [k ].

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115

{

}= W X [k ] = (−1) X [k ]. Hence, ⎧2 X [k ], U [k ] = DFT{u[n]} = DFT{x[n] − x[〈 n − 〉 }= X [k ] − (−1) X [k ] = ⎨ ⎩ 0,

5.22 (a) DFT x[〈 n −

N 〉 2 N

k ( N / 2) N

k

N 2 N

k

for k odd, for k even.

N ⎫ ⎧ (b) V [k ] = DFT{v[n]} = DFT ⎨ x[n] − x[n − ]⎬ = X [k ] + X [k ] = 2 X [k ]. 2 ⎭ ⎩ ( N / 2) n x[n]. Hence, using the circular frequency-shifting (c) y[n] = (−1) n x[n] = W N

property of the DFT given in Table 5.3, we get

{

}

Y [k ] = DFT{ y[n]} = DFT W N( N / 2) n x[n] = X [〈 k −

N 〉 ]. 2 N

5.23 (a) From the circular frequency-shifting property of the DFT given in Table 5.3, we get − m1 n

IDFT{ X [〈 k − m1 〉 N } = W N

−m 2 n

x[n] and IDFT{ X [〈 k − m2 〉 N } = W N

x[n]. Hence,

w[n] = IDFT{W [k ]} = IDFT{αX [〈 k − m1 〉 N } + βX [〈 k − m2 〉 N } − m1n

= αW N

− m2 n

x[n] + βW N

− m1n

x[n] = (αW N

(b) G[k ] = ( X [k ] + (−1) k X [k ]) = 1 2

1 2

− m2 n

+ βW N

(X [k ] + W

) x[n].

)

− ( N / 2) k X [k ] . N

Using the circular

time-shifting property of the DFT given in Table 5.3, we get N 1 g[n] = IDFT{G[k ]} = ⎛⎜ x[n] + x[〈 n − 〉 N ⎞⎟. 2⎝ 2 ⎠ (c) Using the modulation property of the DFT given in Table 5.3, we get y[n] = IDFT{Y [k ]} = N ⋅ x[n] ⋅ x[n] = N ⋅ x 2 [n]. N −1

( N / 2 ) −1

n=0

n=0

5.24 (a) X [2m] = ∑ x[n]W N2mn =

= = =

( N / 2 ) −1

∑

x[n]W N2mn

∑

x[n]W N2 mn +

n=0 ( N / 2 ) −1 n=0 ( N / 2 ) −1

+

∑

x[n]W N2mn +

∑

x[n +

∑

x[n +

N ]W N2mnW Nmn 2

n=0 N / 2 −1 n=0

2 mn ∑ x[n]W N

n= N / 2

N

2m( n + ) N 2 ]W N 2

N / 2 −1

N −1

( N / 2 ) −1 N 2 mn ⎛⎜ x[n] + x[n + N ]⎟⎞ W 2mn = ∑ (x[n] − x[n])W N = 0, 0 ≤ m ≤ − 1. N 2 ⎠ 2 n=0 n=0 ⎝

∑

N −1

(b) X [4l] = ∑ x[n]W N4ln n=0

=

( N / 4 ) −1

∑

n=0

x[n]W N4ln +

( N / 2 ) −1

∑

n= N / 4

x[n]W N4ln +

( 3 N / 4 ) −1

∑

n= N / 2

Not for sale.

x[n]W N4ln +

N −1

4 ln ∑ x[n]W N

n = 3N / 4

116

N −1 4 ⎛ ⎜

4l ( n + ) 4l ( n + ) 4l ( n + ) ⎞⎟ N N 3N 4 ln 4 2 4 = ∑ ⎜ x[n]W N + x[n + ]W N + x[n + ]W N + x[n + ]W N ⎟⎟ 4 2 4 ⎜ n=0 N

N

3N

⎝

⎠

N −1 4

3N N N = ∑ ⎛⎜ x[n] + x[n + ]W Nln + x[n + ]W N2ln + x[n + ]W N3ln ⎞⎟ W N4ln 4 2 4 ⎠ n=0⎝ N −1 4

= ∑ ( x[n] − X [n] + x[n] − x[n]) W N4ln = 0 as W NlN = W N2lN = W N3lN = 1. n=0

N −1

N −1

n=0 N −1

n=0

5.25 (a) X [ N − k ] = ∑ x[n]W N( N − k ) n = ∑ x[n]W N− kn + X * [k ]. N −1

(b) X [0] = ∑ x[n]W N0 ∑ x[n] which is real. n=0 N −1

N 2

n=0

N −1

(c) X [ ] = ∑ x[n]W N( N / 2) n = ∑ (−1) n x[n] which is real. n=0

n=0

N −1

5.26 X [k ] = ∑ x[n]W Nnk . n=0

N −1

(a) X * [k ] = ∑ x * [n]W N− nk . Replacing n by N − n in the summation we obtain n=0

N −1

N −1

n=0

n=0

X * [k ] = ∑ x * [ N − n]W N− ( N − n) k = ∑ x * [ N − n]W Nnk . Thus, DFT{x * [ N − n]} = DFT{x * [〈− n〉 N } = X * [k ]. 1 2

(b) Re{x[n]} = {x[n] + x * [n]}. Taking the DFT of both sides and using the results

of Part (a) we get DFT{Re{x[n]}} =

1 {X [k ] + 2

X * [〈− k 〉 N }.

(c) j Im{x[n]} = {x[n] − x * [n]}. Thus, DFT{ j Im{x[n]}} = 1 2

1 {X [k ] − 2

X * [〈− k 〉 N }.

1

(d) x cs [n] = {x[n] + x * [〈− n〉 N ]}. Using the linearity property and results of Part (b) 2 1

we get DFT{xcs [n]} = { X [k ] + X * [k ]} = Re{ X [k ]}. 2 1

(e) x ca [n] = {x[n] − x * [〈− n〉 N ]}. Using the linearity property and results of Part (b) 2 1 2

we get DFT{xca [n]} = { X [k ] − X * [k ]} = j Im{X [k ]}.

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5.27 Since for a real sequence, x[ n] = x * [ n], taking the DFT of both sides we get X [k ] = X * [〈− k 〉 N ]. This implies Re{ X [k ]} + j Im{X [k ]} = Re{ X [〈− k 〉 N ]} − j Im{X [〈− k 〉 N ]}. Comparing real and imaginary parts we get Re{ X [k ]} = Re{ X [〈− k 〉 N ]} and Im{X [k ]} = − Im{X [〈− k 〉 N ]}.

Also, X [k ] = (Re{ X [k ]}) 2 + (Im{X [k ]}) 2 = (Re{ X [〈− k 〉 N ]}) 2 + (Im{X [〈− k 〉 N ]}) 2 = X [〈− k 〉 N ] and

⎛ − Im{X [〈− k 〉 N ]} ⎞ ⎛ Im{X [k ]} ⎞ ⎟⎟ = − arg{X [〈− k 〉 N ]}. ⎟⎟ = tan −1 ⎜⎜ arg{X [k ]} = tan −1 ⎜⎜ 〈− 〉 Re{ X [ k ]} Re{ X [ k ]} ⎝ ⎠ N ⎝ ⎠ 5.28 (a) x1[〈− n〉 9 ] = {4 3 − 5 1 − 2 − 2 1 − 5 3} = x1[n]. Thus, x1[n] is a circular even sequence and hence, it has a real-valued 9 -point DFT. (b) x2 [〈− n〉 9 ] = {0 − 5 − 1 − 4 3 − 3 4 1 5} = − x2 [n]. Thus, x2 [n] is a circular odd sequence and hence, it has an imaginaryl-valued 9 -point DFT. (c) x3 [〈−n〉 9 ] = {0 − 5 − 1 − 4 3 − 3 4 2 − 5} which is neither equal to

x 3 [n] nor equal to − x 3 [n]. Thus, x3 [ n] has a complex-valued 9 -point DFT. (d) x4 [〈− n〉 9 ] = {− 5 5 − 2 2 4 4 2 − 2 5} = x 4 [n]. Thus, x4 [n] is a circular even sequence and hence, it has a real-valued 9 -point DFT. 5.29 (a) h[n] = g[〈 n − 5〉 8 ]. Hence, H [k ] = W8−5k G[k ] = e j10πk / 8G[k ] = e j 5πk / 4 G[k ]

= {2.6 + j 4.1, e j 5π / 4 (3 − j 2.7), e j 5π / 2 (−4.2 + j1.4), e j15π / 4 (3.5 − j 2.6), e j 5π (0.5), e j 25π / 4 (1.3 + j 4.4), e j15π / 2 (2.4 − j1.6), e j 35π / 4 (−3 + j1.6)}. (b) H [k ] = G[〈 k + 3〉 8 ]. Hence, h[n] = W83 g[n] = e − j 6πn / 8 g[n] = e − j 3πn / 4 g[n]

= {−0.1 − j 0.7, e − j 3π / 4 (1.3 + j ), e − j 3π / 2 (2 + j 0.7), e − j 9π / 4 (1.1 + j 2.2), e − j 3π (−0.8 + j 0.2), e − j15π / 4 (3.4 − j 0.1), e − j 9π / 2 (−1.2 + j 3.1), e − j 21π / 4 ( j1.5)}. 5.30 (a) y[ n] = αg[ n] + βh[ n]. Therefore, N −1

N −1

N −1

n=0

n=0

n=0

Y [k ] = ∑ y[n]W Nnk = α ∑ g[n]W Nnk + β ∑ h[n]W Nnk = αG[k ] + βH [k ]. N −1

(b) x[n] = g[〈 n − no 〉 N ]. Therefore, X [k ] = ∑ g[〈 n − no 〉 N ]W Nnk n=0

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118

n o −1

N −1

n=0

n = no

= ∑ g[ N + n − no ]W Nnk + ∑ g[n − no ]W Nnk

=

N −1

(n + no − N ) k

∑ g[n]W N

n = N − no

−ko n

(c) u[n] = W N

+

N − n −1

(n + no )k

∑ g[no ]W N

n=0

n k N −1

= WN o

n k

nk ∑ g[n]W N = W N o G[k ].

n=0

N −1

N −1

n=0

n=0

(k − k o )n

g[n]. Therefore, U [k ] = ∑ u[n]W Nkn = ∑ g[n]W N

⎧ N −1 (k − k o )n , if k ≥ k o , ⎪⎪ ∑ g[n]W N = 0 n = ⎨ N −1 ⎪ ∑ g[n]W ( N + k − k o ) n , if k < k o . N ⎪⎩ n = 0 if k ≥ k o , ⎧ G[k − k o ], = G[〈 k − k o 〉 N ]. Thus, U [k ] = ⎨ ⎩G[ N + k − k o ], if k < k o , N −1

N −1

(d) h[ n] = G[ n]. Therefore, H [k ] = ∑ h[n]W Nnk = ∑ G[n]W Nnk N −1 N −1

N −1

n=0 N −1 ( k + r ) n

n=0 r =0

r =0

n=0

= ∑ ∑ g[r ]W Nnr W Nkr = ∑ g[r ] ∑ W N

n=0

. The second sum is nonzero only if

k = r = 0 or else if r = N − k and k ≠ 0. Hence, if k = 0, ⎧ Ng[0], H [k ] = ⎨ = Ng[〈− k 〉 N ]. ⎩ Ng[ N − k ], if k > 0, N −1

N −1 N − 1 n=0 m=0

N −1

m=0 N −1

N −1

m =0

n=0

m =0

(e) u[ n] = ∑ g[ m] h[〈− n − m〉 N ]. Therefore, U[ k ] = ∑ ∑ g[ m] h[〈− n − m〉 N ]WNnk

= ∑ g[ m] ∑ h[〈− n − m〉 N ]WNnk = ∑ g[ m] H[ k ]WNmk = H[ k ] G[ k ]. N −1

5.31

∑ g[ n] h * [ n] =

n=0

=

N −1 1 N −1 N − 1 1 N −1 − nk − nk ∑ ∑ G[ k ]WN h * [ n] = ∑ G[ k ] ∑ h * [ n]WN N n=0 k =0 N k =0 n=0

1 N −1 ∑ G[ k ] H * [ k ]. N k =0

N −1⎛ N −1 N −1 ⎞ 5.32 DFT{ r xy [l]} = R xy [l] = ∑ r xy [l]WNlk = ∑ ⎜⎜ ∑ x[ n] y[〈l + n〉 N ⎟⎟ WNlk l=0 ⎝ n=0 l=0 ⎠ N −1 ⎛ N −1 ⎞ N −1 ⎛ N −1 ⎞ = ∑ x[ n]⎜⎜ ∑ y[〈l + n〉 N ]WNlk ⎟⎟ = ∑ x[ n]⎜⎜ ∑ y[ m]WNl( m − n) ⎟⎟ n =0 ⎝ l=0 ⎠ n=0 ⎝ l=0 ⎠

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119

N −1 ⎛ N −1 ⎞ = ∑ x[ n]WN− nk ⎜⎜ ∑ y[ m]WNlm ⎟⎟ = X * [ k ]Y [ k ]. n=0 ⎝ l=0 ⎠

5.33 Note X [k ] is the MN –point DFT of the sequence x e [n] obtained from x[n] by appending it with M ( N − 1) zeros. Thus, the length- MN sequence y[n] is given by M −1

y[ n] = ∑ x e [〈 n − Nl〉 MN ], 0 ≤ n ≤ MN − 1. Taking the MN –DFT of both sides we l =0

⎛ M −1 Nkl ⎞ ⎛ M −1 ⎞ ⎟⎟ X[ k ] = ⎜⎜ ∑ WNkl ⎟⎟ X [ k ]. get Y [ k ] = ⎜⎜ ∑ WMN ⎝ l=0 ⎠ ⎝ l=0 ⎠ 9

5.34 (a) X[0] = ∑ x[ n] = 30. n=0 9

(b) X[5] = ∑ (−1) n x[ n] = 0. n=0

9

(c)

∑ X[ k ] = 10 ⋅ x[0] = −30.

k =0

9 (d) The inverse DFT of e − j 2 πk / 5 X[ k ] is x[〈 n − 2〉 10 ]. Thus, ∑ e − j 2 πk / 5 X[ k ] k =0

= 10 ⋅ x[〈0 − 2〉 10 ] = 10 ⋅ x[8] = −100. 9

9

2

2

(e) From Parseval’s relation, ∑ X[ k ] = 10 ⋅ ∑ x[ n] = 38600. k =0

n=0

5.35 X[7] = X * [〈−7〉12 ] = X * [5] = 2 − j, X[8] = X * [〈−8〉12 ] = X * [ 4] = −3 − j 2, X[9] = X * [〈−9〉12 ] = X * [3] = 6 − j3, X[10] = X * [〈−10〉12 ] = X * [2] = 1 + j12, X[11] = X * [〈−11〉12 ] = X * [1] = 8 + j 2. (a) x[0] =

1 11 ∑ X[ k ] = 4.5, 12 k = 0

(b) x[6] =

1 11 k ∑ (−1) X[ k ] = −0.8333, 12 k = 0

11

(c)

∑ x[ n] = X[0] = 11,

n=0

−4 n −4 n x[ n]. Then, DFT{W12 x[ n]} = X [〈 k − 4〉 12 ]. (d) Let g[ n] = e j 2 πn / 3 x[ n] = W12

11

11

n=0

n=0

Thus, ∑ g[ n] = ∑ e j 2 πn / 3 x[ n] = X[〈0 − 4〉 12 ] = X[8] = −3 − j 2, 11

2

(e) From Parseval’s relation, ∑ x[ n] = n =0

1 11 2 ∑ X[ k ] = 74.8333. 12 k =0

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120

6

5.36 Now, yC [ n] = ∑ g[l] h[〈 n − l〉 7 ]. Hence, l=0

yC [0] = g[0] h[0] + g[1] h[6] + g[2] h[5] + g[3] h[ 4] + g[ 4] h[3] + g[5] h[2] + g[6] h[1], yC [1] = g[0] h[1] + g[1] h[0] + g[2] h[6] + g[3] h[5] + g[ 4] h[ 4] + g[5] h[3] + g[6] h[2], yC [2] = g[0] h[2] + g[1] h[1] + g[2] h[0] + g[3] h[6] + g[ 4] h[5] + g[5] h[ 4] + g[6] h[3], yC [3] = g[0] h[3] + g[1] h[2] + g[2] h[1] + g[3] h[0] + g[ 4] h[6] + g[5] h[5] + g[6] h[ 4], yC [ 4] = g[0] h[ 4] + g[1] h[3] + g[2] h[2] + g[3] h[1] + g[ 4] h[0] + g[5] h[6] + g[6] h[5], yC [5] = g[0] h[5] + g[1] h[ 4] + g[2] h[3] + g[3] h[2] + g[ 4] h[1] + g[5] h[0] + g[6] h[6], yC [6] = g[0] h[6] + g[1] h[5] + g[2] h[ 4] + g[3] h[3] + g[ 4] h[2] + g[5] h[1] + g[6] h[0]. 6

Likewise, y L [ n] = ∑ g[l] h[ n − l]. Hence, l=0

y L [0] = g[0]h[0], y L [1] = g[0]h[1] + g[1]h[0], y L [2] = g[0]h[2] + g[1]h[1] + g[2]h[0], y L [3] = g[0]h[3] + g[1]h[2] + g[2]h[1] + g[3]h[0], y L [ 4] = g[0]h[ 4] + g[1]h[3] + g[2]h[2] + g[3]h[1] + g[ 4]h[0], y L [5] = g[0]h[5] + g[1]h[ 4] + g[2]h[3] + g[3]h[2] + g[ 4]h[1] + g[5]h[0], y L [6] = g[0]h[6] + g[1]h[5] + g[2]h[ 4] + g[3]h[3] + g[ 4]h[2] + g[5]h[1] + g[6]h[0], y L [7] = g[1]h[6] + g[2]h[5] + g[3]h[ 4] + g[ 4]h[3] + g[5]h[2] + g[6]h[1], y L [8] = g[2]h[6] + g[3]h[5] + g[ 4]h[ 4] + g[5]h[3] + g[6]h[2], y L [9] = g[3]h[6] + g[ 4]h[5] + g[5]h[ 4] + g[6]h[3], y L [10] = g[ 4]h[6] + g[5]h[5] + g[6]h[ 4], y L [11] = g[5]h[6] + g[6]h[5], y L [12] = g[6]h[6]. Comparing yC [n] with y L [n] we observe that yC [0] = y L [0] + y L [7], yC [1] = y L [1] + y L [8], yC [2] = y L [2] + y L [9], yC [3] = y L [3] + y L [10], yC [ 4] = y L [ 4] + y L [11], yC [5] = y L [5] + y L [12], yC [6] = y L [6]. 5.37 Since x[n] is a length-9 real sequence, its DFT satisfies X[ k ] = X * [〈− k 〉 9 ]. Therefore, X[1] = X * [〈−1〉 9 ] = X * [8] = −7.7 + j 3.2,

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121

X[3] = X * [〈−3〉 9 ] = X * [6] = 8.6 + j 9.6, X[5] = X * [〈−5〉 9 ] = X * [ 4] = −3.5 − j 5.3, X[7] = X * [〈−7〉 9 ] = X * [2] = 1.2 + j 4.1. 5.38 X[1] = X * [〈−1〉 9 ] = X * [8] = 4.5 − j1.6, X[ 4] = X * [〈−4〉 9 ] = X * [5] = −3.1 − j8.2, X[6] = X * [〈−6〉 9 ] = X * [3] = −7.2 + 4.1, X[7] = X * [〈−7〉 9 ] = X * [2] = 1.2 + j 2.3. 5.39 Since the DFT X [k ] is real-valued, x[n] is a circularly even sequence, i.e., x[ n] = x[〈− n〉12 ]. Therefore, x[1] = x[〈−1〉12 ] = x[11] = −2, x[ 4] = x[〈−4〉12 ] = x[8] = 9.3, x[7] = x[〈−7〉 12 ] = x[5] = 4.1, x[9] = x[〈−9〉 12 ] = x[3] = −3.25, x[10] = x[〈−10〉12 ] = x[2] = 0.7. 5.40 Since the DFT X [k ] is imaginary-valued, x[n] is a circularly odd sequence, i.e., x[ n] = − x[〈− n〉12 ]. Therefore, x[7] = − x[〈−7〉 12 ] = − x[5] = 9.3, x[8] = − x[〈−8〉12 ] = − x[ 4] = −2.87, x[9] = − x[〈−9〉12 ] = − x[3] = −4.1, x[10] = − x[〈−10〉12 ] = − x[2] = 3.25, x[11] = − x[〈−11〉12 ] = − x[1] = −0.7. 5.41 X[ k ] = X * [〈− k 〉174 ] = X * [174 − k ]. X [9] = X * [174 − 9] = X * [165] = −3.4 + j 5.9 ⇒ X [165] = −3.4 − j 5.9. X [51] = X * [174 − 51] = X * [123] = 5 − j1.6 ⇒ X [123] = 5 + j1.6. X [113] = X * [174 − 113] = X * [61] = 8.7 − j 4.9 ⇒ X [61] = 8.7 + j 4.9. X [162 ] = X * [174 − 162 ] = X * [12 ] = 7.1 − j 2.4 ⇒ X [12 ] = 7.1 + j 2.4. X[ k1 ] = 7.1 + j 2.4, X[ k 2 ] = 8.7 + j 4.9, X[ k3 ] = 5 + j1.6, X[ k 4 ] = −3.4 − j 5.9. (a) Comparing these 4 DFT samples with the DFT samples given above we conclude k1 = 12, k2 = 61, k3 = 123, k 4 = 165. (b) dc value of {x[ n]} = X [0] = 11. (c) x[ n] =

1 173 1 − kn − 9n − 51n ( X[0] + 2 Re{X[9]W174 } + 2 Re{X[51]W174 } ∑ X[ k ]W174 = 174 k = 0 174 −87 n −113n −162 n + X [87] W174 + 2 Re{X [113] W174 } + 2 Re{X [162 ] W174 })

Not for sale.

122

173

(d)

2

∑ x[ n] =

n=0

1 173 2 ∑ X[ k ] = 86.0279. 174 k = 0

5.42 X[ k ] = X * [〈− k 〉 126 ] = X * [126 − k ]. X [ 0] = 12.8 + jα. X [13] = X * [126 − 13] = X * [113] = −3.7 + j 2.2 ⇒ X [113] = −3.7 − j 2.2. X [ 51] = X * [126 − 51] = X * [75] = − j1.7 ⇒ X [ 75] = j1.7. X [13] = X * [126 − 13] = X * [113] = −3.7 + j 2.2 ⇒ X [113] = −3.7 − j 2.2. X[ k1 ] = X * [126 − k1 ] = 9.1 − j 5.4 ⇒ X[126 − k1 ] = 9.1 + j 5.4. X[ k2 ] = X * [126 − k2 ] = 6.3 + j 2.3 ⇒ X[126 − k2 ] = 6.3 − j 2.3. X[ k3 ] = X * [126 − k3 ] = γ + j1.7 ⇒ X[126 − k3 ] = γ − j1.7.

X[ k 4 ] = X * [126 − k 4 ] = −3.7 − j 2.2 ⇒ X[126 − k 4 ] = −3.7 + j 2.2. (a) (b) Since x[n] is a real-valued sequence of length 126, X [0] and X[63] must be real. Thus, α = 0 and β = 0. As X[126 − k1 ] and X[108] have the same imaginary part, ε = 9.1 and k1 = 126 − 108 = 18. Likewise, as X[ k2 ] and X [79] have the same real part, δ = −2.3 and k2 = 126 − 79 = 47. Similarly, as X[126 − k3 ] and X[51] have

the same imaginary part, γ = 0 and k3 = 126 − 51 = 75. Finally, as

X[ k 4 ] = X[113], k 4 = 113. (c) dc value of x[n] is X[0] = 12.8. (d) x[ n] =

1 125 1 − kn − 26 πn / 126 X[0] + 2 Re{X[13]W126 } ∑ X[ k ]W126 = 126 k = 0 126

(

− 36 πn / 126 −94 πn / 126 − 63πn + 2 Re{X[18]W126 } + 2 Re{X[ 47]W126 } + X[63]W126 −150πn / 126 − 226πn / 126 + 2 Re{X[75]W126 } + 2 Re{X[113]W126 } .

)

125

(e)

2

∑ x[ n] =

n=0

1 125 2 ∑ X[ k ] = 5.767. 126 k = 0

5.43 Y [ k ] = W3−2 k X [ k ] = W9−6 k X [ k ]. Therefore, y[ n] = x[〈 n − 6〉 9 ]. Thus, y[ 0] = x[3] = 4, y[1] = x[ 4] = −3, y[ 2] = x[ 5] = 5, y[3] = x[6] = −2, y[ 4] = x[ 7] = −2, y[ 5] = x[8] = 4, y[6] = x[ 0] = 3, y[ 7] = x[1] = 5, y[8] = x[ 2] = 1. 5.44 H[ k ] = H * [〈− k 〉 9 ] = X * [9 − k ]. Hence, H [ 5] = H * [ 4] = −6.876 − j11.4883, H [6] = H * [3] = − j8.6603, H [ 7] = H * [ 2] = 6.0346 + j1.957, H [8] = H * [1] = 6.8414 + j 6.0572 .

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123

Now g[ n] = e j 2 πn / 3 h[ n] = e j 6 πn / 9 h[ n] = W9−6 n h[ n]. Therefore, G[ k ] = H[〈 k − 6〉 9 ], 0 ≤ k ≤ 8. Thus, G[0] = H[〈−6〉 9 ] = H[3] = j8.6603, G[1] = H[〈1 − 6〉 9 ] = H[ 4] = −6.876 − j11.4883, G[2] = H[〈2 − 6〉 9 ] = H[5] = H * [ 4] = −6.876 + j11.4883, G[3] = H[〈3 − 6〉 9 ] = H[6] = H * [3] = − j8.6603, G[ 4] = H[〈 4 − 6〉 9 ] = H[7] = H * [2] = 6.0346 + j1.957, G[5] = H[〈5 − 6〉 9 ] = H[8] = H * [1] = 6.8414 + j.0572, G[6] = H[〈6 − 6〉 9 ] = H[0] = 15, G[7] = H[〈7 − 6〉 9 ] = H[1] = 6.8414 − j 6.0572, G[8] = H[〈8 − 6〉 9 ] = H[2] = 6.0346 − j1.957. 5.45 (a) y L [0] = g[0] h[0] = −4, y L [1] = g[0] h[1] + g[1] h[0] = 10, y L [2] = g[0] h[2] + g[1] h[1] + g[2] h[0] = −6, y L [3] = g[0] h[3] + g[1] h[2] + g[2] h[1] = 8, y L [ 4] = g[1] h[3] + g[2] h[2] = 7, y L [5] = g[2] h[3] = −3. (b) yC [0] = g e [0] h[0] + g e [1] h[3] + g e [2] h[2] + g e [3] h[1] = g[ 0]h[ 0] + g[1]h[3] + g[ 2]h[ 2] = 3, yC [1] = g e [0] h[1] + g e [1] h[0] + g e [2] h[3] + g e [3] h[2] = g[0]h[1] + g[1]h[ 0] + g[ 2]h[3] = 7, yC [2] = g e [0] h[2] + g e [1] h[1] + g e [2] h[0] + g e [3] h[3] = g[ 0]h[ 2] + g[1]h[1] + g[2]h[ 0] = −6, yC [3] = g e [0] h[3] + g e [1] h[2] + g e [2] h[1] + g e [3] h[0] = g[0]h[3] + g[1]h[ 2] + g[ 2]h[1] = 8. ⎡Ge [0]⎤ ⎡1 1 1 1 ⎤⎡ 2 ⎤ ⎡ 4 ⎤ ⎢ Ge [1] ⎥ ⎢1 − j − 1 j ⎥ ⎢− 1⎥ ⎢− 1 + j ⎥ = = , (c) ⎢ G [2]⎥ ⎢1 − 1 1 − 1⎥ ⎢ 3 ⎥ ⎢ 6 ⎥ ⎢ e ⎥ ⎢1 j − 1 j ⎥ ⎢ 0 ⎥ ⎢ − 1 − j ⎥ ⎦⎣ ⎦ ⎣ ⎦ ⎣Ge [3]⎦ ⎣ ⎡ H[0]⎤ ⎡1 1 1 1 ⎤ ⎡− 2⎤ ⎡ 3 ⎤ ⎢ H [1] ⎥ ⎢1 − j − 1 j ⎥ ⎢ 4 ⎥ ⎢ − 4 − j 5⎥ ⎢ H[2]⎥ = ⎢1 − 1 1 − 1⎥ ⎢ 2 ⎥ = ⎢ − 3 ⎥. ⎢ H [3]⎥ ⎢⎣1 j − 1 j ⎥⎦ ⎣⎢ − 1 ⎦⎥ ⎢⎣− 4 + j 5⎥⎦ ⎦ ⎣ ⎡YC [0]⎤ ⎡Ge [0]H [0]⎤ ⎡ 12 ⎤ ⎢ YC [1] ⎥ ⎢ Ge [1]H [1] ⎥ ⎢9 + j ⎥ ⎢Y [2]⎥ = ⎢G [2]H[2]⎥ = ⎢ − 18 ⎥. Therefore ⎥ ⎢9 − j ⎥ ⎢ C ⎥ ⎢ e ⎦ ⎣YC [3]⎦ ⎣ Ge [3]H[3]⎦ ⎣

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124

⎡ yC [0]⎤ 1 1 ⎤ ⎡ 12 ⎤ ⎡ 3 ⎤ ⎡1 1 ⎢ yC [1] ⎥ 1 ⎢1 j − 1 − j ⎥ ⎢9 + j ⎥ ⎢ 7 ⎥ ⎢ y [2]⎥ = ⎢1 − 1 1 − 1⎥ ⎢ − 18 ⎥ = ⎢− 6 ⎥. ⎢ C ⎥ 4 ⎢1 − j − 1 j ⎥ ⎢9 − j ⎥ ⎢ 8 ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ yC [3]⎦ (d) g e [ n] = [2, − 1, 3, 0, 0, 0], he [ n] = [− 2, 4, 2, − 1, 0, 0]. yC [0] = g e [0] he [0] + g e [1] he [5] + g e [2] h e [ 4] + g e [3] he [3] + g e [ 4] he [2] + g e [5] he [1]

= g[0]h[0] = −4 = y L [0], yC [1] = g e [0] he [1] + g e [1] he [0] + g e [2] h e [5] + g e [3] he [6] + g e [ 4] he [3] + g e [5] he [2] = g[0]h[1] + g[1]h[0] = 10 = y L [1], yC [2] = g e [0] he [2] + g e [1] he [1] + g e [2] h e [0] + g e [3] he [5] + g e [ 4] he [ 4] + g e [5] he [3] = g[0]h[2] + g[1]h[1] + g[2]h[0] = −6 = y L [2], yC [3] = g e [0] he [3] + g e [1] he [2] + g e [2] h e [1] + g e [3] he [0] + g e [ 4] he [5] + g e [5] he [ 4] = g[0]h[3] + g[1]h[2] + g[2]h[1] = 8 = y L [3], yC [ 4] = g e [0] he [ 4] + g e [1] he [3] + g e [2] h e [2] + g e [3] he [1] + g e [ 4] he [0] + g e [5] he [5] = g[1]h[3] + g[2]h[2] = 7 = y L [ 4], yC [5] = g e [0] he [5] + g e [1] he [ 4] + g e [2] h e [3] + g e [3] he [2] + g e [ 4] he [1] + g e [5] he [0] = g[2]h[3] = −3 = y L [5]. 5.46 We need to show g[ n] O N h[ n] = h[ n] O N g[ n]. Let N −1

x[ n] = g[ n] O N h[ n] = ∑ g[ m ]h[ 〈 n − m〉 N ] and m=0 N −1

n

N −1

m=0 N −1

m = n +1

y[ n] = h[ n] O N g[ n] = ∑ h[ m ]g[ 〈 n − m〉 N ] = ∑ h[ m ]g[ n − m ] + n

= ∑ h[ n − m]g[ m] + m =0

m=0 N −1

∑ h[ m]g[ N + n − m]

∑ h[ N + n − m]g[ m] = ∑ h[〈 n − m〉 N ]g[ m] = x[ n].

m = n +1

m=0

N −1

5.47 (a) y[ n] = x1[ n] O N x 2 [ n] = ∑ x1[ m ] x 2 [ 〈 n − m〉 N ]. Thus, m=0

N −1

N −1

N −1

⎛ N −1 ⎞⎛ N −1 ⎞ ∑ y[ n] = ∑ x1[ m] ∑ x[〈 n − m〉 N ] = ⎜⎜ ∑ x1[ n] ⎟⎟⎜⎜ ∑ x 2 [ n] ⎟⎟. n=0 m =0 n =0 ⎝ n=0 ⎠⎝ n = 0 ⎠

(b) N −1

N −1

N −1

n n ∑ (−1) y[ n] = ∑ x1[ m] ∑ x[〈 n − m〉 N ](−1)

n=0 ⎛ N −1

m=0 n=0 ⎞⎛ m −1 = ⎜⎜ ∑ x1[ m] ⎟⎟⎜⎜ ∑ x 2 [ N + m − n](−1) n ⎝ m =0 ⎠⎝ n = 0

N −1 ⎞ + ∑ x 2 [ m − n](−1) n ⎟⎟. Replacing n by n=m ⎠

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125

N + n − m in the first sum on the right-hand side and by n − m in the second sum on the right-hand side we obtain N −1 N −1 ⎛ N −1 ⎞⎛ m −1 ⎞ n n−N +m + ∑ x 2 [ n](−1) n + m ⎟⎟ ∑ (−1) y[ n] = ⎜⎜ ∑ x1[ m] ⎟⎟⎜⎜ ∑ x 2 [ n](−1) n =0 n=m ⎝ m =0 ⎠⎝ n = 0 ⎠ ⎛ N −1 ⎞⎛ N −1 ⎞ = ⎜⎜ ∑ (−1) n x1[ n] ⎟⎟⎜⎜ ∑ (−1) n x 2 [ n] ⎟⎟. ⎝ n=0 ⎠⎝ n = 0 ⎠

5.48 y[ n] = x[3n], 0 ≤ n ≤

Now, x[ n] =

Y [k ] =

N 3

N −1 3

N −1 3

n=

n=

− 1. Therefore, Y [ k ] = ∑ y[ n]WNnk/ 3 = ∑ x[ 4 n]WNnk/ 3 . N −1

N −1

1 1 − mn − 3mn = ∑ X [ m ] WN ∑ X [ m]WN / 3 . Hence, N m =0 N m=0

N −1 N −1 3

1 N −1

N −1 3

1 ( k − m )n − mn nk ∑ ∑ X [ m ] WN / 3 WN / 3 = ∑ X[ m] ∑ WN / 3 . Since N n =0 m =0 N m=0 n=0

N −1 3

N 2N ⎧N , k + N, ( k − m )n ⎪ , m = k , k + , k + Thus, =⎨3 ∑ WN / 3 3 3 ⎪⎩ 0, n=0 elsewhere. 1 N 2N Y [ k ] = ⎛⎜ X [ k ] + X [ k + ] + X [ k + ] + X [ k + N ] ⎞⎟. 3 3 ⎠ 3⎝ 1 2

5.49 v[ n] = x[ n] + jy[ n]. Hence, X[ k ] = {V [ k ] + V * [〈− k 〉 8 ]} is the 8 –point DFT of x[n]

and Y [ k ] =

1 {V [ k ] − V * [〈− k 〉 8 ]} is the 2j

8 –point DFT of y[n].

V [ k ] = [3 + j 7, − 2 + j 6, 1 − j 5, 4 − j 9, 5 + j 2, 3 − j 2, j 4, − 3 − j8]. V * [〈− k 〉 8 ] = [3 + j 7, − 3 + j8, − j 4, 3 + j 2, 5 − j 2, 4 + j 9, 1 + j 5, − 2 − j 6]. Therefore, 5 1 9 7 7 7 7 5 1 9 X[ k ] = ⎡3 + j 7, − + j 7, + j , − j , 5, + j , − j , − − j 7 ⎤, ⎥⎦ ⎢⎣ 2 2 2 2 2 2 2 2 2 2 1 1 1 11 1 11 1 1 1 1 Y [ k ] = ⎡0, − 1 − j , − − j , − − j , 2, − + j , − + j , − 1 + j ⎤. ⎢⎣ 2 ⎥⎦ 2 2 2 2 2 2 2 2 2 The IDFT of V [k ] obtained using MATLAB is given by Columns 1 through 4 1.3750 - 0.6250i -0.9331 - 0.1187i

-0.8044 + 1.7223i

-1.7500 + 1.2500i

Columns 5 through 8 0.8750 + 2.6250i -1.8169 + 1.1187i

2.5544 - 0.2223i

Not for sale.

3.5000 + 1.2500i

126

The same result is obtained by computing the IDFT x[n] of X [k ] and the IDFT y[n] of Y [k ] using MATLAB and then forming x[ n] + jy[ n]. 5.50 v[ n] = g[ n] + j h[ n] = [2 − j 2, − 1 + j 4, 3 + j 2, − j ]. Therefore, ⎡V [0]⎤ ⎡1 1 1 1 ⎤ ⎡ 2 − j 2 ⎤ ⎡ 4 + j3 ⎤ ⎢ V [1] ⎥ ⎢1 − j − 1 j ⎥ ⎢− 1 + j 4⎥ ⎢ 4 − j 3 ⎥ ⎢V [2]⎥ = ⎢1 − 1 1 − 1⎥ ⎢ 3 + j 2 ⎥ = ⎢ 6 − j 3 ⎥, i.e., ⎢V [3]⎥ ⎢⎣1 j − 1 − j ⎥⎦ ⎢ − j ⎥ ⎢− 6 − j 5⎥ ⎣ ⎦ ⎦ ⎦ ⎣ ⎣ V [ k ] = [4 + j3, 4 − j3, 6 − j3, − 6 − j 5]. Thus, V * [〈− k 〉 4 ] = [4 + j3, − 6 + j 5, 6 + j 3, 4 + j3]. Therefore,

) [

(

1 V [ k ] + V * [〈− k 〉 4 ] = 4, 2 1 H[ k ] = V [ k ] − V * [〈− k 〉 4 ] = 3, 2j

G[ k ] =

) [

(

− 1 + j, 6, − 1 − j ] and

− 4 − j 5, − 3, − 4 + j 5].

5.51 Let p[ n] = IDFT{P[ k ]}. Thus, ⎡ p[0]⎤ 1 1 ⎤ ⎡ − 5 ⎤ ⎡ − 1.25 ⎤ ⎡1 1 ⎢ p[1] ⎥ 1 ⎢1 j − 1 − j ⎥ ⎢− 2 + j 5⎥ ⎢− 4.75⎥ ⎢ p[2]⎥ = ⎢1 − 1 1 − 1⎥ ⎢ 4 ⎥ = ⎢ 0.75 ⎥ ⎥ 4 ⎢1 − j − 1 j ⎥ ⎢− 2 − j 5⎥ ⎣⎢ 0.25 ⎦⎥ ⎢ ⎦⎣ ⎣ ⎦ ⎣ p[3]⎦ ⎡d[0]⎤ 1 1 ⎤⎡ 3 ⎤ ⎡ 1 ⎤ ⎡1 1 ⎢ d[1] ⎥ 1 ⎢1 j − 1 − j ⎥ ⎢4 + j ⎥ ⎢ 2 ⎥ ⎢d[2]⎥ = 4 ⎢1 − 1 1 − 1⎥ ⎢ − 7 ⎥ = ⎢− 3⎥. Therefore, ⎢ d[3]⎥ ⎢⎣1 − j − 1 j ⎥⎦ ⎢⎣4 − j ⎥⎦ ⎢⎣ 3 ⎥⎦ ⎦ ⎣ X ( e jω ) =

− 1.25 − 4.75 e − jω + 0.75 e − j 2 ω + 0.25 e − j 3ω

. 1 + 2 e − j ω − 3 e − j 2 ω + 3 e − j 3ω 5.52 Let p[ n] = IDFT{P[ k ]}. Thus, ⎡ p[0]⎤ 1 1 ⎤ ⎡ 8 ⎤ ⎡ − 1.25 ⎤ ⎡1 1 ⎢ p[1] ⎥ 1 ⎢1 j − 1 − j ⎥ ⎢− 5 − j 6 ⎥ ⎢ 5.75 ⎥ ⎢ p[2]⎥ = ⎢1 − 1 1 − 1⎥ ⎢ − 3 ⎥ = ⎢ 3.75 ⎥ ⎥ 4 ⎢1 − j − 1 j ⎥ ⎢− 5 + j 6 ⎥ ⎢⎣− 0.25⎥⎦ ⎢ [ 3 ] p ⎦⎣ ⎣ ⎦ ⎦ ⎣ ⎡d[0]⎤ 1 1 ⎤ ⎡ 3 ⎤ ⎡ 2.75 ⎤ ⎡1 1 ⎢ d[1] ⎥ 1 ⎢1 j − 1 − j ⎥ ⎢4 + j ⎥ ⎢− 3.75⎥ ⎢d[2]⎥ = 4 ⎢1 − 1 1 − 1⎥ ⎢ − 7 ⎥ = ⎢− 3.25⎥. Therefore, ⎢ d[3]⎥ ⎢⎣1 − j − 1 j ⎥⎦ ⎢⎣4 − j ⎥⎦ ⎣⎢ − 1.75 ⎦⎥ ⎦ ⎣ X (e

jω

)=

− 1.25 + 54.75 e − jω + 3.75 e − j 2 ω − 0.25 e − j 3ω 2.75 − 3.75 e − jω − 3.25 e − j 2ω − 1.75 e − j 3ω N −1

N −1

n=0

n=0

.

5.53 X (e jω ) = ∑ x[ n] e − jωn and Xˆ [ k ] = ∑ x[ n] e − j 2 πkn / M . Now,

Not for sale.

127

xˆ[ n] =

1 M −1 ˆ 1 M −1 N −1 − j 2 πkm / M − nk − nk WM ∑ X [ k ] WM = ∑ ∑ x[ m] e M k =0 M k =0 m =0

N −1 ∞ 1 N −1 − j 2 πk ( m − n ) / M = ∑ x[ n + rM ]. ∑ x[ m] ∑ e M m =0 k =0 r = −∞ Thus, xˆ[ n] is obtained by shifting x[n] in multiples of M and adding the shifted copies. Since the new sequence is obtained by shifting in multiples of hence, to recover the original sequence take any consecutive N samples in the range 0 ≤ n ≤ N − 1 for any value of r. This would be true only if the shifted copies of x[n] did not overlap with each other, that is, if and only if M ≥ N .

=

8

8

n=0

n=0

5.54 (a) X (e jω ) = ∑ x[ n] e − jωn . Therefore, X1[ k ] = ∑ x[ n] e − j 2 πkn / 12 . Hence,

x1[ n] =

1 11 1 11 ⎛ 8 j 2 πkn / 12 j 2 πkm / 12 ⎞⎟ j 2 πkn / 12 = ∑ X[ k ] e ∑ ⎜⎜ ∑ x[ m] e ⎟e 12 k = 0 12 k = 0 ⎝ m = 0 ⎠

11 ∞ 1 8 j 2 πk ( n − m ) / 12 = ∑ x[ n + 12r ] using the results of Problem 5.53. ∑ x[ m] ∑ e 12 m = 0 k =0 r = −∞ Since M = 12 and N = 9, M > N , and hence, x[n] is recoverable from x1[ n]. In fact x1[ n] = {1, − 2, 3, − 4, 5, − 4, 3, − 2, 1, 0, 0, 0}, 0 ≤ n ≤ 11, and x[n] is given by the first 9 samples of x1[ n].

=

8

(b) Here, X 2 [ k ] = ∑ x[ n] e − j 2 πkn / 8 . Hence, n=0

x1[ n] =

7 ⎛ 8 ⎞ 1 7 j 2 πkn / 8 1 = ∑ ⎜⎜ ∑ x[ m] e j 2 πkm / 8 ⎟⎟ e j 2 πkn / 8 ∑ X[ k ] e 8 k =0 8 k =0 ⎝ m =0 ⎠

7 ∞ 1 8 j 2 πk ( n − m ) / 8 = ∑ x[ n + 8r ] using the results of Problem 5.53. ∑ x[ m] ∑ e 8 m =0 k =0 r = −∞ Since M = 8 and N = 9, M < N , and hence, x[n] is not recoverable from x1[ n]. In fact x 2 [ n] = [2, − 2, 3, − 4, 5, − 4, 3, − 2, 2, − 1, 1], 0 ≤ n ≤ 10.

=

1 N −1 − kn ∑ X[ k ]WN . Let n = N − m. Then, N k =0 1 N −1 1 N −1 1 − k( N − m) km x[ m] = = ∑ X [ k ] WN ∑ X[ k ]WN = F {X [ k ]}. Therefore, N k =0 N k =0 N 1 1 F { F {x[ n]}}, or, F { F {x[ n]}} = N ⋅ x[ N − n]. Hence, x[ N − n] = F {X [ k ]} = N N

5.55 x[ n] =

F { F { F { F {x[ n]}}}} = N 2 ⋅ x[ N − n].

Not for sale.

128

100

100

k =0 50

k =0

36

5.56 y[ n] = x[ n] O * h[ n] = ∑ x[ k ] h[ n − k ] = ∑ h[ k ] x[ n − k ] = ∑ h[ k ] x[ n − k ]. k =17

36

u[ n] = x[ n] O N h[ n] = ∑ h[ k ] x[ 〈 n − k 〉 51 ] = ∑ h[ k ] x[ 〈 n − k 〉 51 ]. k =0

k =17

Now, for n ≥ 36, x[〈 n − k 〉 51 ] = x[ n − k ]. Thus, y[ n] = u[ n] for 36 ≤ n ≤ 50. 5.57 (a) Overlap-add method: Since the impulse response is of length and the DFT size to be used is 128, hence, the number of data samples required for each convolution will be 128 − 109 = 19. Thus the total number of DFTs required for the length- 1300 data 1300 ⎤ = 69. Also, the DFT of the impulse response needs to be computed sequence is ⎡ ⎢ 19 ⎥ once. Hence, the total number of DFTs used are = 69 + 1 = 70. The total number of IDFTs used are = 69. (b) Overlap-save method: In this case, since the first 110 − 1 = 109 points are lost, we need to pad the data sequence with 109 zeros for a total length of 1409. Again, each convolution will result in 128 − 109 = 19 correct values. Thus the total number of 1409 ⎤ = 75. Again, 1 DFT is required for the impulse DFTs required for the data are ⎡ ⎢ 19 ⎥ response. The total number of DFTs used are 75 + 1 = 76. The total number of IDFTs used are = 75.

⎧ x[ n / L ], n = 0, L, 2 L,K, ( N − 1)L, 5.58 (a) y[ n] = ⎨ elsewhere. ⎩ 0, NL −1

N −1

N −1

n=0

n=0

n=0

nk nLk = ∑ x[ n]WNL = ∑ x[ n]WNnk . For k ≥ N , let k = k o + rN Y [ k ] = ∑ y[ n]WNL N −1

n( k o + rN )

where ko = 〈 k 〉 N . Then, Y [ k ] = Y [ k o + rN ] = ∑ x[ n]WN n=0

N −1

nk o

= ∑ x[ n]WN n=0

= X[ ko ] = X[〈 k 〉 N ]. (b) Since Y [ k ] = X[〈 k 〉 5 ] for 0 ≤ k ≤ 20, a sketch of Y [k ] is thus as shown below. 5.59 x 0 [ n] = x[2 n + 1] + x[2n], x1[ n] = x[2 n + 1] − x[2n], y0 [ n] = y[2n + 1] + y[2n], and

y1[ n] = y[2n + 1] − y[2n], 0 ≤ n ≤

N 2

− 1. Since x[n] and y[n] are real, symmetric

sequences, it follows that x 0 [ n] and y0 [ n] are real, symmetric sequences, and x1[ n] and y1[ n] are real, antisymmetric sequences. Now, consider the

N 2

–length sequence

u[ n] = x 0 [ n] + y1[ n] + j (x1[ n] + y0 [ n]). Its conjugate sequence is given by u * [ n] = x 0 [ n] + y1[ n] − j (x1[ n] + y0 [ n]). Next, we observe that u[〈− n〉 N / 2 ] = x 0 [〈− n〉 N / 2 ] + y1[〈− n〉 N / 2 ] + j (x1[〈− n〉 N / 2 ] + y0 [〈− n〉 N / 2 ]) Not for sale.

129

= x 0 [ n] − y1[ n] + j (− x1[ n] + y0 [ n]). Its conjugate sequence is given by u * [〈− n〉 N / 2 ] = x 0 [ n] − y1[ n] − j (− x1[ n] + y0 [ n]). By adding the last 4 sequences we get 4 x 0 [ n] = u[ n] + u * [ n] + u[〈− n〉 N / 2 ] + u * [〈− n〉 N / 2 ]. From Table 5.3, if U [ k ] = DFT{u[ n]}, then U * [〈− k 〉 N / 2 ] = DFT{u * [ n]}, U * [ k ] = DFT{u * [〈− n〉 N / 2 ]}, and U[〈− k 〉 N / 2 ] = DFT{u[〈− n〉 N / 2 ]}. Thus, 1 4

X 0 [ k ] = DFT{x 0 [ n]} =

(U[k ] + U * [〈− k〉 N / 2 ] + U[〈− k〉 N / 2 ] + U * [k ]).

Similarly,

j 4 x1[ n] = u[ n] − u * [ n] − u[〈− n〉 N / 2 ] + u * [〈− n〉 N / 2 ]. Hence,

X1[ k ] = DFT{x1[ n]} =

(

)

1 U[ k ] − U * [〈− k 〉 N / 2 ] − U[〈− k 〉 N / 2 ] + U * [ k ] . Likewise, j4

4 y1[ n] = u[ n] − u[〈− n〉 N / 2 ] + u * [ n] − u * [〈− n〉 N / 2 ]. Thus, Y1[ k ] = DFT{y1[ n]} =

1 4

(U[k ] − U[〈− k〉 N / 2 ] + U * [〈− k〉 N / 2 ] − U * [k ]). Finally,

j 4 y0 [ n] = u[ n] + u[〈− n〉 N / 2 ] − u * [ n] − u * [〈− n〉 N / 2 ]. Hence,

Y0 [ k ] = DFT{y0 [ n]} = N −1

(

5.60 X GDFT [ k, a, b] = ∑ x[ n] e n=0

j 1 N −1 x[ n] = ∑ X[ k ] e N k =0 j 1 N −1 N − 1 = ∑ ∑ x[r ] e N k =0 r =0 N −1 j 1 N −1 = ∑ x[r ] ∑ e N r =0 k =0 N −1

∑ e

j

2 π( n − r )( k + b ) N

k =0

)

1 U[ k ] + U[〈− k 〉 N / 2 ] − U * [〈− k 〉 N / 2 ] − U * [ k ] . j4 −j

2 π( n + a )( k + b ) N .

2 π( n + a )( k + b ) N

−j 1 N −1 N − 1 = ∑ ∑ x[r ] e N k =0 r =0

2 π( n + a − r − a )( k + b ) N 2 π( n − r )( k + b ) N

=

2 π( r + a ) k + b ) N

j 1 N −1 N − 1 = ∑ ∑ x[r ] e N k =0 r =0

j

e

2 π( n + a ) k + b ) N

2 π( n − r )( k + b ) N

1 ⋅ x[ n] ⋅ N = x[ n], as from Eq. (5.23) N

N , if n = r , = ⎧⎨ ⎩ 0, otherwise.

5.61 (a) x[ n] = α g[ n] + β h[ n]. Thus, N −1

N −1

πk (2 n +1) ⎞ πk (2 n + 1) ⎞ X DCT [ k ] = ∑ x[ n] cos⎛⎜ ⎟ = ∑ (α g[ n] + β h[ n]) cos⎛⎜ ⎟ ⎝ 2N ⎠ n=0 ⎝ 2N ⎠ n=0

N −1

N −1

πk (2 n +1) ⎞ πk (2 n + 1) ⎞ = ∑ α g[ n] cos⎛⎜ ⎟ + ∑ β h[ n] cos⎛⎜ ⎟ = α G DCT [ k ] + β H DCT [ k ]. ⎝ 2N ⎠ n =0 ⎝ 2N ⎠ n=0

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130

N −1

N −1

πk (2 n + 1) ⎞ πk (2 n + 1) ⎞ (b) G DCT [ k ] = ∑ g[ n] cos⎛⎜ ⎟ ⇒ G *DCT [ k ] = ∑ g * [ n] cos⎛⎜ ⎟. ⎝ 2N ⎠ ⎝ 2N ⎠ n=0 n=0

DCT

Therefore, g * [ n] ↔ G *DCT [ k ]. N , if k = m = 0, ⎧ N −1 πk (2 n +1) ⎞ πm(2 n +1) ⎞ ⎪ (c) Note that ∑ cos⎛⎜ ⎟ cos⎛⎜ ⎟ = ⎨ N/2, if k = m and k ≠ 0, Now, ⎝ 2N ⎠ ⎝ 2N ⎠ ⎪ n=0 0, otherwise. ⎩ g[ n] g * [ n] =

1 N −1 N −1 N2

π(2 n +1)k ⎞ π(2 n +1)k ⎞ * ⎟ cos⎛⎜ ⎟. ∑ α[ k ] α[ m] G DCT [ k ]G DCT [ m] cos⎛⎜ ⎝ 2N ⎠ ⎝ 2N ⎠ k =0 m =0

∑

Thus, N −1

2

∑ g[ n] =

n=0

1 N −1 N − 1 N − 1 N2

π(2 n +1)k ⎞ π(2 n +1)k ⎞ * ⎟ cos⎛⎜ ⎟. ∑ α[ k ] α[ m] G DCT [ k ]G DCT [ m] cos⎛⎜ 2 N 2N ⎝ ⎠ ⎝ ⎠ n=0 k =0 m =0

∑ ∑

Using the orthogonality property mentioned earlier we get N −1

2

∑ g[ n] =

n=0

5.62 (a) HN

where I H N H TN

2 1 N −1 ∑ α[ k ] G DCT [ k ] . 2N k =0

⎡13 ⎢17 =⎢ ⎢13 ⎢ ⎣7 is the ⎡13 = ⎢17 ⎢13 ⎣⎢ 7

13 13 13 ⎤ − 7 − 17⎥⎥ 7 . The matrix H N is orthogonal if H N H TN = cI − 13 − 13 13 ⎥ ⎥ − 17 17 −7⎦ 4 × 4 identity matrix and c is a constant. Now, 13 13 13 ⎤ ⎡13 17 13 7 ⎤ ⎡676 0 0 0 ⎤ 7 0 ⎥. − 13 − 17⎥ = ⎢ 0 676 0 − 7 − 17⎥ ⎢13 7 0 676 0 ⎥ − 13 − 13 13 ⎥ ⎢13 − 7 − 13 17 ⎥ ⎢ 0 0 0 676⎦⎥ − 17 17 − 7 ⎦⎥ ⎣⎢13 − 17 13 − 7 ⎦⎥ ⎣⎢ 0

Hence, the matrix is orthogonal and all its rows have the same L 2 -norm. 1 1 ⎤ ⎡1 1 ⎢2 1 − 1 − 2 ⎥ ⎥. Next, we observe (b) G N = ⎢ ⎢1 − 1 − 1 1 ⎥ ⎢ ⎥ ⎣1 − 2 2 − 1 ⎦ 1 1 ⎤ ⎡4 0 0 0 ⎤ 1 1 ⎤ ⎡1 2 ⎡1 1 1 1 1 2 1 1 2 − − − − ⎢ ⎥ ⎢ ⎥ ⎢0 10 0 0 ⎥ = 1 −1 −1 1 1 −1 −1 − 2 2⎥ = ⎢0 0 4 0 ⎥ which shows that ⎢ ⎥⎢ ⎢⎣1 − 2 2 − 1⎥⎦ ⎢⎣1 − 2 1 − 1⎥⎦ ⎢⎣0 0 0 10⎥⎦ the rows of G N are orthogonal but do not have the same L 2 -norms. G N G TN

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131

1 1 1 ⎤ ⎡1 1 ⎤ = ⎡1 5.63 (a) H 2 = ⎡1 1 ⎤. Now, H t2 H 2 = ⎡1 2 2 ⎢⎣1 − 1⎦⎥ ⎢⎣1 − 1⎦⎥ ⎢⎣0 ⎣⎢1 − 1⎥⎦ H2 ⎤ ⎡H . Thus, H4 = ⎢ 2 ⎣H 2 − H 2 ⎥⎦ ⎡1 t H t2 ⎤ ⎡H 2 H 2 ⎤ 1 ⎡4I 2 4I 2 ⎤ ⎢0 1 t 1 ⎡H 2 H H = ⎢ t = = t ⎥ ⎢H − H 2 ⎥⎦ 4 ⎢⎣4I 2 4I 2 ⎥⎦ ⎢0 4 4 4 4 ⎢H ⎣ 2 − H 2 ⎦⎥ ⎣ 2 ⎢⎣0 1 N

H tN H N =

1 N

⎡H t ⎢ tN / 2 ⎢⎣H N / 2

H tN / 2 ⎤ ⎡H N / 2 ⎥ − H tN / 2 ⎥⎦ ⎢⎣H N / 2

0⎤ = I . 2 1 ⎥⎦

0 1 0 0

0 0 1 0

H N / 2 ⎤ 1 ⎡ NI N / 2 = − H N / 2 ⎥⎦ N ⎢⎣ NI N / 2

0⎤ 0⎥ = I . 4 0⎥ 1⎥⎦ NI N / 2 ⎤ = IN . NI N / 2 ⎥⎦

N −1 1 1 2 (b) From Eq. (5.172), ∑ x[ n] = x t x = ⎛⎜ H tN X Haar ⎞⎟ ⎛⎜ H tN X Haar ⎞⎟ ⎠ ⎠ ⎝N ⎝N n=0 t

=

1 N2

X tHaar H N H tN X Haar =

1 N

X tHaar X Haar =

1 N −1 2 ∑ X Haar [ k ] N k =0

as from Eq.

(5.171) H N H tN + N . N −1 2 πnk ⎞ 2 πnk ⎞ ⎞ 5.64 X DHT [ k ] = ∑ x[ n] ⎛⎜ cos⎛⎜ ⎟ + sin⎛⎜ ⎟ ⎟. Now, ⎝ N ⎠⎠ ⎝ ⎝ N ⎠ n=0

2 πmk ⎞ 2 πmk ⎞ ⎞ X DHT [ k ]⎛⎜ cos⎛⎜ ⎟ + sin⎛⎜ ⎟⎟ N N ⎠⎠ ⎝ ⎠ ⎝ ⎝ N −1

2 πnk ⎞ 2 πnk ⎞ ⎞⎛ 2 πmk ⎞ 2 πmk ⎞ ⎞ = ∑ x[ n] ⎛⎜ cos⎛⎜ ⎟ + sin⎛⎜ ⎟ ⎟⎜ cos⎛⎜ ⎟ + sin⎛⎜ ⎟ ⎟. Therefore, ⎝ N ⎠ ⎠⎝ ⎝ N ⎠ ⎝ N ⎠⎠ ⎝ ⎝ N ⎠ n=0 N −1

2 πmk ⎞ 2 πmk ⎞ ⎞ ⎟ + sin⎛⎜ ⎟⎟ ∑ X DHT [ k ]⎛⎜ cos⎛⎜ ⎝ N ⎠⎠ ⎝ ⎝ N ⎠ k =0 N −1 N −1 2 πnk ⎞ 2 πnk ⎞ ⎞⎛ 2 πmk ⎞ 2 πmk ⎞ ⎞ = ∑ x[ n] ∑ ⎛⎜ cos⎛⎜ ⎟ + sin⎛⎜ ⎟ ⎟⎜ cos⎛⎜ ⎟ + sin⎛⎜ ⎟ ⎟. ⎝ N ⎠ ⎝ N ⎠ ⎠⎝ ⎝ N ⎠ ⎝ N ⎠⎠ n=0 k =0 ⎝

⎧ N , if m = n = 0, ⎪ N / 2, if m = n ≠ 0, It can be shown that =⎨ N / 2, if m = N − n, ⎪ ⎩ 0, otherwise, ⎧ N , if m = n = 0, N −1 π π 2 nk 2 mk ⎞⎟ ⋅ sin⎛⎜ ⎞⎟ = ⎪ N / 2, if m = n ≠ 0, ∑ sin⎛⎜ ⎨ ⎝ N ⎠ ⎝ N ⎠ ⎪− N / 2, if m = N − n, k =0 0, otherwise, ⎩ N −1

2 πnk ⎞ 2 πmk ⎞ ⎟ ⋅ cos⎛⎜ ⎟ ∑ cos⎛⎜ ⎝ N ⎠ ⎝ N ⎠ k =0

N −1

N −1

2 πnk ⎞ 2 πmk ⎞ 2 πnk ⎞ 2 πmk ⎞ ⎟ ⋅ cos⎛⎜ ⎟ = ∑ cos⎛⎜ ⎟ ⋅ sin⎛⎜ ⎟ = 0. ∑ sin⎛⎜ ⎝ N ⎠ ⎝ N ⎠ k =0 ⎝ N ⎠ ⎝ N ⎠ k =0

Hence, x[ m] =

1 N −1 2 πmk ⎞ 2 πmk ⎞ ⎞ ⎟ + sin⎛⎜ ⎟ ⎟. ∑ X DHT [ n] ⎛⎜ cos⎛⎜ ⎝ N ⎠⎠ ⎝ ⎝ N ⎠ N n=0

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132

⎧ x[ n − no + N ], 0 ≤ n ≤ no − 1, 5.65 (a) y[ n] = x[〈 n − no 〉 N ] = ⎨ n o ≤ n ≤ N − 1. ⎩ x[ n − no ], N −1 2 πnk ⎞ 2 πnk ⎞ ⎞ YDHT [ k ] = ∑ y[ n] ⎛⎜ cos⎛⎜ ⎟ + sin⎛⎜ ⎟⎟ ⎝ N ⎠⎠ ⎝ ⎝ N ⎠ n=0 n −1

o 2 πnk ⎞ 2 πnk ⎞ ⎞ N −1 2 πnk ⎞ 2 πnk ⎞ ⎞ = ∑ x[ n − no + N ] ⎛⎜ cos⎛⎜ ⎟ + sin⎛⎜ ⎟ ⎟ + ∑ x[ n − no ] ⎛⎜ cos⎛⎜ ⎟ + sin⎛⎜ ⎟⎟ . ⎝ N ⎠⎠ n=n ⎝ N ⎠⎠ ⎝ ⎝ N ⎠ ⎝ ⎝ N ⎠ n=0 o

Replacing n − no + N by n in the first sum and n − no by n in the second sum we get

YDHT [ k ] =

N −1

2 π( n + n o )k ⎞ 2 π( n + n o )k ⎞ ⎞ x[ n] ⎛⎜ cos⎛⎜ ⎟ + sin⎛⎜ ⎟⎟ N N ⎠ ⎝ ⎠⎠ ⎝ ⎝ n = N − no

∑

n −1

o 2 π( n + n o ) k ⎞ 2 π( n + n o ) k ⎞ ⎞ + ∑ x[ n] ⎛⎜ cos⎛⎜ ⎟ + sin⎛⎜ ⎟⎟ N N ⎠ ⎝ ⎠⎠ ⎝ ⎝ n=0

N −1 2 π( n + n o ) k ⎞ 2 π( n + n o ) k ⎞ ⎞ = ∑ x[ n] ⎛⎜ cos⎛⎜ ⎟ + sin⎛⎜ ⎟⎟ N N ⎠ ⎝ ⎠⎠ ⎝ ⎝ n=0 2 πn o k ⎞ N −1 2 πnk ⎞ 2 πnk ⎞ ⎞ ⎛ = cos⎛⎜ ⎟ + sin⎛⎜ ⎟⎟ ⎟ ∑ x[ n] ⎜ cos⎛⎜ ⎝ N ⎠⎠ ⎝ N ⎠n=0 ⎝ ⎝ N ⎠ 2 πn o k ⎞ N −1 2 πnk ⎞ 2 πnk ⎞ ⎞ ⎛ + sin⎛⎜ ⎟ − sin⎛⎜ ⎟⎟ ⎟ ∑ x[ n] ⎜ cos⎛⎜ ⎝ N ⎠⎠ ⎝ N ⎠n=0 ⎝ ⎝ N ⎠ 2 πn o k ⎞ 2 πn o k ⎞ = cos⎛⎜ [ −k ]. ⎟ X DHT [ k ] + sin⎛⎜ ⎟X ⎝ N ⎠ ⎝ N ⎠ DHT

(b) The N –point DHT of x[〈− n〉 N ] is X DHT[ −k ] . N −1

1 N-1 N −1 ∑ ∑ X DHT [ k ]X DHT [l] × n=0 N 2 k =0 l=0 ⎛ N −1⎛ ⎛ 2 πnk ⎞ 2 πnk ⎞ ⎞⎛ 2 πnl ⎞ 2 πnl ⎞ ⎞ ⎞ ⎜⎜ ∑ ⎜ cos⎜ ⎟ + sin⎛⎜ ⎟ ⎟⎜ cos⎛⎜ ⎟ + sin⎛⎜ ⎟ ⎟ ⎟. ⎝ N ⎠ ⎠⎝ ⎝ N ⎠ ⎝ N ⎠ ⎠ ⎟⎠ ⎝ n=0 ⎝ ⎝ N ⎠ Using the orthogonality property, we observe that the above product is equal to N if N −1 1 N −1 2 k = l and is equal to zero if k ≠ l. Hence, ∑ x 2 [ n] = ∑ (X DHT [ k ]) . N k =0 n=0 (c)

2 ∑ x [ n] =

(

)

(

)

1 2 πnk ⎞ 1 2 πnk ⎞ 5.66 cos⎛⎜ W − nk − WNnk . Therefore, ⎟ = W − nk + WNnk and sin⎛⎜ ⎟= ⎝ N ⎠ 2 N ⎝ N ⎠ 2j N

X DHT [ k ] =

(

)

1 N −1 nk nk − nk − nk ∑ x[ n] WN + WN − j WN + j WN 2 n=0 1 = X[ N − k ] + X[ k ] − j X [ N − k ] + j X[ k ] . 2

(

)

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133

N −1 ⎛ ⎛ 2 πnk ⎞⎟ + sin⎛⎜ 2 πnk ⎞⎟ ⎞ 5.67 y[ n] = x[ n] O N g[ n]. Thus, YDHT [ k ] = ∑ y[ n] ⎜ cos⎜ ⎟ ⎝ N ⎠⎠ ⎝ ⎝ N ⎠ n=0 N −1 N −1 2 πnk ⎞ 2 πnk ⎞ ⎞ = ∑ x[r ] ∑ g[〈 n − r 〉 N ] ⎛⎜ cos⎛⎜ ⎟ ⎟. ⎟ + sin⎛⎜ ⎝ N ⎠⎠ ⎝ ⎝ N ⎠ r =0 n=0

From the results of Problem 5.65 we have N −1 2 πlk ⎞ 2 πlk ⎞ ⎞ YDHT [ k ] = ∑ x[l]⎛⎜ G DHT [ k ] cos⎛⎜ ⎟ + G DHT [〈− k 〉 N ] sin⎛⎜ ⎟⎟ ⎝ N ⎠ ⎝ N ⎠⎠ ⎝ l=0 N −1

N −1

2 πlk ⎞ 2 πlk ⎞ = G DHT [ k ] ∑ x[l] cos⎛⎜ ⎟ + G DHT [〈− k 〉 N ] ∑ x[l] sin⎛⎜ ⎟ ⎝ N ⎠ ⎝ N ⎠ l=0 l =0

= G DHT [ k ] (X DHT [ k ] + X DHT [〈− k 〉 N ]) + G DHT [〈− k 〉 N ] (X DHT [ k ] − X DHT [〈− k 〉 N ]) =

1 2 1 [k] X 2 DHT

(GDHT [k ] + GDHT [〈− k〉 N ]) +

1 2 1 [〈− k 〉 N ] X 2 DHT

(GDHT [k ] − GDHT [〈− k〉 N ]) .

N −1

5.68 (a) y[ n] = ∑ (β1WN− nk + β 2WNnk ) X DCFT [ k ] N −1

k =0 N −1

= ∑ x[ m] ∑ (α1β1WN( m − n)k + α 2 β 2WN− ( m + n)k + α1β 2WN( m + n)k + α1β 2WN− ( m − n)k ) m =0 N −1

k =0

= ∑ x[ m]N ((α1β1 + α 2 β 2 )δ[ m − n] + (α 2 β1 + α1β 2 )(δ[ m + n] + δ[ m + n − N ])) m =0

= N [(α1β1 + α 2 β 2 ) x[ n] + (α 2 β1 + α1β 2 )(x[ n]δ[ n] + x[ N − n](1 − δ[ n])]. If we require y[ n] = x[ n], 0 ≤ n ≤ N − 1, then the following conditions must be satisfied: α 2 β1 + α1β 2 = 0, and N (α1β1 + α 2 β 2 ) = 1.

(5-1) (5-2)

(b) Let α12 − α 22 ≠ 0. Solving for β1 and β2 in Eqs. (5-1) and (5-2) we arrive at α1 − α2 , β2 = . β1 = N (α12 − α 22 ) N (α12 − α 22 ) Then, the inverse DCFT is given by N −1 1 − nk nk x[ n] = ∑ (α1WN − α 2WN ) X DCFT [ k ], 0 ≤ n ≤ N − 1. 2 2 N (α 1 − α 2 ) k = 0 (c) If α1 = α *2 = α re + jα im with α re ≠ 0 and α im ≠ 0, the expression for X DCFT [ k ] reduces to

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N −1 2 πnk ⎞ 2 πnk ⎞ ⎞ X DCFT [ k ] = ∑ ⎛⎜ 2α re cos⎛⎜ ⎟ + 2α im sin⎛⎜ ⎟ ⎟x[ n] ⎝ N ⎠ ⎝ N ⎠⎠ n=0 ⎝ which is real. The inverse DCFT is then given by N −1 1 2 πnk ⎞ 2 πnk ⎞ ⎞ x[ n] = [ k ]. ⎟ + 2α re sin⎛⎜ ⎟ ⎟X ∑ ⎛⎜ 2α im cos⎛⎜ N N ⎠ ⎠ DCFT ⎝ ⎠ ⎝ 4 Nα1α 2 k = 0 ⎝

(d) It can be easily shown that the discrete Hartley transform (DHT) of Eq. (5.192) is a 1

. special case of the real DCFT with α re = α im = 2N 1 1⎤ ⎡1 1 1 1 1 1 1 − − ⎡ ⎤ ⎢ 5.69 (a) H 2 = , H 4 = 1 1 − 1 − 11⎥, and ⎢⎣1 − 1⎥⎦ ⎢ ⎥ ⎢⎣1 − 1 − 1 1 ⎥⎦ 1 1 1 1 1 1 ⎤ ⎡1 1 ⎢1 − 1 1 − 1 1 − 1 1 − 1 ⎥ ⎢1 1 − 1 − 1 1 1 − 1 − 1 ⎥ 1 −1 −1 1 1 −1 −1 1 ⎥ H8 = ⎢ . ⎢1 1 1 1 − 1 − 1 − 1 − 1⎥ ⎢1 − 1 1 − 1 − 1 1 − 1 1 ⎥ ⎢1 1 − 1 − 1 − 1 − 1 1 1⎥ ⎢⎣1 − 1 − 1 1 − 1 1 1 − 1⎥⎦ ⎡H (b) From the structure of H 2 , H 4 , and H 8 , it can be seen that H 4 = ⎢ 2 ⎣H 2 H4 ⎤ ⎡H and H 8 = ⎢ 4 . H − H 4 ⎥⎦ ⎣ 4

H2 ⎤ , − H 2 ⎥⎦

(c) X HT = H N x. Therefore, x = H −N1X HT = N ⋅ H TN X HT = N ⋅ H *N X HT . Hence, N −1

x[ n] = ∑ X HT k =0

l −1

∑ bi ( n ) bi ( k )

[ k ](−1) i =0

, where bi (r ) is the i –th bit in the binary

representation of r. M5.1 (a) N = input('The value of N = '); k = -N:N; y = ones(1,2*N+1); w = 0:2*pi/255:2*pi; Y = freqz(y, 1, w); Ydft = fft(y); n = 0:1:2*N; plot(w/pi,abs(Y),n*2/(2*N+1),abs(Ydft),'o'); xlabel('\omega/\pi'),ylabel('Amplitude');

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(b) Replace the statement k = -N:N; with k = 0:N;, y = ones(1,2*N+1); with y = ones(1,N+1); the statement n = 0:1:2*N; with n = 0:N;, and the statement plot(w/pi,abs(Y),n*2/(2*N+1),abs(Ydft),'o'); with plot(w/pi,abs(Y),n*2/(N+1),abs(Ydft),'o');,in the program of Part (a). (c) Add the statement y = y – abs(k)/N; below the statement y = ones([1,2*N+1]); in the program of Part (a). (d) Replace the statement y = ones(1,2*N+1); with y = N + ones(1,2*N+1) – abs(k); in the program of Part (a). (e) Replace the statement y = ones(1,2*N+1); with y = cos(pi*k/(2*N)); in the program of Part (a).

The plots generated for N = 4 are shown below where the circles denote the DFT samples. (a) (b) 5

10

4 Amplitude

Amplitude

8 6 4

2 1

2 0

3

0 0

0.5

1 ω/ π

1.5

2

0

0.5

1

1.5

2

ω/ π

(d) 4

25

3

20 Amplitude

Amplitude

(c)

2 1 0

15 10 5

0

0.5

1 ω/π

1.5

2

0

0

0.5

1

1.5

2

ω/π

(e)

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6

Amplitude

5 4 3 2 1 0

0

0.5

1

1.5

2

ω/ π

M5.2 The code fragments to be used are as follows:

Y = fft(g).*fft(h); y = iift(Y); (a) y[ n] = g[ n] O 6 h[ n]. The output generated using the above code fragments is y = -6 9 -16 20 -4 45 5 v[ n]. The output generated using the above code fragments is (b) w[ n] = x[ n] O w = Columns 1 through 3 11.0000 +25.0000i -9.0000 +48.0000i

3.0000 +17.0000i

Columns 4 through 5 29.0000 + 0.0000i -10.0000 +12.0000i 5 y[ n]. The output generated using the above code fragments is (c) u[ n] = x[ n] O

u = -23.0000

-69.0000

35.0000

105.0000

73.0000

M5.3 N = 8; % sequence length gamma = 0.5; k = 0:N-1; x = exp(-gamma*k); X = fft(x); % Property 1 X1 = fft(conj(x)); G1 = conj([X(1) X(N:-1:2)]); % Verify X1 = G1 % Property 2 x2 = conj([x(1) x(N:-1:2)]); X2 = fft(x2);

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% Verify X2 = conj(X) % Property 3 x3 = real(x); X3 = fft(x3); G3 = 0.5*(X+conj([X(1) X(N:-1:2)])); % Verify X3 = G3 % Property 4 x4 = j*imag(x); X4 = fft(x4); G4 = 0.5*(X-conj([X(1) X(N:-1:2)])); % Verify X4 = G4 % Property 5 x5 = 0.5*(x+conj([x(1) x(N:-1:2)])); X5 = fft(x5); % Verify X5 = real(X) % Property 6 x6 = 0.5*(x-conj([x(1) x(N:-1:2)])); X6 = fft(x6); % Verify X6 = j*imag(X) M5.4 N = 8; k = 0:N-1; gamma = 0.5; x = exp(-gamma*k); X = fft(x); % Property 1 xpe = 0.5*(x+[x(1) x(N:-1:2)]); xpo = 0.5*(x-[x(1) x(N:-1:2)]); Xpe = fft(xpe); Xpo = fft(xpo); % Verify Xpe = real(X) and Xpo = j*imag(X) % Property 2 X2 = [X(1) X(N:-1:2)]; % Verify X = conj(X2); % real(X) = real(X2) and imag(X) = -imag(X2) % abs(X)= abs(X2) and angle(X) = -angle(X2) M5.5 N = 8; % N is length of the sequence(s) gamma = 0.5; k = 0:N-1; g = exp(-gamma*k); h = cos(pi*k/N); G = fft(g); H=fft(h); % Property 1

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alpha=0.5; beta=0.25; x1 = alpha*g+beta*h; X1 = fft(x1); % Verify X1=alpha*G+beta*H % Property 2 n0 = N/2; % n0 is the amount of shift x2 = [g(n0+1:N) g(1:n0)]; X2 = fft(x2); % Verify X2(k)= exp(-j*k*n0)G(k) % Property 3 k0 = N/2; x3 = exp(-j*2*pi*k0*k/N).*g; X3 = fft(x3); G3 = [G(k0+1:N) G(1:k0)]; % Verify X3=G3 % Property 4 x4 = G; X4 = fft(G); G4 = N*[g(1) g(8:-1:2)]; % This forms N*(g mod(-k)) % Verify X4 = G4; % Property 5 % To calculate circular convolution between % g and h use eqn (3.67) h1 = [h(1) h(N:-1:2)]; T = toeplitz(h',h1); x5 = T*g'; X5 = fft(x5'); % Verify X5 = G.*H % Property 6 x6 = g.*h; X6 = fft(x6); H1 = [H(1) H(N:-1:2)]; T = toeplitz(H.', H1); % .' is the nonconjugate transpose G6 = (1/N)*T*G.'; % Verify G6 = X6.' M5.6 g = input('Type in first sequence = '); h = input('Type in second sequence = '); x = g +i*h; XF = fft(x); XFconj = conj(XF); N = length(g); YF = zeros(N-1,1)'; for k = 1:N-1; YF(k) = XFconj(mod(-k,N)+1);

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end YF1 = [XFconj(1) YF]; GF = (XF + YF1)/2; HF = (XF - YF1)/2; M5.7 x = [-3 5 45 -15 -9 -19 -8 21 -10 23]; XF = fft(x); k =0:9; YF = exp(-i*2*pi*k/5).*XF; output = [XF(1) XF(6) sum(XF) sum(YF)];; disp(output) disp(sum(abs(XF).*abs(XF))

The output data generated by this program is 30 0 -30-0.0000i -100-0.0000i 38600

M5.8 X = [11 8-i*2 1-i*12 6+i*3 -3+i*2 2+i k = 8:12; XF(k)=conj(X(mod(-k+2,12))); XF = [X XF(8:12)]; x = ifft(XF); n = 0:11; y = exp(i*2*pi*n/3).*x; output = [x(1) x(7) sum(x) sum(y)]; disp(output) disp(x.*x)

The output data generated by this program is 4.5000 -0.8333 11.0000 74.8333

15];

-3.0000-2.0000i

M5.9 n=0:255; x = 0.1*n.*exp(-0.03*n); plot(n,x);axis([0 255 0 1.3]); xlabel('n');ylabel('Amplitude'); title('Original signal'); pause z = [zeros(1,50) ones(1,156) zeros(1,50)]; y = 4*rand(1,256)-1; YF = z.*fft(y); yinv = ifft(YF); s = x + yinv; plot(n,s);axis([0 255 -2 4]); xlabel('n');ylabel('Amplitude'); title('Noise corrupted signal'); pause zc = [ones(1,50) zeros(1,156) ones(1,50)]; SF = zc.*fft(s); xr = ifft(SF); plot(n,xr);axis([0 255 0 1.3]); xlabel('n');ylabel('Amplitude');

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title('Signal after Fourier-domain filtering'); Original signal

Noise corrupted signal 4

1.2 3 Amplitude

Amplitude

1 0.8 0.6 0.4

1 0 -1

0.2 0

2

0

50

100

150

200

-2

250

0

50

100

150

200

250

Signal after Fourier-domain filtering 1.2

Amplitude

1 0.8 0.6 0.4 0.2 0

M5.10

0

50

100

150

200

250

function y = overlapsave(x,h) X = length(x); %Length of longer sequence M = length(h); %length of shorter sequence flops(0); if (M > X) %Error condition disp('error'); end %clear all temp = ceil(log2(M)); %Find length of circular convolution N = 2^temp; %zero padding the shorter sequence if(N > M) for i = M+1:N h(i) = 0; end end m = ceil((-N/(N-M+1))); while (m*(N-M+1) <= X) if(((N+m*(N-M+1)) <= X)&((m*(N-M+1)) > 0)) for n = 1:N x1(n) = x(n+m*(N-M+1)); end end if(((m*(N-M+1))<=0)&((N+m*(N-M+1))>=0)) %underflow adjustment

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for n = 1:N x1(n) = 0; end for n = m*(N-M+1):N+m*(N-M+1) if(n > 0) x1(n-m*(N-M+1)) = x(n); end end end if((N+m*(N-M+1)) > X) %overflow adjustment for n = 1:N x1(n) = 0; end for n = 1:(X-m*(N-M+1)) x1(n) =x (m*(N-M+1)+n); end end w1 = circonv1(h,x1); %circular convolution using DFT for i = 1:M-1 y1(i) = 0; end for i = M:N y1(i) = w1(i); end for j = M:N if((j+m*(N-M+1)) < (X+M)) if((j+m*(N-M+1)) > 0) yO(j+m*(N-M+1)) = y1(j); end end end m = m+1; end %disp('Convolution using Overlap Save:'); y = real(yO); function y = circonv1(x1,x2) L1 = length(x1); L2 = length(x2); if L1 ~= L2, error('Sequences of unequal lengths'), end X1 = fft(x1); X2 = fft(x2); X_RES = X1.*X2; y = ifft(X_RES);

The MATLAB program for performing convolution using the overlap-save method is h = [1 1 R = 50;

1]/3;

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d = rand(1,R) - 0.5; m = 0:1:R-1; s = 2*m.*(0.9.^m); x = s + d; %x = [x x x x x x x]; y = overlapsave(x,h); k = 0:R-1; plot(k,x,'r-',k,y(1:R),'b--'); xlabel('Time index n');ylabel('Amplitude'); legend('r-', 's[n]','b--','y[n]');

The output plot generated by the above program is shown below: 8 s[n] y[n]

Amplitude

6 4 2 0

0

10

20 30 Time index n

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40

50

143

Chapter 6 ∞

∞

∞

6.1 X ( z ) = ∑ x[ n] z − n = ∑ x[ n] z − n . Therefore, lim X ( z ) = lim ∑ x[ n] z − n n = −∞

z→∞

n=0 ∞

z →∞ n=0

= lim x[0] + lim ∑ x[ n] z − n = x[0]. z→∞

z → ∞ n =1 ∞

6.2

(a) Z δ{[ n]} = ∑ δ[ n] z − n = δ[0] = 1, which converges everywhere in the z –plane. n = −∞

(b) x[ n] = α n µ[ n]. From Table 6.1, ∞

1

Z {x[ n]} = X ( z ) = ∑ x[ n]z − n =

, z > α . Let g[ n] = nx[ n]. Then, 1 − αz − 1 ∞ dX ( z ) = − ∑ ng[ n]z − n −1 . Hence, Z {g[ n]} = G( z ) = ∑ nx[ n]z − n . Now, dz n = −∞ n = −∞ n = −∞ ∞

z

∞ dX ( z ) dX ( z ) αz − 1 = − ∑ nx[ n]z − n = −G( z ), or, G ( z ) = − z = , z > α. dz dz n = −∞ (1 − αz −1 ) 2

r n jω o n (e − e − jωo n ) µ[ n]. Using the results of Example 2j 6.1 and the linearity property of the z –transform we get ⎞ ⎞ 1 ⎛ 1 ⎛ 1 1 ⎟ ⎟− ⎜ Z {r n sin(ωo n) µ[ n]} = ⎜ 2 j ⎜ 1− r e jωo z −1 ⎟ 2 j ⎜ 1− r e − jωo z −1 ⎟ ⎠ ⎠ ⎝ ⎝

(c) x[ n] = r n sin(ωo n) µ[ n] =

=

6.3

r ( e jω o − e − jω o ) z − 1 2j jω o − jω o − 1 2 −2

1 − r (e

+e

)z

+r z

=

r sin(ωo )z −1 1 − 2r cos(ωo )z

−1

2 −2

+r z

,

∀ z > r.

(a) x1[ n] = α n µ[ n − 2]. Note, x1[ n] is a right-sided sequence. Hence, the ROC of its ∞

∞

z –transform is exterior to a circle. Therefore, X1 ( z ) = ∑ α n µ[ n − 2]z − n = ∑ α n z − n Simplifying we get X1 ( z) =

1 1 − αz

−1

− 1 − αz − 1 =

n = −∞ 2 −2

α z

1 − αz − 1

n=2

whose ROC is given by

z > α. (b) x 2 [ n] = −α n µ[ − n − 3]. Note, x2 [ n] is a left-sided sequence. Hence, the ROC of its z –transform is interior to a circle. Therefore, ∞

−3

∞

∞

n = −∞

n = −∞

m =3

m=3

X 2 ( z) = − ∑ α n µ[ − n − 3]z − n = − ∑ α n z − n = − ∑ α − m z m = − ∑ ( z / α ) m

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Simplifying we get X 2 ( z ) =

(z / α )3 whose ROC is given by z < α . 1 − (z / α )

(c) x 3 [ n] = α n µ[ n + 4]. Note, x3[ n] is a right-sided sequence. Hence, the ROC of its ∞

∞

n = −∞

n = −4

z –transform is exterior to a circle. Therefore, X 3 ( z) = ∑ α n µ[ n + 4]z − n = ∑ α n z − n Simplifying we get X 3 ( z ) =

(α / z ) whose ROC is given by z > α . 1 − (α / z ) −4

(d) x 4 [ n] = α n µ[ − n]. Note, x4 [ n] is a left-sided sequence. Hence, the ROC of its ∞

0

n = −∞

n = −∞

z –transform is interior to a circle. Therefore, X 4 ( z) = ∑ α n µ[ − n]z − n = ∑ α n z − n ∞

1 , z / α < 1. Therefore the ROC of X 4 ( z) is given by z < α . 1 − (z / α)

= ∑ α−mzm = m=0

6.4

Z {(0.4) n µ[ n]} =

1 1 − 0 .4 z − 1

, z > 0.4; Z {( −0.6) n µ[ n]} =

1 1 + 0 .6 z − 1

, z > 0.6;

1

Z {(0.4) n µ[ − n − 1]} = −

, z < 0.4; 1 − 0. 4 z − 1 1 Z {( −0.6) n µ[ − n − 1]} = − , z < 0.6; 1 + 0 .6 z − 1

(a) Z {x1[ n]} =

1 1 − 0 .4 z − 1

(b) Z {x 2 [ n]} = (c) Z {x 3 [ n]} = (d) Z {x 4 [ n]} =

+

1 1 − 0 .4 z − 1 1 1 − 0 .4 z − 1 1 1 − 0 .4 z − 1

1 1 + 0 .6 z − 1

+ + +

=

1 1 + 0 .6 z − 1 1 1 + 0 .6 z − 1 1 1 + 0 .6 z − 1

1 + 0 .2 z − 1 (1 − 0.4 z −1 )(1 + 0.6 z −1 )

= = =

, z > 0 .6 .

1 + 0 .2 z − 1 (1 − 0.4 z −1 )(1 + 0.6 z −1 ) 1 + 0 .2 z − 1 (1 − 0.4 z −1 )(1 + 0.6 z −1 )

, 0 .4 < z < 0 .6 .

, z < 0.4.

1 + 0 .2 z − 1 (1 − 0.4 z −1 )(1 + 0.6 z −1 )

. Since the ROC of

the first term is z < 0.4 and that of the second term is z > 0.6 . Hence, the z – transform of x 4 [ n] does not converge. 6.5

(a) The ROC of Z {x1[ n]} is z > 0.3, the ROC of Z {x 2 [ n]} is z > 0.7, the ROC of Z {x 3 [ n]} is z > 0.4, and the ROC of Z {x 4 [ n]} is z < 0.4. (b) (i) The ROC of Z {y1[ n]} is z > 0.7, (ii) The ROC of Z {y2 [ n]} is z > 0.4,

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(iii) The ROC of Z {y3 [ n]} is 0.3 < z < 0.4, (iv) The ROC of Z {y 4 [ n]} is z > 0.7, (v) The ROC of Z {x 2 [ n]} is z > 0.7, whereas, the ROC of Z {x 4 [ n]} is z < 0.4. Hence, the z -transform of y5 [ n] does not converge. (vi) The ROC of Z {x 3 [ n]} is z > 0.4, whereas, the ROC of Z {x 4 [ n]} is z < 0.4. Hence, the z -transform of y6 [ n] does not converge. 6.6

v[ n] = α

n

= α n µ[ n] + α − n µ[ − n − 1]. Now, Z {α n µ[ n]} =

1 1 − αz − 1

−1

∞

∞

n = −∞

m =1

m=0

, z > α . (See Table

6.1) and Z {α − n µ[ − n − 1]} = ∑ α − n z − n = ∑ α m z m = ∑ α m z m − 1 =

1 −1 1 − αz

αz 1 α (1 − α 2 )z −1 , αz < 1. Therefore, {v[ n]} = V ( z ) = + = 1 − αz 1 − αz −1 z −1 − α (1 − αz −1 )( z −1 − α ) with its ROC given by α < z < 1 / α . =

6.7

(a) x1[ n] = α n µ[ n + 2] + β n µ[ n + 2] with β > α . Note that x1[ n] is a right-sided sequence. Hence, the ROC of its z –transform is exterior to a circle. Now, ∞

∞

n = −2

n=0

Z {α n µ[ n + 2]} = ∑ α n z − n = ∑ α n z − n + (α / z) −1 + (α / z) − 2 =

1 (α / z ) −2 with its ROC given by z > α . Likewise, + ( α / z ) − 1 + (α / z ) − 2 = 1 − (α / z ) 1 − (α / z )

Z {β n µ[ n + 2]} =

( β / z ) −2 with its ROC given by z > β . Hence, 1 − ( β / z)

Z {x1[ n]} = X1 ( z ) = =

(α / z ) − 2 ( β / z ) − 2 + 1 − (α / z ) 1 − ( β / z )

(α −2 + β −2 ) − (αβ −2 + α −2 β )z −1 z − 2 (1 − αz −1 )(1 − βz −1 )

with its ROC given by z > β .

(b) x 2 [ n] = α n µ[ − n − 2] + β n µ[ n − 1] with β > α . Note that x2 [ n] is a two-sided sequence. Now, −2

∞

∞

n = −∞

m=2

m=0

Z {α n µ[ − n − 2]} = ∑ α n z − n = ∑ α − m z m = ∑ ( z / α ) m − 1 − ( z / α ) − ( z / α ) 2 =

(z / α)3 with its ROC given by z < α . Likewise, 1 − (z / α)

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∞

∞

1

n =1

n=0

1 − βz − 1

Z {β n µ[ n − 1]} = ∑ β n z − n = ∑ β n z − n − 1 =

−1 =

βz − 1 1 − βz − 1

with its ROC

given by z > β . Since the two ROCs do not intersect, Z {x 2 [ n]} does not converge. (c) x 3 [ n] = α n µ[ n + 1] + β n µ[ − n − 2] with β > α . Note that x3[ n] is a two-sided sequence. Now, Z {β n µ[− n − 2]} =

=

−2

∞

∞

n = −∞

m=2

m=0

∑ β n z − n = ∑ β − m z m = ∑ ( z / β )m − 1 − ( z / β ) − ( z / β )2

(z / β ) with its ROC given by z < β . Likewise, 1− (z / β ) 3

∞

∞

n =1

n =0

Z {α n µ[n − 1]} = ∑ α n z − n = ∑ α n z − n − 1 =

1 1 − αz −1

−1 =

αz −1 with its ROC given by 1 − αz −1

z > α. 6.8

The denominator factor ( z 2 + 0.3z − 0.18) = ( z + 0.6)( z − 0.3) has poles at z = −0.6 and at z = 0.3, and the factor ( z 2 − 2 z + 4) has poles with a magnitude 2. Hence, the four ROCs are defined by the regions: R 1 : 0 < z < 0.3, R 2 : 0.3 < z < 0.6, R 3 = 0.6 < z < 2, and R 4 : z > 2. The inverse z –transform associated with the ROC R 1 is a left-sided sequence, the inverse z –transforms associated with the ROCs R 2 and R 3 are two-sided sequences, and the inverse z –transform associated with the ROC R 4 is a right-sided sequence.

6.9

X (z ) = Z {x[ n]} with an ROC given by R x . Using the conjugation property of the – transform given in Table 6.2, we observe that Z {x * [ n]} = X * ( z*) whose ROC is 1 2

given by R x . Now, Re{x[ n]} = ( x[ n] + x * [ n]). Hence, Z {Re x[ n]}

=

1 2

(X ( z) + X * ( z)) whose ROC is also R x .

Thus, Z {Im x[ n]} = 6.10

1 2j

Likewise, Im{x[ n]} =

1 ( x[ n] − 2j

x * [ n]).

(X (z) − X * (z)) whose ROC is again R x .

{x[ n]} = {2, 3, − 1, 0, − 4, 3, 1, 2, 4}, − 2 ≤ n ≤ 6. Then, ~ ~ X[ k ] = X ( z) z = e jπk / 3 = X ( z) z = e j 2πk / 6 = X (e jω ) . Note that X [ k ] is a ω = 2πk / 6 periodic sequence with a period 6. Hence, from Eq. (5.49), the inverse of the discrete ∞ t ~ Fourier series X [ k ] is given by x[ n] = ∑ x[ n + 6r ] = x[ n − 6] + x[ n] + x[ n + 6] for r = −∞

0 ≤ n ≤ 5. Let y[ n] = x[ n − 6] + x[ n] + x[ n + 6], − 2 ≤ n ≤ 6. Now,

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{x[ n − 6]} = {0, 0, 0, 0, 0, 0, 2, 3, − 1} and {x[ n + 6] = {1, 2, 4, 0, 0, 0, 0, 0, 0}. Therefore, {y[ n]} = {3, 5, 3, 0, − 4, 3, 3, 5, 3}, − 2 ≤ n ≤ 6. Hence, {~ x[ n]} = {3, 0, − 4, 3, 3, 5}, 0 ≤ n ≤ 5. 6.11

{x[ n]} = {4, 2, − 1, 5, − 3, 1, − 2, 4, 2}, − 6 ≤ n ≤ 2. Then, ~ ~ X[ k ] = X ( z) z = e jπk / 3 = X ( z) z = e j 2πk / 6 = X (e jω ) . Note that X [ k ] is a ω = 2πk / 6

periodic sequence with a period 6. Hence, from Eq. (5.49), the inverse of the discrete ∞ t ~ Fourier series X [ k ] is given by x[ n] = ∑ x[ n + 6r ] = x[ n − 6] + x[ n] + x[ n + 6] for r = −∞

0 ≤ n ≤ 5. Let y[ n] = x[ n − 6] + x[ n] + x[ n + 6], − 6 ≤ n ≤ 5. Now, {x[ n − 6]} = {0, 0, 0, 0, 0, 0, 4, 2, − 1, 5, − 3, 1}, − 6 ≤ n ≤ 5, {x[ n + 6] = {− 2, 4, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0}, − 6 ≤ n ≤ 5. Therefore, {y[ n]} = {2, 6, 1, 5, − 3, 1, 2, 6, 5, − 3, 1}, − 6 ≤ n ≤ 5. x[ n]} = {2, 6, 1, 5, − 3, 1}, 0 ≤ n ≤ 5. Hence, {~ 11

6.12

11

X ( z ) = ∑ x[ n] z − n . X 0 [ k ] = X ( z) z = e (2πk / 9 = ∑ x[ n] e − j 2πkn / 9 , 0 ≤ k ≤ 8. n=0

Therefore, x 0 [ n] =

=

1 8 j 2πkn / 9 ∑ X [k ] e 9 k =0 0

1 8 11 j 2 πk ( n − r ) / 9 ∑ ∑ x[r ] e 9 k =0 r =0

From Eq. (5.11),

=

=

n=0 1 8 ⎛ 11 − j 2 πkr / 9 ⎞⎟ j 2 πkn / 9 ∑ ⎜⎜ ∑ x[r ] e ⎟e 9 k =0 r =0

⎝

9 1 11 j 2 πk ( n − r ) / 9 ∑ x[r ] ∑ e 9 r =0 k =0

1 8 − (r − n ) k ∑ W9 9 k −0

⎠

=

9 1 11 − (r − n ) k . ∑ x[r ] ∑ W9 9 r =0 k =0

⎧1, for r − n = 9i, Hence, =⎨ otherwise. ⎩0,

⎧ x[0] + x[9], for n = 0 i.e., x 0 [ n] = ⎨ otherwise, ⎩ x[ n], {x 0 [ n]} = {20 5 45 − 15 − 9 − 19 − 8 21 − 10}, 0 ≤ n ≤ 8. 6.13

∞

∞

∞

n = −∞

n = −∞

r = −∞

(a) X ( z) = ∑ x[ n] z − n . Hence, X ( z 3 ) = ∑ x[ n] z − 3n = ∑ x[ m / 3] z − m . Define m =3r

⎧ x[ m / 3], m = 0, ± 3, ± 6,K, a new sequence g[ m] = ⎨ We can then express otherwise. ⎩ 0,

Not for sale.

151

∞

X ( z 3 ) = ∑ g[ n] z − n . Thus, the inverse z –transform of X ( z 3 ) is given by g[n]. n = −∞

⎧⎪(−0.5) n / 3 , n = 0, 3, 6,K, For x[ n] = (−0.5) n µ[ n], g[ n] = ⎨ ⎪⎩ 0, otherwise.

(b) Y ( z) = (1 + z −1 ) X ( z 3 ) = X ( z 3 ) + z −1 X ( z 3 ). Therefore,

y[n] = Z −1{Y ( z)} = Z −1 X ( z 3 ) + Z −1 z −1 X ( z 3 ) = g[ n] + g[ n − 1], where g[n] = ⎧⎪(−0.5) n / 3 , n = 0, 3, 6,K, Hence, Z −1 X ( z 3 ) . From Part (a), g[ n] = ⎨ ⎪⎩ 0, otherwise. ⎪⎧(−0.5) ( n −1) / 3 , n = 1, 4, 7,K, Therefore, g[ n − 1] = ⎨ ⎪⎩ 0, otherwise. ⎧ (−0.5) n / 3 , n = 0, 3, 6,K, ⎪⎪ y[ n] = ⎨(−0.5) ( n −1) / 3 , n = 1, 4, 7,K, ⎪ 0, otherwise. ⎪⎩ 6.14

(a) X a (z) = Z {µ[ n] − µ[ n − 5]} =

1 1 − z −1

−

z −5 1 − z −1

=

1 − z −5 1 − z −1

= 1 + z −1 + z −2 + z −3 + z −4 . Since has all poles at the origin, the ROC is the entire z –plane except the point z = 0 , and hence includes the unit circle. On the unit circle,

X a ( z ) z = e j ω = X a ( e j ω ) = 1 + e − j ω + e − j 2 ω + e − j 3ω + e − j 4 ω =

1 − e − j 5ω 1 − e − jω

.

(b) x b [ n] = α n µ[ n] − α n µ[ n − 8], α < 1. From Table 6.1,

z −8

1

X b (z) =

1 − z −8

. The ROC is exterior to the circle at − = 1 − αz − 1 1 − α z − 1 1 − αz − 1 z = α < 1. Hence, the ROC includes the unit circle. On the unit circle,

X b ( z) z = e jω = X b (e jω ) =

1 − e j 8ω 1 − α e jω

.

(c) x c [ n] = (n + 1)α n µ[ n] = nα n µ[ n] + α n µ[ n], α < 1. From Table 6.1,

X c ( z) =

αz − 1

1

1 + αz − 1

. The ROC is exterior to the circle at + = 1 − αz − 1 1 − αz − 1 1 − α z − 1 z = α < 1. Hence, the ROC includes the unit circle. On the unit circle,

X c (e jω ) =

1 + αe jω 1− e

jω

.

Not for sale.

152

N ⎞ 1 − z −(2 N +1) ⎛ 2N (a) Y 1( z ) = ∑ z − n = z N ⎜⎜ ∑ z − n ⎟⎟ = . Y1 ( z) has N poles at n=−N ⎠ z − N (1 − z −1 ) ⎝ n=0 z = 0 and N poles at z = ∞. Hence, the ROC is the entire z –plane excluding the points z = 0 and z = ∞, and includes the unit circle. On the unit circle,

6.15

Y1 ( z) z = e jω

1 sin⎛⎜ ω( N + ) ⎞⎟ 2 ⎠ ⎝ . = Y1 (e jω ) = = sin(ω / 2) e − jNω (1 − e − jω )

1 − e j (2 N + 1)ω

N

(b) Y 2( z) = ∑ z − n =

1 − z −( N + 1)

. Y 2( z) has N poles at z = 0. Hence, the ROC 1 − z −1 is the entire z –plane excluding the point z = 0. On the unit circle, N +1 ⎞ ω⎟ sin⎛⎜ j ( N + 1)ω 1− e 2 ⎝ ⎠. jω − Nω / 2 Y2 ( z ) z = e jω = Y2 (e ) = =e − jω sin(ω / 2) 1− e n ⎧⎪ (c) y3 [ n] = ⎨1 − N , − N ≤ n ≤ N , Now, y3 [ n] = y0 [ n] O * y 0 [ n] where ⎪⎩ 0, otherwise. ⎧⎪1, − N ≤ n ≤ N , (1 − z −( N + 1) ) 2 2 y 0 [ n] = ⎨ . 2 2 Therefore, Y3 ( z ) = Y0 ( z ) = ⎪⎩0, z − N (1 − z −1 ) 2 otherwise. n=0

Y3 ( z ) has

N N poles at z = 0 and poles at z = ∞. Hence, the ROC is the entire z – 2 2

plane excluding the points z = 0 and z = ∞, and includes the unit circle. On the unit N +1 ⎞ ⎞ sin 2 ⎛⎜ ω⎛⎜ ⎟⎟ ⎝ ⎝ 2 ⎠⎠ . circle, Y3 (e jω ) = Y02 (e jω ) = sin 2 (ω / 2)

⎧N + 1 − n , − N ≤ n ≤ N , = y1 [ n] + N ⋅ y 3 [ n], where y1[ n] is the (d) y 4 [ n] = ⎨ 0, otherwise, ⎩ sequence of Part (a) and y 3 [ n] is the sequence of Part (c). Therefore, Y4 ( z ) = Y1 ( z ) + N ⋅ Y3 ( z ) =

1 − z −(2 N + 1)

+N

(1 − z −( N +1) )2

. Since the ROCs of z − N (1 − z −1 )2 z − N (1 − z −1 ) both Y1( z) and Y3 ( z ) include the unit circle, the ROC of Y4 ( z) also includes the unit N +1 ⎞ ⎞ 1 ⎟⎟ sin⎛⎜ ω( N + ) ⎞⎟ sin 2 ⎛⎜ ω⎛⎜ 2 ⎠ ⎝ 2 ⎠⎠ ⎝ ⎝ jω + . circle. On the unit circle, Y4 (e ) = sin(ω / 2) sin 2 (ω / 2)

⎧cos(πn / 2 N ), − N ≤ n ≤ N , Therefore, (e) y f [ n] = ⎨ 0, otherwise. ⎩

Not for sale.

153

1 N − j (πn / 2 N ) − n 1 N j (πn / 2 N ) − n z + z ∑e ∑e 2 n=−N 2 n=−N e j (π / 2) z N ⎛⎜ 1 − e − j (2 N + 1)(π / 2 N ) z − (2 N + 1) ⎞⎟ = ⎜ ⎟ 2 1 − e − j (π / 2 N ) z −1 ⎝ ⎠ e − j (π / 2) z N ⎛⎜ 1 − e j (2 N +1)(π / 2 N ) z − (2 N +1) ⎞⎟ + . j (π / 2 N ) −1 ⎜ ⎟ 2 e z − 1 ⎝ ⎠ Y f (z) has N poles at z = 0 and N poles at z = ∞. Hence, the ROC is the entire z – Y f ( z) =

plane excluding the points z = 0 and z = ∞, and includes the unit circle. On the unit π ⎞⎛ π ⎞⎛ 1 ⎞ 1 ⎞ sin⎛⎜ ⎛⎜ ω − sin⎛⎜ ⎛⎜ ω + ⎟⎜ N + ⎞⎟ ⎟ ⎟⎜ N + ⎞⎟ ⎟ 2 N 2 2 N 2 ⎠⎠ 1 1 ⎝ ⎠ ⎝ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠+ ⎝ circle, Y f (e jω ) = . π ⎞ ⎞ π ⎞ ⎞ 2 2 ⎛ ⎛ ⎛ ⎛ sin⎜ ⎜ ω − sin⎜ ⎜ ω + ⎟ / 2⎟ ⎟ / 2⎟ 2N ⎠ ⎠ 2N ⎠ ⎠ ⎝⎝ ⎝⎝ 6.16

X ( z ) = −4 z 3 + 5 z 2 + z − 2 − 3 z − 1 + 2 z −3 , Y ( z ) = 6 z − 3 − z − 1 + 8 z − 3 + 7 z −4 − 2 z − 5 , W ( z ) = 3 z − 2 + 2 z − 3 + 2 z − 4 − z −5 − 2 z − 7 + 5 z −8 . (a) U ( z ) = X ( z )Y ( z ) = (−4 z 3 + 5z 2 + z − 2 − 3z −1 + 2 z −3 )(6 z − 3 − z −1 + 8 z −3 + 7z −4 − 2 z −5 ) = −24 z 4 + 42 z 3 − 5z 2 − 20 z − 45 + 23z −1 + 66 z −2 − 25z −3 − 42 z −4 − 17 z −5 + 22 z −6 + 14 z −7 − 4 z −8 . Hence, {u[ n]} = {− 24, 42, − 5, − 20, − 45, 23, 66, − 25, − 42, − 17, 22, 14, − 4}, − 4 ≤ n ≤ 8.

(b) V ( z ) = X ( z )W ( z )

= (−4 z 3 + 5z 2 + z − 2 − 3z −1 + 2 z −3 )(3z −2 + 2 z −3 + 2 z −4 − z −5 − 2 z −7 + 5z −8 ) = −12 z + 7 + 5z −1 + 10 z −2 − 16 z −3 − 3z −4 − 28 z −5 + 30 z −6 + 13z −7 − 6 z −8 − 15 z −4 − 4 z −10 + 10 z −11 . Hence, {v[ n]} = {− 12, 7, 5, 10, − 16, − 3, − 28, 30, 13, − 6, − 15, − 4, 10}, − 1 ≤ n ≤ 11.

(c) G ( z ) = W ( z )Y ( z )

= (3z −2 + 2 z −3 + 2 z −4 − z −5 − 2 z −7 + 5z −8 )(6 z − 3 − z −1 + 8 z −3 + 7z −4 − 2 z −5 ) = 18 z −1 + 3z −2 + 3z −3 − 14 z −4 + 25z −5 + 26 z −6 + 60 z −7 − 11z −8 − 16 z −9 − 14 z −10 + 26 z −11 + 39 z −12 − 10 z −13 . Hence, {g[ n]} = {18, 3, 3, − 14, 25, 26, 60, − 11, − 16, − 14, 26, 39, − 10}, 1 ≤ n ≤ 13.

6.17

2 N − 1 ⎛ N −1 N −1⎛ N −1 ⎞ ⎞ YL ( z) = ∑ ⎜⎜ ∑ x[ m]h[ n − m] ⎟⎟ z − n and YC ( z) = ∑ ⎜⎜ ∑ x[ m]h[〈 n − m〉 N ]⎟⎟ z − n . n=0 ⎝ m =0 n=0 ⎝ m=0 ⎠ ⎠

Not for sale.

154

Now, YL (z) can be rewritten as N −1⎛ N −1 2 N −1⎛ N −1 ⎞ ⎞ YL ( z ) = ∑ ⎜⎜ ∑ x[ m]h[ n − m] ⎟⎟z − n + ∑ ⎜⎜ ∑ x[ m]h[ n − m] ⎟⎟z − n n=0 ⎝ m =0 n= N ⎝ m =0 ⎠ ⎠ N −1⎛ N −1 N −1⎛ N −1 ⎞ ⎞ = ∑ ⎜⎜ ∑ x[ m]h[ n − m] ⎟⎟z − n + ∑ ⎜⎜ ∑ x[ m]h[ k − m − N ] ⎟⎟z − ( k − N ) . Therefore, n=0 ⎝ m =0 k =0 ⎝ m=0 ⎠ ⎠ N −1⎛ N −1 N −1⎛ N −1 ⎞ ⎞ 〈YL ( z)〉 − N = ∑ ⎜⎜ ∑ x[ m]h[ n − m] ⎟⎟z − n + ∑ ⎜⎜ ∑ x[ m]h[〈 k − m〉 N ] ⎟⎟z − k (z −1) n=0 ⎝ m =0 k =0 ⎝ m=0 ⎠ ⎠ N −1⎛ N −1 ⎞ = ∑ ⎜⎜ ∑ x[ m]h[〈 n − m〉 N ] ⎟⎟z − n = YC ( z ). n=0 ⎝ m=0 ⎠

6.18

G( z) = 2 − z −1 + 3z −2 , H ( z) = −2 + 4 z −1 + 2 z −2 − z −3 . Now, Y L ( z) = G( z)H ( z) = (2 − z −1 + 3z −2 )(−2 + 4 z −1 + 2 z −2 − z −3 ) = −4 + 10 z −1 − 6 z −2 + 8 z −3 + 7 z −4 − 3z −5 . Therefore, y L [ n] = {− 4 6 2 12 5 − 3}, 0 ≤ n ≤ 5. Using the MATLAB statement y = conv([2 -1 3], [-2 4 2 -1]); we obtain y = -4 10 -6 8 7 -3 which is seen to be the same result as given above.

YC ( z) = 〈YC ( z)〉

= 〈−4 + 10 z −1 − 6 z −2 + 8z −3 + 7z −4 − 3z −5 〉

( z − 4 −1) −2 −3

( z − 4 −1)

= −4 + 10 z −1 − 6 z + 8 z + 7 − 3z −1 = 3 + 7 z −1 − 6 z −2 + 8 z −3 . Therefore, yC [ n] = {1 3 2 12}, 0 ≤ n ≤ 3. Using the MATLAB statement y = circonv([2 -1 3 0],[-2 4 2 -1]); we obtain y = 3 7 -6 8 which is seen to be the same result as given above.

6.19

G( z) =

ρl =

P( z) P( z) = . The residue ρl of G ( z ) at the pole is given by D( z ) (1 − λl z −1 ) R( z )

P( z) . Now, R( z ) z = λ l

D' ( z ) =

dD( z) dz

−1

=

d[(1 − λ l z −1 ) R( z)] dz

−1

= − λl R( z ) + (1 − λl z −1 )

D' ( z) z = λl = − λl R( z) z = λ . Therefore, ρl = − λl l

Not for sale.

dR( z ) dz −1

. Hence,

P( z) . D' ( z ) z = λ l

155

ρ1 ρ2 3z 3z −1 = = + , ( z + 0.6)( z − 0.3) (1 + 0.6 z −1 )(1 − 0.3z −1 ) 1 + 0.6 z −1 1 − 0.3z −1 3 3 10 3 3 10 where ρ1 = = = − , ρ2 = = = . z − 0 . 3 z = − 0. 6 − 0 . 9 z + 0.6 z = 0.3 0.9 3 3

6.20 (a) X a ( z ) =

10 / 3

Therefore, X a ( z ) = −

10 / 3

+

. 1 + 0 .6 z − 1 1 − 0 .3 z − 1 There are three ROCs - R 1 : z < 0.3, R 2 : 0.3 < z < 0.6, R 3 : z > 0.6. The inverse z –transform associated with the ROC R1 is a left-sided sequence:

Z −1{X a ( z)} = x a [ n] =

10 ⎛ n ⎜ ( −0.6) 3 ⎝

− (0.3) n ⎞⎟µ[ −n − 1]. ⎠

The inverse z –transform associated with the ROC R 2 is a two-sided sequence: Z −1{X a ( z)} = x a [ n] = − (−0.6) n µ[ −n − 1] + (0.3) n µ[ n]. 3 3 The inverse z –transform associated with the ROC R 3 is a right-sided sequence: 10

10

(

)

Z −1{X a ( z)} = x a [ n] = − (−0.6) n + (0.3) n µ[ n]. 3 10

(b) X b ( z ) =

3z −1 + 0.1z −2 + 0.87 z −3 (1 + 0.6 z

−1

K = X b (0) = 0, ρ1 =

γ2 = γ1 =

)(1 − 0.3z

−1 2

=K+

)

(1 − 0.3z −1 ) 2

1 + 0.6 z −1

1 + 0.6 z −1

−

+

γ1 1 − 0 .3 z

−1

+

γ2

(1 − 0.3z −1 ) 2

= 2.7279, z = −0.6

= 0.6190, z = 0.3 −2

1 d ⎛⎜ 3z −1 + 0.1z + 0.87 z − 3 ⎞⎟ ⋅ ⎟ − 0.3 dz −1 ⎜⎝ 1 + 0 .6 z − 1 ⎠

X b ( z) =

1 + 0 .6 z

−1

3z −1 + 0.1z − 2 + 0.87z − 3

3z −1 + 0.1z − 2 + 0.87z − 3

2.7279

ρ1

0.3469 1 − 0.3z −1

+

= −0.3469. Hence, z = 0. 3

0.6190 (1 − 0.3z −1 ) 2

.

There are three ROCs - R 1 : z < 0.3, R 2 : 0.3 < z < 0.6, R 3 : z > 0.6. The inverse z –transform associated with the ROC R1 is a left-sided sequence: Z −1{ X b ( z )} = xb [n] = 2.7279(−0.6) n µ[−n − 1]

+ (− 0.3469 + 0.6190(n + 1))(0.3)n µ[ −n − 1]. The inverse z –transform associated with the ROC R 2 is a two-sided sequence: Z −1{ X b ( z )} = xb [ n] = 2.7279( −0.6) n µ[ − n − 1]

+ (− 0.3469 + 0.6190(n + 1))(0.3) n µ[ n]. The inverse z –transform associated with the ROC R 3 is a right-sided sequence: Not for sale.

156

.

Z −1{X b ( z )} = x b [ n] = 2.7279(−0.6) n µ[ n] + (− 0.3469 + 0.6190( n + 1))(0.3) n µ[ n].

G( z ) =

6.21

P (∞ ) P( z ) p0 + p1 z −1 + L + p M z − M . Thus, G(∞ ) = . Now, a partial= − 1 − N D(∞ ) D( z ) 1 + d1 z + L + d N z ρl

N fraction expansion of G ( z ) in z −1 is given by G( z) = ∑

l =1 1 − λl z

−1

, from which we obtain

N p G (∞ ) = ∑ ρ l = 0 . d0 l =1

1

6.22 H ( z ) =

, z > r > 0. By using partial-fraction expansion we write 1 − 2r cos θz −1 + r 2 z − 2 ⎞ ⎛ ⎞ e jθ e − jθ e jθ e − jθ 1 1 ⎛⎜ ⎟. ⎜ ⎟ = − H (z) = − jθ jθ ⎜ jθ −1 − jθ −1 ⎟ 2 sin θ ⎜ jθ − 1 − jθ − 1 ⎟ 1 − re z ⎠ z ⎠ (e − e − j ) ⎝ 1 − re z 1 − re ⎝ 1 − re z 1 r n ⎛⎜ e jθ ( n + 1) − e − jθ ( n + 1) ⎞⎟ Thus, h[ n] = r n e jθ e jnθ µ[ n] − r n e − jθ e − jnθ µ[ n] = µ[ n] ⎟ j 2 sin θ sin θ ⎜⎝ 2j ⎠

(

=

)

r n sin ((n + 1)θ ) µ[ n]. sin θ ∞

−1

∞

∞

∞

n = −∞

n = −∞

m =1

m =1

m =0

6.23 (a) X ( z ) = ∑ α n µ[ −n − 1]z − n = ∑ α n z − n = ∑ α − m z m = ∑ ( z / α) m = ∑ ( z / α) m − 1 =

1 z/α = , z < α. 1 − ( z / α ) 1 − αz −1

(b) Using the differentiation property, we obtain from Part (a), Z{nx[ n]} = − z =

α z −1 (1 − αz

−1 2

dX ( z ) α z −1 = , z < α . Therefore, Z {y[ n]} = Z{nx[ n] + x[ n]} dz (1 − αz −1 ) 2 +

)

1 1 − αz

−1

=

1 (1 − αz −1 ) 2

, z < α. ∞

6.24 (a) Expanding X1 ( z) in a power series we get X1 ( z) = ∑ z − 3n , z > 1. Thus, n=0

⎧1, if n = 3k and n ≥ 0, x1[ n] = ⎨ Alternately, using partial-fraction expansion we get otherwise. ⎩0,

X1 ( z ) =

1 1 − z −3

=

1 3

1 − z −1

+

1 3 1 1+ ( 2

+j

3 −1 )z 2

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+

1 3 1 1+ ( − 2

j

3 −1 )z 2

. Therefore,

157

1 1 1 x1[ n] = µ[ n] + ⎛⎜ − − j 3 3⎝ 2

=

1 3

3⎞ 1⎛ 1 ⎟ µ[ n] + 3 ⎜ − 2 + 2 ⎠ ⎝

µ[ n] + e − j 2πn / 3 µ[ n] + e j 2 πn / 3 µ[ n] = 1 3

1 3

3⎞ ⎟ µ[ n] 2 ⎠

j

1 3

2 3

µ[ n] + cos(2πn / 3) µ[ n]. Thus,

⎧1, if n = 3k and n ≥ 0, x1[ n] = ⎨ otherwise. ⎩0, ∞

(b) Expanding X 2 ( z) in a power series we get X 2 ( z ) = ∑ z − 4 n , z > 1. Thus, n=0

⎧1, if n = 4k and n ≥ 0, Alternately, using partial-fraction expansion we get x 2 [ n] = ⎨ otherwise. ⎩0, X 2 (z) =

x 2 [ n] =

1 4

1 − z −1 1 4

+

1 4

1 + z −1

+

1 4 1 2

1+ ( + j

3 −1 )z 2

1 1 1 µ[ n] + (−1) n µ[ n] + ⎛⎜ − − j 4 4⎝ 2

+

1 4

1+ (

1 − 2

j

3 −1 )z 2

3⎞ 1 1 µ[ n] + ⎛⎜ − + ⎟ 2 ⎠ 4⎝ 2

j

. Thus,

3⎞ ⎟ µ[ n] 2 ⎠

=

1 4

µ[ n] + ( −1) n µ[ n] + e − j 2 πn / 3 µ[ n] + e j 2 πn / 3 µ[ n]

=

1 4

⎧1, if n = 4k and n ≥ 0, 1 1 µ[ n] + ( −1) n µ[ n] + cos(2πn / 3) µ[ n]. Thus, x 2 [ n] = ⎨ 4 2 otherwise. ⎩0,

1 4

1 4

1 4

6.25 (a) X1 ( z) = log(1 − αz −1 ), z > α . Expanding X1( z) in a power series we get

X 1 ( z ) = − αz

−1

∞ αn α 2 z − 2 α 3 z −3 − − −K = − ∑ z − n . Therefore, 2 3 n =1 n

αn µ[ n − 1]. n ⎛ α − z −1 ⎞ ⎟ = log 1 − (αz ) −1 , z < α . Expanding X 2 ( z) in a power series (b) X 2 ( z) = log⎜ ⎜ α ⎟ ⎝ ⎠

x1[ n] = −

(

we get X 2 ( z ) = −(α z ) −1 −

)

∞ (α z ) − n ( αz ) − 2 ( αz ) − 3 − −K = − ∑ . Therefore, 2 3 n n =1

α −n µ[ n − 1]. n ⎛ ⎞ 1 ⎟, z > α . Expanding X 3 ( z ) in a power series we get (c) X 3 ( z ) = log⎜⎜ −1 ⎟ ⎝ 1 − αz ⎠

x 2 [ n] = −

X 3 ( z ) = αz − 1 +

∞ αn α 2 z −2 α 3 z −3 αn + K= ∑ z − n . Therefore, x3 [ n] = µ[ n − 1]. 2 3 n n =1 n

Not for sale.

158

(

)

⎛ α ⎞ ⎟ = − log 1 − (αz) −1 , z < α . Expanding X 4 ( z) in a power (d) X 4 ( z) = log⎜⎜ −1 ⎟ ⎝α−z ⎠ series we get X 4 ( z) = (α z ) −1 +

x 4 [ n] = 6.26 H ( z ) =

α −n µ[ n − 1]. n z − 1 + 1 .7 z − 2 (1 − 0.3z −1 )(1 + 0.5z −1 )

∞ (α z ) − n ( α z ) − 2 ( αz ) − 3 . Therefore, + −K = ∑ 2 3 n n =1

=k+

ρ1

1 − 0 .3 z − 1

+

ρ2

1 + 0 .5 z − 1

, where k = H (0) = −

34 , 3

z −1 + 1.7z − 2 25 , ρ2 = = 3. 1 − 0.3z −1 z = −0.5 1 + 0.5z −1 z = 0.3 3 The statement [r,p,k]=residuez([0 1 1.7],conv([1 -0.3],[1 0.6]); yields r = 3.0000 8.3333 ρ1 =

z −1 + 1.7z − 2

=

p = -0.5000 0.3000 k = -11.3333 34 25 / 3 3 . Hence, its inverse z –transform is given by + + 3 1 − 0 .3 z − 1 1 + 0 .5 z − 1 34 25 h[ n] = − δ[ n] + (0.3) n µ[ n] + 3(−0.5) n µ[ n]. 3 3

Thus, H ( z ) = −

∞

∞

n = −∞

n = −∞

6.27 G( z ) = Z{g[ n] = ∑ g[ n]z − n with a ROC given by R g and H ( z ) = Z{h[ n] = ∑ h[ n]z − n with a ROC given by R h . ∞

∞

n = −∞

n = −∞

(a) G * ( z ) = ∑ g * [ n]( z*) − n and G * ( z*) = ∑ g * [ n]z − n . Therefore, Z{g * [ n]} = G * ( z*) with a ROC given by R g .

(b) Replace n by − m in the sum defining G ( z ). Then ∞

∑ g[ − m]z

m = −∞

m

=

∞

∑ g[ − m](1 / z)

m = −∞

−m

= G(1 / z). Thus, Z{g[ − n]} = G (1 / z ). Since z has

been replaced by 1 / z, the ROC of is given by 1 / R g .

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159

(c) Let y[ n] = αg[ n] + βh[ n]. Then Y ( z ) = Z{αg[ n] + βh[ n] = αZ{g[ n] + βZ{h[ n] = αG ( z ) + βH ( z ). In this case Y ( z ) will converge wherever both G ( z ) and H ( z ) converge. Hence the ROC of Y ( z ) is R g ∩ Rh . (d) y[ n] = g[ n − no ]. Hence, ∞

∞

Y ( z ) = ∑ y[ n]z − n = ∑ g[ n − n 0 ]z − n = =z

n = −∞ − no ∞

∑ g[ n]z

n = −∞

−m

m = −∞

∞

∑ y[ m]z

− (m + n o )

m = −∞

= z − n o G( z ). In this case, the ROC of Y ( z ) is same as that of G ( z )

except for the possible addition or elimination of the point z = 0 or z = ∞ (due to the factor of z − n o ). ∞

∞

∞

n = −∞

n = −∞

n = −∞

(e) y[ n] = α n g[ n]. Hence, Y ( z) = ∑ y[ n]z − n = ∑ αg[ n]z − n = ∑ g[ n]( z / α ) − n = G ( z / α ). The ROC of Y ( z ) is α R g . ∞

∞

(f) y[ n] = n g[ n]. Hence, Y ( z) = ∑ y[ n]z − n = ∑ ng[ n]z − n . Now, n = −∞

∞

n = −∞

∞

dG( z) = − ∑ ng[ n]z − n −1 . Hence, dz n = −∞ n = −∞ ∞ dG( z) dG( z ) . In this case, the ROC of −z = ∑ ng[ n]z − n −1 . Thus, Y(z) = Z{ng[ n]} = − dz dz n = −∞ Y ( z ) is same as that of G ( z ) except possibly the point z = 0 or z = ∞ . G( z ) = ∑ g[ n]z − n . Thus,

6.28 From Eq. (6.111), for N = 3, we get ⎡1 z −1 z − 2 ⎤ 0 0 ⎢ ⎥ −1 D 3 = ⎢1 z1 z1− 2 ⎥. The determinant of D 3 is given by ⎢1 z −1 z − 2 ⎥ 2 2 ⎦⎥ ⎣⎢ 1 z 0−1 det(D 3 ) = 1 z1−1 1 z 2−1

z 0− 2

1

z 2− 2

0 z 2−1 − z 0−1

z1− 2 = 0 z1−1 − z 0−1

= ( z1−1 − z 0−1 )( z 2−1 − z 0−1 ) =

z 0−1

1 z1−1 + z 0−1 1

z 2−1

+ z 0−1

z 0− 2

z −1 − z −1 z1− 2 − z 0− 2 = 1−1 0−1 z2 − z0 z 2− 2 − z 0− 2

z1− 2 − z 0− 2 z 2− 2 − z 0− 2

= ( z1−1 − z 0−1 )( z 2−1 − z 0−1 )( z 2−1 − z1−1 )

−1 −1 ∏ ( z k − z l ).

2≥k ≥l≥0

From Eq. (6.111), for N = 4, we get

Not for sale.

160

1 z 0−1 D4 =

1 z1−1 1 z 2−1 1 z3−1

z 0− 2 z1− 2 z 2− 2 z3− 2

1 z 0−1 1 z1−1 det(D 4 ) = 1 z 2−1 1 z3−1 z1−1 − z 0−1

= z 2−1 − z 0−1 z3−1 − z 0−1

z 0− 3

z1− 3 . The determinant of D 4 is given by z 2− 3 z3− 3 z 0− 2 z1− 2 z 2− 2 z3− 2

z1− 2 − z 0− 2

z 2− 2 − z 0− 2 z3− 2 − z 0− 2

z 0− 3

z 0−1

1

z 0− 2

z1− 3 0 z1−1 − z 0−1 = z 2− 3 0 z 2−1 − z 0−1 z 3− 3 0 z3−1 − z 0−1

z1− 2 − z 0− 2 z 2− 2 − z 0− 2 z3− 2 − z 0− 2

z 0− 3

z1− 3 − z 0− 3 z 2− 3 − z 0− 3 z3− 3 − z 0− 3

z1− 3 − z 0− 2

z 2− 3 − z 0− 2 z3− 3 − z 0− 2 1 z1−1 + z 0−1

z1− 2 + z1−1 z 0−1 + z 0− 2

= ( z1−1 − z 0−1 )( z 2−1 − z 0−1 )( z3−1 − z 0−1 ) 1 z 2−1 + z 0−1 1 z 3−1 + z 0−1

z 2− 2 + z 2−1 z 0−1 + z 0− 2 z 3− 2 + z3−1 z 0−1 + z 0− 2

1 z1−1 + z 0−1

z1− 2 + z1−1 z 0−1 + z 0− 2

= ( z1−1 − z 0−1 )( z 2−1 − z 0−1 )( z3−1 − z 0−1 ) 0 z 2−1 − z 0−1 0 z3−1 − z 0−1

( z 2−1 − z 0−1 )( z 2−1 + z1−1 + z 0−1 )

( z3−1 − z 0−1 )( z3−1 + z1−1 + z 0−1 )

z −1 − z −1 ( z 2−1 − z1−1 )( z 2−1 + z1−1 + z 0−1 ) = ( z1−1 − z 0−1 )( z 2−1 − z 0−1 )( z3−1 − z 0−1 ) 2−1 1−1 z3 − z1 ( z3−1 − z1−1 )( z3−1 + z1−1 + z 0−1 ) = ( z1−1 − z 0−1 )( z 2−1 − z 0−1 )( z 3−1 − z 0−1 )( z 2−1 − z1−1 )( z 3−1 − z1−1 )( z 3−1 − z 2−1 ) =

−1 −1 ∏ ( z k − z l ) . Hence, in the general case, det( D N ) =

3≥ k ≥ l ≥ 0

−1 −1 ∏ ( z k − z l ). It follows

N − 1≥ k ≥ l ≥ 0

from this expression that the determinant det(D N ) is non-zero, i.e., D N is non-singular, if the sampling points z k are distinct. 6.29 X ( z) = 1 − 2 z −1 + 3z −2 − 4 z −3 . Thus, 1 X NDFT [ 0] = X ( z 0 ) = 1 − 2⎛⎜ − ⎞⎟ ⎝ 2⎠

−1

1 + 3⎛⎜ − ⎞⎟ ⎝ 2⎠ X NDFT [1] = X ( z1 ) = 1 − 2 + 3 − 4 = −2, 1 X NDFT [ 2] = X ( z 2 ) = 1 − 2⎛⎜ ⎞⎟ ⎝2⎠ 1 X NDFT [3] = X ( z 3 ) = 1 − 2⎛⎜ ⎞⎟ ⎝ 3⎠

−1

−1

1 + 3⎛⎜ ⎞⎟ ⎝2⎠ 1 + 3⎛⎜ ⎞⎟ ⎝ 3⎠

−2

−2

−2

1 − 4⎛⎜ − ⎞⎟ ⎝ 2⎠

1 − 4⎛⎜ ⎞⎟ ⎝2⎠ 1 − 4⎛⎜ ⎞⎟ ⎝ 3⎠

Not for sale.

−3

−3

−3

= 1 + 4 + 3 ⋅ 4 + 4 ⋅ 8 = 49,

= 1 − 4 + 3 ⋅ 4 − 4 ⋅ 8 = −23, = 1 − 2 ⋅ 3 + 3 ⋅ 9 − 4 ⋅ 27 = −86.

161

I 0 ( z ) = (1 − z −1 )(1 − z −1 )(1 − z −1 ) = 1 −

11 −1 1 1 z + z − 2 − z − 3 ⇒ I 0 (− ) = 10, 6 6 2 1 1 −1 1 −1 1 −1 1 −1 1 − 2 1 −3 I1 ( z ) = (1 + z )(1 − z )(1 − z ) = 1 − z − z + z ⇒ I1 (1) = , 2 2 3 3 4 12 2 1 2

1 3

2 1 1 5 1 1 1 I 2 ( z ) = (1 + z −1 )(1 − z −1 )(1 − z −1 ) = 1 − z −1 − z − 2 + z − 3 ⇒ I 2 ( ) = − , 2 3 6 3 6 2 3 5 1 1 1 1 1 I 3 ( z ) = (1 + z −1 )(1 − z −1 )(1 − z −1 ) = 1 − z −1 − z − 2 + z − 3 ⇒ I 3 ( ) = . 2 2 4 4 3 2 49 −2 − 23 − 86 Therefore, X ( z ) = I 0 ( z) + I1 ( z ) + I 2 (z) + I 3 ( z) 10 1/ 2 − 2/3 5/2 = 4.9 I 0 ( z ) − 4 I1 ( z ) + 34.5I 2 ( z ) − 34.4 I 3 ( z ) = 1 − 2 z −1 + 3z −2 − 4 z −3 . ∞

ˆ 6.30 (a) X ( z ) = ∑ x[ n]z − n . Let Xˆ ( z ) = log( X ( z )) ⇒ X ( z ) = e X ( z ) . Thus,

X ( e jω ) = e (b) xˆ[ n] =

n = −∞ Xˆ (e jω )

.

(

)

1 π jω jωn jω − jω ). ∫ log X (e ) e dω. If x[n] is real, then X (e ) = X * (e 2π − π

(

)

(

)

Therefore, log X (e jω ) = log X * (e − j jω ) . .

xˆ * [ n] =

(

)

(

)

1 π 1 π jω − jωn − jω dω = ) e − jωn dω ∫ log X * (e ) e ∫ log X (e 2π − π 2π − π

=

(

)

1 π jω jωn ∫ log X (e ) e dω = xˆ[ n]. 2π − π

(

xˆ[ n] + xˆ[ − n] 1 π jω (c) xˆ ev [ n] = = ∫ log X (e ) 2 2π − π

= Similarly, xˆ ev [ n] =

(

)

)

⎛ e jωn + e − jωn ⎞ ⎜ ⎟ dω ⎜ ⎟ 2 ⎝ ⎠

1 π jω ∫ log X (e ) cos(ωn)dω. 2π − π

(

xˆ[ n] − xˆ[ − n] 1 π jω = ∫ log X (e ) 2 2π − π

=

π

(

)

)⎛⎜⎜ e ⎝

j ωn

− e − jωn ⎞⎟ dω ⎟ 2 ⎠

j jω ∫ log X (e ) sin(ωn)dω. 2π − π

6.31 x[ n] = aδ[ n] + bδ[ n − 1]. Thus, X ( z ) = Z{x[ n]} = a + bz −1. Also, ∞

n

(b / a ) − n z . Xˆ ( z) = log(a + bz −1 ) = log(a ) + log(1 + b / az −1 ) = log(a) + ∑ (−1) n −1 n n = −∞

Not for sale.

162

log( a ), if n = 0, ⎧ ⎪ n (b / a ) ⎪ Therefore, xˆ[ n] = ⎨(−1) n −1 , for n > 0, n ⎪ 0, otherwise. ⎪⎩ Nγ

Nα

Nβ

Nδ

6.32 (a) Xˆ ( z ) = log( K ) + ∑ log(1 − α k z −1 ) + ∑ log(1 − γ k z ) − ∑ log(1 − β k z −1 ) − ∑ log(1 − δ k z ) k =1 n Nα ∞ α k −n

= log( K ) − ∑ ∑

k =1 n =1 n

z

k =1 k =1 n n N Nλ ∞ γ Nδ ∞ δ n β ∞ β k n k −n k n

− ∑ ∑

k =1 n =1 n

z − ∑ ∑

k =1 n =1 n

z

− ∑ ∑

k =1 n =1 n

k =1

z .

⎧ ⎪ log( K ), n = 0, ⎪ n n N Nα ∞ α ⎪⎪ β ∞ β Thus, xˆ[ n] = ⎨ ∑ ∑ k − ∑ ∑ k , n > 0, ⎪ k =1 n =1 n k = 1 n =1 n ⎪ Nγ ∞ γ−n Nδ ∞ δ −n ⎪ ∑ ∑ k − ∑ ∑ k , n < 0. k =1 n =1 n ⎩⎪k =1 n =1 n (b) xˆ[ n] < N

r

n

as n → ∞, where r is the maximum value of α k , β k , γ k , and δ k for all n values of k, and N is a constant. Thus, xˆ[ n] is a decaying bounded sequence as n → ∞.

(c) From Part (a) if α k = β k = 0, then xˆ[ n] = 0 for all n > 0, and is thus anti-causal. (d) If γ k = δ k = 0, then xˆ[ n] = 0 for all n < 0, and is thus causal. 6.33 If X ( z ) has no poles and zeros on the unit circle, then from Part (b) of Problem 6.32, γ k = δ k = 0, then xˆ[ n] = 0 for all n < 0. dXˆ ( z) 1 dX ( z) dXˆ ( z ) dX ( z ) = . Thus, z Xˆ ( z ) = log (X ( z )). Therefore, . = zX ( z ) dz X ( z) dz dz dz n

Taking the inverse –transform we get nx[ n] = ∑ k xˆ[ k ]x[ n − k ], n ≠ 0. Or, k =0

n −1 k

x[ n] = ∑

k =0 n

xˆ[ k ]x[ n − k ] + xˆ[ n]x[0]. Hence, xˆ[ n] =

x[ n] n −1 ⎛ k ⎞ xˆ[ k ]x[ n − k ] , n ≠ 0. − ∑ ⎜ ⎟ x[0] k = 0 ⎝ n ⎠ x[0]

For n = 0, xˆ[0] = Xˆ ( z ) z = ∞ = X ( z ) z = ∞ = log( x[0]). Thus,

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163

⎧ ⎪ 0, n < 0, ⎪ xˆ[ n] = ⎨ log( x[0]), n = 0, ⎪ x[ n] n −1 ⎛ k ⎞ xˆ[ k ]x[ n − k ] − ∑ ⎜ ⎟ , n > 0. ⎪ x[0] ⎩ x[0] n = 0 ⎝ n ⎠

6.34

| H(e j5ω ) |

0 0.02π

0.06π

_π 5

π

3π __ 5

ω

6.35 Given the real part of a real, stable transfer function ∑ N a cos(iω) A(e jω ) , H re (e jω ) = i = 0 i = ∑ iN= 0 bi cos(iω) B(e jω ) the problem is to determine the transfer function H ( z ) =

(6-1) −i P( z ) ∑ iN= 0 pi z . = D( z ) ∑ iN= 0 pi z − i

(a) H re (e jω ) = [ H (e jω ) + H * (e jω )] = [ H (e jω ) + H (e − jω )] 2 2 1

1

= [ H ( z ) + H ( z −1 )] 1 2

z = e jω

.

Substituting H ( z ) = P ( z ) / D( z ) in the above we get

H re (e jω ) =

1 P ( z ) D( z − 1 ) + P ( z − 1 ) D ( z ) 2 D( z ) D( z − 1 )

,

(6-2)

z = e jω

which is Eq. (6.117). (b) Comparing Eqs. (6-1) and (6-2) we get B(e jω ) = D( z ) D( z −1 )

A(e jω ) = [ P( z) D( z −1 ) + P( z −1 ) D( z)] 1 2

z = e jω

z = e jω

,

(6-3)

.

(6-4)

N

Now, D ( z ) is of the form D( z ) = Kz − N ∏ ( z − z i ), i =1 jω

where the zi ' s are the roots of B( z ) = B(e

)

e jω = z

(6-5) inside the unit circle and K is a scalar

constant. Putting ω = 0 in Eq. (6-3) we get B(1) = [ D(1)]2 , or Hence, K = B(1) / ∏ iN=1 (1 − zi ). Not for sale.

B(1) = K ∏ iN=1 (1 − zi ). (6-6)

164

(c) By analytic continuation, Eq. (6-4) yields A( z ) = [ P( z ) D( z −1 ) + P( z −1 ) D( z )]. (6-7) 1 2

Substituting A( z ) =

1 N i −i ∑ ai ( z + z ) and the polynomial forms of P ( z ) and D ( z ), we get 2 i =0

N N N N ⎛ N i −i − i ⎞⎛ i⎞ ⎛ i ⎞⎛ −i ⎞ ∑ ai ( z + z ) = ⎜⎜ ∑ p i z ⎟⎟⎜⎜ ∑ d i z ⎟⎟ + ⎜⎜ ∑ p i z ⎟⎟⎜⎜ ∑ d i z ⎟⎟ and equating the coefficients i =0 ⎠ ⎠⎝ i = 0 ⎠ ⎝ i =0 ⎠⎝ i = 0 ⎝i =0

of ( z i + z −i ) on both sides, we arrive at a set of N equations which can be solved for the numerator coefficients pi of H ( z ). For the given example, i.e., H re (e) =

1 + cos(ω) + cos(2ω) , we observe 17 − 8 + cos(2ω)

A( z ) = 1 + ( z + z −1 ) + ( z 2 + z − 2 ). 1 2

1 2

Also, B( z ) = 17 − 4( z 2 + z −2 ), which has roots at z = ±

1 2

(6-8)

and z = ±2. Hence,

D( z ) = Kz − 2 ( z − )( z + ) = K ( z 2 − )z − 2 . 1 2

1 4

1 2

(6-9)

Also, from Eq. (6-6) we have K = 17 − 8 /(1 − ) = 4, so that D( z ) = 4 − z −2 . 1 4

Substituting Eqs. (6-8), (6-9) and P ( z ) = p0 + p1 z −1 + p2 z −2 in Eq. (6-7) we get

[

]

1 + ( z + z −1 ) + ( z 2 + z − 2 ) = ( p0 + p1 z −1 + p2 z − 2 )(4 − z 2 ) + ( p0 + p1 z + p2 z 2 )(4 − z − 2 ) . 2 2 1

1

Equating the coefficients of ( z i + z −i ) / 2, 0 ≤ i ≤ 2, on both sides we get 4 p0 − p2 = 1, 3 p1 = 1, 4 p2 − p0 = 1. Solving these equations we then arrive at

p0 = p1 = p2 = 1 / 3. Therefore, H ( z ) =

1 + z −1 + z −2 3(4 − z − 2 )

.

d M + d M − 1 z −1 + L + d 1 z − M + 1 + z − M

⇒ A(1) = 1 and A( −1) = −1 if M is odd. 1 + d1 z −1 + L + d M −1 z − M + 1 + d M z − M In which case, G (1) = H (1) and G ( −1) = H (−1). If is even, then G (1) = H (1) and G ( −1) = H (1). .

6.36 A( z ) =

6.37 H ( z ) = H1 ( z ) H 2 ( z ) + H 3 ( z ) = (1.2 + 3.3z −1 + 0.7 z −2 )(−4.1 − 2.5z −1 + 0.9 z −2 ) + 2.3 + 4.3z −1 + 0.8 z −2 = −2.62 − 12.23z −1 − 9.24 z −2 + 1.22 z −3 + 0.63z −4 .

6.38 (a) (1 − 0.1z −1 + 0.14 z −2 + 0.49z −3 )Y ( z ) = (5 + 9.5z −1 + 1.4 z −2 − 24 z −3 ) X ( z ) ⇒

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165

Y (z) 5 + 9.5z −1 + 1.4 z −2 − 24 z −3 = . Using Program 6_1.m we factorize H ( z ) X ( z) 1 − 0.1z −1 + 0.14 z − 2 + 0.49z − 3 and develop is pole-zero plot shown below: H ( z) =

Numerator factors 1.00000000000000 3.10000000000000 1.00000000000000 -1.20000000000000

4.00000000000000 0

Denominator factors 1.00000000000000 -0.80000000000000 1.00000000000000 0.70000000000000

0.70000000000000 0

Gain constant 5 The factored form of H ( z ) is thus H ( z ) =

5(1 + 3.1z −1 + 4 z −2 )(1 − 1.2 z −1 ) (1 − 0.81z −1 + 0.7 z − 2 )(1 + 0.7 z −1 )

Imaginary Part

1 0.5 0 -0.5 -1 -1

0 Real Part

1

As all poles are inside the unit circle, H ( z ) is BIBO stable. (b) (1 − 0 . 5 z − 1 + 0 . 1 z − 2 + 0 . 3 z − 3 − 0 . 0936 z − 4 )Y ( z )

= (5 + 16.5z −1 + 14.7z −2 − 22.4 z −3 − 33.6 z −4 ) X ( z ) ⇒

Y ( z) 5 + 16.5z −1 + 14.7z −2 − 22.4 z −3 − 33.6 z −4 . Using Program 6_1.m we = X ( z) 1 − 0.5z −1 + 0.1z − 2 + 0.3z − 3 − 0.0936z − 4 factorize H ( z ) and develop is pole-zero plot shown below: H (z) =

Numerator factors 1.00000000000000 1.00000000000000 1.00000000000000

-1.20000000000000 3.10000000000001 1.39999999999999

0 4.00000000000001 0

Denominator factors 1.00000000000000 1.00000000000000 1.00000000000000

0.59950918226500 -0.80131282790906 -0.29819635435594

0 0.52021728677142 0

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166

Gain constant 5 The factored form of H ( z ) is thus 5(1 − 1.2 z −1 )(1 + 3.1z −1 + 4 z −2 )(1 + 1.4 z −1 )

H ( z) =

(1 + 0.5995 z −1 )(1 − 0.801313 z −1 + 0.520217 z − 2 )(1 − 0.2982 z −1 ) As all poles are inside the unit circle, H ( z ) is BIBO stable.

Imaginary Part

1 0.5 0 -0.5 -1 -2

-1

0 Real Part

1

6.39 A partial-fraction expansion of H ( z ) in z −1 using the M-file residuez yields

H ( z) = −

1.21212 1 − 0.4 z −1

+

2.21212(1 − 0.81781z −1 ) 1 + 0.5z −1 + 0.3z − 2

. Comparing the denominator of the

quadratic factor with 1 − 2r cos(ω o ) z −1 + r 2 z −2 we get r = 0.3 = 0.54772 and 0.5 cos(ωo ) = − , or ωo = 2.04478. Hence, from Table 6.1 we have 2 0.3

h[ n] = −1.21212(0.4) n µ[ n] + ( 0.3 ) n cos(2.04478n)µ[ n]. 6.40 (a) A partial-fraction expansion of H ( z ) in z −1 using the M-file residuez yields 5 2 H ( z ) = −2 + − . Hence, from Table 6.1 we have −1 1 + 0 .6 z 1 − 0 .3 z − 1

h[ n] = −2δ[ n] + 5(−0.6) n µ[ n] − 2(0.3) n µ[ n]. (b) x[ n] = 2.1(0.4) n µ[ n] + 0.3(−0.3) n µ[ n]. Its z –transform is thus given by X ( z) =

2 .1 1 − 0 .4 z − 1

+

0 .3 1 + 0 .3 z − 1

=

2.4 + 0.51z −1 (1 − 0.4 z −1 )(1 + 0.3z −1 )

, z > 0.4. The z –transform of

⎡ ⎤ ⎡1 − 3.3z −1 + 0.36 z − 2 ⎤ 2.4 + 0.51z −1 the output is then given by Y ( z ) = ⎢ ⎥⋅⎢ ⎥. ⎢⎣ (1 − 0.4 z −1 )(1 + 0.3z −1 ) ⎥⎦ ⎢⎣1 + 0.3z −1 − 0.18 z − 2 ⎥⎦ A partial-fraction expansion of Y ( z ) in z −1 using the M-file residuez yields

Not for sale.

167

Y ( z) =

9.3 1 + 0.6z −1

−

(

16.8

+

1 − 0 .4 z − 1

12.3

−

1 − 0.3z −1

2.4 1 + 0.3z −1

, z > 0.6. Hence, from Table 6.1

)

we have y[ n] = 9.3(−0.6) n − 16.8(0.4) n + 12.3(0.3) n − 2.4(−0.3) n µ[ n]. 6.41 (a) H ( z ) = Z{h[ n]} =

Y ( z) = H ( z) X ( z) =

1 1 + 0 .4 z − 1 1

, z > 0.4, X ( z ) = Z{x[ n]} =

(1 + 0.4 z −1 )(1 − 0.2 z −1 )

using the M-file residuez yields Y ( z ) = 2 3

1 1 − 0 .2 z − 1

, z > 0.4. Thus,

, z > 0.4. A partial-fraction expansion of 2/3

1 + 0 .4 z

−1

+

1/ 3 1 − 0 .2 z − 1

. Hence, from Table 6.1

1 3

y[ n] = (−0.4) n µ[ n] + (0.2) n µ[ n]. (b) H ( z ) = Z{h[ n]} =

Y ( z) = H ( z) X ( z) =

1 1 + 0 .2 z 1

−1

(1 + 0.2 z −1 ) 2

, z > 0.2, X ( z ) = Z{x[ n]} =

1 1 + 0 .2 z − 1

, z > 0.4. Thus,

, z > 0.2. Hence, from Table 6.1,

y[ n] = (n + 1)(−0.2) n µ[ n]. 6.42 Y ( z ) = Z{y[ n]} =

2 1 + 0 .3 z

−1

, z > 0.3, X ( z ) = Z{x[ n]} =

4 1 − 0 .6 z − 1

, z > 0.2. Thus,

Y ( z ) 0.5(1 − 0.6 z −1 ) = , z > 0.3. A partial-fraction expansion of using the M-file X ( z) 1 + 0.3z −1 1 .5 residuez yields H ( z ) = −1 + . Hence, from Table 6.1, 1 + 0 .3 z − 1 H (z) =

h[ n] = − δ[ n] + 1.5(−0.3) n µ[ n]. 6.43 (a) Taking the –transform of both sides of the difference equation we get Y (z) 2 Y ( z ) = 0.2 z −1Y ( z ) + 0.08z −2Y ( z ) + 2 X ( z ). Hence, H ( z ) = = . X ( z ) 1 − 0.2 z −1 − 0.08 z − 2 (b) A partial-fraction expansion of using the M-file residuez yields 4/3 2/3 H ( z) = + . Hence, from Table 6.1, −1 1 − 0 .4 z 1 + 0 .2 z − 1 4 3

2 3

h[ n] = (0.4) n µ[ n] + (−0.2) n µ[ n]. (c) Now S ( z ) = Z{s[ n]} = H ( z ) ⋅ Z{µ[ n]} =

Not for sale.

2 (1 − 0.2 z −1 − 0.08z − 2 )(1 − z −1 )

.

168

A partial-fraction expansion of using the M-file residuez yields 2.7778 0.8889 0.1111 S ( z) = − + . Hence, from Table 6.1, −1 −1 1− z 1 − 0 .4 z 1 + 0 .2 z − 1

s[ n] = 2.7778 µ[ n] − 0.8889(0.4) n µ[ n] + 0.1111(−0.2) n µ[ n]. 6.44 H ( z) =

1 − z −2 1 − (1 + α) cos(ωc )z −1 + αz − 2

. Thus,

1 − e j 2ω

H ( e jω ) =

. 1 − (1 + α) cos(ωc ) e − jω + αe − j 2ω 2(1 − cos 2ω) H ( e jω ) 2 = 2 2 1 + (cos ω c ) (1 + α ) + α 2 + 2α cos 2ω − 2 cos ω c (1 + α ) 2 cos ω

=

4 sin 2 ω 2

2

2

(1 + α) (cos ω − cos ωc ) + (1 − α) sin ω

. Note that H (e jω ) 2 is maximum when

cos ω = cos ωc , i.e, ω = ωc . Then H (e jω c ) 2 =

4 sin 2 ωc (1 − α) 2 sin 2 ωc

=

4 (1 − α) 2

, and hence,

H (e jω c ) = 2 /(1 − α ).

6.45 H ( z ) = 1 − αz −R ⇒ H (e jω ) = 1 − αe − jωR . Then, H (e jω = 1 + α 2 − 2α cos(ωR). H(e jω is maximum when cos( ωR ) = −1 and is minimum when cos(ωR ) = 1. The

maximum value of H (e jω ) is 1 + α , and the minimum value is 1 − α . H(e jω has R peaks and R dips in the range 0 ≤ ω < 2π. 2 πk (2 k + 1)π and the dips are located at ω = ω k = The peaks are located at ω = ω k = , R R 0 ≤ k ≤ R − 1. Phase Response 1

1.5

0.5

Phase, radians

Amplitude

Magnitude Response 2

1 0.5 0

0

0.5

(

6.46 G (e jω ) = H (e jω )

1 ω/π

1.5

2

0 -0.5 -1

0

0.5

1

1.5

2

ω/ π

)3 = (1 − αe jωR )3 . Not for sale.

169

M −1

6.47 G(e jω ) = ∑ α n e − jωn = n=0

1 − α M e − jωM 1 − αe

− jω

. Note that G (e jω ) = H (e jω ) for

1− αM . Hence, to make the dc value of the 1− α magnitude response equal to unity, the impulse response should be multiplied by a constant αn =

1 , M

0 ≤ n ≤ M − 1. Now, G(e j 0 ) =

K = (1 − α ) /(1 − α M ) .

6.48 Y (e jω ) = X (e jω ) + αe − jωR Y (e jω ). Hence, H (e jω ) =

Y ( e jω )

=

1

. Maximum X ( e jω ) 1 − α e − jωR 1 1 . There are R peaks and dips value of H (e jω ) is and the minimum value is 1− α 1+ α

in the range 0 ≤ ω ≤ 2π. The locations of the peaks and dips are given by α 1 − αe − jωR = 1 ± α or, e − jωR = ± . The locations of the peaks are given by α (2π + 1)k 2πk and the locations of the dips are given by ω = ωk = , ω = ωk = R R

0 ≤ k ≤ R − 1. Plots of the magnitude and phase responses of H (e jω ) for α = 0.8 and R = 6 are shown below: Phase Response 1

4

0.5

Phase, radians

Amplitude

Magnitude Response 5

3 2 1 0

0

0.5

jω

1 ω/π

2

-0.5 -1

0

0.5

1

1.5

2

ω/ π

b0 + b1e − jω + b2 e − j 2ω

(b0 e jω + b2 e − jω ) + b1

= 1 + a1e − jω + a 2 e − j 2ω (e jω + a 2 e − jω ) + a1 b + (b0 + b2 ) cos ω + j (b0 − b2 ) sin ω . Therefore, = 1 a1 + (1 + a 2 ) cos ω + j (1 − a 2 ) sin ω

6.49 A(e

A(e

)=

1.5

0

jω

)

2

[ b1 + (b0 + b2 )]2 cos 2 ω + (b0 − b2 ) 2 sin 2 ω = 1. Hence, at ω = 0, we have = [a1 + (1 + a2 )]2 cos 2 ω + (1 − a2 ) 2 sin 2 ω

b1 + (b0 + b2 ) = ±[ a1 + (1 + a 2 )], and at ω = π / 2, we have b0 − b2 = ± (1 − a 2 ). Solution #1: Consider b0 − b2 = 1 − a 2 . Choose b 0 = 1, − b 2 = 1 − a 2 , and b 2 = a 2 . Substituting these values in b1 + (b0 + b2 ) = ±[ a1 + (1 + a 2 )], we get b1 = a1 . In this case,

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170

A(e

jω

)=

1 + a1e − jω + a 2 e − j 2ω

1 + a1e − jω + a 2 e − j 2ω

= 1, a trivial solution.

Solution #2: Consider b0 − b2 = a 2 − 1. Choose b0 = a 2 and b2 = 1. Substituting these values in b1 + (b0 + b2 ) = ±[ a1 + (1 + a 2 )], we get b1 = a1 . In this case,

A(e jω ) =

a 2 + a1e − jω + e − j 2ω . 1 + a1e − jω + a 2 e − j 2ω

6.50 From Eq. (2.20), the input-output relation of a factor-of-2 up-sampler is given by ⎧ x[ n / 2], n = 0, ± 2, ± 4,K x u [ n] = ⎨ otherwise. ⎩ 0, The DTFT of x u [n] is therefore given by ∞

Y (e jω ) = ∑ y[ n]e − jωn = n = −∞

∞

∑ x[ n / 2]e

− jωn

n = −∞ n even

=

∞

∑ x[ m]e

− j 2ωm

m = −∞

= X (e j 2ω ) where X (e jω )

is the DTFT of x[n]. 1 . Thus, 1 − 0.5 cos ω + j 0.5 sin ω 1 − 0.5e 1 1 H (e ± jπ / 4 ) = = = 0.6512 m j1.1907. 1 − 0.5 cos(± π / 4) + j 0.5 sin(± π / 4) 0.6464 ± j 0.3536

6.51 H (e jω ) =

1

− jω

=

Therefore, H (e ± jπ / 4 ) = 1.3572 and arg{H (e ± jπ / 4 )} = θ (± jπ / 4) = m1.0703. Now, for an input x[ n] = sin(ω o n) µ[ n], the steady-state output is given by y[ n] = H (e jωo ) sin (ωo n + θ (ωo )) which for ωo = π / 4 reduces to π π y[ n] = H (e jπ / 4 ) sin⎛⎜ n + θ (π / 4) ⎞⎟ = 1.3572 sin⎛⎜ n − 1.0703 ⎞⎟. 4 ⎝ ⎠ ⎝4 ⎠

6.52 To guarantee the stability of G ( z ), the transformation z → F ( z ) should be such that the unit circle remains inside the ROC after the mapping. If the points inside the unit circle after the mapping remains inside the unit circle, G ( z ) will be causal and stable. On the other hand, if the points inside the unit circle after the mapping move outside the unit circle, G ( z ) will be stable but anti-causal. For example, the mapping z → − z will ensure that G ( z ) will be causal and stable, whereas, the mapping z → z −1 will result in a G ( z ) that is stable, but anti-causal. 6.53

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171

K

6.54

∞

−1 n ∑ h[ n] = 0.95 ∑ h[ n] . Since H ( z) = 1 /(1 − βz ), h[ n] = ( β ) µ[ n]. Thus, 2

n=0

1− β

2

n=0

2K

1− β

2

=

0.95 1− β

2

. Solving this equation for we get K = 0.5

log(0.05) . log( α )

6.55 Let the output of the predictor of Figure P6.3(a) be denoted by E ( z ). Then analysis of this structure yields E ( z) = P( z)[U ( z) + E ( z)] and U ( z ) = X ( z ) − E ( z ). From the first equation P( z) we have E ( z ) = U ( z ) which when substituted in the second equation yields 1 − P( z ) U ( z) H (z) = = 1 − P ( z). X (z) Analyzing Figure P6.3(b) we get Y ( z ) = V ( z ) + P ( z )Y ( z ) which leads to 1 Y ( z) , which is seen to be the inverse of H ( z ). G( z) = = V ( z ) 1 − P( z) 1 For P ( z) = h1 z −1 , H ( z ) = 1 − h1 z −1 and G ( z ) = . Similarly, for 1 − h1 z −1 1 P( z ) = h1 z −1 + h2 z −2 , H ( z) = 1 − h1 z −1 − h2 z −2 and G ( z ) = . −1 1 − h1 z − h2 z − 2 6.56 Y ( z) = [H 0 ( z)F0 ( z) − H 0 (− z )F0 (− z)]X ( z). Since the output is a delayed replica of the input, we must have H 0 ( z ) F0 ( z ) − H 0 ( − z ) F0 ( − z ) = z − r . But, H 0 ( z ) = 1 + αz −1 . Hence, (1 + αz −1 ) F0 ( z ) − (1 − αz −1 )F0 ( − z ) = z −r . Let F0 ( z ) = a 0 + a1 z −1 . This implies, 2( a 0 α + a1 ) z −1 = z −r .

The solution is therefore, r = 1 and 2(a 0 α + a1 ) = 1. One possible solution is thus

a 0 = 1 / 2 and a1 = 1 / 4. Hence, F0 ( z ) = 0.25(1 + z −1 ). 6.57 H1 ( z ) = Z{h1[ n]} = 1.2 +

0 .5

−

0 .6

1 + 0 .5 z − 1 1 − 0 .2 z − 1 function of the inverse transform is thus

=

1.1 − 0.04 z −1 − 0.12 z −2 1 + 0.3z −1 − 0.1z − 2

. The transfer

1 1 + 0.3z −1 − 0.1z −2 (1 + 0.5z −1 )(1 − 0.2 z −1 ) = . As both = H1 ( z ) 1.1 − 0.04 z −1 − 0.12 z − 2 1.1(1 − 0.349 z −1 )(1 + 0.3126 z −1 ) poles are inside the unit circle, H 2 ( z) is stable and causal with an ROC z > 0.349. A H 2 ( z) =

partial-fraction expansion in obtained using the M-file residuez is

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172

0.498 0.42224464 1 + − . Hence, 1.2 1 − 0.34897 z −1 1 + 0.31261z −1 1 δ[ n] + 0.498(0.34897) n µ[ n] − 0.42224464(−0.31261) n µ[ n]. 1 .2

H 2 (z) =

h2 [ n] =

6.58 Now, H ( z ) z = e jω = H (e jω ) = H (e jω ) e jθ(ω) . Denote H ' ( z) =

dH ( z ) . dz

From the above we get ln H ( z ) z = e jω = ln H (e jω ) + jθ(ω). Therefore, jω d H (e jω ) / dω H ' ( z) dz dθ(ω) d H (e ) / dω = − jτ g (ω). ⋅ = +j H ( z) dω z = e jω dω H ( e jω ) H (e jω )

(6-A)

d H (e jω ) / dω H ' ( z) −j Hence, τ g (ω) = − z . H ( z ) z = e jω H ( e jω )

(6-B)

Replacing by in Eq, (6-A) we arrive at

d H (e jω ) / dω ' ( ) H z τ g (ω) = − z −1 +j . H ( z ) z = e jω H (e jω ) Adding Eqs. (6-B) and (6-C), and making use of the notation T ( z) = z

τ g (ω) = −

T ( z ) + T ( z −1 ) 2

(6-C)

H ' ( z) we finally get H ( z)

. z = e jω

M6.1 (a) The output data generated by Program 6_1 is as follows: Numerator factors 1.00000000000000 -2.10000000000000 5.00000000000000 1.00000000000000 -0.40000000000000 0.90000000000000 Denominator factors 1.00000000000000 2.00000000000000 1.00000000000000 -0.20000000000000

4.99999999999999 0.40000000000001

Gain constant 0.50000000000000 Hence, G1 ( z ) =

0.5(1 − 2.1z −1 + 5z −2 )(1 + 2 z −1 + 0.9 z −2 )

(1 + 2 z −1 + 5z − 2 )(1 − 0.2 z −1 + 0.4 z − 2 ) The pole-zero plot of G1 ( z) is given below:

Not for sale.

.

173

Imaginary Part

2 1 0 -1 -2 -3

-2

-1

0 1 Real Part

2

3

There are 3 ROCs associated with G1 ( z) : R1 : z < 0.4 , R 2 : 0.4 < z < 5 , and R 3 : z > 5. The inverse z –transform corresponding to the ROC R1 is a leftsided sequence, the inverse z –transform corresponding to the ROC R 2 is a two-sided sequence, and the inverse z –transform corresponding to the ROC R 3 is a right-sided sequence. (b) The output data generated by Program 6_1 is as follows: Numerator factors 1.00000000000000 1.20000000000000 3.99999999999999 1.00000000000000 -0.50000000000000 0.90000000000001 Denominator factors 1.00000000000000 2.10000000000000 1.00000000000000 0.60000000000003 1.00000000000000 0.39999999999997

4.00000000000001 0 0

Gain constant 1 Hence, G2 ( z ) =

(1 + 1.2 z −1 + 4 z −2 )(1 − 0.5z −1 + 0.9 z −2 )

(1 + 2.1z −1 + 4 z − 2 )(1 + 0.6 z −1 )(1 + 0.4 z −1 ) The pole-zero plot of G2 ( z) is given below:

Imaginary Part

2 1 0 -1 -2 -2

-1

0 1 Real Part

2

There are 4 ROCs associated with G2 ( z) : R1 : z < 0.4, R 2 : 0.4 < z < 0.6,

Not for sale.

174

R 3 : 0.6 < z < 2, and R 4 : z > 2. The inverse z –transform corresponding to the ROC R1 is a left-sided sequence, the inverse z –transform corresponding to the ROC R 2 is a two-sided sequence, the inverse z –transform corresponding to the ROC R 3 is a twosided sequence, and the inverse z –transform corresponding to the ROC R 4 is a rightsided sequence. M6.2 (a) The output data generated by Program 6_3 is as follows: Residues -3.33333333333333 3.33333333333333 Poles -0.60000000000000 0.30000000000000 Constants 0 Hence, the partial-fraction expansion of X a (z ) is given by 10 / 3 10 / 3 X a ( z) = − + . The z –transform has poles at z = −0.6 and at 1 + 0 .6 z − 1 1 − 0 .3 z − 1 z = 0.3. Thus, it is associated with ROCs as given in the solution of Problem 6.20 which also shows their corresponding inverse z –transform. (b) The output data generated by Program 6_3 is as follows: Residues Columns 1 through 2 2.33333333333333 -3.66666666666667 + 0.00000008829151i Column 3 4.33333333333333 - 0.00000008829151i Poles Columns 1 through 2 -0.60000000000000 0.30000000000000 - 0.00000000722385i Column 3 0.30000000000000 + 0.00000000722385i Constants Hence, the partial-fraction expansion of X b (z) is given by 2.3333 3.66666 4.33333 . The z –transform has two poles X b ( z) = − + 1 + 0.6 z −1 1 − 0.3z −1 (1 − 0.3z −1 ) 2 at z = −0.6 and one at z = 0.3. Thus, it is associated with ROCs as given in the solution of Problem 6.20 which also shows their corresponding inverse z –transform. M6.3 (a) X1 ( z) = 3 −

4 5+ z

−1

−

7 6+z

−1

= 3−

4/5 1 + (1 / 5)z

Not for sale.

−1

−

7/6 1 + (1 / 6)z −1

.

175

The output data generated by Program 6_4 is as follows: Numerator polynomial coefficients 1.03333333333333 0.73333333333333 0.1000000000000 Denominator polynomial coefficients 1.00000000000000 0.36666666666667 Hence, X1 ( z) =

31 + 22 z

−1

+ 3z

−1

+ z −2

30 + 11z

.

1.4 + z −1

3

(b) X 2 ( z) = −2.5 +

0.03333333333333

−2

− 1 + 0.4 z −1 1 + 0.6 z − 2 3 0.7 − j 6454972243679 0.7 + j 6454972243679 = −2.5 + − . − 1 + 0.4 z −1 1 − j 0.774596669z −1 1 + j 0.774596669z −1 The output data generated by Program 6_4 is as follows: Numerator polynomial coefficients -0.9000 -2.5600 -0.1000 -0.6000 Denominator polynomial coefficients 1.0000 0.4000 0.6000 0.2400 Hence, X b ( z) = −

0.9 + 2.56 z −1 + 0.1z −2 + 0.6 z −3 1.0 + 0.4 z −1 + 0.6 z − 2 + 0.24 z − 3 5

(c) X 3 ( z) =

6

+

+

.

−4

1 + 0.64 z −1 4 + 2 z −1 (4 + 2 z −1 ) 2 5 1.5 − 0.25 = + + . 1 + 0.64 z −1 1 + 0.5z −1 (1 + 0.5z −1 ) 2 The output data generated by Program 6_4 is as follows: Numerator polynomial coefficients 6.2500 6.5500 1.7300 0 Denominator polynomial coefficients 1.0000 1.6400 0.8900 0.1600 Hence, X 3 ( z ) =

6.25 + 6.55z −1 + 1.73z −2 1 + 1.64 z −1 + 0.89z − 2 + 0.16 z − 3

(d) X 4 ( z) = −5 +

2

+

.

z −1

4 + 3z −1 4 + 3z −1 + 0.9z − 2 0.5 j 0.4303 j 0.4303 . = −5 + − + 1 + 0.75z −1 1 + (0.3750 - j0.2905)z −1 1 + (0.3750 + j0.2905)z −1 The output data generated by Program 6_4 is as follows: Numerator polynomial coefficients Not for sale.

176

-4.5000

-6.8750

-3.6375

-0.8438

Denominator polynomial coefficients 1.0000 1.5000 0.7875 0.1688 Hence, X 4 ( z) = −

4.5 + 6.875z −1 + 3.6375z −2 + 0.84375z −3 1 + 1.5z −1 + 0.7875z − 2 + 0.16875z − 3

.

M6.4 (a) The inverse z –transform of X a (z ) from it partial-fraction expansion form is thus 4

7

x a [ n] = 3δ[ n] − (−1 / 5) n µ[ n] − (−1 / 6) n µ[ n]. The first 10 samples of xa [n] obtained 5 6 by evaluating this expression in MATAB are given by Columns 1 through 4 1.0333333333 0.3544444444 -0.0644074074 0.0118012345 Columns 5 through 8 -0.0021802057 0.0004060343

-0.0000762057

0.0000144076

Columns 9 through 10 -0.0000027426 0.0000005254 The first 10 samples of the inverse z–transform of the rational form of X a (z ) obtaining using the M-file impz are identical to the samples given above. (b) The inverse z –transform of X b (z) from it partial-fraction expansion form is thus x b [ n] = −2.5δ[ n] + 3( −0.4) n µ[ n] − (0.7 − j 0.6454972243 679)( j 0.7745966692 41) n µ[ n]

− (0.7 + j 0.6454972243679)(− j 0.774596669241) n µ[ n] . The first 10 samples of x b [n] obtained by evaluating this expression in MATAB are given by Columns 1 through 4 -0.900000000 -2.200000000

1.320000000

0.408000000

Columns 5 through 8 -0.427200000 -0.390720000

0.314688000

0.211084800

Columns 9 through 10 -0.179473920 -0.130386432 The first 10 samples of the inverse z–transform of the rational form of X b (z) obtaining using the M-file impz are identical to the samples given above. (c) The inverse z –transform of X c (z ) from it partial-fraction expansion form is thus x c [ n] = 5( −0.64) n µ[ n] + 1.5(−0.5) n µ[ n] − 0.25( n + 1)( −0.5) n µ[ n].

The first 10 samples of x c [n] obtained by evaluating this expression in MATAB are given by Not for sale.

177

Columns 1 through 5 6.250000000 -3.700000000 0.854485800

2.235500000

-1.373220000

Columns 6 through 10 -0.536870912 0.3396911337 -0.215996076 0.137807801 -0.0881188675 The first 10 samples of the inverse z–transform of the rational form of X c (z ) obtaining using the M-file impz are identical to the samples given above. (d) The inverse z –transform of X d (z ) from it partial-fraction expansion form is thus

x 4 [ n] = −5 + 0.5(−0.75) n µ[ n] − j 0.430331483(−0.375 + j 0.29047375) n µ[ n] + j 0.430331483(−0.375 − j 0.29047375) n µ[ n]. The first 10 samples of x d [n] obtained by evaluating this expression in MATAB are given by Columns 1 through 5 4.50000000000000 -0.12499999991919 0.09374999993940 0.12656249997773 0.13710937500069 Columns 6 through 10 -0.12181640625721 0.09610839844318 0.05192523193410 -0.03790274963350

-0.07136938476820

The first 10 samples of the inverse z–transform of the rational form of X d (z ) obtaining using the M-file impz are identical to the samples given above. M6.5 To verify using MATLAB that H 2 ( z) =

H1 ( z ) =

1 + 0.3z −1 − 0.1z −2 1.1 − 0.04 z −1 − 0.12 z − 2

is the inverse of

1.1 − 0.04 z −1 − 0.12 z −2

, we determine the first 20 samples of h1[ n] and h2 [ n] , 1 + 0.3z −1 − 0.1z − 2 and then form the convolution of these two sequences using the M-file conv. The first samples of the convolution result are as follows: Columns 1 through 9 1.0000 0 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 -0.0000 Columns 10 through 18 0.0000 -0.0000 -0.0000 -0.0000 -0.0000

0.0000 -0.0000

0.0000

0.0000

Columns 19 through 21 -0.0000 -0.0000 -0.0000 M6.6 % As an example try a sequence x = 0:24; % calculate the actual uniform dft

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178

% % % % %

and then use these uniform samples with this ndft program to get the original sequence back [X,w] = freqz(x,1,25,’whole’); use freq = X and points = exp(j*w)

freq = input(‘The sample values = ‘); points = input(‘Frequencies at which samples are taken = ‘); L = 1; len = length(points); val = zeros(size(1,len)); L = poly(points); for k = 1:len if (freq(k) ~= 0) xx = [1 –points(k)]; [yy, rr] = deconv(L,xx); F(k,:) = yy; Down = polyval(yy,points(k))*(points(k))*(points(k)^(-len+1)); F(k,:) = freq(k)/down*yy; val = val+F(k,:); end end coeff = val;

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179

Chapter 7 7.1

⎧⎪ 1 , 0 ≤ n ≤ M − 1, Impulse response of the moving average filter is: h[ n] = ⎨ M ⎪⎩ 0, otherwise. Its frequency response is: H (e jω ) =

1 ⎛⎜ 1 − e − jMω ⎞⎟ 1 sin( Mω / 2) − jMω / 2 = ⋅ e . M ⎜⎝ 1 − e − jω ⎟⎠ M sin(ω / 2)

Now, for a BR transfer function, H (e jω ) ≤ 1, ∀ω. For the moving-average filter, H ( e jω ) =

1 sin( Mω / 2) . We shall show by induction that M sin(ω / 2)

1 sin( Mω / 2) ≤ 1, ∀ω. Now, for M sin(ω / 2) M = 2, H (e jω ) =

1 sin(ω) 1 2 sin(ω / 2) cos(ω / 2) = = cos(ω / 2) ≤ 1, ∀ω. 2 sin(ω / 2) 2 sin(ω / 2)

We assume next that

1 sin( mω / 2) ≤ 1 for 1 ≤ m ≤ M − 1. We can express m sin(ω / 2)

1 sin( Mω / 2) 1 sin (( M − 1)ω / 2 ) cos(ω / 2) + cos(( M − 1)ω / 2 )sin(ω / 2) = M sin(ω / 2) M sin(ω / 2)

⎤ 1 ⎡ sin(( M − 1)ω / 2 ) cos(ω / 2) + cos(( M − 1)ω / 2 ) ⎥. Now, cos(mω / 2 ) ≤ 1 for ⎢ M⎣ sin(ω / 2) ⎦ 1 sin( Mω / 2) 1 all values of m. Hence, ≤ [M − 1 + 1] ≤ 1. M sin(ω / 2) M ≤

7.2

A( z ) =

1 − d1* z (1 − d1* z )(1 − d1z*) 2 . Therefore, . A( z ) = A( z ) A * ( z ) = z − d1 ( z − d1 )( z * − d1* ) 2

1 − A( z ) = 2

=

( z − d1 )( z * − d1* ) − (1 − d1* z )(1 − d1z*) ( z − d1 )( z * − d1* )

2

2

2

z + d 1 − d1 z * − d1* z − 1 − d 1 z + d1z * + d1* z ( z − d1 )( z * − d1* )

⎧ > 0, if z > 1,

2⎪

2

=

2

( z − 1)(1 − d 1 ) z −d1

2

. Hence,

⎧ < 1, if z > 1,

2⎪

1 − A1 ( z ) ⎨= 0, if z = 1, Thus, A1 ( z ) ⎨= 1, if z = 1, Thus, Eq. (7.20) holds ⎪ < 0, if z < 1. ⎪ > 1, if z < 1. ⎩ ⎩ for any first-order allpass function. A higher-order allpass function can br factored into a product of first-order allpass functions. Since, Eq. (7.20) holds true each of these factors individually, hence, it also holds true for the product.

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7.3

An m –th order stable, real allpass transfer function can be expressed as a product

1 − di* z . If di is z − di 1 − di z . Now, complex, then A(z ) has another factor of the form Ai' ( z ) = z − di* of first-order allpass transfer functions of the form Ai ( z) =

Ai (e jω ) =

1 − d *i e jω e

jω

− di

= e − jω

Then, Ai (e jω ) = e − jω

(1 − d *i e jω )(1 − d *i e jω ) − jω

(1 − di e )(1 − d *i e jω ) − jθ jω 2

(1 − α e

e

)

[1 + α 2 − 2α cos(θ − ω)]

{

. Let di = di e jθ = α e jθ .

. Therefore,

}

{

}

arg{Ai (e jω )} = − ω + arg (1 − α e − jθ e jω ) 2 = − ω + 2 arg (1 − α e − jθ e jω )

⎡ α sin(θ − ω) ⎤ = − ω + 2 tan −1 ⎢ ⎥. We can show similarly, ⎣1 − α cos(θ − ω) ⎦ ⎡ α sin(θ + ω) ⎤ arg{ Ai' (e jω )} = − ω + 2 tan −1 ⎢ ⎥. ⎣1 − α cos(θ + ω) ⎦ ⎡ α sin ω ⎤ If d i = α is real, then arg{Ai (e jω )} = − ω + 2 tan −1 ⎢ ⎥. Now for real di , ⎣1 − α cos ω ⎦ arg{A i (e j 0 )} − arg{A i (e jπ )} = {−0 + 2 tan −1 (−0)} − {− π + 2 tan −1 ( −0)} = π. For

complex di , arg{A i (e j 0 )} + arg{ A i' (e j 0 )} − arg{A i (e jπ )} − arg{ A i' (e jπ )} = ⎡ α sin θ ⎤ ⎡ − α sin θ ⎤ = −0 + 2 tan −1 ⎢ − 0 + 2 tan −1 ⎢ ⎥ ⎥ ⎣ 1 − α cos θ ⎦ ⎣ 1 − α cos θ ⎦ ⎡ α sin θ ⎤ ⎡ α sin θ ⎤ + π − 2 tan −1 ⎢ + π − 2 tan −1 ⎢ ⎥ ⎥ = 2π. Now, ⎣ 1 + α cos θ ⎦ ⎣ 1 + α cos θ ⎦ d arg{A(e jω )} . Therefore, τ ( ω) = − dω

(

π

π

0

0

)

jω j0 jπ ∫ τ (ω)dω = − ∫ d[arg{A(e )}] = arg{A(e ) − arg{A(e ). Since m

π

i =1

0

arg{A(e jω ) = ∑ arg{Ai (e jω ), it follows then ∫ τ (ω)dω = mπ. 7.4

h[ n] = a1δ[ n − 2] − a2 δ[ n − 1] − a3δ[ n] + a 4 δ[ n + 1] − a5δ[ n + 2]. Therefore, H (e jω ) = a1e j 2 ω − a 2 e jω − a3 + a 4 e − jω − a 5 e −2 ω

= a1 (cos 2ω + j sin 2ω) − a1 (cos ω + j sin ω) − a3 + a 4 (cos ω − j sin ω) − a3 + a 4 (cos ω − j sin ω) − a5 (cos 2ω − j sin 2ω) = (a1 − a5 ) cos 2ω − (a2 − a4 ) cos ω − a3 + j ((a1 + a5 ) sin 2ω − (a2 + a 4 ) sin ω). To

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have a zero-phase frequency response, the imaginary part of must be equal to zero for all values of ω. Hence, for a zero-phase response, a1 = − a5 and a2 = − a4 . 7.5

v[ n] = x[ − n] O * h[ n], and u[ n] = v[ − n] = x[ n] O * h[ − n]. Hence, jω jω jω y[ n] = (h[ n] + h[ − n]) O * x[ n]. Therefore, G (e ) = H (e ) + H * (e ). Thus, the

equivalent frequency response is real and has zero phase. 7.6

Now, AM ( z) = ±

z − M D*M (1 / z*) , where DM ( z)

D M ( z ) = 1 + d1z −1 + d 2 z −2 + L + d M z − M . 2

AM ( z ) = AM ( z ) A*M (1 / z*) =

Consider the first-order factor 1 + az −1 . Its square-magnitude function is given by (1 + az −1 )(1 + az )

z = e jω

= (1 + a 2 ) + 2 a cos ω. We thus rewrite the given square-

magnitude function as

H (e

jω

2

) = H ( z)H ( z

−1

)

z = e jω

Therefore, H ( z )H ( z −1 ) =

=

9[1.0625 + 0.25( z + z −1 )][1.49 − 0.7( z + z −1 )] [1.36 + 0.6( z + z −1 )][1.64 + 0.8( z + z −1 )]

9(1 + 0.25z )(1 + 0.25z −1 )(1 − 0.7 z )(1 − 0.7 z −1 )

(1 + 0.6 z )(1 + 0.6 z −1 )(1 + 0.8 z )(1 + 0.8 z −1 ) As can be seen, there are 4 possible causal, stable transfer functions:

(i) H ( z ) =

3(1 + 0.25z −1 )(1 − 0.7 z −1 )

(iii) H ( z ) =

(1 + 0.6 z −1 )(1 + 0.8 z −1 )

3(1 + 0.25z )(1 − 0.7z ) (1 + 0.6 z −1 )(1 + 0.8z −1 )

,

, (ii) H ( z ) =

(iv)

.

3(1 + 0.25z )(1 − 0.7 z −1 ) (1 + 0.6 z −1 )(1 + 0.8 z −1 )

H (z) =

. z = e jω

,

3(1 + 0.25z −1 )(1 − 0.7 z ) (1 + 0.6 z −1 )(1 + 0.8 z −1 )

.

The zero locations of the four FIR transfer functions are given below: (ii)

(i) 1

1.5 Imaginary Part

Imaginary Part

7.7

D *M (1 / z*) D M ( z ) ⋅ = 1. D M ( z ) D *M (1 / z*)

0.5 0 -0.5

1 0.5 0 -0.5 -1 -1.5

-1 -1

-0.5

0 0.5 Real Part

-4

1

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-2 -1 Real Part

0

1

179

(iii)

(iv) 1

1

Imaginary Part

Imaginary Part

1.5 0.5 0 -0.5 -1

0.5 0 -0.5

-1.5 -1 -4

-3

-2 -1 Real Part

0

1

-1

-0.5

0 0.5 Real Part

1

1.5

The zeros of the transfer function of a linear-phase FIR filter exhibits mirror-image symmetry with respect to the unit circle. The zeros of the transfer function of a minimum-phase FIR filter are all inside the unit circle and that of a maximum-phase FIR filter are outside the unit circle. (a) The transfer functions of (iii) and (iv) have linear-phase as their zeros exhibit mirror-image symmetry. (b) The transfer function of (i) is minimum-phase as its zeros are inside the unit circle. (c) The transfer function of (ii) is maximum-phase as its zeros are inside the unit circle. 7.9

G( z ) = 6(1 + 1.7 z −1 − 2 z −2 )(1 − 0.5z −1 ) = 6(1 + 2.5z −1 )(1 − 0.8 z −1 )(1 − 0.5z −1 ) = 6 + 7.2 z −1 − 17.1 z −2 + 6 z −3 . (a) Other transfer functions having the same magnitude responses are: ⎛ 2.5 + z −1 ⎞ ⎟⎟ (i) G2 ( z ) = 6(1 + 2.5z −1 )(1 − 0.8z −1 )(1 − 0.5z −1 )⎜⎜ ⎝ 1+ 2.5z −1 ⎠ = 6(2.5 + z −1 )(1 − 0.8 z −1 )(1 − 0.5z −1 ) = 15 − 13.5 z −1 − 1.8 z −2 + 2.4 z −3 . ⎛ − 0.8 + z −1 ⎞ (ii) G3 ( z ) = 6(1 + 2.5z −1 )(1 − 0.8z −1 )(1 − 0.5z −1 )⎜⎜ ⎟⎟ ⎝ 1− 0.8 z −1 ⎠ = 6(1 + 2.5z −1 )(−0.8 + z −1 )(1 − 0.5z −1 ) = −4.8 − 3.6 z −1 + 18 z −2 − 7.56 z −3 . (iii)

⎛ − 0.5 + z −1 ⎞ ⎟⎟ G4 ( z ) = 6(1 + 2.5z −1 )(1 − 0.8 z −1 )(1 − 0.5z −1 )⎜⎜ ⎝ 1− 0.5z −1 ⎠

= 6(1 + 2.5z −1 )(1 − 0.8 z −1 )(−0.5 + z −1 ) = −3 + 0.9 z −1 + 16.28 z −2 − 124 z −3 . (iv)

⎛ 2.5 + z −1 ⎞⎛ − 0.8 + z −1 ⎞ G5 ( z ) = 6(1 + 2.5z −1 )(1 − 0.8 z −1 )(1 − 0.5z −1 )⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎝ 1+ 2.5z −1 ⎠⎝ 1− 0.8 z −1 ⎠

= 6(2.5 + z −1 )(−0.8 + z −1 )(1 − 0.5z −1 ) = −12 + 16.2 z −1 + 0.9 z −2 − 3 z −3 . Not for sale

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(v)

⎛ − 0.8 + z −1 ⎞⎛ − 0.5 + z −1 ⎞ G6 ( z ) = 6(1 + 2.5z −1 )(1 − 0.8z −1 )(1 − 0.5z −1 )⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎝ 1− 0.8 z −1 ⎠⎝ 1− 0.5z −1 ⎠

= 6(1 + 2.5z −1 )(−0.8 + z −1 )(−0.5 + z −1 ) = 2.4 − 1.8 z −1 − 13.5 z −2 + 15 z −3 . ⎛ 2.5 + z −1 ⎞⎛ − 0.5 + z −1 ⎞ (vi) G7 ( z ) = 6(1 + 2.5z −1 )(1 − 0.8z −1 )(1 − 0.5z −1 )⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎝ 1+ 2.5z −1 ⎠⎝ 1− 0.5z −1 ⎠ = 6(2.5 + z −1 )(1 − 0.8 z −1 )(−0.5 + z −1 ) = −7.5 + 18 z −1 − 3.6 z −2 − 4.8 z −3 . ⎛ 2.5 + z −1 ⎞⎛ − 0.8 + z −1 ⎞⎛ − 0.5 + z −1 ⎞ (vii) G8 ( z ) = 6(1 + 2.5z −1 )(1 − 0.8z −1 )(1 − 0.5z −1 )⎜⎜ ⎟⎟⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎝ 1+ 2.5z −1 ⎠⎝ 1− 0.8 z −1 ⎠⎝ 1− 0.5z −1 ⎠ = 6(1 + 2.5z −1 )(−0.8 + z −1 )(−0.5 + z −1 ) = −6 − 17.1 z −1 + 7.2 z −2 + 6 z −3 . (b) The minimum phase filter is G2 ( z), as all its zeros inside the unit circle. Likewise, the maximum phase filter is G6 ( z ), as all its zeros outside the unit circle. (c) The partial energy of impulse responses of each of the above transfer functions for different values of k are given by: k=0

k=1

k=2

k=3

G1(z)

36

87.84

380.25

416.25

G2(z)

225

407.25

410.49

416.25

G3(z)

23.04

36

360

416.25

G4(z)

9

9.81

272.25

416.25

G5(z)

144

406.44

407.25

416.25

G6(z)

5.76

9

191.25

416.25

G7(z)

56.25

380.25

393.25

416.25

G8(z)

36.00

328.41

380.25

416.25

The partial energy remains the same for values of k ≥ 3. From the above table it n

2

n

2

can be seen that ∑ g k [ m] ≤ ∑ g2 [ m] , and ∞

2

m=0 ∞

m=0

2

∑ g k [ m] = ∑ g2 [ m] = 416.25, 1 ≤ k ≤ 8, where G2 ( z) is the minimum

m=0

m=0

n

2

n

2

phase transfer function. Likewise, ∑ g k [ m] ≥ ∑ g6 [ m] , 1 ≤ k ≤ 8, where m=0

m=0

G6 ( z ) is the maximum phase transfer function. 7.9 H1 ( z ) = 1 − 0.5z −1 + 0.8z −2 − 0.4 z −3 + 0.25z −4 − 0.125z −5 + 0.2 z −6 − 0.1z −7

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= (1 − 0.8z −2 )(1 + z −1 + 0.5z −2 )(1 − z −1 + 0.5z −2 )(1 − 0.5z −1 ). H 2 ( z ) = 0.5 + 0.25z −1 + 0.4 z −2 − 0.425z −3 + 0.75z −4 − 0.75z −5 + 0.6 z −6 − 0.2 z −7 = (1 − 0.8 z −2 )(0.5 + z −1 + z −2 )(1 − z −1 + 0.5z −2 )(1 − 0.5z −1 ). H 3 ( z ) = −0.25 + 0.25z −1 + 0.175 z −2 + 0.7 z −3 − 0.45z −4 + 0.9 z −5 − 0.6 z −6 + 0.4 z −7

= (1 − 0.8 z −2 )(0.5 + z −1 + z −2 )(1 − z −1 + 0.5z −2 )(−0.5 + z −1 ). H 4 ( z ) = −0.5 + z −1 − 0.4 z −2 + 0.8 z −3 − 0.125z −4 + 0.25z −5 − 0.1z −6 + 0.2 z −7 = (1 − 0.8 z −2 )(1 + z −1 + 0.5z −2 )(1 − z −1 + 0.5z −2 )(−0.5 + z −1 ). H 5 ( z ) = −0.1 + 0.2 z −1 − 0.125 z −2 + 0.25z −3 − 0.4 z −4 + 0.8 z −5 − 0.5z −6 + z −7

= (−0.8 + z −2 )(0.5 + z −1 + z −2 )(0.5 − z −1 + z −2 )(−0.5 + z −1 ) = z −7 H1 ( z −1 ). Each factor of H1 ( z) has roots inside the unit circle. Hence, H1 ( z) is a minimumphase transfer function. Since H 5 ( z ) is a mirror-image of H1 ( z) , it has all zeros outside the unit circle and is thus a maximum-phase transfer function. There are 11 other length- FIR filters having the same magnitude responses as that of the above filters. 7.11 H (e jω ) = a1 + a 2 e − jω + a3 e − j 2 ω + a 4 e − j 3ω + a 5 e − j 4 ω + a6 e − j 5ω

[

= e − j 5ω / 2 (a1e j 5ω / 2 + a6 e − j 5ω / 2 ) + (a 2 e j 3ω / 2 + a 5 e − j 3ω / 2 ) + (a3 e jω / 2 + a 4 e − jω / 2 ) −j

5ω 2 2[( a

5ω j (a1 − a6 ) sin⎛⎜ ⎞⎟ + ⎝ 2 ⎠ ω ω 3ω 3ω + (a2 + a5 ) cos⎛⎜ ⎞⎟ + j (a2 − a5 ) sin⎛⎜ ⎞⎟ + (a3 + a 4 ) cos⎛⎜ ⎞⎟ + j (a3 − a 4 ) sin⎛⎜ ⎞⎟] ⎝2⎠ ⎝2⎠ ⎝ 2 ⎠ ⎝ 2 ⎠

=e

⎛ 5ω ⎞ 1 + a6 ) cos⎜ 2 ⎟ + ⎝ ⎠

It follows from the above that H (e jω ) will have linear-phase, i.e., constant group delay, if the imaginary parts inside the square brackets are zero. Hence, for a constant group delay we must have a1 = a6 , a2 = a5 , and a3 = a 4 . 7.12 The frequency response of the LTI discrete-time system is given by H (e jω ) = a1e jkω − a2 e j ( k −1)ω + a 2 e j ( k − 3)ω − a1e j ( k − 4)ω = e − j (2 − k )ω [ a1 (e j 2ω − e − j 2ω ) + a2 (e jω − e − jω )] = − je

− j ( 2 − k )ω

2[ a1 sin(2ω) + a2 sin ω] = − e

π 2

− j ( 2 − k − )ω

2[ a1 sin(2ω) + a 2 sin ω].

π 2

So, for k = 2 − , the system will have a frequency response that is a real function of ω .

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]

7.13 H 1(e jω ) = α + β e − jω and H 2 (e jω ) = H (e jω ) = H 1(e jω )H 2 (e jω ) =

1 1 + γ e − jω

α + β e − jω

1 − γe − jω

. Thus,

.

α + β e − jω α + β e jω α 2 + β 2 + 2αβ cos ω ⋅ = = K2 2 − jω jω 1 − γe 1 − γe 1 + γ − 2 γ cos ω

2

H ( e j ω ) = H ( e jω ) H ( e − j ω ) =

if α 2 + β 2 = K 2 (1 + γ 2 ) and αβ = − K 2 γ, i.e., (α + β ) 2 = K 2 (1 − γ ) 2 . α

7.14 Y (e jω ) = X (e jω ) e j arg X (e

jω )

. Hence, H (e jω ) =

Y ( e jω ) X (e jω )

= X ( e jω )

(α −1)

. Since

H (e jω ) is a real function of ω, it has zero phase.

(

)

7.15 H (e jω ) = h[0](1 + e − j 2ω ) + h[1]e − jω = e − jω h[0](e jω + e − jω ) + h[1]

= e − jω (2h[0] cos ω + h[1]). Thus, we require H (e j 0.3 ) = 2h[ 0] cos(0.3) + h[1] = 1 and H (e j 0.6 ) = 2h[ 0] cos(0.6) + h[1] = 0. Solving these two equations we get h[0] = 3.8461 and h[1] = −6.3487. 7.16 (a) H (e jω ) = h[0] + h[1]e − jω − h[1]e − j 3ω − h[0]e − j 4ω

(

= e − j 2ω h[0](e j 2ω − e − j 2ω ) + h[1](e jω − e − jω )

)

= j 2e − j 2ω (h[0] sin(2ω) + h[1] sin(ω)). Therefore, H (e jω ) = 2(h[ 0] sin( 2ω) + h[1] sin( ω)). Thus, H (e j 0.3 ) = 2(h[0] sin(0.6) + h[1] sin(0.3)) = 0.3 and H (e j 0.6 ) = 2(h[ 0] sin(1.2) + h[1] sin(0.6)) = 0.8. Solving these two equations we get

h[1] = −0.9873 and h[0] = 2.5686. (b) H (e jω ) = j 2e − j 2ω (2.5686 sin(2ω) − 0.9873 sin(ω)). The plot of the magnitude and phase responses are shown below. 7.17 (a) H (e jω ) = h[0] + h[1]e − jω − h[1]e − j 2ω − h[0]e − j 3ω

(

= e − j 3ω / 2 h[0](e j 3ω / 2 − e − j 3ω / 2 ) + h[1](e jω / 2 − e − jω / 2 ) = j 2e − j 3ω / 2 (h[0] sin(3ω / 2) + h[1] sin(ω / 2)). Therefore,

)

H (e jω ) = 2(h[ 0] sin(3ω / 2) + h[1] sin(ω / 2)). Thus,

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H (e j 0.25 ) = 2(h[ 0] sin(0.375) + h[1] sin( 0.125)) = 0.2 and H (e j 0.8 ) = 2(h[0] sin(1.2) + h[1] sin(0.4)) = 0.8. Solving these two equations we get

h[1] = 6.2573 and h[0] = −1.8569. (b) H (e jω ) = j 2e − j 3ω / 2 (− 1.8569 sin(3ω / 2) + 6.2573 sin(ω / 2)). 7.18 (a) H (e jω ) = h[0] + h[1]e − jω + h[2]e − j 2 ω + h[1]e − j 3ω + h[0]e − j 4ω

(

)

= e − j 2ω h[0](e j 2ω + e − j 2ω ) + h[1](e jω + e − jω ) + h[2] =e

− j 2ω

(2h[0] cos(2ω) + 2h[1] cos ω + h[2]) .

Therefore,

H (e jω ) = 2h[ 0] cos(2ω) + 2h[1] cos ω + h[ 2]. Thus,

H (e j 0.3 ) = 2 h[0] cos(0.6) + 2 h[1] cos(0.3) + h[2] = 1, H (e j 0.5 ) = 2 h[0] cos(1.0) + 2 h[1] cos(0.5) + h[2] = 0, and H (e j 0.8 ) = 2 h[0] cos(1.6) + 2 h[1] cos(0.8) + h[2] = 1. Solving these three equations

we get h[0] = 17.7761, h[1] = −58.7339 , and h[2] = 83.8786 . (b) H (e jω ) = e − j 2ω (35.5522 cos(2ω) − 117.4677 cos ω + 83.8786 ). The plot of the magnitude and phase responses are shown below. 4 Phase, in radians

2

Magnitude

1.5 1 0.5 0

0

0.2

0.4

0.6

0.8

1

ω

2 0 -2 -4

0

0.5

1

1.5

2

2.5

3

ω

7.19 H (e j 0 ) = 13, H (e j 3π / 4 ) = −3 − j 4, and H (e jπ ) = −3. Using the symmetry property of the DTFT of a real sequence, we observe H (e jπ / 4 ) = H * (e j 3π / 4 ) = −3 + j 4. Thus, the 4 –point DFT H [k ] of the length- 4

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sequence {h[n]} is given by {H[ k ]} = {13, − 3 + j 4, − 3, − 3 − j 4}. Its 4 –point 1 1 ⎤ ⎡ 13 ⎤ ⎡ 1 ⎤ ⎡h[0]⎤ ⎡1 1 ⎢ h[1] ⎥ ⎢1 j − 1 − j ⎥ ⎢ − 3 + j 4⎥ ⎢ 2 ⎥ ⎥= 1⎢ ⎥⎢ ⎥ = ⎢ ⎥. inverse DFT is thus given by ⎢ ⎢h[2]⎥ 4 ⎢1 − 1 1 − 1⎥ ⎢ − 3 ⎥ ⎢ 4 ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ h[3]⎦ ⎣1 − j − 1 j ⎦ ⎣ − 3 − j 4 ⎦ ⎣6 ⎦ Hence, H ( z ) = 1 + 2 z −1 + 4 z −2 + 6 z −3 . 7.20 Now, for a real, anti-symmetric sequence {h[ n]} of even length, H (e j 0 ) = 0. Using the symmetry property of the DTFT of a real sequence, we observe also H (e jπ / 4 ) = H * (e j 3π / 4 ) = −5 + j 5. Hence, the 4 –point DFT of the length- 4 sequence {h[ n]} is given by {H[ k ]} = {0, − 5 + j 5, 20, − 5 − j 5}. Its 4 –point ⎡h[0]⎤ 1 1 ⎤ ⎡ 0 ⎤ ⎡ 2 .5 ⎤ ⎡1 1 ⎢ h[1] ⎥ 1 ⎢1 j − 1 − j ⎥ ⎢− 5 + j 5⎥ ⎢− 7.5⎥ inverse DFT is thus given by ⎢ = . = h[2]⎥ 4 ⎢1 − 1 1 − 1⎥ ⎢ 20 ⎥ ⎢− 2.5⎥ ⎢ h[3]⎥ ⎢⎣1 − j − 1 j ⎥⎦ ⎢⎣ − 5 − j 5⎥⎦ ⎢⎣ 7.5 ⎥⎦ ⎦ ⎣ Hence, H ( z ) = 2.5 − 7.5 z −1 − 2.5 z −2 + 7.5 z −3 . 7.21 (a) H A (e jω ) = 0.5 − e − jω + 0.5e − j 2ω = e − jω (cos ω − 1) and H B (e jω ) = 0.5 + e − jω + 0.5e − j 2ω = e − jω (cos ω + 1). Hence, H A (e jω ) = cos ω − 1 and H B (e jω ) = cos ω + 1 . Plots of these two magnitude

functions is given below: jω

jω

HB(e )

2

2

1.5

1.5 Magnitude

Magnitude

HA(e )

1

0.5

0.5 0

1

0

0.2

0.4

0.6

0.8

1

0

0

0.2

0.4

0.6

0.8

1

ω/ π

ω/ π

As can be seen from the above plots, h A [n] is a highpass filter and hB [n] is a lowpass filter. (b) hC [ n] = ( −1) n h A [ n] = h B [ n]. Hence, H C (e jω ) = H A (e j (ω + π ) ) = H B (e jω ). HC (e jω ) is a shifted version of H A (e jω ) shifted by π radians.

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7.22 h[ n] = h[ N − n]. Thus, N −1

N −1

N −1

n=0

n=0

k =0

H ( z ) = ∑ h[ n]z − n = ∑ h[ N − n]z − n = ∑ h[ k ]z − ( N − k ) = z − N H ( z −1 ). As H ( z ) has zeros in a mirror-image symmetry in the z –plane, G ( z ) = 1 / H ( z ) will have poles outside the unit circle making it unstable.

7.23 (a) Since H ( z ) is a minimum-phase FIR transfer function, any other FIR transfer function G ( z ) with the same magnitude response as that of H ( z ) can be expressed as G ( z ) = H ( z ) A( z ) where A( z ) is an allpass function. Now, g[ 0] = lim G ( z ). z→∞

Thus, g[0] = | lim G( z) |= | lim H ( z) A( z) |= | lim H ( z) | ⋅ | lim A( z ) | z→∞

z →∞

z →∞

z→∞

≤ | lim H ( z ) |, as | lim A( z ) | < 1 (see Eq. (7.20)). Hence, g[0] ≤ h[0] . z→∞

z→∞

(b) If λ1 is a zero of H ( z ), then λ1 < 1, since H ( z ) is a minimum-phase causal stable transfer function with all zeros inside the unit circle. Let H ( z ) = B( z )(1 − λ1 z −1 ). It follows that is also a minimum-phase causal transfer function. ⎛ λ* − z −1 ⎞ ⎟. If Now, consider the transfer function F ( z) = B( z )( λ*1 − z −1 ) = H ( z )⎜ 1 ⎜ 1 − λ z −1 ⎟ 1 ⎝ ⎠ h[ n], b[ n], and f [ n] denote, respectively, the inverse –transform of H ( z ), B( z ), and F ( z ), then we get ⎧⎪ b[0], n = 0, ⎧ λ*1b[0], n = 0, h[ n] = ⎨ and f [ n] = ⎨ * ⎪⎩ λ1b[ n] − b[ n − 1], n ≥ 1. ⎩b[ n] − λ1b[ n − 1], n ≥ 1, Consider m

2

m

2

2

ε = ∑ h[ n] − ∑ f [ n] = b[0] − λ*1 n=0 2

h[ n] = 2

f [ n] = Hence, ε

2

n=0

2

m

2

m

2

b[0] + ∑ h[ n] − ∑ f [ n] . Now,

n =1 n =1 2 2 2 b[ n] + λ1 b[ n − 1] − λ1 b[ n − 1]b * [ n] − λ*1 b * [ n − 1]b[ n], and 2 2 2 λ1 b[ n] + b[ n − 1] − λ1 b[ n − 1]b * [ n] − λ*1b * [ n − 1]b[ n]. m 2 2 2 2 2 2 = b[0] − λ1 b[0] + ∑ ⎛⎜ b[ n] − λ1 b[ n − 1] ⎞⎟ ⎝ ⎠ n =1 m 2 2 2 2 2 − ∑ ⎛⎜ λ1 b[ n] − b[ n − 1] ⎞⎟ = ⎛⎜ 1 − λ1 ⎞⎟ b[ m] . ⎝ ⎠ ⎝ ⎠ n =1 m m m m 2 2 2

2

Since λ1 < 1, ε > 0, i.e., ∑ h[ n] > ∑ f [ n] . Hence, ∑ h[ n] ≥ ∑ g[ n] . n=0

n=0

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n=0

186

7.24 (a) H1 ( z ) =

z 3 + 3z 2 + 2 z + 7 (2 z + 3)( z 2 + 0.5 z + 0.8)

=

( z + 3.0867)( z 2 − 0.0867 z + 2.2678) (2 z + 3)( z 2 + 0.5 z + 0.8)

has a pole at

3 2

z = − , which is outside the unit circle. Hence, H1 ( z) is not a stable transfer function. To arrive at a stable transfer function with an identical magnitude 2z+3

resulting in response, we multiply H1 ( z) with the allpass function 2 + 3z z 3 + 3z 2 + 2 z + 7

2z + 3 z 3 + 3z 2 + 2 z + 7 . = (2 z + 3)( z 2 + 0.5 z + 0.8) 2 + 3 z (2 + 3 z )( z 2 + 0.5 z + 0.8)

(b) H 2 ( z ) =

⋅

4 z 3 − 2 z 2 + 5z − 6 (1.5 z 2 + 3 z − 5)( z 2 − 0.3 z + 0.7)

=

4( z 2 + 0.4181+ 1.6338)( z − 0.9181) 1.5( z + 3.0817)( z −1.0817)( z 2 − 0.3 z + 0.7)

has

two poles outside the unit circle at z = −3.0817 and z = 1.0817 . Hence, H 2 ( z) is not a stable transfer function. To arrive at a stable transfer function with an identical magnitude response, we multiply H 2 ( z) with the allpass function z 2 − 0.3 z + 0.7 0 .7 z 2 − 0 .3 z + 1

=

resulting in

4 z 3 − 2 z 2 + 5z − 6

⋅

z 2 − 0.3 z + 0.7

(1.5z 2 + 3z − 5)( z 2 − 0.3z + 0.7) 0.7 z 2 − 0.3 z + 1

4 z 3 − 2 z 2 + 5z − 6 (5 z 2 + 3 z −1.5)( z 2 − 0.3z + 0.7)

.

7.25 The transfer function of the simplest notch filter is given by H ( z ) = (1 − e jωo z −1 )(1 − e − jωo z −1 ) = 1 − 2 cos(ωo ) z −1 + z − 2 . In the steady-state, the output for an input x[ n] = cos(ωo n) is given by y[ n] = H (e where θ (ωo ) = arg[ H (e

jω o

jω o

) cos(ωo n + θ (ωo ))

)].

(a) Comparing H ( z ) = 1 − z −1 + z −2 with H ( z ) as given above we conclude cos(ωo ) = , or ωo = . Here, H1 (e jω o ) = 1 − e − jω o + e − j 2ω o 2 3 π

1

= 1 − e − jπ / 3 + e − j 2π / 3 = 1 − (0.5 + j 0.866) + (−0.5 + j 0.866) = 0. Hence, y[ n] = H1 (e jωo ) cos(ωo n + θ (ωo )) = 0. (b) Comparing H 2 ( z ) = 1 − 0.8 z −1 + z −2 with H ( z ) as given above we conclude cos(ωo ) = 0.4 or ωo = 0.369π. Here, H 2 (e jω o ) = 1 − e − jω o + e − j 2ω o = 1 − (0.32 + j 0.7332 ) + (−0.68 + j 0.7332 ) = 0. Hence, y[ n] = H 2 (e

jω o

) cos(ωo n + θ (ωo )) = 0. .

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(c) Comparing H 3 ( z ) = 1 − 1.6 z −1 + z −2 with H ( z ) as given above we conclude cos(ωo ) = 0.8 or ωo = 0.2048π. Here, H 2 (e jω o ) = 1 − e − jω o + e − j 2ω o = 1 − (1.2891 + j 0.9599 ) + (0.2801 + j 0.9599 ) = 0. Hence, y[ n] = H 3 (e

jω o

) cos(ωo n + θ (ωo )) = 0.

Y (z) Y (z) 7.26 From the figure, H 0 ( z ) = 0 = G L ( z )G H ( z 2 ), H1 ( z ) = 1 = G H ( z )G H ( z 2 ), X ( z) X ( z) Y ( z) Y ( z) H 2 ( z) = 2 = G H ( z )G L ( z 2 ), H 3 ( z ) = 3 = G L ( z )G L ( z 2 ). X ( z) X (z) For the magnitude responses of G L (z) and G H (z) shown below GH(e jω)

G L(e jω)

1

1

0

π/4

0

π/2

π

ω

0

π/2

0 2

3π/4

π

ω

the magnitude responses of G L ( z 2 ) and G H ( z ) are as shown below G L(e

j2ω

GH(e j2ω )

)

1

1

0

π/4

0

π/2

π

ω

0

π/2

0

3π/4

π

ω

Hence, the magnitude responses of the casacaded filters H i ( z ), 0 ≤ i ≤ 3, are as indicated below: H 0(e jω )

H 1(e jω ) 1

1

0

0

π/4

π/2

π

ω

π/2

0

3π/4

π

ω

H 3(e jω )

H 2(e jω )

1

1

0

0

0

π/4

π/2

π

ω

0

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0

π/4

π/2

π

ω

188

⎧1, 0 ≤ ω < ωc , 7.27 G(e jω ) = H LP (e j (π − ω) ). Since, H LP (e jω ) = ⎨ ⎩0, ωc ≤ ω < π, ⎧0, 0 ≤ ω < π − ωc , G ( e jω ) = ⎨ Hence, G(e jω ) is a highpass filter with a cutoff − ≤ < 1 , π ω ω π . c ⎩ frequency given by ω o = π − ω c . Also, since G( z) = H LP (− z), we have g[ n] = (−1) n hLP [ n]. 7.28 G( z ) = H LP (e jωo z ) + H LP (e − jωo z ). Hence, g[ n] = hLP [ n]e − jωo n + hLP [ n]e jωo n = 2 h LP [ n ] cos( ω o n ). Thus, G ( z ) is a real coefficient bandpass filter with a center frequency at ωo and a passband width of 2ω p . 7.29 F ( z ) = H LP (e jωo z ) + H LP (e − jωo z ) + H LP ( z ). Hence, f [ n] = (1 + 2 cos(ωo n))hLP [ n]. Thus, F ( z ) is a real coefficient bandstop filter with a center frequency at ωo and a stopband width of (π − 3ω p ) / 2. 7.30 V(z)

+

X(z)

hLP[n]

U(z)

+

Y(z)

( _1)n

( _1)n

From the above figure, we get V ( z) = X (− z), U ( z) = H LP ( z) X (− z), and Y ( z) Y ( z) = U (− z) = H LP (− z) X ( z). Hence, H eq ( z ) = = H LP (− z ), which is the X (z) highpass filter of Problem 7.27. 7.31 2

V 0(z) cos(ωo n) +

cos(ωo n)

+

U1 (z) 2

hLP[n]

Y(z)

V 1(z)

+

X(z)

hLP[n]

+

+

U0 (z)

sin(ωo n)

sin(ωo n)

From the above figure, we have u0 [ n] = 2 x[ n] cos(ωo n) = x[ n] e jω o n + x[ n] e − jω o n or, U 0 (e jω ) = X (e j (ω + ω o ) ) + X (e j (ω − ω o ) ). Likewise, U1 (e jω ) = j X (e j (ω + ω o ) ) − j X (e j (ω − ω o ) ). We also have

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V0 ( z ) = H LP ( z ) X ( z e jω o ) + H LP ( z ) X ( z e − jω o ), and

V1 ( z ) = j H LP ( z ) X ( z e jω o ) − j H LP ( z ) X ( z e − jω o ). Therefore, Y ( z) =

1 2

(V (z e 0

jω o

) (V (z e

) + V0 ( z e − jω o ) + j

1 2

1

jω o

)

) − V1 ( z e − jω o ) , which after

simplification yields

{H LP (z e jωo ) X (z e j 2ωo ) + H LP (z e jωo ) X (z) + H LP (z e − jωo ) X (z) 1 + H LP ( z e − jω o ) X ( z e − j 2ω o )} − {H LP ( z e jω o ) X ( z e j 2ω o ) − H LP ( z e jω o ) X ( z ) 4 − jω o − jω o + H LP ( z e ) X ( z ) + H LP ( z e ) X ( z e − j 2ω o )}. Hence,

Y ( z) =

1 4

Y ( z) =

1 2

{H

H eq ( z ) =

1 2

LP ( z e

{H

jω o

LP ( z e

}

) + H LP ( z e − jω o ) X ( z ). Therefore,

jω o

}

) + H LP ( z e − jω o ) . Thus, the structure of Figure P7.4

implements the bandpass filter of Problem 7.28.

7.32

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|H 0(e jω )|

π 0 __ π _ __ M M |H 1(e jω )|

0

2π __ M

π

2π

π

2π

π

2π

π

2π

ω

ω

|H 2(e jω )|

0

4π __ M

ω

|HM _ 1(e jω )|

_ 2π __ M

0

2( M _ 1)π _____ M

ω

The output of the k –th filter is a bandpass signal occupying a bandwidth of 2π / M and centered at ω = kπ / M. In general, the k –th filter H k (z ) has a complex impulse response generating a complex output sequence. To realize a real coefficient bandpass filter, one can combine the outputs of H k (z ) and H M − k (z ). 7.33 H 0 ( z ) = (1 + z −1 ). Thus, H 0 (e jω ) = cos(ω / 2). Now, G ( z ) = (H 0 ( z )) M . 1 2

2

Hence, G(e jω ) = H 0 (e jω )

2M

= (cos(ω / 2))2 M . The 3 –dB cutoff frequency

ωc of G ( z ) is thus given by (cos(ωc / 2))2 M = . Hence, ωc = 2 cos −1 (2 −1 / 2 M ). 1 2

2

7.34 H1 ( z ) = (1 − z −1 ). Thus, H1 (e jω ) = sin 2 (ω / 2). Let, F ( z ) = (H 0 ( z )) M . Then, 2 1

2

F (e jω ) = (sin(ω / 2))2 M . The 3 –dB cutoff frequency ωc of F ( z ) is thus given

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by (sin(ωc / 2))2 M = , which yields, ωc = 2 sin −1 (2 −1 / 2 M ). 2 1

7.35 H LP ( z ) =

1 − α ⎛⎜ 1 + z −1 ⎞⎟ . Note that H LP (z) is stable if α < 1. Now, 2 ⎜⎝ 1 − αz −1 ⎟⎠

1 − sin(ωc ) cos 2 (ωc / 2) + sin 2 (ωc / 2) − 2 sin(ωc / 2) cos(ωc / 2) α= = cos(ωc ) cos 2 (ωc / 2) − sin 2 (ωc / 2)

=

cos(ωc / 2) − sin(ωc / 2) 1 − tan(ωc / 2) = . cos(ωc / 2) + sin(ωc / 2) 1 + tan(ωc / 2)

(7-a)

If 0 ≤ ω < π, then tan(ωc / 2) ≥ 0. Hence, α < 1. 7.36 From Eq. (7-a), α =

1 − tan(ωc / 2) 1− α . Hence, tan(ωc / 2) = . 1+ α 1 + tan(ωc / 2)

1 − sin(0.6) = 0.5275. Substituting this value of α cos(0.6) ⎛ ⎞ 1 + z −1 ⎟. in Eq.(7.71), we arrive at H LP ( z ) = 0.2363⎜ ⎜ 1 − 0.5275 z −1 ⎟ ⎝ ⎠

7.37 (a) From Eq.(7.73b), we get α =

1 − sin(0.45) = 0.0787. Substituting this value of cos(0.45) ⎛ ⎞ 1 + z −1 ⎟. α in Eq.(7.71), we arrive at H LP ( z ) = 0.4607⎜ ⎜ 1 − 0.0787 z −1 ⎟ ⎝ ⎠ 1 + α ⎛⎜ 1 − z −1 ⎞⎟ 1 + α ⎛⎜ 1 − e − jω ⎞⎟ 7.38 H HP ( z ) = . Thus, H HP (e jω ) = . Thus, 2 ⎜⎝ 1 − αz −1 ⎟⎠ 2 ⎜⎝ 1 − αe − jω ⎟⎠ 2 ⎛ 1 + α ⎞2 ⎛ 2 − 2 cos(ω) ⎞⎟ H HP (e jω ) = ⎜ . At –dB cutoff frequency ωc , ⎟ ⎜⎜ ⎝ 2 ⎠ ⎝ 1 + α 2 − 2α cos(ω) ⎟⎠ 2 1 2α H HP (e jωc ) = which yields cos(ωc ) = . 2 1 + α2 1 − sin(0.6) 7.39 (a) From Eq.(7.73b), we get α = = 0.5275. Substituting this value of cos(0.6) ⎛ 1 + z −1 ⎞ ⎟. α in Eq.(7.74), we arrive at H HP ( z ) = 0.7637⎜ ⎜ 1 − 0.5275z −1 ⎟ ⎝ ⎠ 1 − sin(0.45) (b) From Eq.(7.73b), we get α = = 0.0787. Substituting this value of cos(0.45) (b) From Eq.(7.73b), we get α =

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⎛ 1 + z −1 ⎞ ⎟. α in Eq.(7.74), we arrive at H HP ( z ) = 0.5394⎜ ⎜ 1 − 0.0787 z −1 ⎟ ⎝ ⎠ 7.40 H ( z) =

1 − z −1 1 − kz

−1

(1 − cos ω) 2 + sin 2 ω

2

. Hence, H (e jω ) =

2

Now, H (e jπ ) =

4 (1 + k ) 2

2

2

2

(1 − k cos ω) + k sin ω

=

2 − 2 cos ω 1 + k 2 − 2 k cos ω

. Thus, the scaled transfer function is given by

1 + k ⎛⎜ 1 − z −1 ⎞⎟ . A plot of the magnitude responses of the scaled transfer H (z) = 2 ⎜⎝ 1 − kz −1 ⎟⎠

function for k = 0.95, 0.9, and − 0.5 are given below: 1

←k=0.95 ←k=0.9

Magnitude

0.8

k=-0.5→

0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/pi

7.41 H BP ( z ) =

⎞ 1 − α ⎛⎜ 1 − z−2 ⎟. Thus, 2 ⎜⎝ 1 − β (1 + α )z −1 + αz − 2 ⎟⎠

H BP (e jω ) = H BP (e jω ) =

2

⎞ 1 − α ⎛⎜ 1 − e − j 2ω ⎟. Hence, 2 ⎜⎝ 1 − β (1 + α )e − jω + αe − j 2ω ⎟⎠ =

(1 − α ) 2 2(1 − 2 cos(2ω)) ⋅ 2 1 + β 2 (1 + α ) 2 + α 2 + 2α cos(2ω) − 2 β (1 + α ) 2 cos(ω)

(1 − α) 2 sin 2 ω (1 + α) 2 (cos ω − β) 2 + (1 − α ) 2 sin 2 ω

. At the center frequency ω o ,

2

H BP (e jωo ) = 1. Hence, (cosω o − β ) 2 = 0 or cosω o = β. 2

At the 3 –dB bandedges ω1 and ω2 , H BP (e jωi ) = , i = 1, 2. This imples (1 + α ) 2 (cos ωi − β) 2 = (1 − α ) 2 sin 2 ωi ,

1 2

(7-c)

⎛1+ α ⎞ or, sin ωi = ±⎜ ⎟ (cos ωi − β), i = 1, 2. Since, ω1 < ωo < ω2 , sin ω1 must have ⎝1− α ⎠ positive sign and sin ω2 must have negative sign, because otherwise, sin ω2 < 0

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.

for ω2 in (0, π). Now, Eq. (7-c) can be rewritten as 2(1 + α 2 ) cos 2 ωi − 2β(1 + α) 2 cos ωi + β 2 (1 + α) 2 − (1 − α 2 ) = 0. Hence,

(1 + α) 2

β 2 (1 + α) 2 − (1 − α) 2 , and (cos ω1 )(cos ω2 ) = . 2(1 + α 2 ) 1 + α2 Denote ω3dB = ω2 − ω1. Then cos ω3dB = cos ω2 cos ω1 + sin ω2 sin ω1

cos ω1 + cos ω2 = β

2

(

)

2α ⎛1+ α ⎞ 2 . = cos ω2 cos ω1 − ⎜ ⎟ cos ω2 cos ω1 + β (cos ω2 + cos ω1 ) = ⎝1− α ⎠ 1 + α2

7.42 (a) Using Eq.(7.76), we get β = cos( 0.55π) = −0.1564 . Next, from Eq.(7.78), we 2α get = cos(0.25π) = 0.7071, or, equivalently, 0.7071α 2 − 2α + 0.7071 = 0. 2 1+ α Solution of this quadratic equation yields α = 2.4142 and α = 0.4142. Substituting α = 2.4142 and β = −0.1564 . in Eq.(7.77) we arrive at the denominator polynomial of the transfer function H BP (z) as D( z ) = 1 + 0.5340 z −1 + 2.4142 z −2 . Comparing with Eq. (7.136) we note d1 = 0.5340 and d 2 = 2.4142. Since the condition of Eq. (7.139) is not satisfied, the corresponding H BP (z) is unstable. Substituting α = 0.4142 and β = −0.1564 . in Eq.(7.77) we arrive at the denominator polynomial of the transfer function H BP (z) as D( z ) = 1 + 0.2212 z −1 + 0.4142 z −2 . Comparing with Eq. (7.136) we note d1 = 0.2212 and d 2 = 0.4142. Since the conditions of Eqs. (7.139) and (7.141) are satisfied, the corresponding H BP (z) is a stable transfer function. Hence, the desired transfer function is H BP ( z) =

0.2929(1 − z −2 )

1 + 0.2212 z −1 + 0.4142 z − 2

.

(b) Using Eq.(7.76), we get β = cos( 0.3π) = 0.5878 . Next, from Eq.(7.78), we get 2α = cos(0.3π) = 0.5878, or, equivalently, 0.5878 α 2 − 2α + 0.5878 = 0. 2 1+ α Solution of this quadratic equation yields α = 3.0766 and α = 0.3249. Substituting α = 3.0766 and β = 0.5878. in Eq.(7.77) we arrive at the denominator polynomial of the transfer function H BP (z) as D( z ) = 1 + 2.3968 z −1 + 0.788 z −2 . Comparing with Eq. (7.136) we note d1 = 2.3968 and d 2 = 0.788. Since the condition of Eq. (7.141) is not satisfied, the corresponding H BP (z) is unstable. Substituting α = 0.3249 and β = 0.5878. in Eq.(7.77) we arrive at the denominator polynomial of the transfer function H BP (z) as D( z ) = 1 + 0.7788 z −1 + 0.3249z −2 . Not for sale

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Comparing with Eq. (7.136) we note d1 = 0.2212 and d 2 = 0.4142. Since the conditions of Eqs. (7.139) and (7.141) are satisfied, the corresponding H BP (z) is a stable transfer function. Hence, the desired transfer function is

0.3376(1 − z −2 )

H BP ( z ) =

. 1 + 0.7788z −1 + 0.3249z − 2 1 + α ⎛⎜ 1 − 2βz −1 + z − 2 7.43 H BS ( z ) = 2 ⎜⎝ 1 − β(1 + α)z −1 + αz − 2 H BS (e =

jω

⎞ ⎟. Thus, ⎟ ⎠

2

2 + 4β 2 − 8β cos(ω) + 2 cos(2ω) ⎛1+ α ⎞ ) =⎜ ⎟ ⋅ ⎝ 2 ⎠ 1 + β 2 (1 + α) 2 + α 2 + 2α cos(2ω) − 2β(1 + α) 2 cos(ω) 2

(1 + α) 2 (cos ω − β) 2 (1 + α) 2 (cos ω − β) 2 + (1 − α ) 2 sin 2 ω

. At the center frequency ω o ,

2

H BS (e jωo ) = 0. Hence, (cosω o − β ) 2 = 0 or cosω o = β. At the 3 –dB 2

bandedges ω1 and ω2 , H BP (e jωi ) = , i = 1, 2. This leads to Eq. (7-c) given in 2 the solution of Problem 7.41. Hence, as in the solution of Problem 7.41, 2α . ω3dB = 1 + α2 1

7.44 (a) Using Eq.(7.76), we get β = cos( 0.35π) = 0.454 . Next, from Eq.(7.78), we get 2α = cos(0.2 π) = 0.809, or, equivalently, 0.809 α 2 − 2α + 0.809 = 0. Solution 2 1+ α of this quadratic equation yields α = 1.9627 and α = 0.3249. Substituting α = 1.9627 and β = 0.454 in Eq.(7.80) we arrive at the denominator polynomial of the transfer function H BS (z ) as D( z ) = 1 − 1.3451 z −1 + 1.9627 z −2 . Since the condition of Eq. (7.139) is not satisfied, the corresponding H BS (z ) is unstable. Substituting α = 0.3249 and β = 0.454 in Eq.(7.80) we arrive at the denominator polynomial of the transfer function H BS (z ) as D( z ) = 1 + 0.6015 z −1 + 0.3249 z −2 . Since the conditions of Eqs. (7.139) and (7.141) are satisfied, the corresponding H BS (z ) is a stable transfer function. Hence, the desired transfer function is H BS ( z ) =

0.6624(1 − 0.908 z −1 + z −2 ) 1 + 0.6015 z −1 + 0.3249 z − 2

.

(b) Using Eq.(7.76), we get β = cos( 0.6 π) = −0.309. Next, from Eq.(7.78), we get 2α = cos(0.15π) = 0.891, or, equivalently, 0.891α 2 − 2α + 0.891 = 0. Solution 2 1+ α Not for sale

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of this quadratic equation yields α = 1.6319 and α = 0.6128 . Substituting α = 1.6319 and β = −0.309 in Eq.(7.80) we arrive at the denominator polynomial of the transfer function H BS (z ) as D( z ) = 1 + 2.345 z −1 + 1.6319 z −2 . Since the condition of Eq. (7.139) is not satisfied, the corresponding H BS (z ) is unstable. Substituting α = 0.6128 and β = −0.309 in Eq.(7.80) we arrive at the denominator polynomial of the transfer function H BS (z ) as D( z ) = 1 + 0.4974 z −1 + 0.6128 z −2 . Since the conditions of Eqs. (7.139) and (7.141) are satisfied, the corresponding H BS (z ) is a stable transfer function. Hence, the desired transfer function is H BS ( z ) = 7.45

0.564(1 + 0.618 z −1 + z −2 ) 1 + 0.4984 z −1 + 0.6128 z − 2

(1 − α) 2 (1 + cos ωc ) 2

2(1 + α − 2α cos ωc )

.

= 2 −1 / K . Let C = 2 ( K −1) / K . Simplifying the first equation

we then get α 2 (cosωc +1 − C ) − 2α(1 + cosωc −C cosωc ) + 1 + cosωc −C = 0. Solving the quadratic equation for α we obtain α=

= =

2(1 + (1 − C ) cos ωc ) ± 4(1 + (1 − C ) cos ωc )2 − 4(1 + cos ωc − C ) 2

(1 + (1 − C ) cos ωc ) ±

2(1 + cos ωc − C )

(2 + 2 cos ωc − C − C cos ωc )(C (1 − cos ωc ) ) 1 + cos ωc − C

1 + (1 − C ) cos ωc ± sin ωc 2C − C 2 . For stability we require α < 1, hence the 1 + cos ωc − C

desired solution is α =

1 + (1 − C ) cos ωc − sin ωc 2C − C 2 . 1 + cos ωc − C 2

2 ⎛ 1 + α ⎞ 2 1 − e − jω 1 + α ⎛⎜ 1 − z −1 ⎞⎟ . H HP (e jω ) = ⎜ 7.46 H HP ( z ) = . ⎟ 2 ⎜⎝ 1 − αz −1 ⎟⎠ ⎝ 2 ⎠ 1 − α e − jω

H HP (e

jω

)

2K

⎛1+ α ⎞ =⎜ ⎟ ⎝ 2 ⎠

2K

1 − e − jω 1 − αe

2K

− jω

the –dB cutoff frequency ωc , H HP (e jω c )

⎛1+ α ⎞ =⎜ ⎟ ⎝ 2 ⎠ 2K

2K

2 K (1 − cos ω) K 2

(1 + α − 2α cos ω)

K

. At

= . Let C = 2 ( K −1) / K . 1 2

Simplifying the above equation we get α 2 (1 − cosωc −C ) − 2α(1 − cosωc +C cosωc ) + 1 − cosωc −C = 0. Solving the quadratic equation for α we obtain

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α=

− 2(1 − cos ωc + C cos ωc ) ± 2 (1 − cos ωc + C cos ωc ) 2 − (1 − cos ωc − C ) 2 2(1 − cos ωc − C )

.

For stability we require α < 1, hence the desired solution is

sin ωc 2C − C 2 − (1 − cos ωc + C cos ωc ) . α= 1 − cos ωc − C K 1 7.47 (a) Analyzing Figure P7.6(a) we get Y ( z ) = ⎡ (1 + A1 ( z )) + (1 − A1 ( z ))⎤ X ( z ). ⎢⎣ 2 ⎥⎦ 2 Y (z) ⎛ 1 + K ⎞ ⎛ 1 − K ⎞ Hence, H ( z ) = =⎜ ⎟+⎜ ⎟A1 ( z ). X (z) ⎝ 2 ⎠ ⎝ 2 ⎠ ⎛1+ K ⎞ ⎛1− K ⎞ (b) Analyzing Figure P7.6(b) we get Y ( z ) = ⎜ ⎟ X ( z) + ⎜ ⎟A1 ( z ) X ( z ). ⎝ 2 ⎠ ⎝ 2 ⎠ Y (z) ⎛ 1 + K ⎞ ⎛ 1 − K ⎞ Hence, H ( z ) = =⎜ ⎟+⎜ ⎟A1 ( z ). X (z) ⎝ 2 ⎠ ⎝ 2 ⎠ K 7.48 (a) Analyzing Figure P7.6(a) we get Y ( z ) = X ( z ) + (1 + A1 ( z ))X ( z ). 2 Y ( z) ⎛ 1 + K ⎞ ⎛ K ⎞ Therefore, H ( z ) = =⎜ ⎟ + ⎜ ⎟A1 ( z ). X ( z) ⎝ 2 ⎠ ⎝ 2 ⎠ K (b) Analyzing Figure P7.6(b) we get Y ( z ) = X ( z ) + (1 − A1 ( z ))X ( z ). Therefore, 2 Y ( z) ⎛ 1 + K ⎞ ⎛ K ⎞ =⎜ H ( z) = ⎟ − ⎜ ⎟A1 ( z ). X ( z) ⎝ 2 ⎠ ⎝ 2 ⎠

7.49 H (e

jω

⎧1, ω p1 ≤ ω ≤ ω p2 , ⎪ ) = ⎨0, G(e jω ) = H (e j ( π − ω) ), This implies that the 0 ≤ ω ≤ ω s1 , ⎪0, ω s 2 ≤ ω < π. ⎩

frequency response of H ( − z ) is a shifted version of the frequency response of H ( z ) , shifted by π radians. Therefore,

⎧1, π − ω p2 ≤ ω ≤ π − ω p1, ⎪ ) = ⎨0, G ( e jω ) = H ( e Hence, H (− z ) is also a 0 ≤ ω ≤ π − ωs 2 , ⎪0, π − ωs1 ≤ ω < π. ⎩ bandpass filter with passband edges at π − ω p2 and π − ω p1 , and stopband edges at j ( π − ω)

π − ω s 2 and π − ω s1 with π − ωs 2 < π − ω p1 < π − ω p1 < π − ωs1. 1 − α ⎛⎜ 1 + z −1 ⎞⎟ 1 − α ⎛⎜ 1 − z −1 ⎞⎟ , G HP ( z) = . Let β = − α. Then, 7.50 H LP ( z) = 2 ⎜⎝ 1 − αz −1 ⎟⎠ 2 ⎜⎝ 1 + αz −1 ⎟⎠

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G HP ( z) =

1 + β ⎛⎜ 1 − z −1 ⎞⎟ . Therefore, ωc = cos −1 ( β ) = cos −1 (− α ). − 1 ⎜ ⎟ 2 ⎝ 1 − βz ⎠

7.51 The magnitude responses of H ( z ), H (− z), H ( z 3 ), and H (− z 3 ) are shown below: |H(e jω )|

|H(_ e jω )|

1

1

π __ 3

0

2π __ 3

π

ω

0

|H(_ e j3ω )|

|H(e j3ω )| 1

π __ 3

2π __ 3

ω

π

1

0 π__

9

π __ 3

5π __ 7π __ 2π __ 9 3 9

ω

π

0

2π __ 4π __ π __ 9 3 9

2π __ 3

8π __ π 9

ω

The magnitude responses of H ( z)H ( z 3 ), H (− z) H ( z 3 ), and H ( z)H (− z 3 ) are shown below: |H(_ e jω )H(e j3ω )|

|H(e jω )H(e j3ω )|

1

|H(e jω )H(_ e j3ω)|

1

1

0 π__

9

π __ 3

2π __ 3

π

ω

0

π __ 3

2π __ 7π __ 3 9

π

ω

0

2π __ __ π 9 3

2π __ 3

π

ω

( 7.52 From Eq. (7.49) we observe that the amplitude response H (ω) of a Type 1 FIR ( transfer function is a function of cos( ωn). Thus, H (ω + 2πk ) will be a function of cos((ω + 2πk )n ) = cos(ωn + 2πkn) = cos(ωn) cos(2πkn) − sin(ωn) sin(2πkn) = cos(ωn). ( Hence, H (ω) is a periodic function in ω with a period 2π. ( Likewise, from Eq. (7.53) we observe that the amplitude response H (ω) of a Type ( 3 FIR transfer function is a function of sin( ωn). Thus, H (ω + 2πk ) will be a function of sin((ω + 2πk )n ) = sin(ωn + 2πkn) = sin(ωn) cos(2πkn) ( + cos( ωn) sin( 2πkn) = sin( ωn) . Hence, H (ω) is a periodic function in ω with a period 2π. ( Next, from Eq. (7.52) we observe that the amplitude response H (ω) of a Type 2 ( 1 FIR transfer function is a function of cos⎛⎜ ω(n − ) ⎞⎟. Thus, H (ω + 4πk ) will be a 2 ⎠ ⎝ function of 1 1 1 1 1 cos⎛⎜ (ω + 4πk )(n − ) ⎞⎟ = cos⎛⎜ ω(n − ) + 4πk (n − ) ⎞⎟ = cos⎛⎜ ω(n − ) ⎞⎟ cos⎛⎜ 4πk (n − ) ⎞⎟ 2 ⎠ 2 2 ⎠ 2 ⎠ 2 ⎠ ⎝ ⎝ ⎝ ⎝

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1 1 1 1 − sin⎛⎜ ω(n − ) ⎞⎟ sin⎛⎜ 4πk (n − ) ⎞⎟ = cos⎛⎜ ω(n − ) ⎞⎟ as cos⎛⎜ 4πk (n − ) ⎞⎟ = 1 and 2 ⎠ 2 ⎠ 2 ⎠ 2 ⎠ ⎝ ⎝ ⎝ ⎝ ( 1 sin⎛⎜ 4πk (n − ) ⎞⎟ = 0. Hence, H (ω) is a periodic function in ω with a period 4π. 2 ⎠ ⎝

( Finally, from Eq. (7.54) we observe that the amplitude response H (ω) of a Type ( 1 4 FIR transfer function is a function of sin⎛⎜ ω(n − ) ⎞⎟. Thus, H (ω + 4πk ) will be a 2 ⎠ ⎝ function of 1 1 1 1 1 sin⎛⎜ (ω + 4πk )(n − ) ⎞⎟ = sin⎛⎜ ω(n − ) + 4πk (n − ) ⎞⎟ = sin⎛⎜ ω(n − ) ⎞⎟ cos⎛⎜ 4πk (n − ) ⎞⎟ 2 ⎠ 2 2 ⎠ 2 ⎠ 2 ⎠ ⎝ ⎝ ⎝ ⎝ 1 1 1 1 + cos⎛⎜ ω(n − ) ⎞⎟ sin⎛⎜ 4πk (n − ) ⎞⎟ = sin⎛⎜ ω(n − ) ⎞⎟ as cos⎛⎜ 4πk (n − ) ⎞⎟ = 1 and 2 2 2 2 ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ( 1 sin⎛⎜ 4πk (n − ) ⎞⎟ = 0. Hence, H (ω) is a periodic function in ω with a period 4π. 2 ⎠ ⎝ 7.53 The remaining zeros are at: z 5 =

1 1 = = 1.25; z 6 = z 2* = j; z 7 = z 3* = 2 + j 2; z1 0.8

1 1 = = 0.25 + j 0.25; z 9 = z8* = 0.25 − j 0.25; z10 = z *4 = −0.5 − j 0.3; z3 2 − j 2 1 1 0 .5 0 .3 0 .5 0 .3 z11 = = =− +j , z12 = z11* = − −j . z 4 − 0 .5 + j 0 .3 0.34 0.34 0.34 0.34

z8 =

12

H1 ( z) = ∏ (1 − z k z −1 ) = 1 − 2.6088z −1 + 1.7576 z − 2 + 11.0226 z − 3 + 5.6432 z − 4 k =1 −5

− 24.2166 z

+ 9.7711z −6 − 24.2166 z −7 + 5.6432 z −8 + 11.0226 z −9 + 1.7576 z −10

− 2.6088 z −11 + z −12 .

7.54 The remaining zeros are at: z 4 =

1 1 = = 0.3226, z 5 = z 2* = −2 − j 4, z1 3.1

1 1 = = −0.1 − j 0.2, z 7 = −0.1 − j 0.2, z2 − 2 + j 4 1 1 z8 = z 3* = 0.8 − j 0.4, z 9 = = = 1 − j 0.5, z10 = 1 + j 0.5, z11 = −1. z 3 0 .8 + j 0 .4 z6 =

11

H 2 ( z) = ∏ (1 − z k z −1 ) = 1 − 1.8226 z −1 + 7.1039z − 2 − 79.4635z − 3 +182.19z − 4 k =1

− 111.2306 z −5 −111.2306 z −6 +182.19 z −7 −79.4635 z −8 +7.1039 z −9 −1.8226 z −10 + z −11 .

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7.55 The remaining zeros are at: z 4 = z1* = 0.1 + j 0.599, 1 1 z5 = = = 0.2711 + j 1.6242, z 6 = z 5* = 0.2711 - j 1.6242, z1 0.1 − j 0.599 1 z 7 = z 2* = − 0 . 3 − j 0 . 4 , z8 = = −1.2 − j 1.6, z 9 = z8* = −1.2 + j 1.6, z2 1 z 3 = 2, z10 = = 0.5, z10 = 1, z11 = −1. z3 H 3 ( z ) = 1 − 0 .2423 z − 1 + 1 .0076 z − 2 − 6 .5294 z − 3 + 1 .3338 z − 4 − 17 .2533 z − 5 + 17.2533 z −7 − 1.3338 z −8 + 6.5294 z −9 − 1.0076 z −10 + 0.2423 z −11 − z −12 .

7.56 The remaining zeros are at: z 4 = 2.2 − j 3.4, z 5 =

1 = 0.1341 − j 0.2073, z1

z 6 = z *5 = 0.1341 + j 0.2073, z 2 = 0 . 6 + j 0 . 9 , z 7 = z 2* = 0 . 6 − j 0 . 9 , z8 = z 9 = z8* = 0.5128 + j 0.7692, z 3 = −0.5, z10 =

1 = 0.5128 − j 0.7692, z2

1 = −2, z11 = 1. z3

H 4 ( z) = 1 − 5.3939z −1 + 19.446 z −2 − 5.0838z −3 − 51.8577z −4 + 119.7073z −5 − 119.7073 z −6 + 51.8577 z −7 + 5.0838 z −8 − 19.446 z −9 + 5.3939 z −10 − z −11 .

7.57 The magnitude responses of H ( z M ) (solid line) and F1 ( z ) (dashed line) are shown below: |F1(e jω)|

1

0

ω __p M

ω__s M

|H(e jMω )|

_ω 2π s __ M

2π __ M

ω

4π __ M

Hence, G1 ( z) = H ( z M )F1 ( z ) is a lowpass filter with a unity passband magnitude, passband edge at ω p / M and stopband edge at ωs / M . The magnitude responses of H ( z M ) (solid line) and F2 ( z) (dashed line) are shown below:

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1

|F2(e jω)|

|H(e jMω )|

0

ω__s M

ω __p M

_ω +ωs 2π 2π 2π __ s __ __ M M 2π _ ω M 2π + ω __p __p M M

_ω 4π s __ M

ω

4π __ M

Hence, G2 ( z) = H ( z M )F2 ( z ) is a bandpass filter with a unity passband magnitude, passband edges at (2π − ω p ) / M and (2π + ω p ) / M , and stopband edges at

(2π − ωs ) / M and (2π + ωs ) / M . N

N

n=0

n=0

7.58 H ( z) = ∑ h[ n] z − n and H (e jω ) = ∑ h[ n] e − jωn . The frequency response will exhibit generalized linear phase if it can be expressed in the form H (e jω ) ( ( = H (ω)e − jαωe − jβ , where H (ω), the amplitude function, is a real function of ω and α and β are constants. We need to examine the case when the order N is even and when N is odd separately. Without any loss of generality, assume first 5

N = 5. Then H ( z ) = ∑ h[ n] z − n , and n=0 − jω

H (e jω ) = h[0] + h[1] e

+ h[2] e − j 2ω + h[3] e − j 3ω + h[ 4] e − j 4ω + h[5] e − j 5ω

= e − j 5ω / 2 h[ 0]e j 5ω / 2 + h[5]e − j 5ω / 2 + h[1] e j 3ω / 2 + h[ 4] e − j 3ω / 2 + h[ 2] e jω / 2

(

+ h[3] e − j 3ω / 2 = e − j 5ω / 2 (h[ 0] + h[ 5]) cos( 5ω / 2) + (h[1] + h[ 4]) cos(3ω / 2)

)

[

+ (h[ 2] + h[3]) cos(3ω / 2) + je − j 5ω / 2 (h[ 0] − h[ 5]) sin( 5ω / 2) + (h[1] − h[ 4]) sin(3ω / 2)

] + (h[2] − h[3]) cos(3ω / 2)] .

[

It follows then that if h[ n] = h[ 5 − n], 0 ≤ n ≤ 5, we

( have H (e jω ) = e − j 5ω / 2 H (ω), where ( H (ω) = 2h[0] cos(5ω / 2) + 2h[1] cos(3ω / 2) + 2h[2] cos(ω / 2), which is a real function of and as a result, H (e jω ) has generalized phase. Alternately, if h[ n] = − h[ 5 − n], 0 ≤ n ≤ 5, then we have ( ( H (e jω ) = je − j 5ω / 2 H (ω) = e − j 5ω / 2 e jπ H (ω), where, ( H (ω) = 2h[0] sin(5ω / 2) + 2h[1] sin(3ω / 2) + 2h[2] sin(ω / 2), which is a real function of and as a result, H (e jω ) has generalized phase. 6

Next, assume N = 6. Then H ( z) = ∑ h[ n] z − n , and H (e jω ) = h[0] + h[1] e − jω n=0

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+ h[2] e − j 2 ω + h[3] e − j 3ω + h[ 4] e − j 4ω + h[5] e − j 5ω + h[6] e − j 6ω = e − j 3ω (h[ 0] + h[6]) cos(3ω) + (h[1] + h[5]) cos( 2ω) + (h[ 2] + h[ 4]) cos(ω) + h[3]

[ + je − j 3ω[(h[ 0] − h[6]) sin(3ω) + (h[1] − h[ 5]) sin( 2ω) + (h[ 2] − h[ 4]) sin( ω)].

]

( Hence, it follows that if h[ n] = h[6 − n], 0 ≤ n ≤ 6, then H (e jω ) = e − j 3ω H (ω), ( where H (ω) = 2h[0] cos(3ω) + 2h[1] cos(2ω) + 2h[2] cos(ω) + h[3], which is a real function of and as a result, H (e jω ) has generalized phase. Alternately, if h[ n] = − h[6 − n], 0 ≤ n ≤ 5, then we have ( ( H (e jω ) = je − j 3ω H (ω) = e − j 3ω e jπ H (ω), where, ( H (ω) = 2h[0] sin(3ω) + 2h[1] sin(2ω) + 2h[2] sin(ω), which is a real function of and as a result, H (e jω ) has generalized phase. 7.59 Type 1: Type 2: Type 3: Type 4:

h[ n] = {a h[ n] = {a h[ n] = {a h[ n] = {a

−b −b −b −b

−c −c −c −c

d d d d

e e e e

d − c − b a}. e d − c − b a}. 0 − e − d c b − a}. − e − d c b − a}.

7.60 (a) Type 1: {h[ n]} = {1 − 3 − 4 6 8 6 − 4 − 3 1}. Hence,

H ( z ) = 1 − 3 z − 1 − 4 z − 2 + 6 z −3 + 8 z − 4 + 6 z − 5 − 4 z − 6 − 3 z − 7 + z −8 . The zero plot obtained using the M-file zplane is shown below:

Imaginary Part

2 1 8

0 -1 -2 -1

0

1 2 Real Part

3

It can be seen from the above that a complex-conjugate zero pair on the unit circle appear singly and real zeros appear in mirror-image symmetry. There are no zeros at z = 1. (b) Type 2: {h[ n]} = {1 − 3 − 4 6 8 8 6 − 4 − 3 1}. Hence,

H ( z ) = 1 − 3 z − 1 − 4 z − 2 + 6 z − 3 + 8 z −4 + 8 z − 5 + 6 z − 6 − 4 z − 7 − 3 z − 8 + z − 9 . The zero plot obtained using the M-file zplane is shown below:

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1.5

Imaginary Part

1 0.5 0

3

9

-0.5 -1 -1.5 -1

0

1 Real Part

2

3

From the above zero plot it can be seen that complex-conjugate zero pairs on the unit circle appear singly and real zerosappear in mirror-image symmetry. There are 3 zeros at z = −1. (c) Type 3: {h[ n]} = {1 − 3 − 4 6 8 0 − 8 − 6 4 3 − 1}. Hence,

H ( z) = 1 − 3z −1 − 4 z −2 + 6 z −3 + 8z −4 − 8z −6 − 6 z −7 + 4 z −8 + 3z −9 − z −10 . The zero plot obtained using the M-file zplane is shown below: 1.5 Imaginary Part

1 0.5 10

0 -0.5 -1 -1.5 -1

0

1 2 Real Part

3

From the above zero plot it can be seen that complex-conjugate zero pairs on the unit circle appear singly and real zeros appear in mirror-image symmetry. There is one zero at z = −1. and one zero at z = 1. (d) Type 4: {h[ n]} = {1 − 3 − 4 6 8 − 8 − 6 4 3 − 1}. Hence,

H ( z ) = 1 − 3 z − 1 − 4 z − 2 + 6 z − 3 + 8 z −4 + 8 z − 5 + 6 z − 6 − 4 z − 7 − 3 z − 8 + z − 9 . The zero plot obtained using the M-file zplane is shown below:

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1.5 Imaginary Part

1 0.5 9

0 -0.5 -1 -1.5 -1

0

1 2 Real Part

3

From the above zero plot it can be seen that complex-conjugate zeros appear in mirror-image symmetry, a complex-conjugate zero pair on the unit circle appear singly, and real zeros appear in mirror-image symmetry. There is one zero at z = 1. 7.61 H1 ( z) is of Type 1 and hence, it has a symmetric impulse response of odd length 2 N + 1. Let α be the constant term of H1 ( z). Then, the coefficient of the highest power of z −1 of H1 ( z) is also α. H 2 ( z) is of Type 2 and hence, it has a symmetric impulse response of even length 2M. Let β be the constant term of H 2 ( z). Then, the coefficient of the highest power of z −1 of H 2 ( z) is also β . H 3 ( z) is of Type 3 and hence, it has an anti-symmetric impulse response of odd length 2 R + 1. Let γ be the constant term of H 3 ( z). Then, the coefficient of the highest power of z −1 of H 3 ( z) is − γ. H 4 ( z) is of Type 4 and hence, it has an anti-symmetric impulse response of even length 2K . Let δ be the constant term of H 4 ( z). Then, the coefficient of the highest power of z −1 of H 4 ( z) is − δ. (a) The length of H1 ( z)H1 ( z) is (2 N + 1) + (2 N + 1) − 1 = 4 N + 1 which is odd. The constant term of H1 ( z)H1 ( z) is α 2 and the coefficient of the highest power of z −1 of H1 ( z)H1 ( z) is also α 2 . Hence, H1 ( z)H1 ( z) is of Type 1.

(b) The length of H1 ( z)H 2 ( z) is (2 N + 1) + (2 M ) − 1 = 2( N + M ) which is even. The constant term of H1 ( z)H 2 ( z) is αβ and the coefficient of the highest power of z −1 of H1 ( z)H 2 ( z ) is also αβ . Hence, H1 ( z )H 2 ( z ) is of Type 2.

(c) The length of H1 ( z)H 3 ( z) is (2 N + 1) + (2 R + 1) − 1 = 2( N + R ) + 1 which is odd. The constant term of H1 ( z)H 3 ( z) is αγ and the coefficient of the highest power of z −1 of H1 ( z)H 3 ( z) is − αγ. . Hence, H1 ( z)H 3 ( z) is of Type 3.

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(d) The length of H1 ( z)H 4 ( z) is (2 N + 1) + (2 K ) − 1 = 2( N + K ) which is even. The constant term of H1 ( z)H 4 ( z) is αδ and the coefficient of the highest power of z −1 of H1 ( z)H 4 ( z) is − αδ . Hence, H1 ( z)H 4 ( z) is of Type 4.

(e) The length of H 2 ( z)H 2 ( z) is (2 M ) + (2 M ) − 1 = 4 M − 1 which is odd. The constant term of H 2 ( z)H 2 ( z) is β 2 and the coefficient of the highest power of z −1 of H 2 ( z)H 2 ( z) is also β 2 . Hence, H 2 ( z)H 2 ( z) is of Type 1.

(f) The length of H 3 ( z)H 3 ( z ) is (2 R + 1) + (2 R + 1) − 1 = 4 R + 1 which is odd. The constant term of H 3 ( z)H 3 ( z ) is γ 2 and the coefficient of the highest power of z −1 of H 3 ( z)H 3 ( z ) is also γ 2 . Hence, H 3 ( z)H 3 ( z ) is of Type 1.

(g) The length of H 4 ( z)H 4 ( z) is (2 K ) + (2 K ) − 1 = 4 K − 1 which is odd. The constant term of H 4 ( z)H 4 ( z) is δ 2 and the coefficient of the highest power of z −1 of H 4 ( z)H 4 ( z) is also δ 2 . Hence, H 4 ( z)H 4 ( z) is of Type 1.

(h) The length of H 2 ( z)H 3 ( z ) is (2 M ) + (2 R + 1) − 1 = 2( M + R ) which is even. The constant term of H 2 ( z)H 3 ( z ) is βγ and the coefficient of the highest power of z −1 of H 2 ( z)H 3 ( z ) is − βγ . Hence, H 2 ( z)H 3 ( z ) is of Type 4.

(i) The length of H 3 ( z)H 4 ( z) is (2 R + 1) + (2 K ) − 1 = 2( R + K ) which is even. The constant term of H 3 ( z)H 4 ( z) is γδ and the coefficient of the highest power of z −1 of H 3 ( z)H 4 ( z) is also γδ . Hence, H 3 ( z)H 4 ( z) is of Type 2.

7.62 (a) F 1( z ) = 2.1 − 3.5z −1 + 4.2 z −2 = 2.1(1 − 1.667z −1 + 2 z −2 ). F1( z) has complex conjugate zeros at z = 0.8333 ± j1.1426. To generate a linear-phase transfer function H ( z ) , we need to multiply F1( z) with the factor F 2 ( z) which has complex-conjugate zeros situated in the z –plane with a mirror-image symmetry with respect to the zeros of F1( z) . Hence, F 2 ( z) = 2 − 1.667z −1 + z −2 , resulting in

H ( z ) = F1 ( z )F2 ( z) = 2.1(2 − 5z −1 + 7.7778z −2 − 5z −3 + 2 z −4 ) = 4.2 − 10.5z −1 + 16.3333 z −2 − 10.5z −3 + 4.2 z −4 .

(b) F 1( z) = 1.4 + 5.2 z −1 − 2.2 z −2 + 3.3z −3

= 1.4(1 + 3.7143z −1 − 1.5714 z −2 + 2.3571z −3 ). F1( z) has complex conjugate zeros

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at z = 0.2524 ± j 0.7035 . and a real zero at − 4.2192. To generate a linear-phase transfer function H ( z ) , we need to multiply F1( z) with the factor F 2 ( z) which has situated in the z –plane with a mirror-image symmetry with respect to the zeros of

F1( z) . Hence, F 2 ( z ) = 2.3571 − 1.5714 z −1 + 3.7143z −2 + z −3 , resulting in H ( z) = F1 ( z)F2 ( z) = 1.4(2.3571 + 7.1837z −1 − 5.8265z −2 + 22.8214 z −3 − 5.8265z −4 + 7.1837z −5 + 2.3571z −6 ) = 3.3 + 10.0571z −1 − 8.1671z −2 + 31.95z −3 − 8.1571z −4 + 10.0571z −5 + 3.3z −6 . N

7.63 We rewrite the polynomial H ( z ) in the form f ( z) = Kz − N ∏ ( z − λi ) . Its i =1

'

logarithmic differential is given by

z

f ( z) N N 1 or =− +∑ f ( z) z i =1 z − λi

N N f ' ( z) z 1 . = −N + ∑ = −N + ∑ f ( z) z − λ 1 − ( λ / z ) i i i =1 i =1

For any | z |> max | λi | we can expand the above in a Taylor series as i N

[

]

f ′( z) = − N + ∑ 1 + (λi / z) + (λi / z)2 + (λi / z)3 + ... , or f ( z) i =1 f ′(z) z = − N + N + (S1 / z) + (S2 / z)2 + (S3 / z)3 + ... f ( z) z

where S m =

N

∑ λi

m

(1)

.

i =1

Now set f ( z ) = h[0] + h[1]z −1 + h[2]z −2 + ... + h[ N ]z − N . So that

z

f ′( z) h[1]z −1 + 2h[2]z −2 + ... + Nh[ N ]z − N . =− f ( z) h[0] + h[1]z −1 + h[2]z−2 + ... + h[ N ]z − N

(2)

Equations (1) and (2) pertain to the same quantity hence identically the right hand sides are the same. Thus the following convolution holds true

{h[0],

h[1], h[2], K, h[ N ]}O * {0, S1 , S 2 , S 3 , K} = −{h[1], 2h[ 2], K, Nh[ n]}

Hence the Newton Identities. n1

n2

i =1

i =1

7.64 The root moments of Sm of H ( z) = K ∏ (1 − αi z −1 )∏ (1 − βi z −1 ) are defined as

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Sm =

n1

∑ α im +

i =1

n2

∑ βim .

For real H ( z ), the complex roots occur in conjugate pairs

i =1

and hence, their corresponding powers are also in this form, thereby making their sum entirely real. (a) If H ( z ) is minimum-phase, then βi = 0 for all i, and αi < 1. Therefore as m → ∞, S m will decrease exponentially. n1

n2

i =1

i =1

(b) Write ln H ( z ) = ln K + ∑ ln(1 − αi z −1 ) + ∑ ln(1 − βi z −1 ) and then expand ln H ( z ) in a Laurent series. The second summation can be re-expressed as n2

∑ ln(1 − βi z

−1

)=

i =1

ln H ( z) = ln K +

n2

⎛

i =1

⎝

∑ ln( − βi z −1 )⎜⎜ 1 −

z βi

⎞ ⎟⎟. Hence, ⎠

n2

n1

n2

i =1

i =1

i =1

z ∑ ln(− βi ) − n 2 ln z + ∑ (1 − αi z −1 ) + ∑ (1 − β ). Now, βi appear i

n2

in complex conjugate pairs, hence, ln K + ∑ ln(− βi ) can be written as ln K1 where i =1

K1 is real. ⎡ ⎤ (α z −1 )2 (αi z −1 )3 + + L⎥ and However, ln(1 − αi z −1 ) = − ⎢αi z −1 + i 2 3 ⎢⎣ ⎥⎦ ⎡ z ( z / β ) 2 ( z / β )3 ⎤ ⎛ z ⎞ i i ⎜ ⎟ + + L⎥. These two Taylor series expansions are ln⎜ 1 − ⎟ = − ⎢ + βi ⎠ 2 3 ⎝ ⎣⎢ βi ⎦⎥ valid since αi z −1 < 1 and z / βi < 1 on z = 1. Thus,

ln H ( z ) = ln K1 −

N2 n1 n1 ∞ ⎛ S N1 ⎞ N2 m m z − m + S − m z m ⎟ where S N1 = ⎜ α and S = βi− m . ∑⎜ ∑ ∑ m i −m ⎟ m m =1 ⎝ m i =1 i =1 ⎠

( On z = 1 we have H ( z ) z = e jω = H (ω) e jθ (ω) and therefore,

⎛ ∞ S N1 − jmω ∞ S N 2 jmω ⎞ ( ⎟ + (− jn ω). ln H (ω) + jθ (ω) = ln K1 − ⎜ ∑ m e + ∑ −m e 2 ⎜ m =1 m ⎟ m m =1 ⎝ ⎠ On equating real and imaginary parts of the equation we arrive at ∞ ⎛ S N1 + S N 2 ⎞ ( − m ⎟ cos(mω), ln H (ω) = ln K1 − ∑ ⎜ m ⎜ ⎟ m m =1 ⎝ ⎠ ∞ ⎛ S N1 − S N 2 ⎞ − m ⎟ sin(mω). θ (ω) = − n2 ω + ∑ ⎜ m ⎜ ⎟ m m =1 ⎝ ⎠

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(c) From the expression for the phase as given above, it can be seen that the second term contributes non-linear components to the phase, whilst the only term which is linear is the first (− n2 ω). Thus to have linear phase we must have (1) n2 ≠ 0 and (2) N1 Sm = S −Nm2 . The second condition means that the zeros of H ( z ) outside the unit circle must be the same in number as those inside the unit circle and the zeros outside must be at locations of the zeros inside the unit circle.

d + e − jω d1e jω / 2 + e − jω / 2 d + z −1 7.65 (a) A1 ( z ) = 1 . Thus, A1 (e jω ) = 1 = 1 + d1e − jω e jω / 2 + d1e − jω / 2 1 + d1z −1 α e jβ = = e j 2 β , where αe jβ = d1e jω / 2 + e − jω / 2 ) − jβ

αe

= (d1 + 1) cos(ω / 2) + j(d1 − 1) sin(ω / 2). Therefore, phase is given by ⎛ 1 − d1 ⎞ θ (ω) = 2 β = −2 tan −1 ⎜⎜ tan(ω / 2) ⎟⎟. Now, for small values of x, tan( x ) ≅ x and ⎝ 1 + d1 ⎠ tan −1 ( x ) ≅ x. Hence, the approximate expression for the phase at low frequencies

⎛ 1 − d1 ⎞ ω ⎛ 1 − d1 ⎞ ⎟⎟ = −⎜⎜ ⎟⎟ ω. Therefore, the approximate is given by θ (ω) ≅ −2⎜⎜ ⎝ 1 + d1 ⎠ 2 ⎝ 1 + d1 ⎠ θ ( ω) 1 − d1 expression for the phase delay is given by τ p (ω) = − samples. =δ≅ ω 1 + d1 (b) For δ = 0.5 samples, d1 =

1− δ 1+ δ

=

0 .5 1.5

=

1 . 3

Then, A1 ( z) =

1 3

+ z −1

1 + z −1 1 3

. Thus, the

2 1− d 2 θ ( ω) = ⎛⎜ 1 tan(ω / 2) ⎞⎟ = (0.5 tan(ω / 2)). exact phase delay is given by τ p (ω) = − ω ω ⎝ 1+ d1 ω ⎠ For a sampling rate of 20 kHz, the normalized angular frequency equivalent to 1

kHz is ω o =

τ p (ω o ) =

10 3 20×10 3

2 ωo

=

1 20

= 0.05. The exact phase delay at ωo is thus

(0.5 tan(ωo / 2) =

2 (0.5 tan(0.025) = 0.500078 samples, which is 0.05

seen to be very close to the desired phase delay of 0.5 samples. Phase delay, in samples

Phase delay, in samples

1 0.9 0.8 0.7 0.6 0.5

0

0.2

0.4

0.6

0.8

1

0.55 0.5 0.45 0.4

ω/π

0

0.05

0.1

0.15

0.2

ω/ π

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d2 + d1e − jω + e − j 2 ω d2 e jω + d1 + e − jω 7.66 A2 (e ) = = 1 + d1e − jω + d2 e − j 2ω e jω + d1 + d2 e − jω (d + d2 cos ω + cos ω) + j (d2 sin ω + sin ω) (d1 + (d2 + 1) cos ω) + j (d2 − 1) sin ω = . = 1 (d1 + d2 cos ω + cos ω) + j (sin ω − d2 sin ω) (d1 + (d2 + 1) cos ω) − j (d2 − 1) sin ω ( d 2 −1) sin ω ⎞ Therefore, θ(ω) = 2 tan −1 ⎛⎜ ⎟. Now, ⎝ d1 + ( d 2 +1) cos ω ⎠ θ(ω) 2 ⎛ ( d 2 −1) sin ω ⎞ . For ω ≅ 0, sin ω = ω and τ p (ω) = − = − tan −1 ⎜ ⎟ ω ω ⎝ d1 + ( d 2 +1) cos ω ⎠ ( d −1)ω ⎞ 2 −1 cos ω = 1. Then, τ p (ω) = − tan −1 ⎛⎜ 2 ⎟. Also, for x ≅ 0, tan x ≅ x. ω d + ( d + 1 ) 2 ⎝ 1 ⎠ 2( d 2 −1) 2 ( d 2 −1)ω 2−δ ⎞ =− . Now, substituting d1 = 2⎛⎜ Hence, τ p (ω) = − ⎟ and ω d1 + ( d 2 +1) d1 + d 2 +1 ⎝ 1+ δ ⎠ jω

d2 =

2( d 2 −1) (2 − δ)(1− δ) , we can easily show that − = δ. (2 + δ)(1+ δ) d1 + d 2 +1

7.67 Since G ( z ) is non-minimum phase but causal, it will have some zeros outside the unit circle. Let z = α be one such zero. We can then write G( z) = P( z)(1 − αz −1 ) −1 ⎞ ⎛ 1− αz −1 ⎞ ⎛ ⎟. Note that ⎜ 1− αz ⎟ = P( z)(−α * + z −1 )⎜⎜ ⎜ ( −α * + z −1 ) ⎟ is a stable first-order −1 ⎟ ⎠ ⎝ ( −α * + z ) ⎠ ⎝ allpass function. If we carry out this operation for all zeros of G ( z ) that are outside the unit circle, we can write G ( z ) = H ( z ) A( z ) where H ( z ) will have all zeros inside the unit circle and will thus be a minimum phase function and will be a product of stable first-order allpass functions, and hence an allpass function.

(3z − 2.1)( z 2 + 2.5z + 5) . In order to correct for magnitude distortion we ( z − 0.65)( z + 0.48) require the transfer function G ( z ) to satisfy the following property 1 G(e jω ) = . Hence, one possible solution is jω H (e )

7.68 H ( z ) =

Gd ( z ) =

1 ( z − 0.65)( z + 0.48) . Note that the coefficients of the pole = H ( z) (3z − 2.1)(z 2 + 2.5z + 5)

factor ( z 2 + 2.5z + 5) in the denominator of Gd (z ) do not satisfy the condition of Eq. (7.139) and hence, has roots outside the unit circle making Gd (z ) unstable. To develop a stable transfer function with magnitude response same as Gd (z ) , we multiply it with the stable allpass function

z 2 + 2.5z + 5

5z 2 + 2.5z + 1

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resulting in the transfer

209

( z − 0.65)( z + 0.48)

function G( z) =

(3z − 2.1)(5z 2 + 2.5z + 1)

which is the desired stable solution

satisfying the condition G (e jω ) H (e jω ) = 1. 7.69 (a) G ( z ) = H ( z ) A( z ), where A( z ) is an allpass function. Then, g[ 0] = lim G ( z ). z→∞

Hence, g[0] = | lim G( z) | = | lim H ( z ) A( z) | = | lim H ( z) | | lim A( z) | ≤ | lim H ( z) | z →∞

z →∞

z →∞

z →∞

z →∞

≤ | lim H ( z ) | because | lim A( z ) | < 1 because of Property 2 of stable allpass z→∞

z→∞

function (see Eq. (7.20)). Hence, g[0] ≤ h[0] . (b) If λ1 is a zero of H ( z ), then λ1 < 1, since H ( z ), is a minimum-phase causal stable transfer function. As H ( z ) has all zeros inside the unit circle, we can write

H ( z ) = B( z)(1 − λ1z −1 ). It follows that B( z ) is also a minimum-phase causal transfer function. ⎛ λ* − z −1 ⎞ ⎟. If Now consider the transfer function F ( z ) = B( z)( λ*1 − z −1 ) = H ( z)⎜ 1 ⎜ 1 − λ z −1 ⎟ 1 ⎝ ⎠ h[ n], b[ n], and f [ n] denote, respectively, the inverse z –transforms of H ( z ), B( z ),

b[0], n = 0, ⎧ and and F ( z ), then we get h[ n] = ⎨ ⎩b[ n] − λ1b[ n − 1], n ≥ 1, ⎧⎪ λ*1b[0], n = 0, f [ n] = ⎨ * ⎪⎩ λ1b[ n] − b[ n − 1], n ≥ 1. m

2

m

2

2

2

m

2

2

m

2

Consider ε = ∑ h[ n] − ∑ f [ n] = b[0] − λ*1 b[0] + ∑ h[ n] − ∑ f [ n] . n=0

2

n=0 2

2

n =1

n =1

2

Now, h[ n] = b[ n] + λ1 b[ n − 1] − λ1b[ n − 1]b * [ n] − λ*1b * [ n − 1]b[ n], and 2

2

2

2

f [ n] = λ1 b[ n] + b[ n − 1] − λ1b[ n − 1]b * [ n] − λ*1b * [ n − 1]b[ n]. Hence, m m 2 2 2 2 2 2 2 2 2 ε = b[0] − λ1 b[0] + ∑ ⎛⎜ b[ n] + λ1 b[ n − 1] ⎞⎟ − ∑ ⎛⎜ λ1 b[ n] − b[ n − 1] ⎞⎟ ⎝ ⎠ ⎝ ⎠ n =1 n =1 2

m

2

2

m

2

= (1 − λ1 ) b[ m] . Since λ1 < 1, ε > 0, i.e., ∑ h[ n] > ∑ f [ n] . Hence, n=0

m

2

m

n=0

2

∑ h[ n] ≥ ∑ g[ n] .

n=0

n=0

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(2 z + 3)(4 z − 1) 3 has a zero at z = which is outside the unit circle and is 2 ( z + 0.4)( z − 0.6) thus a non-minimum phase transfer function. To develop a minimum phase transfer

7.70 H ( z) =

function G ( z ) such that G (e jω ) = H (e jω ) , we multiply with an allpass function

3z + 2 (3z + 2)(4 z − 1) and arrive at G( z) = which is minimum phase. ( z + 0.4)( z − 0.6) 2z + 3 The first 5 impulse response of H ( z ) are {h[ n]} = {8 11.6 1.24 3.032 0.904}, 0 ≤ n ≤ 4, and The first 5 impulse response of G ( z ) are {g[ n]} = {12 7.4 2.36 2.248 1.016}, 0 ≤ n ≤ 4. m

0

m

Σ h[ n]

n=0 m

Σ g[ n]

2 2

n= 0

64

1

2

3

4

198.56 200.0976 209.2906 210.1078

144 198.76 204.3296 209.3831 210.4154 m

m

2

2

It follows from the above that ∑ g[ n] > ∑ h[ n] n=0

n=0

for m ≥ 1.

7.71 See Example 7.14. (a) H BS ( z) = (1 + z − 2 )2 . Thus, H BP ( z ) = z − 2 − (1 + z − 2 )2 = − (1 − z − 2 )2 . 4 4 4 1

1

1

(b) H BS ( z) = (1 + z − 2 )(−1 + 6 z − 2 − z − 4 ). Thus, 16 1

H BP ( z ) = z − 4 − =

1 (1 + z − 2 )(−1 + 6 z − 2 16

− z−4 ) =

1 (1 − 4 z − 2 16

+ 6 z − 4 − 4 z − 6 + z −8 )

1 −2 4 (1 − z ) . 16

(c) H BS ( z ) = (1 + z − 2 )2 (−3 + 14 z − 2 − 3z − 4 ). Thus, 32 1

H BP ( z) = z − 4 −

1 (1 + z − 2 )2 (−3 + 14 z − 2 32

− 3z − 4 ) =

1 (3 − 8z − 2 32

+ 10 z − 4 − 8 z − 6 + 3z − 8 ).

7.72 H 0 ( z ) = A0 ( z ) + A1 ( z), and H1 ( z) = A0 ( z) − A1 ( z), where A0 ( z) and A1 ( z) are allpass functions of orders M and N , respectively. Hence, the orders of H 0 ( z)

z − M D0 ( z −1 ) and and H1 ( z) are M + N . Now, we can write A0 ( z) = D0 ( z)

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A1 ( z) =

z − N D1 ( z −1 ) . Then, D1 ( z )

H0 (z) =

P( z) z − M D0 ( z −1 ) D1 ( z ) + z − N D0 ( z) D1 ( z −1 ) = and D( z ) D0 ( z) D1 ( z )

H1 ( z) =

Q( z) z − M D0 ( z −1 ) D1 ( z) − z − N D0 ( z) D1 ( z −1 ) . Since P (z ) is of degree = D( z ) D0 ( z ) D1 ( z )

[

M + N and z − ( M + N ) P ( z −1 ) = z − ( M + N ) z M D0 ( z ) D1 ( z −1 ) + z N D0 ( z −1 ) D1 ( z )

]

= z − M D0 ( z −1 ) D1 ( z ) + z − N D0 ( z ) D1 ( z −1 ) = P ( z ). Hence, P (z ) is symmetric. Similarly, one can show that Q(z ) is anti-symmetric. 1

1

7.73 H 0 ( z) = [ A0 ( z) + A1 ( z )] and H1 ( z) = [ A0 ( z) − A1 ( z)]. Thus, 2 2 H 0 ( z ) H 0 ( z −1 ) + H1 ( z ) H1 ( z −1 ) = [ A0 ( z) + A1 ( z)][ A0 ( z −1 ) + A1 ( z −1 )] 1 4 −1

+ [ A0 ( z ) − A1 ( z )][ A0 ( z −1 ) − A1 ( z )] 4 1

= [ A0 ( z ) A0 ( z −1 ) + A0 ( z) A1 ( z −1 ) + A1 ( z) A0 ( z −1 ) + A1 ( z) A1 ( z −1 )] 4 1

+ [ A0 ( z) A0 ( z −1 ) − A0 ( z) A1 ( z −1 ) − A1 ( z ) A0 ( z −1 ) + A1 ( z) A1 ( z −1 )] 4 1

2

2

= [ A0 ( z) A0 ( z −1 ) + A1 ( z ) A1 ( z −1 )] = 1. Thus, H 0 (e jω ) + H1 (e jω ) = 1 1 2

implying that H 0 ( z) and H1( z) form a power-complementary pair. 2

7.74 H 0 (e jω ) =

{A0 (e jω ) A0* (e jω ) + A1(e jω ) A0* (e jω ) + A0 (e jω ) A1* (e jω ) + A1 (e jω ) A1* (e jω )}. Since A0 ( z) and A1 ( z) are allpass functions, 1 4

A0 (e jω ) = e jφ0 (ω) and A1 (e jω ) = e jφ1 (ω) . Therefore, 2

H 0 ( e jω ) =

{

}

1 2 + e j (φ0 (ω) − φ1 (ω)) + + e − j (φ0 (ω) − φ1 (ω)) ≤ 1 as maximum values of 4

and e − j (φ0 (ω) − φ1 (ω)) are 1. H 0 ( z) is stable since A0 ( z) and A1 ( z) are stable allpass functions. Hence, H 0 ( z) is BR. e

j ( φ0 ( ω)− φ1( ω))

1 M −1 1 M −1M −1 −1 −1 ∑ Ak ( z). Thus, H ( z)H ( z ) = 2 ∑ ∑ Ak ( z) Ar ( z ). Hence, M k =0 M k =0 r =0 2 M − 1 M − 1 1 j (φ (ω) − φr (ω)) H ( e jω ) = ≤ 1. Also, H (z ) is stable since ∑ ∑ e k 2 M r =0 k =0 Ai ( z ), 0 ≤ i ≤ M − 1, are stable allpass functions. Hence, H (z ) is BR.

7.75 H ( z) =

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1

1

7.76 H BP ( z) = [1 − A( z)] and H BS ( z) = [1 + A( z)] where 2 2 A( z ) =

α − β (1 + α )z −1 + z −2

is an allpass function. Note 1 − β (1 + α )z −1 + αz − 2 H BP ( z ) + H BS ( z ) = A( z ) and from the solution of Problem 7.73, 2

2

H BP (e jω ) + H BS (e jω ) = 1. Hence, H BP (z) and H BS (z) are doublycomplementary pair. 7.77 H ( z )H ( z −1 ) + H (− z)H (− z −1 ) = K . On the unit circle, this reduces to

H (e jω )H (e − jω ) + H (−e jω )H (−e − jω ) = K , or, equivalently, 2

2

H (e jω ) + H (−e jω ) = K , as H (z ) is a real-coefficient transfer function. Now, 2

2

H (−e jω ) = H (e j (π + ω) ) . Hence, for ω = π / 2, the power-symmetric condition 2

2

reduces to H (e jπ / 2 ) + H (e j (π + π / 2) ) = K . Since H (z ) is a real-coefficient transfer function, H (e jω ) 2

2

is an even function of ω, and thus, 2

2

H (e jπ / 2 ) = H (e j (2π − π / 2) ) . As a result, 2 H (e jπ / 2 ) = K , from which we obtain 10 log10 2 + 20 log10 H (e jπ / 2 ) = 10 log10 K , or, 20 log10 H (e jπ / 2 ) = 10 log10 K − 3 dB.

7.78 H ( z ) = A0 ( z 2 ) + z −1 A1 ( z 2 ). Therefore, H ( z )H ( z −1 ) + H ( − z ) H ( − z −1 ) = [ A0 ( z 2 ) + z −1 A1 ( z 2 )][ A0 ( z −2 ) + zA1 ( z −2 )]

+ [ A0 (( − z )2 ) − z −1 A1(( − z )2 )] [ A0 (( − z )2 ) − z A1(( − z )2 )] = A0 ( z 2 ) A0 ( z −2 ) + z −1 A1 ( z 2 ) A0 ( z −2 ) + z A0 ( z 2 ) A1 ( z −2 ) + A1 ( z 2 ) A1 ( z −2 )

+ A0 ( z 2 ) A0 ( z −2 ) − z −1 A1 ( z 2 ) A0 ( z −2 ) − z A0 ( z 2 ) A1 ( z −2 ) + A1 ( z 2 ) A1 ( z −2 ) = 4, as A0 ( z 2 ) A0 ( z −2 ) = A1 ( z 2 ) A1 ( z −2 ) = 1.

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7.79 H ( z) =

− 0.1 + 0.5z −1 + 0.05z −2 + 0.05z −3 + 0.5z −4 − 0.1z −5

where A( z) =

1 + 0.1z

−2

− 0.2 + 0.1z −1 + z −2 1 + 0.1z −1 − 0.2 z − 2

− 0.2 z

−4

= [ A( z 2 ) + z −1 ], 1 2

is a stable allpass function. Thus,

H ( z )H ( z −1 ) + H (− z )H (− z −1 ) = [ A( z 2 ) + z −1 ][ A( z − 2 ) + z ] 1 4

+ [ A( z 2 ) − z −1 ][ A( z − 2 ) − z] = 1. 1 4

7.80 (a) H a ( z ) = 1 − 2 z −1 + 4.5z −2 + 6 z −3 + z −4 + 0.5z −5 . Thus, H a ( z )H a ( z −1 )

= (1 − 2 z −1 + 4.5z −2 + 6 z −3 + z −4 + 0.5z −5 )(1 − 2 z + 4.5z 2 + 6 z 3 + z 4 + 0.5z 5 ) = 0.5z 5 + 6.25z 3 + 22.5z + 62.5 + 22.5z −1 + 6.25 z −3 + 0.5z −5 .

Next, we compute H a ( − z ) H a (− z −1 ) =

= (1 + 2 z −1 + 4.5z −2 − 6 z −3 + z −4 − 0.5z −5 )(1 + 2 z + 4.5z 2 − 6 z 3 + z 4 − 0.5z 5 ) = −0.5z 5 − 6.25 z 3 − 22.5z + 62.5 − 22.5z −1 − 6.25 z −3 − 0.5z −5 .

Hence, H a ( z ) H a ( z −1 ) + H a ( − z ) H a ( − z −1 ) = 125. (b) H b ( z) = 1 + z −1 + z − 2 − z − 4 + 2 z − 5 . Thus, H a ( z )H a ( z −1 ) 4 2 1 15 15 1 H b ( z)H b ( z −1 ) = ⎛⎜1 + z −1 + z − 2 − z − 4 + 2 z − 5 ⎞⎟⎛⎜1 + z + z 2 − z 4 + 2 z 5 ⎞⎟ 2 4 4 2 ⎝ ⎠⎝ ⎠ 1

15

= 2 z 5 + 7 z 3 + 0.375 z + 20.3125 + 0.375 z −1 + 7 z −3 + 2 z −5 .

Next, we compute H b (− z )H b ( − z −1 ) =

= (1 − 0.5z −1 + 3.75z −2 − z −4 − 2 z −5 )(1 − 0.5z + 3.75z 2 − z 4 − 2 z 5 ) = −2 z 5 − 7 z 3 − 0.375 z + 20.3125 − 0.375 z −1 − 7 z −3 − 2 z −5 .

Hence, H a ( z )H a ( z −1 ) + H a ( − z ) H a (− z −1 ) = 40.625. 7.81 H ( z )H ( z −1 ) = a 2 (1 + bz −1 )(1 + bz) = a 2 bz + a 2 (1 + b 2 ) + a 2 bz −1 = cz + d + cz −1 . Thus, c = a 2 b and d = a 2 (1 + b 2 ). Now, H ( z)H ( z −1 ) + H (− z)H (− z −1 ) cz + d + cz −1 − cz + d − cz −1 = 2 d. Therefore, 2 d = 2 a 2 (1 + b 2 ) = 1. This 1

condition is satisfied by a = 1 2

2(1+ b 2 )

b = −1, a = , and b = 10, a =

1 10

1 2

. For b = 1, a = . Other solutions include

.

Since H (z ) is a first-order causal FIR transfer function, G( z) = − z −1 H (− z −1 ) is also a first-order causal FIR transfer function. Now,

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H ( z )H ( z −1 ) + G( z)G( z −1 ) = H ( z )H ( z −1 ) + [ − z −1 H (− z −1 )][− zH (− z )] = H ( z )H ( z −1 ) + H (− z )H (− z −1 ) = 1. Hence, H (z ) and G (z ) are power complementary. 7.82 H ( z)H ( z −1 ) = (cz + d + cz −1 )[ d 2 z 2 + d1 (1 + d 2 )z + (1 + d12 + d 22 ) + d1 (1 + d 2 )z −1 + d 2 z −2 ] Thus, H ( z)H ( z −1 ) + H (− z)H (− z −1 ) = 2[cd1 (1 + d 2 )z 2 + dd 2 z 2 + 2cd1 (1 + d 2 )

+ d (1 + d12 + d 22 ) + dd 2 z −2 + cd1 (1 + d 2 )z −2 ] = 1. Hence, we require dd2 + cd1 (1 + d 2 ) = 0 and 2cd1 (1 + d 2 ) + + d (1 + d12 + d 22 ) = 1. Solving these two d2 equations we arrive at c = , and d 2 (1 + d 2 )(2d 2 − 1 − d12 − d 22 ) 1 1 d=− . For d1 = d2 = 1, we get c = − and d = 1. 2 2 d 2 − 1 − d12 − d 22 Since H (z ) is a third-order causal FIR transfer function, G( z) = − z −3 H (− z −1 ) is also a third-order causal FIR transfer function. Now,

H ( z )H ( z −1 ) + G( z )G( z −1 ) = H ( z )H ( z −1 ) + [ − z −3 H (− z −1 )][ − z 3 H (− z)] = H ( z )H ( z −1 ) + H (− z )H (− z −1 ) = 1. Hence, H (z ) and G (z ) are power complementary. 1

1

7.83 H 0 ( z) = [ A 0 ( z) + A1 ( z )] and H1 ( z) = [ A 0 ( z ) − A1 ( z)], where A0 ( z ) and A1 ( z) 2 2 are stable allpass transfer functions. From these two equations we obtain H 0 ( z) + H1 ( z) = A0 ( z ), and H 0 ( z) − H1 ( z ) = A1 ( z). Moreover, we have 2

2

H 0 (e jω ) + H1 (e jω ) = 1. Choose G 0 ( z ) = H 02 ( z ) and G1 ( z) = − H12 ( z). 2

2

Hence, G 0 (e jω ) + G1 (e jω ) = H 0 (e jω ) + H 1(e jω ) = 1 . 7.84 (a) H1 ( z) = (1 + 3z −1 ). Therefore, H1 ( z)H1 ( z −1 ) = (1 + 3z −1 )(1 + 3z) ⇒ 4 16 1

2

H 1 ( e jω ) =

1

2⎞ d ⎛ 10 + 6 cos ω 3 ⎜⎜ H1 (e jω ) ⎟⎟ = − sin ω < 0 for 0 ≤ ω < π. . Thus, 8 dω ⎝ 16 ⎠

Thus, H1 (e jω ) is a montonically decreasing function of ω. The maximum value of H1 (e jω ) = 1 is at ω = 0, and the minimum value is at ω = π. Hence, H1 ( z) is BR.

(1 − 1.2 z −1 ). Therefore, H 2 ( z)H 2 ( z −1 ) = (1 − 1.2 z −1 )(1 − 1.2 z) ⇒ (b) H 2 ( z) = 2.2 4.84 1

1

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2

H 2 ( e jω ) =

2⎞ d ⎛ 2.44 − 2.4 cos ω 2.4 ⎜⎜ H 2 (e jω ) ⎟⎟ = sin ω > 0 for . Thus, dω ⎝ 4.84 ⎠ 4.84

0 ≤ ω < π. Thus, H 2 (e jω ) is a montonically increasing function of ω. The maximum value of H 2 (e jω ) = 1 is at ω = π, and the minimum value is at ω = 0. Hence, H 2 ( z) is BR.

1 + α z −1 (1 + αz −1 )(1 − β z −1 ) G z ( ) = (c) H 3 ( z ) = and = G1 ( z )G 2 ( z ), where 1 1+ α (1 + α)(1 + β) G2 ( z) =

2 1 + α 2 + 2α cos ω 1 − β z −1 . Now, G1 (e jω ) = . Thus, 1+ β (1 + α) 2

2⎞ d ⎛ 2α sin ω ⎜⎜ G1 (e jω ) ⎟⎟ = − < 0 for 0 ≤ ω < π as α > 0. As a result, G1 (e jω ) is 2 dω ⎝ (1 + α) ⎠

a monotonically decreasing decreasing function of ω. The maximum value of G1 (e jω ) = 1 is at ω = 0, and the minimum value is at ω = π. Hence, G1 ( z) is BR. 2

Likewise, G 2 (e jω ) =

1 + β 2 − 2β cos ω (1 + β) 2

. Thus,

2 ⎞ 2β sin ω d ⎛ ⎜⎜ G2 (e jω ) ⎟⎟ = >0 dω ⎝ ⎠ (1 + β) 2

for 0 ≤ ω < π as β > 0. Thus, H 2 (e jω ) is a montonically increasing function of

ω. The maximum value of G 2 (e jω ) = 1 is at ω = π, and the minimum value is at ω = 0. Hence, G2 ( z) is BR. Therefore, H 3 ( z ) = G3 ( z )G2 ( z) is also BR. (1 − 0.3z −1 )(1 + 0.2 z −1 )(1 − 0.5z −1 ) ⎛⎜ 1 − 0.3z −1 ⎞⎟⎛⎜ 1 + 0.2 z −1 ⎞⎟⎛⎜ 1 − 0.5z −1 ⎞⎟ = . ⎜ ⎟⎜ ⎟⎜ ⎟ 2.34 1 . 3 1 . 2 1 . 5 ⎝ ⎠⎝ ⎠⎝ ⎠ Since each individual factor on the right-hand side is BR, H 4 ( z) is BR. (d) H 4 ( z ) =

7.85 (a) H1 ( z) =

2.6 + 2.6 z −1 4.2 + z −1

A0 ( z) = 1 and A1 ( z ) =

=

1 ⎛⎜ 1 + 4.2 z −1 ⎞⎟ 1 1+ = ( A0 ( z ) + A1 ( z )), where 2 ⎜⎝ 4.2 + z −1 ⎟⎠ 2

1 + 4.2 z −1

are stable allpass transfer functions. Therefore, 4.2 + z −1 H1 ( z) is BR (See solution of Problem 7.74).

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(b) H 2 ( z) =

1.6 − 1.6 z −1

1 ⎛⎜ 1 + 4.2 z −1 ⎞⎟ 1 = 1− = ( A ( z) − A1 ( z)), where −1 −1 ⎟ 2 0 ⎜ 2 4.2 + z 4.2 + z ⎠ ⎝

A0 ( z) = 1 and A1 ( z ) =

1 + 4.2 z −1

are stable allpass transfer functions. Therefore, 4.2 + z −1 H 2 ( z) is BR (See solution of Problem 7.74). (c) H 3 ( z ) =

0.1(1 − z − 2 ) 1 + 0.4 z −1 + 0.8z − 2

1 ⎛⎜ 0.8 + 0.4 z −1 + z − 2 1− 2 ⎜⎝ 1 + 0.4 z −1 + 0.8z − 2

=

⎞ ⎟ ⎟ ⎠

0.8 + 0.4 z −1 + z −2 1 are stable = ( A0 ( z ) − A1 ( z )) , where A0 ( z) = 1 and A1 ( z) = 2 1 + 0.4 z −1 + 0.8z − 2 allpass transfer functions. Therefore, H 3 ( z) is BR (See solution of Problem 7.74). (d) H 4 ( z ) =

4.5 + 2 z −1 + 4.5z − 2 5 + 2 z −1 + 4 z − 2

where A0 ( z) = 1 and A1 ( z ) =

=

1 ⎛⎜ 4 + 2 z −1 + 5z − 2 1+ 2 ⎜⎝ 5 + 2 z −1 + 4 z − 2

⎞ 1 ⎟ = ( A ( z) + A ( z)) , 1 ⎟ 2 0 ⎠

4 + 2 z −1 + 5z −2

are stable allpass transfer functions. 5 + 2 z −1 + 4 z − 2 Therefore, H 4 ( z) is BR (See solution of Problem 7.74).

7.86 Since A1 ( z) and A2 ( z) are LBR, A1 (e jω ) = 1 and A2 (e jω ) = 1. Thus,

(

)

⎛ ⎞ 1 ⎟ = A e − jφ 2 (ω) . A1 (e jω ) = e jφ1 (ω) and A2 (e jω ) = e jφ 2 (ω) . Now, A1 ⎜ 1 ⎜ A ( e jω ) ⎟ ⎝ 2 ⎠ ⎛ 1 ⎞ ⎟⎟ is LBR. Thus, A1 e − jφ 2 (ω) = 1. Thus, A1 ⎜⎜ ⎝ A2 ( z) ⎠

(

)

⎛ G ( e jω ) + α ⎞ ⎛ jφ(ω) + α ⎞ ⎛ G( z) + α ⎞ ⎟ = e jω ⎜ e ⎟ 7.87 F ( z) = z⎜⎜ ⎟⎟. Thus, F (e jω ) = e jω ⎜ j ω j φ ( ω ) ⎜ ⎟ ⎜ ⎟ ⎝ 1 + αG ( z ) ⎠ ⎝ 1 + αG ( e ) ⎠ ⎝ 1 + αe ⎠ since G (z ) is LBR. Therefore, F (e

=

(cos(φ(ω) + α )2 + (sin(φ(ω))2 (1 + α cos(φ(ω))2 + (α sin(φ(ω))2

of F (z ). Then G( z ) z = λ =

=

jω

)

2

=

e jφ(ω) + α

2

1 + αe jφ(ω)

1 + 2α cos(φ(ω) + α 2 1 + 2α cos(φ(ω) + α 2

= 1. Let z = λ be a pole

F ( z ) − αz 1 = − , or, G(λ) = 1 / α . If α < 1, then α z − αF ( z ) z = λ

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G(λ) > 1, which is satisfied by the LBR G (z ) if λ < 1. Hence, F (z ) is LBR. The order of F (z ) is same as that of G (z ). G (z ) can be realized in the form of a two-pair constrained by the transfer function F (z ) as shown below: Y2

X1 G(z)

F(z) Y1

X2

To this end, we express G (z ) in terms of F (z ) arriving at G( z) =

− α + z −1 F ( z )

=

C + D F (z) , where A, B, C , and D are the chain A + B F (z)

1 − α z −1 F ( z ) parameters of the two-pair. Comparing the above two expressions we get A = 1,

B = −α z −1 , C = −α, and D = z −1 . The corresponding transfer parameters are given by t11 = −α, t 21 = 1, t12 = (1 − α 2 )z −1 and t 22 = α z −1. 1 ⎞ . Now, A(z ) being LBR, A(e jω ) = e jφ(ω) . Thus, 7.88 Let F ( z) = G⎛⎜ ⎟ ⎝ A( z ) ⎠ ⎛ 1 ⎞ ⎟ = G(e − jφ(ω) ). Since G (z ) is a BR function, G (e jω ) ≤ 1. F (e jω ) = G⎜⎜ ⎟ j ω ⎝ A(e ) ⎠ ⎛ 1 ⎞ ⎟ ≤ 1. Hence, F (e jω ) = G ⎜⎜ jω ⎟ ⎝ A(e ) ⎠ Let z = ξ be a pole of F (z ) . Hence, F (z ) will be a BR function if ξ < 1. Let

z = λ be a pole of G (z ). Then this pole is mapped to the location z = ξ of F (z ) by 1 1 1 > 1 because of Eq. = λ, or A(ξ) = . Hence, A(ξ) = the relation A( z ) z = ξ λ λ (7.20). This implies, λ < 1. Thus, G (z ) is a BR function. 7.89 (a) H ( z) =

2.6(1 + z −1 ) 4.2 + z −1

H ( z) + G( z) =

, G( z ) =

2.6(1 + z −1 ) 4.2 + z

−1

+

H ( z)H ( z −1 ) + G( z )G( z −1 ) = =

1.6(1 − z −1 ) 4.2 + z −1

1.6(1 − z −1 ) 4.2 + z

−1

=

. Now,

4.2 + z −1 4.2 + z

−1

= 1. Next,

) 2.6(1 + z) 1.6(1 − z −1 ) 1.6(1 − z) ⋅ + ⋅ 4.2 + z 4.2 + z 4.2 + z −1 4.2 + z −1

2.6(1 + z

−1

(6.76 z + 13.52 + 6.76 z −1 ) + (−2.56 z + 5.12 − 2.56 z −1 ) 4.2 z −1 + 18.64 + 4.2 z −1

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=

4.2 z −1 + 18.64 + 4.2 z −1 4.2 z −1 + 18.64 + 4.2 z −1

218

= 1.

2

2

Thus, H (e jω ) + G(e jω ) = 1. Hence, H (z ) and G (z ) are both allpasscomplementary and power complementary. As a result, they are doubly complementary. (b) H ( z ) =

0.1(1 − z −2 ) 1 + 0 .4 z − 1 + 0 .8 z − 2

H ( z ) + G( z ) =

, G( z) =

0.1(1 − z −2 ) 1 + 0.4 z −1 + 0.8 z − 2

+

0 .9 + 0 .4 z −1 + 0 .9 z −2 1 + 0 .4 z − 1 + 0 .8 z − 2

0.9 + 0.4 z −1 + 0.9 z −2 1 + 0 .4 z − 1 + 0 .8 z − 2

=

. Now, 0 .8 + 0 .4 z −1 + z −2

1 + 0.4 z −1 + 0.8 z − 2

implying that and are allpass complementary. Next, H ( z )H ( z −1 ) + G ( z)G( z −1 ) =

0.1(1 − z −2 )

0.1(1 − z 2 )

1 + 0.4 z −1 + 0.8 z − 2 1 + 0.4 z + 0.8 z 2 =

=

⋅

+

0.9 + 0.4 z −1 + 0.9 z −2 0.9 + 0.4 z + 0.9 z 2 ⋅ 1 + 0 .4 z − 1 + 0 .8 z − 2 1 + 0.4 z + 0.8 z 2

( −0.01 z 2 + 0.02 − 0.01 z −2 ) + (0.81 z 2 + 0.72 z + 1.78 + 0.72 z −1 + 0.81 z −2 ) 0.8 z 2 + 0.72 z + 1.8 + 0.72 z −1 + 0.8 z − 2

0.8 z 2 + 0.72 z + 1.8 + 0.72 z −1 + 0.8 z −2

= 1. Hence H (z ) and G (z ) are also 0.8 z 2 + 0.72 z + 1.8 + 0.72 z −1 + 0.8 z − 2 power complementary. As a result, they are doubly complementary.

2(2 + z −1 + 2 z −2 )

1 3 + 2 z −1 + 5z −2 = [1 + A( z)] where A( z ) = is an 7.90 (a) H a ( z) = 2 5 + 2 z −1 + 3 z − 2 5 + 2 z −1 + 3 z − 2 allpass function. Hence, the power complementary transfer function of H a (z) is given by G a ( z ) =

(b) H b ( z ) =

−1 −2 ⎤ −2 ⎡ 1 [1 − A(z)] = 1 ⎢1 − 3 + 2 z −1 + 5z − 2 ⎥ = 1 −−1z − 2 . 2 2 ⎢⎣ 5 + 2 z + 3z ⎥⎦ 5 + 2 z + 3z

3 + 7 .5 z −1 + 7 .5 z −2 + 3 z −3 8 + 8 z −1 + 4 z − 2 + z − 3

=

3 + 7 .5 z −1 + 7 .5 z − 2 + 3 z −3 2(1 + 0.5z −1 )(4 + 2 z −1 + z − 2 )

1 + 2 z −1 + 4 z −2 0.5 + z −1 1 and A1 ( z ) = are = [A0 ( z ) + A1 ( z )] where A0 ( z ) = 2 4 + 2 z −1 + z − 2 1 + 0.5z −1 allpass functions. Hence, the power complementary transfer function of H b (z) is 1 − 1 − 2.5z −1 + 2.5z −2 + z −3 . given by Gb ( z ) = [A0 ( z) − A1 ( z )] = 2 8 + 8 z −1 + 4 z − 2 + z − 3 7.91 From Eq. (7.126) we have X1 = AY2 + BX 2 , Y1 = CY2 + DX 2 . From the first 1 B equation, Y2 = X1 − X 2 . Substituting this in the second equation we get A A

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AD − BC C B ⎛1 ⎞ X 2 . Comparing the last two Y1 = C ⎜ X1 − X 2 ⎟ + DX 2 = X1 + A A A ⎝A ⎠ C AD − BC B 1 equations with Eq. (7.123) we arrive at t11 = , t12 = , t 21 = , t 22 = − . A A A A

From Eq. (7.123) we have Y1 = t11 X1 + t12 X 2 , Y2 = t 21 X1 + t 22 X 2 . From the t 1 Y2 . Substituting this in the first second equation we get X1 = − 22 X 2 + t 21 t 21 equation we get ⎛ t ⎞ t t t −t t 1 Y1 = t11 ⎜⎜ − 22 X 2 + Y2 ⎟⎟ + t12 X 2 = 11 Y2 + 12 21 11 22 X 2 . Comparing the t 21 ⎠ t 21 t 21 ⎝ t 21 last two equations with Eq. (7.126) we arrive at A =

t , B = − 22 , t 21 t 21

1

t t t −t t C = 11 , D = 12 21 11 22 . t 21 t 21 7.92 From Eq. (7.128a) we note t12 =

AD − BC = 1.

AD − BC 1 and t 21 = . Hence, t12 = t 21 imply A A

⎡ "⎤ ⎡Y ' ⎤ ⎡k (1 − k 2 )z −1 ⎤ ⎡ X ' ⎤ ⎡Y " ⎤ ⎡k (1 − k 2 )z −1 ⎤ ⎡ X " ⎤ 1 1 , 1 = 2 2 1 , where X1 = 7.93 ⎢ 1' ⎥ = ⎢ 1 ⎥ ⎢ ⎢ "⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ − k 2 z −1 ⎥⎦ ⎢⎣ X 2" ⎥⎦ − k1 z −1 ⎥⎦ ⎢⎣ X 2' ⎥⎦ ⎢⎣Y2" ⎥⎦ ⎢⎣ 1 ⎢⎣ X 2 ⎥⎦ ⎢⎣Y2 ⎥⎦ ⎢⎣ 1 ⎡Y ' ⎤ ⎢ 1' ⎥ . The transfer matrices of the two two-pairs are given by ⎣⎢Y2 ⎦⎥ ⎡k ⎡k (1 − k 2 )z −1 ⎤ 1 2 τ1= ⎢ 1 and τ = ⎥ 2 ⎢ −1 − k1 z ⎥⎦ ⎢⎣ 1 ⎢⎣ 1

(1 − k 22 )z −1 ⎤ ⎥. The corresponding chain − k 2 z −1 ⎥⎦

⎡1 matrices are obtained using Eq. (7.128b) and are given by Γ 1 = ⎢ ⎣⎢k1

k1 z −1 ⎤ ⎥ and z −1 ⎦⎥

⎡ 1 k z −1 ⎤ 2 Γ2= ⎢ ⎥. Therefore, the chain matrix of the Γ-cascade is given by z −1 ⎥⎦ ⎢⎣k 2 ⎡ 1 k z −1 ⎤ ⎡ 1 k z −1 ⎤ ⎡1 + k k z −1 k z −1 + k z −1 ⎤ 1 2 1 2 2 1 Γ1Γ 2 = ⎢ . Next using Eq. = −1 ⎥ ⎢ −1 ⎥ ⎢ −1 −1 −2 ⎥ k z k z k k z k k z z + + 2 1 2 ⎦⎥ ⎦⎥ ⎣⎢ 1 ⎦⎥ ⎣⎢ 2 ⎣⎢ 1 (7.128a) we arrive at the transfer matrix of the Γ-cascade as

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⎡ k + k z −1 2 ⎢ 1 ⎢1 + k1k 2 z −1 τ =⎢ 1 ⎢ ⎢ 1 + k k z −1 1 2 ⎣

z − 2 (1 − k12 )(1 − k 22 ) ⎤ ⎥ ⎥ 1 + k1k 2 z −1 . −1 −1 ⎥ z (k 2 + k1 z ) ⎥ ⎥ 1 + k1k 2 z −1 ⎦

⎡ Y ' ⎤ ⎡X " ⎤ k2 z −1 ⎤ ⎡ Y2" ⎤ , where ⎢ 2' ⎥ = ⎢ 1" ⎥. The −1 ⎥ ⎢ " ⎥ z ⎥⎦ ⎢⎣ X 2 ⎥⎦ ⎢⎣ X 2 ⎥⎦ ⎢⎣Y1 ⎥⎦ ⎡ 1 k z −1 ⎤ 1 chain matrices of the two two-pairs are given by Γ 1 = ⎢ and −1 ⎥ ⎣⎢k1 z ⎦⎥

⎡X ' ⎤ ⎡ 1 7.94 ⎢ 1' ⎥ = ⎢ ⎢⎣ Y1 ⎥⎦ ⎢⎣k1

⎡1 Γ2= ⎢ ⎢⎣k 2

k1 z −1 ⎤ ⎡ Y2' ⎤ ⎥⎢ ⎥ z −1 ⎥⎦ ⎢⎣ X 2' ⎥⎦

⎡X " ⎤ ⎡ 1 ⎢ 1" ⎥ = ⎢ ⎢⎣Y1 ⎥⎦ ⎢⎣k 2

k 2 z −1 ⎤ ⎥. The corresponding transfer matrices are obtained using Eq. z −1 ⎥⎦

⎡k (1 − k 2 )z −1 ⎤ ⎡k (1 − k 2 )z −1 ⎤ 1 2 2 (7.128a) and are given by τ 1 = ⎢ 1 and τ = ⎥. The ⎥ 2 ⎢ −1 − k 2 z −1 ⎦⎥ − k1 z ⎦⎥ ⎣⎢ 1 ⎣⎢ 1 transfer matrix τ of the τ-cascade is therefore given by ⎡k (1 − k 2 )z −1 ⎤ ⎡k (1 − k 2 )z −1 ⎤ 2 1 1 τ 2 τ1= ⎢ 2 ⎥ −1 ⎥ ⎢ − k2 z − k1z −1 ⎥⎦ ⎥⎦ ⎢⎣ 1 ⎢⎣ 1 ⎡k k + z −1 (1 − k 2 ) z −1k (1 − k 2 ) − z − 2 k (1 − k 2 ) ⎤ 2 2 1 1 2 . Using Eq. (7.128b) we thus =⎢ 1 2 −1 −1 2 −1 2 −2 ⎥ k k z z k z k k z ( 1 ) ( 1 ) − − + − 1 2 1 2 1 ⎦⎥ ⎣⎢ arrive at the chain matrix of the τ-cascade as ⎡ − (k1k 2 z −1 + 1 − k12 )z −1 ⎤ 1 ⎢ ⎥ ⎢ ⎥ k1 − k 2 z −1 k1 − k 2 z −1 Γ= ⎢ ⎥. −1 2 −2 + − k k z k ( 1 ) z 1 2 2 ⎢ ⎥ −1 −1 ⎢ ⎥ − − k k z k k z 1 2 1 2 ⎣ ⎦ 7.95 (a) Analyzing Figure P7.10(a) we obtain Y2 = X1 − k m z −1 X 2 and 2 −1 Y1 = k m Y2 + z −1 X 2 = k m ( X1 − k m z −1 X 2 ) z −1 X 2 = k m X1 + (1 − k m )z X 2 . 2 −1 ⎤ ⎡k (1 − k m )z m = Hence, the transfer matrix is given by τ ⎢ ⎥. Using Eq. (7.128b) − k m z −1 ⎦⎥ ⎣⎢ 1

⎡1 we then arrive at the chain matrix Γ = ⎢ ⎢⎣ k m

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k m z −1 ⎤ ⎥. z −1 ⎥⎦

221

(b) Labeling V1 the output variable of the top left adder connected to input X1 we then analyze Figure P7.10(b) and obtain V1 = k m ( X 1− z −1 X 2 ), Y1 = V1 + X 1 = (1 + k m ) X 1− k m z −1 X 2 , and Y2 = V1 + z −1 X 2 = k m X1 + (1 − k m ) z −1 X 2 . Hence, ⎡1 + k − k m z −1 ⎤ m the transfer matrix of the two-pair is given by τ = ⎢ ⎥. Using (1 − k m )z −1 ⎦⎥ ⎣⎢ k m

⎡ 1 ⎢ Eq. (7.128b) we then arrive at the chain matrix Γ = ⎢ k m ⎢1 − k m ⎢ ⎣ km 7.96 Solving G( z) =

−

(1 − k m )z −1 ⎤ ⎥ km ⎥. ⎥ 1 − ⎥ km ⎦

k + z −1G( z ) for H (z ) we get H ( z) = m . z −1[1 − k m H ( z)] 1 + k m z −1G( z)

H ( z) − k m

For the constrained two-pair H ( z ) =

C + D ⋅ G( z) . Comparing the last two A + B ⋅ G( z)

equations we thus get C = k m , D = z −1 , A = 1, B = k m z −1 . Substituting these values of the chain parameters in Eq. (7.128a) we get C AD − BC 1 B 2 t11 = = k m , t12 = = z −1 (1 − k m ), t 21 = = 1, t 22 = − = − k m z −1 . A A A A 7.97 (a) H ( z ) =

α + z −1G ( z )

1 + αz −1G( z )

. For the constrained two-pair H ( z ) =

C + D ⋅ G( z) . A + B ⋅ G( z)

Comparing the last two equations we thus get C = α, D = z −1 , A = 1, B = αz −1 . (b) 7.98 From the results of Problem 7.95, Part (a), we observe that the chain matrix of the ⎡ 1 k z −1 ⎤ i i -th lattice two-pair is given by Γ i = ⎢ ⎥, i = 1, 2, 3. Thus, the chain matrix z −1 ⎥⎦ ⎢⎣ ki of the cascade of the three lattice two-pairs is given by ⎡ 1 k z −1 ⎤ ⎡ 1 k z −1 ⎤ ⎡ 1 k3 z −1 ⎤ 1 2 Γ cascade = ⎢ ⎥ ⎥⎢ −1 ⎥ ⎢ z −1 ⎦⎥ ⎣⎢ k3 z −1 ⎦⎥ ⎦⎥ ⎣⎢ k 2 ⎣⎢ k1 z

⎡ 1 + k k z −1 + k z −1 (k + k z −1 ) k3 z −1 + k 2 z −1 + k1 z −1 (k 2 k3 + z − 2 ) ⎤ 2 3 1 2 3 =⎢ . −1 −1 −1 −1 −1 −1 −2 ⎥ ⎢⎣ k1 (1 + k 2 k3 z ) + z (k 2 + k3 z ) k1 (k3 z + k 2 z ) + z (k 2 k3 + z )⎥⎦ C+D From Eq. (7.135a) we obtain A3 ( z ) = A+ B

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222

1 + (k2 k3 + k1k2 + k3 )z −1 + (k1k3 + k 2 + k1k 2 k3 )z −2 + k1 z −3 which is seen to be k1 + (k1k3 + k 2 + k1k 2 k3 )z −1 + (k 2 k3 + k1k 2 + k3 )z − 2 + z − 3 an allpass function. =

7.99 Let

D( z) = 1 + d1 z −1 + d 2 z −2 = (1 − λ1 z −1 )(1 − λ2 z −1 ) = 1 − ( λ1 + λ2 )z −1 + λ1 λ2 z −2 . Thus, d 2 = λ1 λ2 and d1 = −( λ1 + λ2 ). For stability, λi < 1, i = 1, 2. As a result, d 2 = λ1 λ2 < 1. Case 1: Complex poles: d 2 > 0. In this case, λ2 = λ*1 . Now, λ1 , λ2 =

− d1 ± d12 − 4d 2

. Hence, λ1 and λ2 will be complex, if d12 < 4d 2 . In

2 d j 4d 2 − d12 . Thus, this case, λ1 = − 1 + 2 2 Consequently, if the poles are complex and circle.

λ1

2

=

1 4

(d12 + 4d2 − d12 ) = d2 < 1.

d 2 < 1, then they are inside the unit

Case 2: Real poles. In this case we get − 1 < λi < 1, i = 1, 2. Since, λi < 1, it follows then d1 < λ1 + λ2 < 2. Now, − 1 <

− d1 ± d12 − 4 d 2 2

< 1, or

± d12 − 4 d 2 < 2 + d1 . It is not possible to satisfy the inequality on the right hand side with a minus sign in front of the square root as it would imply then d1 < −2. Therefore,

d12 − 4d 2 < 2 + d1 , or d12 − 4 d 2 < 4 + d12 + 4d1 , or − d1 < 1 + d 2 . Similarly,

− d1 ± d12 − 4d 2

(7-x)

< −1, or ± d12 − 4d 2 > −2 + d1 . Again it is not

2 possible to satisfy the inequality on the right hand side with a plus sign in front of the square root as it would imply then d1 > 2. Therefore, − d12 − 4 d 2 > −2 + d1 , or

d12 − 4d 2 < 2 − d1 , or d12 − 4 d 2 < 4 + d12 − 4d1 , or equivalently, d1 < 1 + d 2 .

(7-y)

Combining Eqs. (7-x) and (7-y) we get d1 < 1 + d 2 . 7.100 (a) D a ( z ) = 4(1 + 0.75z −1 + 0.25z −2 ) ⇒ d 1= 0.75, d 2 = 0.5. Since

d 2 = 0.5 < 1 and 1 + d 2 = 1.5, d1 = 0.75 < 1 + d 2 . Hence, both roots of Da (z ) are inside the unit circle.

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(b) Db ( z ) = 2(1 + 0.5z −1 + 0.5z −2 ) ⇒ d 1= 0.5, d 2 = 0.5. Since d 2 = 0.5 < 1 and 1 + d 2 = 1.5, d1 = 0.5 < 1 + d 2 . 1 + d 2 = 1.5, d1 = 0.75 < 1 + d 2 . Hence, both roots of Db (z ) are inside the unit circle. (c) Dc ( z) = 3(1 + z −1 − z − 2 ) ⇒ d 1= , d 2 = − . Since d 2 = > 1, at least 3 3 3 3 3 one root of Dc (z ) is outside the unit circle. 4

4

4

4

4

(d) Dd ( z) = 3(1 − z −1 − z − 2 ) ⇒ d 1= − , d 2 = − . Since d 2 = < 1 and 3 3 3 6 6 1

1

1

1

1

1

2

1 + d 2 = , d1 = < 1 + d 2 . Hence, both roots of Dd (z) are inside the unit circle. 3 3 7.101 (a) A3 ( z) =

0.25 + 0.5z −1 + 0.75z −2 + z −3 1 + 0.75z −1 + 0.5z − 2 + 0.25z − 3

(7.148) we arrive at A2 ( z ) =

+ z −1 + z − 2

1 3

2 3

1

. Here, k 2 = < 1. Continuing this 1 2 3 1 + z −1 + z − 2

0.5 + z

process we get A1 ( z) =

. Note k3 = 0.25 < 1. Using Eq.

3 −1

1 + 0.5z −1

3

. Finally, k1 = 0.5 < 1. Since ki < 1, i = 1, 2, 3,

H a (z) is stable. (b) A3 ( z) =

− − z −1 + z − 2 + z − 3 1 3

2 3 2 −1 z 3

1+

arrive at A2 ( z) =

−

2 3 2 −2 1 −3 z − z 3 3 −1 −2

− 0.5 + 0.5z

+z

1 + 0.5z −1 − 0.5z − 2

process we get A1 ( z ) =

1 + z −1 1 + z −1

. Note k3 =

1 3

< 1. Using Eq. (7.148) we

. Here, k 2 = 0.5 < 1. Continuing this

. Finally, k1 = 1. Since k1 is not less than 1 ,

H b (z) is not stable.

(c) A4 ( z) =

− + z −1 + z − 2 + z − 3 + z − 4 1 6

1+

1 2 2 −1 z 3

we get A3 ( z) =

+

2 3 2 −2 z 3

+

2 3 1 −3 1 − 4 z − z 6 2 −1

0.6286 + 0.8z

+ 0.7714z −2 + z −3

1 + 0.7714z −1 + 0.8z − 2 + 0.6286z − 3

Using Eq. (7.148) we next get A2 ( z) =

1

. Note k 4 = < 1. Using Eq. (7.148) 6

. Note k3 = 0.6286 < 1.

0.5209 + 0.444z −1 + z −2

1 + 0.444z −1 + 0.5209z − 2

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. Here,

224

k 2 = 0.209 < 1. Continuing this process we get A1 ( z) =

0.2919 + z −1

1 + 0.2919z −1 k1 = 0.2919 < 1. Since ki < 1, i = 1, 2, 3, 4, H c (z ) is stable. (d) A4 ( z) =

0.2 + 0.4 z −1 + 0.6 z − 2 + 0.8 z − 3 + z − 4 1 + 0.8z −1 + 0.6 z − 2 + 0.4 z

Eq. (7.148) we get A3 ( z) =

2 3 −3

+ 0.2 z − 4

. Note k 4 = 0.2 < 1. Using

0.25 + 0.5z −1 + 0.75z −2 + z −3

. Note k3 = 0.25 < 1.

1 + 0.75z −1 + 0.5z − 2 + 0.25z − 3

Using Eq. (7.148) we next get A2 ( z ) = Continuing this process we get A1 ( z) =

1 3

+ z −1 + z − 2

1+

2 3

2 −1 1 − 2 z + z 3 3 −1

0.5 + z

1 + 0.5z −1

. Finally,

1

. Here, k 2 = < 1. 3

. Finally, k1 = 0.5 < 1. Since

ki < 1, i = 1, 2, 3, 4, H d (z) is stable. (e) A5 ( z) =

0.1 + 0.2 z −1

−1

+ 0.3z −2 + 0.5z −3 + 0.7z −4 + z −5

1 + 0.7z + 0.5z − 2 + 0.3z − 3 + 0.2 z − 4 + 0.1z − 5 Using Eq. (7.148) we get A4 ( z ) =

. Note k 5 = 0.1 < 1.

0.1313 + 0.2525z −1 + 0.4747 z −2 + 0.6869z −3 + z −4

1 + 0.6869z −1 + 0.4747 z − 2 + 0.2525z − 3 + 0.1313z − 4 k 4 = 01313 < 1. Using Eq. (7.148) we next get

A3 ( z) =

0.1652 + 0.4196z −1 + 0.6652z −2 + z −3

1 + 0.6652z −1 + 0.4196z − 2 + 0.1652z − 3

Eq. (7.148) we next get A2 ( z ) =

. Note

. Note k3 = 0.1652 < 1. Using

0.3185 + 0.6126z −1 + z −2

1 + 0.6126z −1 + 0.3185z − 2


. Here,

0.4646 + z −1

1 + 0.4646z −1 k1 = 0.4646 < 1. Since ki < 1, i = 1, 2, 3, 4, 5, H e (z) is stable.

7.102

(a) A5 ( z ) =

0.1 + 0.2 z −1

−1

+ 0.4 z −2 + 0.6 z −3 + 0.8z −4 + z −5

1 + 0.8z + 0.6 z − 2 + 0.4 z − 3 + 0.2 z − 4 + 0.1z − 5 k 5 = 0.1 < 1. Using Eq. (7.148) we get

A4 ( z ) =

0.1212 + 0.3434z −1 + 0.5657 z −2 + 0.7879z −3 + z −4

1 + 0.7879z −1 + 0.5657 z − 2 + 0.3434z − 3 + 0.1212z − 4 k 4 = 0.1211 < 1. Using Eq. (7.148) we next get

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. Finally,

. Note

. Note

225

A3 ( z) =

0.2516 + 0.5045z −1 + 0.7574z −2 + z −3

1 + 0.7574z −1 + 0.5045z − 2 + 0.2516z − 3

Eq. (7.148) we next get A2 ( z ) =

. Note k3 = 0.2516 < 1. Using

0.3351 + 0.6730z −1 + z −2

1 + 0.6730z −1 + 0.3351z − 2

. Here,

0.5041 + z −1

k 2 = 0.3351 < 1. Continuing this process we get A1 ( z ) =

. Finally, 1 + 0.5041z −1 k1 = 0.5041 < 1. Since ki < 1, i = 1, 2, 3, 4, 5, D a (z) has all roots inside the unit circle.. (b) A5 ( z) =

− 0.25 + 0.5z −1 + 0.625z −2 + 0.75z −3 + 0.875z −4 + z −5

1 + 0.875z −1 + 0.75z − 2 + 0.625z − 3 + 0.5z − 4 − 0.25z − 5 k 5 = 0.25 < 1. Using Eq. (7.148) we get

A4 ( z ) =

0.7667 + 0.8667 z −1 + 0.9667 z −2 + 1.0667 z −3 + z −4

1 + 1.0667 z −1 + 0.9667 z − 2 + 0.8667 z − 3 + 0.7667 z − 4 k 4 = 0.7667 < 1. Using Eq. (7.148) we next get

A3 ( z ) =

0.1186 + 0.5472z −1 + 0.9757 z −2 + z −3

1 + 0.9757z −1 + 0.5472z − 2 + 0.1186z − 3

Eq. (7.148) we next get A2 ( z ) =

. Note

. Note

. Note k3 = 0.1186 < 1. Using

0.4376 + 0.9238z −1 + z −2

1 + 0.9238z −1 + 0.4376z − 2


. Here,

0.6426 + z −1

. Finally, 1 + 0.6426z −1 k1 = 0.6426 < 1. Since ki < 1, i = 1, 2, 3, 4, 5, D b (z) has all roots inside the unit circle.. M7.1 The MATLAB code fragments used to simulate the FIR filter are b = [3.8461 -6.3487 3.8461];zi = [0 0]; n=0:49;x1 = cos(0.3*n);x2=cos(0.6*n); y = filter(b,1,x1+x2,zi); The plot generated by the above program is shown below: 2

y[n] x 1[n] x 2[n]

Amplitude

1.5 1 0.5 0 -0.5 -1

0

10

20 30 Time index n

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40

226

M7.2 The MATLAB code fragments used to simulate the FIR filter are: b =[17.7761 -58.7339 83.8786 -58.7339 17.7761]; n = 0:49; x1 = cos(0.3*n);x2=cos(0.5*n);x3=cos(0.8*n); y = filter(b,1,x1+x2+x3); The plot generated by the above program is shown below: 3

Amplitude

2 1 0 -1 -2 -3

y[n] x1[n]+x3[n] 0

10

20 30 Time index n

40

M7.3 The gain response of H (z ) is shown below: 20 0

Gain, dB

-20 -40 -60 -80 -100

0

0.2

0.4

0.6

0.8

1

ω/ π

M7.4 (a) The MATLAB code fragments used to evaluate H ( z)H ( z −1 ) + H (− z)H (− z −1 ) is shown below: n = [1 -2 3.5];n = [n fliplr(n)]; d = [8 0 -2 0 -1 0]; k = 0:5;n1 = (-1).^k.*n; a = conv(n,fliplr(n))+conv(n1,fliplr(n1)); b = conv(d,fliplr(d)); The numerator and denominator of are H ( z)H ( z −1 ) + H (− z)H (− z −1 ) given by the vectors a and b : a = 0 -8 0 -14 0 69 0 -14 0 -8 0 b = 0 -8 0 -14 0 69 0 -14 0 -8 0 Hence H ( z )H ( z −1 ) + H (− z)H (− z −1 ) = 1 verifying the power-complementary property of H (z ) and H ( − z ).

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(b) The MATLAB code fragments used to evaluate H ( z)H ( z −1 ) + H (− z)H (− z −1 ) is shown below: n = [1 1.5 5.25 7.25];n = [n fliplr(n)]; d = [12 0 13 0 4.5 0 0.5 0]; k = 0:7;n1 = (-1).^k.*n; a = conv(n,fliplr(n))+conv(n1,fliplr(n1)); b = conv(d,fliplr(d)); The numerator and denominator of are H ( z)H ( z −1 ) + H (− z)H (− z −1 ) given by the vectors a and b : a = Columns 1 through 10 0 6.0000 0 60.5000 0 216.7500 0 333.5000 0 216.7500 Columns 11 through 15 0 60.5000 0 6.0000 0 b = Columns 1 through 10 0 6.0000 0 60.5000 0 216.7500 0 333.5000 0 216.7500 Columns 11 through 15 0 60.5000 0 6.0000 0 Hence H ( z )H ( z −1 ) + H (− z)H (− z −1 ) = 1 verifying the power-complementary property of H (z ) and H ( − z ). M7.5 The magnitude and phase responses of H (z ) are shown below: 4

phase, radians

Magnitude

1 0.8 0.6 0.4

2 0 -2

0.2 0

0

0.2

0.4

0.6

0.8

1

-4

0

0.2

0.4

ω/π

0.6

0.8

1

ω/π

From the magnitude response plot given above it can be seen that represents a highpass filter. The difference equation representation of H (z ) is given by y[ n] + 0.7074 y[ n − 1] + 0.7976 y[ n − 2] + 0.2004 y[ n − 3] = 0.2031x[ n] − 0.2588 x[ n − 1] + 0.2588 x[ n − 2] − 0.2031x[ n − 3]. M7.6 The magnitude and phase responses of H (z ) are shown below:

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5

4

phase, radians

Magnitude

4 3 2 1 0

0

0.2

0.4

0.6

0.8

2 0 -2 -4

1

0

ω/π

0.2

0.4

0.6

0.8

1

ω/π

From the magnitude response plot given above it can be seen that represents a bandpass filter. The difference equation representation of H (z ) is given by y[ n] + 0.6402 y[n − 1] + 1.7497 y[ n − 2] + 0.5354 y[ n − 3] + 0.7015 y[ n − 4] = 0.2031x[ n] − 0.2588 x[ n − 1] + 0.2588 x[ n − 2] − 0.2031x[ n − 3]. M7.7 Here K = 5. Using Eq. (7.85) we obtain first C = 1.7411. Then using Eq. (7.84) we obtain α = −0.3779. From Eq. (7.71) we get the transfer function of the lowpass filter as H LP ( z ) =

0.6889(1 + z -1 )

. A plot of the gain response of a 1 − 0.3779 z −1 cascade of 5 lowpass filters is shown below: Passband Details 5

-1 -2 Gain, in dB

Gain, in dB

0 -5 -10 -15 -20 -25 -30

-3 -4 -5

0

0.2

0.4

0.6

0.8

1

ω/π

0.4 0.42 0.44 0.46 0.48 0.5 0.52 ω/π

M7.8 From the solution of Problem 7.46 we observe

sin ωc 2C − C 2 − (1 − cos ωc + C cosωc ) , where C = 2 ( K −1) / K . α= 1 − cos ωc − C Substituting K = 6 in the second equation we get C = 1.7818. Next, substituting this value of C and ωc = 0.4 π in the first equation we arrive at α = 0.5946. From Eq. (7.74) we get the transfer function of the lowpass filter as H HP ( z ) =

0.7973(1 − z -1 )

. A plot of the gain response of a cascade of 6 1 − 0.5946 z −1 highpass filters is shown below:

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Passband Details 5

-2 -4 Gain, in dB

Gain, in dB

0 -5 -10 -15 -20

-8 -10

-25 -30

-6

-12 0

0.2

0.4

0.6

0.8

0.2

1

0.3

0.4 ω/ π

ω/ π

0.5

0.6

M7.9 Using Eq. (7.73b) we obtain α = −0.1584. Sunstituting this value of α in Eqs.

0.4208(1 − z -1 ) . and H HP ( z ) = 1 − 0.1584 z −1 1 − 0.1584 z −1 Plots of the magnitude responses of H LP (z ) and H HP (z ) along with the plot of the magnitude response of H LP ( z ) + H HP ( z ) and plot of

(7.71) and (7.74) we get H LP ( z ) =

2

2

H LP (e jω ) + H HP(e jω )

0.5792(1 + z -1 )

are shown below verifying the doubly complementary

property of H LP (z ) and H HP (z ). 1

jω

HLP(z)

HHP(z) Magniitude

Magniitude

0.8 0.6 0.4 0.2 0

jω

|HLP(e )+HHP(e )|

1 0.8 0.6 0.4 0.2

0

0.2

0.4

0.6

0.8

0

1

0

0.2

0.4

ω/ π jω 2

Square Magniitude

0.6

0.8

1

ω/ π jω 2

|HLP(e )| +|HHP(e )|

1 0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/ π

M7.10 From Eq. (7.78) we arrive at the quadratic equation α 2 − 2.8284 α + 1 = 0 whose solution yields α = 0.4142 and α = 2.4142. A stable bandpass and bandstop transfer function requires α < 1. Hence we choose α = 0.4142. Next, from Eq.

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230

(7.76) we get β = 0.3090. Substituting these two parameters in Eqs. (7.77) and (7.80) we obtain H BP ( z ) = H BS ( z ) =

0.2929(1 − z −2 ) 1 − 0.4370 z −1 + 0.4142 z − 2

and

0.7071(1 − 0.6180 z -1 + z −2 )

. Plots of the magnitude responses of 1 − 0.4370 z −1 + 0.4142 z − 2 H BP (z ) and H BS (z ) along with the plot of the magnitude response of 2

H BP ( z ) + H BS ( z ) and plot of H BP(e jω ) + H BS (e jω )

2

are shown below

verifying the doubly complementary property of H BP (z ) and H BS (z ).

0.8 0.6 0.4 0.2 0

jω

|HBP(e )+HBS(e )|

1 Magniitude

Magniitude

jω

HBS(z)

HBP(z)

1

0.8 0.6 0.4 0.2

0

0.2

0.4

0.6

0.8

0

1

0

0.2

0.4

ω/ π jω 2

Square Magniitude

0.6

0.8

1

ω/ π jω 2

|HBP(e )| +|HBS(e )|

1 0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/ π

M7.11 From Eq. (7.78) we arrive at the quadratic equation α 2 − 2.4721α + 1 = 0 whose solution yields α = 0.5095 and α = 1.9626. A stable bandpass transfer function requires α < 1. Hence we choose α = 0.5095. Next, from Eq. (7.76) we get β = −0.3090. Substituting these two parameters in Eqs. (7.77) we obtain

H BP ( z ) =

0.2453(1 − z −2 )

1 + 0.4665 z −1 + 0.5095 z − 2 H BP (z ) is shown below:

. A plot of the magnitude response of

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231

Magniitude

1 0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/ π

M7.12 From Eq. (7.78) we arrive at the quadratic equation α 2 − 2.4721α + 1 = 0 whose solution yields α = 0.5095 and α = 1.9626. A stable bandstop transfer function requires α < 1. Hence we choose α = 0.5095. Next, from Eq. (7.80) we get β = −0.3090. Substituting these two parameters in Eqs. (7.77) we obtain

H BS ( z ) =

0.7548(1 + 0.6180 z −1 + z −2 )

1 + 0.4665 z −1 + 0.5095 z − 2 H BS (z ) is shown below:

. A plot of the magnitude response of

Magniitude

1 0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/ π

M7.13 A plot of the magnitude response of H BP ( z ) + H BS ( z ) and plot of H BP(e jω ) + H BS (e jω )

2

2

are shown below verifying the doubly complementary property of

H BP (z ) and H BS (z ).

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232

j

Square Magniitude

Magniitude

j

0.8 0.6 0.4 0.2 0

0

0.2

|HBP(e ω )| +|HBS(e ω )| j

|HBP(e ω )+HBS(e ω )|

1

0.4

0.6

0.8

1

j

2

0.8 0.6 0.4 0.2 0

1

2

0

0.2

0.4

ω/ π

0.6

0.8

1

ω/ π

M7.14 (a) jω

jω 2

Square Magnitude

Magnitude

jω

|H(e )+G(e )|

1 0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

0.8 0.6 0.4 0.2 0

1

jω 2

|H(e )| +|G(e )|

1

0

0.2

0.4

ω/π

0.6

A plot of the magnitude response of H ( z ) + G ( z ) and plot of H (e jω )

+ G ( e jω )

0.8

1

ω/π

2

2

are shown above verifying the doubly complementary property of H (z )

and G (z ). (b) jω

jω

jω 2

|H(e )+G(e )| Square Magnitude

Magnitude

0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

jω 2

|H(e )| +|G(e )|

1

0.8

1

ω/π

1 0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

A plot of the magnitude response of H ( z ) + G ( z ) and plot of H (e jω )

+ G ( e jω )

0.8

1

ω/π

2

2

are shown above verifying the doubly complementary property of H (z )

and G (z ). Not for sale

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M7.15 (a) The pole-zero plot obtained using the M-file zplane is shown below: Ha(z) 1 0.5

0.8 Magnitude

Imaginary Part

1

0 -0.5

0.6 0.4 0.2

-1 -1.5

-1

-0.5

0 0.5 Real Part

1

0

1.5

0

0.2

0.4

0.6

0.8

1

ω/π

It can be seen from the above pole-zero plot that the two poles of H a (z ) are inside the unit circle and hence H a (z ) is stable. From the magnitude response plot given above, we observe that H a (e jω ) ≤ 1 and hence, H a (z ) is a BR function. (b) The pole-zero plot obtained using the M-file zplane is shown below: Hb(z)

1

0.8 Magnitude

Imaginary Part

1 0.5 0 -0.5

0.6 0.4 0.2

-1 -1.5

-1

-0.5

0 0.5 Real Part

1

1.5

0

0

0.2

0.4

0.6

0.8

1

ω/π

It can be seen from the above pole-zero plot that the three poles of H b (z ) are inside the unit circle and hence H b (z ) is stable. From the magnitude response plot given above, we observe that H a (e jω ) ≤ 1 and hence, H a (z ) is a BR function. M7.16 (a) The power-complementary transfer function G a ( z) = Q( z) / D( z ) to the transfer function H a ( z ) = P( z ) / D( z) satisfy the relation G a ( z )G a ( z −1 ) = 1 − H a ( z ) H a ( z −1 ), or equivalently the relation Q( z )Q( z −1 ) = D( z ) D( z −1 )

− P( z)P ( z −1 ). Here, D( z ) = 1 + 0.4 z −1 + 0.6 z −2 and P( z) = 0.8 + 0.4 z −1 + 0.8z −2 . (b)

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M7.17 (a) The pole-zero plot of H a (z ) obtained using the M-file zplane is shown below. From this plot it can be seen that all 3 poles of H a (z ) are inside the unit circle, and hence, H a (z ) is a stable transfer function. Ha(z)

Imaginary Part

2 1 0 -1 -2 -2

0 Real Part

2

(b) The pole-zero plot of H b (z ) obtained using the M-file zplane is shown below. From this plot it can be seen that one pole of H b (z ) is on the unit circle, and hence, H b (z ) is an unstable transfer function. Hb(z)

Imaginary Part

3 2 1 0 -1 -2 -3 -2

0

2 4 Real Part

6

8

(c) The pole-zero plot of H c (z ) obtained using the M-file zplane is shown below. From this plot it can be seen that all 4 poles of H c (z ) are inside the unit circle, and hence, H c (z ) is a stable transfer function. Hc (z)

Imaginary Part

2 1 0 -1 -2 -3

-2

-1

0 1 Real Part

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2

3

235

(d) The pole-zero plot of H d (z ) obtained using the M-file zplane is shown below. From this plot it can be seen that all 4 poles of H d (z ) are inside the unit circle, and hence, H d (z ) is a stable transfer function. Hd(z)

Imaginary Part

1 0.5 4

0 -0.5 -1 -1.5

-1

-0.5

0 0.5 Real Part

1

1.5

(e) The pole-zero plot of H e (z ) obtained using the M-file zplane is shown below. From this plot it can be seen that all 5 poles of H e (z ) are inside the unit circle, and hence, H e (z ) is a stable transfer function. He(z)

Imaginary Part

1 0.5 5

0 -0.5 -1 -1.5

-1

-0.5

0 0.5 Real Part

1

1.5

M7.18 (a) The output data generated by running Program 7_2 are: The stability test parameters are 0.2500 0.3333 0.5000 stable = 1 Hence, H a (z ) is a stable transfer function. (b) The output data generated by running Program 7_2 are: The stability test parameters are -0.3333 -0.5000 1.0000 stable = 0

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Hence, H b (z ) is an unstable transfer function. (c) The output data generated by running Program 7_2 are: The stability test parameters are -0.1667 0.6286 0.5209 0.2919 stable = 1 Hence, H c (z ) is a stable transfer function. (d) The output data generated by running Program 7_2 are: The stability test parameters are 0.2000 0.2500 0.3333 0.5000 stable = 1 Hence, H d (z ) is a stable transfer function. (e) The output data generated by running Program 7_2 are: The stability test parameters are 0.1000 0.1313 0.1652 0.3185 stable = 1

0.4646

Hence, H e (z ) is a stable transfer function. M7.19 (a) The output data generated by running Program 7_2 are: The stability test parameters are 0.1000 0.1212 0.2516 0.3351 stable = 1

0.5041

Hence, all roots of Da (z ) are inside the unit circle. (b) The output data generated by running Program 7_2 are: (c) The stability test parameters are -0.2500 0.7667 0.1186 0.4376 0.6426 stable = 1 Hence, all roots of Db (z ) are inside the unit circle. Not for sale

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k M7.20 H ( z ) = − + ⎛⎜ 1 + 4 ⎝

k ⎞ −1 ⎟z 4⎠

− z−2. k 4

3 k=2 Magnitude

2.5 2 k=1 1.5 1

0

0.2

0.4

0.6

0.8

1

ω/π

k k k k ⎞ k k ⎛ M7.21 H ( z ) = − 2 + 1 z −1 + ⎜ 1 − 1 + 2 ⎟ z − 2 + 1 z − 3 − 2 z − 4 . 4 4 4 4 ⎠ 4 4 ⎝ 5 k 1=2, k 2=-4

Magnitude

4 3 2

k 1=1, k 2=-2

1 0

0

0.2

0.4

0.6

0.8

1

ω/π

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Chapter 8 8.1

Analyzing Figure P8.1, we get w[n] = A( X [n] + CDu[n]) and y[n] = C ( ABx[n] + u[n]). A direct implementation of these two equations leads to the structure shown below which has no delay-free loop. A w[n]

x[n] AB CD

y[n]

8.2

C

u[n]

(a) Figure P8.2(a) with input and output nodes removed and delays removed is as shown below. There are two delay-free loops in the reduced structure shown below. Analysis of this structure yields W3

W1 α1

k1

U2

k2 W2 U 3

U1

(1): W1 = W3 + U 2 , (2): W2 = α1W1 + U1 , (3): W3 = k1W3 + k 2W2 + U 2 . From Eq. (3) we get (4): (1 − k1 )W3 − k 2W2 = U 2 , and from Eq. (1) we get W3 = W1 − U 2 . Substituting the last equation in Eq. (4) we get (1 − k1 )W1 − k 2W2 = (1 − k1 )U 2 + U 3 and from Eq. (2) we have − α1W1 + W2 = U1. k 1 − k1 1 U 2 + U 3 and Solving the last two equations we arrive at W1 = 2 U 1 + ∆ ∆ ∆ α (1 − k1 ) α 1 − k1 W2 = U1 + 1 U 2 + 1 U 3 . A realization based on the above two ∆ ∆ ∆ equations after insertion of the input and output nodes and the delays is shown below which has no delay-free loops.

Y (z)

W1 (1_ k1)/∆ U2 k 2 /∆

α2 X(z)

_1

z

(1_ k1)/∆ U1

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1/∆

α1 W2

_ z1

U3

239

(b) Figure P8.2(b) with input and output nodes removed and delays removed is as shown below. There is one delay-free loop in the reduced structure shown below. Analysis of this structure yields δ

U1

W1 U2

W2

β U3

(1): W1 = U1 + δW2 , and (2): W2 = U 2 + βW1 + U 3 . Substituting Eq. (1) in Eq. (2) β 1 1 we get after some algebra (3): W2 = U 1 + U 2 + U 3 , where ∆ = 1 − βδ. ∆ ∆ ∆ δ βδ δ Substituting Eq. (3) in Eq. (1) we get (5): W1 = U 1 + U 1 + U 2 + U 3 . A ∆ ∆ ∆ realization based on Eqs. (3) and (5) after insertion of the input and output nodes and the delays is shown below which has no delay-free loop. δ

X(z)

U1

1/∆ W 1

_1

z

β

1/∆

U2 _ z 1

U3

W2

Y(z)

γ

8.3 Analysis yields Y ( z ) = G ( z )( X ( z ) − C ( z )Y ( z ) ). Hence, Y ( z) G( z) 2 H ( z) = = = . The overall transfer function has X ( z ) 1 + G ( z )C ( z ) (1 + 2 K ) + 3 z −1 a pole at z = −

3 3 . Thus, the system is stable if < 1, or, K > 1 or 1 + 2K 1 + 2K

K < −2. 8.4 From the results of Problem 8.3, we have H ( z ) =

G( z) which can be 1 + G ( z )C ( z )

G( z) − H ( z) . Substituting the expressions for G (z ) and G( z)H ( z) H (z ) in this expression we get

solved yielding C ( z ) =

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C ( z) = −

z −1 (7.0 + 9.1z −1 + 6.87 z −2 + 1.84 z −3 + 0.04 z −4 )

. Pole-zero plots of 0.04(1 + 1.8 z −1 + 1.46 z − 2 + 0.56 z − 3 + 0.08 z − 4 ) G ( z ), C ( z ), and H (z ) obtained using zplane are shown below: H(z) 1

0.5

0.5

Imaginary Part

Imaginary Part

G(z) 1

0 -0.5 -1

0 -0.5 -1

-1.5

-1

-0.5

0 0.5 Real Part

1

1.5

-1.5

-1

-0.5

0 0.5 Real Part

1

1.5

C(z)

Imaginary Part

1 0.5 2

0 -0.5 -1 -1.5

-1

-0.5

0 0.5 Real Part

1

1.5

Ac can be seen from the above ploe-zero plots, G ( z ), C ( z ), and H (z ) are stable transfer functions. 8.5

The structure with internal variables is shown below. Analysis of this structure yields K _

_

z1

X(z)

z1

W 1(z) β

Y(z) α

_1

W 2(z)

_1 _

z1

_

z1

W1 ( z ) = K X ( z ) − z −2Y ( z ), W2 ( z ) = − Kz −1 X ( z ) + z −1Y ( z ), and Y ( z ) = β W1 ( z ) + αW2 ( z ) + Kz −2 X ( z ). Substituting the first two equations into the third, and rearranging we get (1 − αz −1 + βz −2 )Y ( z ) = K (β − αz −1 + z −2 ) X ( z ).

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Hence, the transfer function is given by H ( z ) =

K (β − αz −1 + z −2 )

. 1 − αz −1 + β z − 2 (a) Since the structure employs 2 delays and has a second order transfer function, it is a canonic structure.

(b) and (c) We form ⎛ β − α z −1 + z − 2 H ( z ) H ( z −1 ) = K 2 ⎜ ⎜ 1 − α z −1 + β z − 2 ⎝

⎞⎛ β z − 2 − αz −1 + 1 ⎞ ⎟ = K 2 . Therefore, ⎟⎜ ⎟⎜ z − 2 − αz −1 + β ⎟ ⎠ ⎠⎝

H (e jω ) = K for all values of ω. Hence, H (e jω ) = 1 if K = 1.

8.6

The structure with internal variables is shown below. Analysis of this structure yields

+ _ z 1

W 1(z)

_k

Y(z) 2

_

k2 _ z 1

+

W2 (z)

k1

+ γ1

γ2 X(z)

W1 ( z ) = γ 2 X ( z ) + k 2Y ( z ) + z −1W2 ( z ), W2 ( z ) = γ1 X ( z ) − k1Y ( z ), and Y ( z ) = z −1W1 ( z ) − k 2 z −1W2 ( z ). Substituting the second equation in the first and rearranging we get W1 ( z ) = ( γ 2 + γ1 z −1 ) X ( z ) + (k 2 − k1 z −1 )Y ( z ). Substituting this equation and the second equation in the third equation we get after some rearrangement [1 − (k1 + 1)k 2 z −1 + k1 z −2 ]Y ( z ) = [( γ 2 − k 2 γ1 ) z −1 + γ1 z −2 ] X ( z ). ( γ 2 − k 2 γ 1 ) z −1 + γ 1 z −2 Y ( z) Hence, the transfer function is given by H ( z ) = = . X ( z ) 1 − (k1 + 1) k 2 z −1 + k1 z − 2 For stability, we must have k1 < 1 and (1 + k1 )k 2 < 1 + k1 , or k 2 < 1. 8.7

The structure with internal variables is shown on next page. Analysis of this structure yields S 0 ( z ) = X ( z ) − S1 ( z ), S1 ( z ) = z −1 S 2 ( z ) + z −1 S 5 ( z ), S 2 ( z ) = a3 S 0 ( z ) − z −1 S 2 ( z ), S 3 ( z ) = a3 S 5 ( z ) − z −1 S 3 ( z ), S 4 ( z ) = S 0 ( z ) + z −1 S 3 ( z ), S 5 ( z ) = a1 S 4 ( z ) − z −1 S 5 ( z ), and Y ( z ) = a 4 S 4 ( z ). Eliminating S 0 ( z ), S1 ( z ), S 2 ( z ), S 3 ( z ), S 4 ( z ) and S 5 (z ) from these equations we get after some algebra

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S 0(z)

X(z)

S 4(z)

a4

a3

Y(z)

a1 _

1 S2 (z) _ 1 z _1 _ S 3(z) z1

_1

S 5(z)

_1

_1

z

a2 S 1(z)

a 4 (1 + z −1 ) 3 Y ( z) H ( z) = = . X ( z ) 1 + (3 + a1 + a3 ) z −1 + (3 + 2 a1 + 2a3 − a1a2 ) z − 2 + (1 + a1 + a3 − a1a2 − a1a2 a3 ) z − 3 8.8

The structure with internal variables is shown below. Analysis of this structure yields _1

X(z)

W 1(z)

W 3(z) _1

z

R1 (z) α1

β1

α0

_

z1

β2

α2

β3

R 2(z)

α3 _

z1

W2 (z)

Y(z)

_1

Y ( z ) = α 0 X ( z ) + β1 z −1 R1 ( z ), W1 ( z ) = X ( z ) − β 2 z −1 R2 ( z ),

R1 ( z ) = W1 ( z ) + α1β1 z −1 R1 ( z ), W2 ( z ) = β1 z −1 R1 ( z ) − β 3 z −1W3 ( z ), R2 ( z ) = W2 ( z ) + α 2 β 2 z −1 R2 ( z ), W3 ( z ) = β 2 z −1 R2 ( z ) + α 3β 3 z −1W3 ( z ).

From the third equation we get W1 ( z ) = (1 − α1β1 z −1 ) R1 ( z ). From the sixth equation we get W3 ( z ) =

β 2 z −1 R2 ( z ) α 3β 3 z

−1

. From the fifth equation we get R2 ( z ) =

W2 ( z )

1 − α 2 β 2 z −1

.

Rewriting the fourth equation we get W2 ( z ) + β 3 z −1W3 ( z ) = β1 z −1 R1 ( z ), in which we substitute the expressions for W3 (z ) and W1 (z ) resulting in after some algebra R2 ( z ) =

β1 z −1 (1 − α 3β 3 z −1 )

β 2 β 3 z −1 + (1 − α 2 β 2 z −1 )(1 − α 3β 3 z −1 )

R1 ( z ). Substituting this expression

for R2 (z ) and the expression for W1 (z ) in the second equation we arrive at

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R1 ( z ) =

β 2 β 3 z −2 + (1 − α 2 β 2 z −1 )(1 − α 3β 3 z −1 )

β 2 β 3 z − 2 (1 − α1β1 z −1 ) + (1 − α1β1 z −1 )(1 − α 2 β 2 z −1 )(1 − α 3β 3 z −1 )

X ( z ).

β1β 2 z − 2 (1 − α 3β 3 z −1 ) Substituting the above in the first equation we finally arrive at H ( z) =

β1β 2 β 3 z −3 + β1 z −1 (1 − α 2 β 2 z −1 )(1 − α 3β 3 z −1 ) Y ( z) = α0 + . X ( z) (1 − α1β1 z −1 )(1 − α 2 β 2 z −1 )(1 − α 3β 3 z −1 ) + β 2 β 3 z − 2 (1 − α1β1 z −1 ) β1β 2 z − 2 (1 − α 3β 3 z −1 )

8.9 The structure with internal variables is shown below. Analysis of this structure yields U(z) _ k1

W(z) X(z) k1

(3): Y ( z ) =

Y(z)

k2

1 z–

(1): W ( z) = X ( z) + k1Y ( z), (2): U ( z ) =

1 z–

1 1 − z −1

W ( z ) + k 2Y ( z ), and

− k1

U ( z ). Substituting Eq. (2) in Eq. (3) we get 1 − z −1 − k1 k k − k1 ⎛ 1 ⎞ (4): Y ( z ) = ⎜⎜ W ( z ) − 1 2 Y ( z ). W ( z ) + k 2Y ( z ) ⎟⎟ = 1 1 − − − 1 2 1− z ⎝1− z 1 − z −1 ⎠ (1 − z ) Substituting Eq. (1) in Eq. (4) we then get k1 k k Y (z) = − [ X ( z ) + k1Y ( z )] − 1 2 Y ( z ) (1 − z −1 ) 2 1 − z −1 =−

k1

(1 − z −1 ) 2

X ( z) −

⎡ k1 + k 2 − k 2 z −1 ⎤ ⎢ ⎥Y ( z ), or, 1 − z −1 ⎣⎢ 1 − z −1 ⎦⎥

k1

⎡ k1 (k1 + k 2 − k 2 z −1 ) ⎤ k1 X ( z ). Hence, ⎢1 + ⎥Y ( z ) = − − 1 2 (1 − z ) (1 − z −1 ) 2 ⎢⎣ ⎥⎦ k1 Y (z) . H (z) = =− X ( z) [1 + k1 (k1 + k 2 )] − (2 + k1k 2 )z −1 + z − 2

8.10

(a) A direct form realization of H ( z ) = 1 − 3.6 z −1 + 5.4 z −2 − 4.32 z −3 + 1.944 z −4

− 0.4666z −5 + 0.0467z −6 and its transformed structure are shown on next page:

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_ z 1

_ z 1

x[n]

_ 3.6

_ z 1

+ z

_

5.4

+

–1

z _

0.0467

z

0.4666

–1

z _

1.944

0.0467

–1

z

–1

z

–1

y[n]

_ 3.6

5.4

4.32

y[n]

+

+

+

+

–1

_ 0.4666

1.944

4.32

_ z 1

_ z 1

_ z 1

x[n]

(b) A cascade realization of H ( z ) = (1 − 0.6 z −1 )(1 − 0.6 z −1 )(1 − 0.6 z −1 )(1 − 0.6 z −1 )(1 − 0.6 z −1 )(1 − 0.6 z −1 ) is shown below: x[n]

+ _1

+

+

_1

z

_1

z _

_

0.6

z

_1

_

0.6

z _

0.6

y[n]

_1

z

z _

0.6

+

+

+ _1

_

0.6

0.6

(c) A cascade realization of H ( z ) = (1 − 1.2 z −1 + 0.36 z −2 )(1 − 1.2 z −1 + 0.36 z −2 )(1 − 1.2 z −1 + 0.36 z −2 ) is shown below: x[n]

+

+

+ _1

z

z

z _ 0.2

_ 0.2

+

_ 0.2

+ z

z

z

0.36

0.36

+

_1

_1

_1

y[n]

_1

_1

0.36

(d) A cascade realization of H ( z ) = (1 − 1.8 z −1 + 1.08 z −2 − 0.216 z −3 )(1 − 1.8 z −1 + 1.08 z −2 − 0.216 z −3 ) is shown below: y[n] x[n] + + _ z 1

_

_ z 1

1.8

_

+

_1

+

_1

z

z 1.08

1.08

+ _ z 1

1.8

_ 0.216

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+ _ z 1

_ 0.216

245

(e) A cascade realization of H ( z ) = (1 − 0.6 z −1 )(1 − 0.6 z −1 )(1 − 1.2 z −1 + 0.36 z −2 )(1 − 1.2 z −1 + 0.36 z −2 ) is shown below: x[n]

+ _1

z

_ 0.6

+

+

+

z

z

z

_ 0.2

_ 0.6

y[n]

_1

_1

_1

_ 0.2

+

+

_1

_1

z

z

0.36

0.36

All realizations given above require 6 multipliers and 6 two-input adders. 8.11 (a) A four branch polyphase decomposition is given by H ( z) = E 0 ( z 4 ) + E1 ( z 4 )z −1 + E 2 ( z 4 )z −2 + E 3 ( z 4 )z −3 , where E 0 ( z ) = h[0] + h[ 4]z −1 + h[8] z −2 , E1 ( z ) = h[1] + h[5] z −1 + h[9] z −2 , E 2 ( z ) = h[ 2] + h[6]z −1, E 3 ( z) = h[3] + h[7]z −1.

(b) A canonic four-branch polyphase realization of the above transfer function is shown below: h[3] h[7]

+ _ z 1

h[2] h[6]

x[n]

_ z 4

_ z 4

+

+ _ z 1

h[9] h[5]

+

h[1]

_ z 1

h[8] h[4] h[0]

+

+

+

y[n]

8.12 (a) A three branch polyphase decomposition is given by H ( z) = E 0 ( z 3 ) + E1 ( z 3 )z −1 + E 2 ( z 3 )z −2 , where

E 0 ( z) = h[0] + h[3]z −1 + h[6]z −2 + h[9]z −3 , E1 ( z ) = h[1] + h[ 4]z −1 + h[7]z −2 , E 2 ( z ) = h[ 2] + h[ 5]z −1 + h[8]z −2 .

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(b) A canonic three branch polyphase realization of the above transfer function is shown below: h[2] h[5]

+

h[8] _ z 1 _3

_3

_3

z

z

z

h[7] h[4]

+

+

h[1] _ z 1

h[9]

h[6]

+

h[3]

+

+

h[0]

8.13 (a) A two branch polyphase decomposition is given by H ( z) = E 0 ( z 2 ) + E1 ( z 2 )z −1 ,

where E 0 ( z) = h[0] + h[2]z −1 + h[ 4]z −2 + h[6]z −3 + h[8]z −4 , E1 ( z ) = h[1] + h[3]z −1 + h[ 5]z −2 + h[ 7]z −3 + h[9]z −4 .

(b) A canonic two branch polyphase realization of the above transfer function is shown below: + + + + _ z 2

_ z 2

_ z 2

h[5]

h[3]

h[9]

h[7]

+

h[1]

h[8]

+

_ z 2

h[6]

+

x[n]

h[4]

h[2]

_ z 1

+

h[0]

y[n]

+ 8.14

_ z 1

x[n]

_ z 1

_ z 1

+

+

+ _1

_1

_ z 1

_1

_ z 1

_ z 1

h[1]

h[0]

+

h[2]

+

y[n]

8.15

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_ z 1

_ z 1

_ z 1

+

+

_ z 1

_ z 1

h[1]

h[0]

_1

_1

_ z 1

_ z 1

+

+

_1

_1

h[3]

h[2]

+

+

+

8.16 G ( z ) = z − N / 2 − H ( z ). A canonic realization of both G ( z ) and H ( z ) is shown below for N = 6. _ _ _ + G(z) z 1 z 1 z 1

+

+

+ _ z 1

_ z 1

h[1]

h[0]

_ z 1

h[2]

h[3]

+

+

_1

+

H(z)

8.17 Without any loss of generality, assume M = 5 which means N = 11. In this case, the transfer function is given by H ( z )

[

]

= z − 5 h[ 5] + h[ 4]( z + z −1 ) + h[3]( z 2 + z − 2 ) + h[ 2]( z 3 + z − 3 ) + h[1]( z 4 + z − 4 ) + h[ 0]( z 5 + z − 5 ) .

Now, the recursion relation for the Chebyshev polynomial is given by Tr ( x ) = 2 x Tr −1 ( x ) − Tr − 2 ( x ), r ≥ 2 with T0 ( x ) = 1 and T1 ( x ) = x. Hence,

T2 ( x ) = 2 x T1 ( x ) − T0 ( x ) = 2 x 2 − 1, T3 ( x) = 2 x T2 ( x ) − T1 ( x ) = 2 x(2 x 2 − 1) − x = 4 x 3 − 3 x, T4 ( x ) = 2 x T3 ( x ) − T2 ( x ) = 2 x(4 x 3 − 3 x) − (2 x 2 − 1) = 8 x 4 − 8 x 2 + 1, T5 ( x ) = 2 x T4 ( x ) − T3 ( x ) = 2 x(8 x 4 − 8 x 2 + 1) − (4 x 3 − 3 x) = 16 x 5 − 20 x 3 + 5 x. We can thus rewrite the expression inside the square brackets given above as ⎛ z + z −1 ⎞ ⎛ z + z −1 ⎞ ⎛ z + z −1 ⎞ ⎛ z + z −1 ⎞ ⎛ z + z −1 ⎞ h[ 5] + 2h[ 4]T1 ⎜ ⎟ ⎟ + 2h[ 0]T5 ⎜ ⎟ + 2h[1]T4 ⎜ ⎟ + 2h[2]T3 ⎜ ⎟ + 2h[3]T2 ⎜ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎡ ⎛ z + z −1 ⎞ 3 ⎡ ⎛ z + z −1 ⎞ 2 ⎤ ⎛ z + z −1 ⎞ ⎤ ⎛ z + z −1 ⎞ = h[5] + 2h[ 4]⎜ ⎟⎥ ⎟ − 3⎜ ⎟ − 1⎥ + 2h[2]⎢ 4⎜ ⎟ + 2h[3]⎢2⎜ ⎝ 2 ⎠ ⎥⎦ ⎝ 2 ⎠ ⎢⎣ ⎝ 2 ⎠ ⎥⎦ ⎢⎣ ⎝ 2 ⎠ 2 3 ⎤ ⎡ ⎛ z + z −1 ⎞ 4 ⎡ ⎛ z + z −1 ⎞ 5 ⎛ z + z −1 ⎞ ⎛ z + z −1 ⎞ ⎛ z + z −1 ⎞ ⎤ ⎥ ⎢ + 2h[1]⎢8⎜ − + + − 8 1 2 h [ 0 ] 16 20 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ + 5⎜ ⎟⎥ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦ ⎥⎦ ⎢⎣ ⎝ 2 ⎠ ⎢⎣ ⎝ 2 ⎠ 5

n

⎛ z + z −1 ⎞ = a[ n]⎜ ⎟ , where a[ 0] = h[ 5] − 2h[3] + 2h[1], a[1] = 2h[ 4] − 6h[ 2] + 10 h[ 0], 2 ⎠ ⎝ n=0

∑

a[ 2] = 4h[3] − 16 h[1], a[3] = 8h[ 2] − 40 h[1], a[ 4] = 16 h[1], and a[5] = 32h[0]. A realization of

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H ( z) =

z−5

2

n

5

⎛ 1+ z −2 ⎞ − 3 ⎛ z + z −1 ⎞ ⎛ 1+ z − 2 ⎞ − 4 a[ n]⎜ ⎟ z ⎟ = a[0]z − 5 + a[1]⎜ ⎟ z + a[2]⎜ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ n=0

∑

5

4

3

⎛ 1+ z − 2 ⎞ ⎛ 1 + z − 2 ⎞ −1 ⎛ 1+ z − 2 ⎞ − 2 + a[3]⎜ ⎟ is shown below: ⎟ z + a[5]⎜ ⎟ z + a[ 4]⎜ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ _

_

1+z 2 ______ 2

x[n]

_ z 1

+

a[2] _ z 1

1+z 2 ______ 2

+

a[5]

a[4]

a[3] _ z 1

+

_

1+z 2 ______ 2

1+z 2 ______ 2

a[1]

a[0]

_

_

1+z 2 ______ 2

_ z 1

_ z 1

+

+

y[n]

P( z ) P1( z) P 2 ( z) P 3 ( z) . Assume all zeros of P ( z ) and D ( z ) = ⋅ ⋅ D( z ) D1 ( z) D2 ( z) D3 ( z) are complex. Note that the numerator of the first stage can be one of the 3 factors, P1 ( z), P2 ( z), and P3 ( z ). Likewise, the numerator of the second stage can be one of the remaining 2 factors, and the numerator of the third stage is the remaining factor. Similarly, that the denominator of the first stage can be one of the 3 factors, D1 ( z), D2 ( z), and D3 ( z ). Likewise, the denominator of the second stage can be one of the remaining 2 factors, and the denominator of the third stage is the remaining factor. Hence, there are (3! ) 2 = 36 different types of cascade realizations.

8.18 Consider H ( z) =

If the zeros of P ( z ) and D ( z ) are all real, then P ( z ) has 6 real zeros and D ( z ) has 6 real zeros. In this case, then there are (6! ) 2 = 518400 different types of cascade realizations. 8.19 H ( z ) =

K

∏ Dii (z). i =1

P (z)

⎛K⎞ Here, the numerator of the first stage can be chosen in ⎜ ⎟ ways, ⎝1⎠

⎛ K −1 ⎞ the numerator of the second stage can be chosen in ⎜ ⎟ ways, and so on, until there ⎝ 1 ⎠ is only one possible choice for the numerator of the K –th stage. Likewise, the K denominator of the first stage can be chosen in ⎛⎜ ⎞⎟ ways, the denominator of the ⎝1⎠ ⎛ K − 1⎞ second stage can be chosen in ⎜ ⎟ ways, and so on, until there is only one possible ⎝ 1 ⎠ choice for the denominator of the K –th stage. Hence, the total number of possible 2

2

2

2

2

K K −1⎞ ⎛ K − 2 ⎞ ⎛ 2 ⎞ ⎛ 1⎞ cascade realizations are equal to ⎛⎜ ⎞⎟ ⎛⎜ ⎟ ⎜ ⎟ L ⎜ ⎟ ⎜ ⎟ = ( K! ) 2 . 1 1 1 ⎝ 1 ⎠ ⎝ 1⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

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8.20 A canonic direct form II realization of H ( z ) =

3 − 5 z −2 + 3 z −3

1 − 4 z −1 + 3 z − 2 + 6 z − 4 on the left. Its transposed realization is shown below on the right. x[n]

2

+

+

y[n]

2

x[n]

4

+

_ z 1

_5

3

_5

+

_3

+ _ z 1

_ z 1

3

3

_

4

+ _ z 1

_

y[n]

+ _ z 1

_ z 1

+

is shown below

+ _ z 1

_ z 1

_6

6

8.21 (a) A cascade canonic realization of H1( z) based on the decomposition

⎛ z −1 + 3 z − 2 H 1 ( z ) = 0.3⎜ ⎜ 1 − 1.6 z −1 + 2.1z − 2 ⎝ x[n]

⎞⎛ 1 − 0.8 z −1 ⎞ ⎟ is shown below: ⎟⎜ ⎟⎜ 1 − 0.75 z −1 ⎟ ⎠ ⎠⎝

0.3 z

–1

–1

z _

0.75

y[n] 0.8

1.6

–1

z

_ 2.1

3

An alternate cascade canonic realization of H1( z) based on the decomposition ⎛ z −1 − 0.8 z − 2 H 1 ( z ) = 0.3⎜ ⎜ 1 − 1.6 z −1 + 2.1z − 2 ⎝ x[n]

⎞⎛ 1 + 3 z −1 ⎞ ⎟ is shown below: ⎟⎜ ⎟⎜ 1 − 0.75 z −1 ⎟ ⎠ ⎠⎝

0.3 z–1

–1

z

y[n] 0.75

3

1.6 _ 2.1

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–1

z

_

0.8

250

(b) A cascade canonic realization of H 2 ( z) based on the decomposition ⎞ ⎛ 3.1 + 0.85z −1 ⎞⎛ 1 − 4.5z −1 ⎞⎛ 3 − 0 .5 z − 2 ⎟ is shown below: ⎟⎜ ⎟⎜ H 2 ( z ) = ⎜⎜ −1 ⎟⎜ 1 + 0.2 z −1 ⎟⎜ 1 + 0.5z −1 + 0.1z − 2 ⎟ 1 0 . 15 z − ⎠ ⎠⎝ ⎠⎝ ⎝ 3.1

3

x[n] z–1

z–1

_

_ 0.2

0.85

0.15

–1

z

y[n] _ 0.5

4.5

z

–1 _ 0.5

_ 0.1

An alternate cascade canonic realization of H 2 ( z) based on the decomposition ⎛ 1 − 4.5z −1 ⎞⎛ 3.1 + 0.85z −1 ⎞⎛ ⎞ 3 − 0 .5 z − 2 ⎟⎜ ⎟⎜ ⎟ is shown below: H 2 ( z ) = ⎜⎜ ⎟ ⎜ ⎟ ⎜ ⎟ − 1 − 1 − 1 − 2 ⎝ 1 − 0.15z ⎠⎝ 1 + 0.2 z ⎠⎝ 1 + 0.5z + 0.1z ⎠ 3.1

3

x[n] –1

z–1

z

_

0.15

4.5

–1

z

y[n]

_ 0.2

_ 0.5

0.85 _

–1

z

_ 0.5

0.1

(c) The factored form of the denominator of H 3 ( z ) is given by: (1 + 3.1z −1 )(1 − 2.0246 z −1 )(1 − 0.3754 z −1 ). A cascade canonic realization of H 3 ( z ) based on the decomposition ⎛ 1 .5 + 0 .7 z −1 ⎞ ⎛ ⎞ 5.2 − 8.4 z −1 + 7 z − 2 ⎟⎜ ⎟ H 3 ( z ) = ⎜⎜ −1 ⎟⎜ (1 − 2.0246 z −1 )(1 − 0.3754 z −1 ) ⎟ + 1 3 . 1 z ⎝ ⎠⎝ ⎠ − 1 − 1 − 2 ⎞ ⎛ 1.5 + 0.7 z ⎞⎛ 5.2 − 8.4 z + 7 z ⎟ is shown below: ⎟⎜ =⎜ ⎜ 1 + 3.1z −1 ⎟⎜ 1 − 2.4 z −1 + 0.76 z − 2 ⎟ ⎠ ⎠⎝ ⎝ 1.5

5.2

x[n]

z _ 3.1

–1

2.4 0.7

z

–1

z _ 0.76

_ 8.4

y[n]

–1 7.0

An alternate cascade canonic realization of H 3 ( z ) based on the decomposition ⎛ 1 .5 + 0 . 7 z − 1 ⎞ ⎛ ⎞ 5 .2 − 8 .4 z − 1 + 7 z − 2 ⎟⎜ ⎟ H 3 ( z ) = ⎜⎜ −1 ⎟⎜ (1 + 3.1z −1 )(1 − 0.3754 z −1 ) ⎟ 1 2 . 0246 z − ⎝ ⎠⎝ ⎠

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⎛ 1.5 + 0.7 z −1 ⎞⎛ ⎞ 5 .2 − 8 .4 z − 1 + 7 z − 2 ⎟⎜ ⎟ is shown below: = ⎜⎜ − 1 ⎟⎜ 1 + 2.7246 z −1 − 1.1637 z − 1 ⎟ − 1 2 . 0246 z ⎝ ⎠⎝ ⎠ 1.5

5.2

x[n]

z

–1

_ 2.7246

2.024

0.7

z–1

z

_ 8.4

y[n]

–1 7.0

1.1637

8.22 (a) A partial fraction expansion of H1( z) in z −1 obtained using the M-file residuez

is given by H1 ( z ) = 0.4571 +

− 0.0513

+

− 0.4059 + 0.9878z −1

. The Parallel Form I 1 − 0.75z −1 1 − 1.6 z −1 + 2.1z − 2 realization based on this expansion is shown on the next page on the left side. 0.4571

x[n]

_ 0.0513

+

+

y[n]

x[n]

+

_ z 1

_ z 1

0.75 _

0.0385

+

y[n]

0.4059

+

+

+

_ z 1

+

_

0.75

_ z 1

1.6

+

0.9878

0.3385

1.6

_ z 1

+ _ z 1

_ 2.1

_ 2.1

_ 0.8523

A partial fraction expansion of H1( z) in z obtained using the M-file residue is given by H1 ( z ) =

− 0.0385z −1

+

0.3385z −1 − 0.8523z −2

. The Parallel Form II 1 − 0.75z −1 1 − 1.6 z −1 + 2.1z − 2 realization based on this expansion is shown above on the right side. (b) A partial fraction expansion of H 2 ( z) in z −1 obtained using the M-file residuez 146.71 238.6 261.4905 + 79.8813z −1 + + . The is given by H 2 ( z ) = −637.5 + 1 + 0.2 z −1 1 − 0.15z −1 1 + 0.5z −1 + 0.1z − 2 Parallel Form I realization based on this expansion is shown on the next page on the left side.

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146.71

+

+ _ z 1

_ z 1

_ 0.2

x[n]

0.15 238.6

+

+

y[n]

x[n]

+

_1

_ z 1

z 0.15

_

637.5

+

+ _

+

0.5

+

y[n]

9.3 _ z 1

261.4905

_ 50.8639

_ 0.5

79.8813

+

+

_ z 1

_1

z _

_ 29.3414

_ 0.2

+ _ z 1

35.7904

0.1

_ 26.1491

_ 0.1

A partial fraction expansion of H 2 ( z) in z obtained using residue is given by − 29.3414 z −1 35.7904 z −1 − 50.8639z −1 − 26.1491z −2 + + . The H 2 ( z ) = 9.3 + 1 + 0.2 z −1 1 − 0.15z −1 1 + 0.5z −1 + 0.1z − 2 Parallel Form II realization based on this expansion is shown above on the right side. (c) A partial fraction expansion of H 3 ( z ) in z −1 obtained using residuez is given 5.939 2.4696 − 2.6884 + + . The Parallel Form I by H 3 ( z ) = 2.0798 + 1 + 3.1z −1 1 − 0.0246 z −1 1 − 0.3754 z −1 realization based on this expansion is shown below on the left side. 2.0798

7.8

5.939

+

+

_ z 1

_ 3.1

_ 3.1

x[n]

_ z 1

2.4696

+

+

y[n]

x[n]

_ 18.4109

+

_1

z

+ _ z 1

0.0246

0.0246

y[n]

5.0

_ 2.6884

+

+ _1

_ z 1

z

0.3754

_

1.0092

0.3754

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A partial fraction expansion of H 3 ( z ) in z obtained using residue is given by − 18.4109z −1 5.0 z −1 − 1.0092 z −1 + + H 3 ( z ) = 7.8 + . The Parallel Form II 1 + 3.1z −1 1 − 0.0246 z −1 1 − 0.3754 z −1 realization based on this expansion is shown on the previous page on the right side. 8.23 A cascade realization of based on the factored form given by ⎛ ⎞⎛ 0.44 + 0.362 z −1 + 0.02 z − 2 ⎞ z −1 ⎟⎜ ⎟ is shown below: H ( z ) = ⎜⎜ ⎟ − 1 + 0 .5 z − 2 ⎟ ⎜ −1 + − 1 0 . 8 z 1 0 . 4 z ⎝ ⎠⎝ ⎠ x[n]

+

0.44

+

_

+

_ z 1

8.24 (a) H ( z ) =

y[n]

_ z 1

0.8

0.362

0.4 _ z 1

_

+

+

_ z 1

0.5

0.02

2 − 0.3z −1 0.4 + z −1 3 2.4 + 5.64 z −1 − 0.9z −2 ⋅ ⋅ = . 1 + 0.5z −1 1 + 0.4 z −1 1 + 0.5z −1 1 + 1.4 z −1 + 0.65z − 2 + 0.1z − 3

(b) y[ n] = 2.4 x[ n] + 5.64 x[ n − 1] − 0.9 x[ n − 2] − 1.4 y[ n − 1] − 0.65 y[ n − 2] − 0.1y[ n − 3]. (c) A cascade realization of H (z ) is shown below: x[n]

2

+

+

0.4

+

_ z 1 _ 0.5

+

3

+ _ z 1

_ z 1 _

_

0.3

0.4

y[n]

_ 0.5

(d) A partial-fraction expansion of H ( z ) in z −1 obtained using residuez is given by − 62.4 − 277.2 342 + + H ( z) = . The Parallel Form I realization − 1 − 1 2 1 + 0.5z (1 + 0.5z ) 1 + 0.4 z −1 based on this expansion is shown on the next page. (e) The inverse z –transform of the partial-fraction of H ( z ) given in Part (d) yields h[ n] = 342(−0.5) n µ[ n] − 62.4( n + 1)( −0.5) n µ[ n] − 277.2( −0.4) n µ[ n].

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_

+

277.2

_ z 1 _ 0.4

342

+

x[n]

+

y[n]

, X ( z) =

1

_ z 1 _ 0.5 _

+

62.4

_ z 1

+

_1

_ 0.25

8.25 (a) Y ( z ) =

1 1 − 0.4 z −1

−

0.3z −1 1 − 0.4 z −1

=

_ z 1

1 − 0.3z −1 1 − 0.4 z −1

1 − 0.2 z −1 Y ( z ) (1 − 0.3z −1 )(1 − 0.2 z −1 ) 1 − 0.5z −1 + 0.06 z −2 = = H (z) = . X (z) 1 − 0.4 z −1 1 − 0.4 z −1

. Thus,

(b) y[ n] = x[ n] − 0.5 x[ n − 1] + 0.06 x[ n − 2] + 0.4 y[ n − 1].

x[n]

+

_ 0.5

_ z 1

+

+

(c) y[n]

0.06 _ z 1

0.4

A partial-fraction expansion of H ( z ) in z −1 using residuez is given by 0.125 whose realization yields the Parallel Form I H ( z ) = 0.875z −1 − 0.15 + 1 − 0.4 z −1 structure as indicated on the next page. (d)

(e) The inverse z -transform of H ( z ) yields h[ n] = 0.875δ[ n − 1] − 0.15δ[ n] + 0.125(0.4) n µ[ n].

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_1

+

x[n]

z 0.4

_ z 1

(f) X ( z ) =

1

−

0.4 z −1

=

_

0.15

+

0.125 y[n]

0.875

1 − 0.4 z −1

. Therefore, Y ( z ) = H ( z ) X ( z ) 1 − 0.3z −1 1 − 0.3z −1 ⎛ (1 − 0.3z −1 )(1 − 0.2 z −1 ) ⎞⎛ 1 − 0.4 z −1 ⎞ ⎟⎜ ⎟ = 1 − 0.2 z −1 , whose inverse z -transform = ⎜⎜ ⎟ ⎜ ⎟ − 1 − 1 1 − 0 .4 z ⎝ ⎠ ⎝ 1 − 0 .3 z ⎠ yields y[ n] = δ[ n] − 0.2δ[ n − 1]. 1 − 0.3z −1

8.26 Figure P8.12 can be seen to be a Parallel Form II structure. A partial-fraction 2 z 2 − 3.2 z − 18.6 in z obtained using the M-file residue is expansion of H ( z ) = z 2 − 2.6 z − 1.2 5 z −1 − 3z −1 . Comparing the coefficients of the given by H ( z ) = 2 + + 1 − 3z −1 1 + 0.4 z −1 expansion with the corresponding multiplier coefficients in Figure P8.12 we conclude that the multiplier coefficient − 5 in the feed forward path should be 5 and the multiplier coefficient − 0.5 in the feedback path should be replaced with − 0.4. 8.27 Figure P8.13 can be seen to be a Parallel Form I structure. A partial-fraction expansion 4 z 2 − 5.6 z in z −1 obtained using the M-file residuez is given by of H ( z ) = 2 z + 0.2 z − 0.08 ⎛ − 1.5 ⎞ 12 −8 1 ⎟⎟. Comparing the + = −8⎜⎜ H (z) = + 1 + 0.4 z −1 1 − 0.2 z −1 ⎝ 1 + 0.4 z −1 1 − 0.2 z −1 ⎠ coefficients of the expansion with the corresponding multiplier coefficients in Figure P8.12 we conclude that the multiplier coefficient A has a value equal to − 8 and the multiplier coefficient B has a value equal to − 0.4. 8.28 (a) The difference equation corresponding to the transfer function H ( z ) =

=

Y ( z) X ( z)

(1 + α1 + α 2 )(1 + 2 z −1 + z −2 )

is given by 1 − α1z −1 + α2 z − 2 y[ n] + α1 y[ n − 1] − α2 y[ n − 2] = (1 + α1 + α2 )( x[ n] + 2 x[ n − 1] + x[ n − 2]), which can be rewritten as y[ n] = ( x[ n] + 2 x[ n − 1] + x[ n − 2]) + α1 ( y[ n − 1] − x[ n] − 2 x[ n − 1] − x[ n − 2]) − α2 ( y[ n − 2] − x[ n] − 2 x[ n − 1] − x[ n − 2]). Denoting w[ n] = x[ n] + 2 x[ n − 1] + x[ n − 2], the difference equation representation becomes

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y[ n] = w[ n] + α1 ( y[ n − 1] − w[ n]) − α2 ( y[ n − 2] − w[ n]). A realization of H (z ) based on the last two equations is as indicated below: w[n]

_ z 1

α1

_

1

2

+

+

_1

z

y[n]

+

+

+

x[n]

_1

z

_α

_

_ z 1

2

1

+ An interchange of the two stages leads to an equivalent realization shown below: x[n]

+

+

+

α1

_

1

+

_1

z

_α

_1

_1

z

_1

2

2

y[n]

+

_1

z

z

+ Finally, by delay sharing the above structure reduces to the canonic realization shown below: x[n] y[n] + + + α1

_

1

+ _

_α

_ z 1

2

+

_1

2

z

1

+ (b) The difference equation corresponding to the transfer function H ( z ) =

=

Y ( z) X ( z)

(1 − α2 )(1 − z −2 )

is given by 1 − α1z −1 + α2 z − 2 y[ n] − α1 y[ n − 1] + α2 y[ n − 2] = (1 − α2 )( x[ n] − x[ n − 2]) which can be rewritten as y[ n] = α1 y[ n − 1] − α2 y[ n − 2] + x[ n] − α2 x[ n] − x[ n − 2] + α2 x[ n − 2] = α1 y[ n − 1] − α2 ( x[ n] − x[ n − 2] + y[ n − 2]) + ( x[ n] − x[ n − 2]). Denoting w[ n] = x[ n] − x[ n − 2], we can rewrite the last equation as y[ n] = α1 y[ n − 1] − α2 (w[ n] + y[ n − 2]) + w[ n]. A realization of H (z ) based on the last two equations is as shown on next page.

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+

x[n] _ z 1

w[n] _α

_1

+

+ 2

y[n] α1 _ 1 z

+

_ z 1

_ z 1

_

1

An interchange of the two stages leads to an equivalent realization shown below: x[n]

_α

+

+ 2

+

α1 _ 1 z

z

_1

+ _ z 1

_

1

y[n]

_1

_ z 1

Finally, by delay sharing the above structure reduces to a canonic realization as shown below: x[n]

_α

+

+ 2

+

α1 _ 1 z

y[n]

_1

+ _ z 1

_1

8.29 (a)

+

+

X1 α1

α2

+ α3

_ z 1

Y1

α4

_ z 1

(b) +

+

X1 α4

Y1 α2

_ z 1

α1

+

α3

_ z 1

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258

C + D ⋅ H N −1 ( z ) Y , 8.30 (a) From the structure of Figure P8.14 it follows that H N ( z ) = 1 = X1 A + B ⋅ H N − 1 ( z ) C − A ⋅ H N ( z) . Substituting the expression for H N ( z ) we from which we get H N −1 ( z ) = B ⋅ H N (z) − D N N C ⎛⎜ 1 + di z − i ⎞⎟ − A⎛⎜ pi ⎞⎟ = 1 = 0 i i ⎠ ⎝ ⎠ then arrive at H N −1 ( z ) = ⎝ N N B⎛⎜ pi ⎞⎟ − D⎛⎜ 1 + d z − i ⎞⎟ i =1 i ⎝ ⎠ ⎝ i=0 ⎠

∑

∑

∑

=

∑

(C − Ap 0 ) + (Cd1 − Ap1 )z −1 + K + (Cd N −1 − Ap N −1 )z − N + 1 + (Cd N − Ap N )z − N ( Bp 0 − D ) + ( Bp1 − Dd1 )z −1 + L + ( Bp N −1 − Dd N −1 )z − N + 1 + ( Bp N − Dd N )z − N

.

Substituting the values A = 1, B = d N z −1 , C = p0 , and D = p N z −1 , we get H N −1 ( z ) =

( p 0 − p 0 ) + ( p 0 d 1 − p1 ) z − 1 + L + ( p 0 d N − 1 − p N − 1 ) z − N + 1 + ( p 0 d N − p N ) z − N (d N p0 − p N )z −1 + (d N p1 − p N d1 )z − 2 + L + (d N p N −1 − p N d N −1 )z − N + (d N p N − p N d N )z − N −1

=

( p 0 d 1 − p1 ) + L + ( p 0 d N − 1 − p N − 1 ) z − N + 2 + ( p 0 d N − p N ) z − N + 1

(d N p0 − p N ) + (d N p1 − p N d1 )z −1 + L + (d N p N −1 − p N d N −1 )z − N + 1

p ' + p1' z −1 + L + p 'N − 2 z − N + 2 + p 'N −1 z − N + 1 where = 0 ' −2z− N +2 + dN ' −1 z − N + 1 1 + d1' z −1 + L + d N

p k' =

p k d k +1 − pk +1 p d − pN dk , 0 ≤ k ≤ N − 1, and d k' = k N , 1 ≤ k ≤ N − 1. p0 d N − p N p0 d N − p N

(b) From the chain parameters, we obtain for the first stage t 11=

C = p0 , A

AD − BC 1 B = ( p N − p0 d N )z −1 , t 21= = 1, t 22 = − = − d N z −1 . The A A A corresponding input-output relations of the two-pair are given by Y1 = p 0 X1 + ( p N − p 0 d N )z −1 X 2 = p 0 ( X1 − d N z −1 X 2 ) + p N z −1 X 2 , t12 =

Y2 = X1 − d N z −1 X 2 . Substituting the second equation into the first we rewrite it as Y1 = p0Y2 + p N z −1 X 2 . A realization of the two-pair based on the last two equations is therefore as indicated below: X1

Y2 _d

N

p

0

Y1

p

z–1

X2

N

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259

(c) Except for the first stage, all other stages require 2 multipliers. Hence, the total number of multipliers needed to implement an N –th order transfer function H N (z) is 2 N + 1. The total number of 2-input adders required is 2N , while the overall realization is canonic requiring N delays. 8.31 (a) From G3 ( z) = H1 ( z) =

0.3z −1 + 0.66 z −2 − 0.72 z −3

, using Eq. (8.132a) and 1 − 2.35z −1 + 3.3z − 2 − 1.575z − 3 − 0.4167 − 0.9167z −1 + z −2 . Repeating the procedure we get (8.132b) we get G2 ( z ) = 1 − 3.0063z −1 + 1.8562 z − 2 − 1.2232 + z −1 . From G 3 ( z), G 2 ( z ), and G1( z), we arrive at the cascaded lattice G1( z) = 1 − 0.7357z −1 realization shown below: _1.8562

1.575

_ _

_1.2232

0.4167

z–1

0.72

(b) From G 4 ( z ) = H 2 ( z ) =

0.7357

z

–1

z

–1

9.3 − 39.3z −1 − 13.025z −2 + 6.55z −3 + 1.9125z −4

, using Eq. 1 + 0.55z −1 + 0.095z − 2 − 0.01z − 3 − 0.003z − 4 − 22.8896 − 7.1679z −1 + 3.4235z −2 + z −3 . (8.132a) and (8.132b) we get G3 ( z) = 1 + 0.4813z −1 + 0.0735z − 2 + 0.00027z − 3 3.8259 + 5.0744 z −1 + z −2 and Repeating the procedure twice we get G2 ( z) = 1 + 0.4803z −1 + 0.0721z − 2 4.4704 + z −1 . From G 4 ( z ), G 3 ( z), G 2 ( z), and G1( z), we arrive at the cascaded G1( z) = 1 + 0.1579z −1 lattice realization shown below: 0.003 9.3 1.9125

_

_ 0.0721

0.00027

_

22.8896 –1 z

(c) From G3 ( z ) = H 3 ( z ) =

4.7915 0.9179

3.8259 z–1

z–1

z–1

7.8 − 8.96 z −1 + 4.62 z −2 + 4.92 z −3

, using Eq. (8.132a) and 1 + 0.7 z −1 − 6.68z − 2 + 2.356 z − 3 1.07 − 4.209z −1 + z −2 . Repeating the procedure we get (8.132b) we get G2 ( z) = 1 − 1.8209z −1 + 3.2364 z − 2

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260

G1( z) =

0.9179 + z −1

. From G 3 ( z), G 2 ( z ), and G1( z), we arrive at the cascaded lattice 1 − 4.7915z −1 realization shown below: _ 2.356

_3.2364

4.7915

7.8

1.07

0.9179

z

4.92

z–1

–1

z

–1

8.32 When H N (z) is an allpass transfer function of the form H N ( z ) = AN ( z ) =

d N + d N −1 z −1 + L + d 1z − N + 1 + z − N , then from Eq. (8.132a), the 1 + d 1z −1 + L + d N −1 z − N + 1 + d N z − N

numerator coefficients of H N −1 ( z ) are given by − p k + 1 d N d k + 1 − d N − k −1 p d = , and from Eq. (8.132b) the denominator p k' = 0 k + 1 2 −1 p0 d N − p N dN coefficients of H N −1 ( z) are given by p N − k −1d N − d N − k − 1 d k + 1d N − d N − k − 1 ' dN = = pk' , implying H N −1 ( z) is an − k −1 = 2 −1 p0 d N − pN dN allpass transfer function of order N − 1. Since here p N = 1 and p0 = d N , the lattice structure of Problem 8.30 then reduces to the lattice structure employed in the Gray-Markel realization procedure. 8.33 (a) Consider the realization of Type 1B allpass structure. From its transfer parameters given in Eq. (8.50b) we arrive at Y1 = z −1 X1 + (1 + z −1 ) X 2 = z −1 ( X1 + X 2 ) + X 2 , and Y2 = (1 − z −1 ) X1 − z −1 X 2 = X1 − z −1 ( X1 + X 2 ). A realization of the two-pair based on these two equations is as shown below which leads to the structure of Figure 8.24(b).

X1

+

_1

_ z 1

X2

+ Y2

+

Y1

(b) From the transfer parameters of Type 1At allpass given in Eq. (8.50c) we obtain Y1 = z −1 X1 + X 2 , and Y2 = (1 − z −2 ) X1 − z −1 X 2 = X1 − z −1 ( z −1 X1 + X 2 ) = X1 − z −1Y1 . A realization of the two-pair based on these two equations is as shown on next page which leads to the structure of Figure 8.24(c).

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261

_ z 1

X1

+

Y1

_1

X2

z

Y2

_1

+

(c) From the transfer parameters of Type 1At allpass given in Eq. (8.50d) we obtain Y1 = z −1 X 1+ (1 − z −1 ) X 2 = z −1 ( X1 − X 2 ) + X 2 , and Y2 = (1 + z −1 ) X1 − z −1 X 2

= X1 + z −1 ( X1 − X 2 ). A realization of the two-pair based on these two equations is as shown below which leads to the structure of Figure 8.24(d). _ z 1

+

X1

+

_1

Y2

X2

+

Y1

8.34 (a) A cascade connection of 4 Type 1A first-order allpass networks is shown below: _ z 1

+

_ z 1

+

+

+

_1

_1

a

_ z 1

_ z 1

+

+

_1 _ z 1

b

_ z 1

+

+

_1 _ z 1

c

_ z 1

d

Simple block diagram manipulation of the above structure leads to: _ z 1

+ _1

+

_ z 1

+

_ z 1

+

z

c

b _1

+

+

_1

z

z

a

_ z 1

+

_1

_1

_ z 1

+

d

_1

_1

Finally, by delay sharing between adjacent allpass sections we arrive at the following equivalent realization requiring now 5 delays compared to 8 delays in the direct realization shown on the previous page. +

+

_ z 1

+

+

_ z 1

+

+

_ z 1

_1 _ z 1

a _1

+

d

c

b

_ z 1

+

_1

_1

(b) A cascade connection of 4 Type 1At first-order allpass networks is shown below which requires 8 delays. _ _ _ _ z 1 + z 1 + z 1 + z 1 + a

+

_ z 1 _1

b

+

_ z 1 _1

Not for sale

c

+

_ z 1 _1

d

+

_ z 1 _1

262

By delay sharing between adjacent allpass sections we arrive at the following equivalent realization requiring now 5 delays. _ _ _ _ + + z 1 z 1 z 1 + z 1 + a _1

_ z 1

_1

_1

_1

+

+

+

+

d

c

b

8.35 The structure of Figure P8.15 with internal variables labeled is shown below: –1 –1

X(z)

W(z)

z

a

U(z)

Y(z) –1

Its analysis yields (1): W ( z ) = X ( z ) − z −1U ( z ), (2): U ( z ) = a W ( z ) + z −1U ( z ), and (3): a W ( z). Substituting Eq. Y ( z ) = −W ( z ) + U ( z ). From Eq. (2) we obtain (4): U ( z ) = 1 − z −1 ⎛ 1 − a − z −1 ⎞ ⎟W ( z ). Substituting Eq. (4) in Eq. (1) we get (4) in Eq. (3) we get (5): Y ( z ) = −⎜⎜ −1 ⎟ − 1 z ⎝ ⎠ ⎛ 1 − a − z −1 + a z −1 ⎞ ⎟W ( z ). Finally, from Eqs. (5) and (6) we arrive at (6) : X ( z ) = ⎜ ⎜ ⎟ −1 1 z − ⎝ ⎠ − 1 Y ( z) − (1 − a ) + z . H (z) = = X ( z) 1 − (1 − a)z −1 8.36 We realize A2 ( z ) =

d1d 2 + d1 z −1 + z −2 in the form of a constrained three-pair as 1 + + d1 z −1 + d1d 2 z − 2

indicated below: X1

Y1

t11 t12 t13

Y2

t21 t22 t23 t31 t32 t33

_d

X 3 d2

X2

Y3

⎡ Y1 ⎤ ⎡ t 11 t 12 From the above figure, we have ⎢⎢Y2 ⎥⎥ = ⎢⎢t 21 t 22 ⎢⎣Y3 ⎥⎦ ⎢⎣t 31 t 32

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1

t 13 ⎤ ⎡ X1 ⎤ t 23 ⎥⎥ ⎢⎢ X 2 ⎥⎥, and X 2 = − d1Y 2, t 33 ⎥⎦ ⎢⎣ X 3 ⎥⎦

263

Y N ( z) , where X 3 = d 2Y 3. From these equations, we get after some algebra A2 ( z ) = 1 = X1 D ( z ) D( z) = 1 + d1t 22 − d 2 t 33 + d1d 2 (t 23 t 32 − t 22 t 33 ), and N ( z) = t11 + d1 (t11t 22 − t12 t 21 ) − d 2 (t11t 33 − t13 t 31 ) + d1d 2 {t 21 (t12 t 33 − t13 t 32 ) + t 31 (t 22 t13 − t12 t 23 ) + t11 (t 23 t 32 − t 22 t 33 )}. Comparing the denominator of the desired allpass transfer function with D (z ) we get

t 22 = z −1 , t 33 = 0, t 23 t 32 = z −2 . Next, comparing the numerator of the desired allpass transfer function with N (z ) we get t11 = z −2 , t12 t 21 = z −1 ( z −2 − 1), t13 t 31 = 0, and t 32 (t11t 23 − t 21t13 ) + t 31 (t 22 t13 − t12 t 23 ) = 1. Substituting the appropriate transfer parameters from the previous equations into the last equation we simplify it to t13 t 21t 32 + t 31t12 t 23 = z −4 − 1. Since t13 t 31 = 0, either t13 = 0, or t 31 = 0. (Both cannot be simultaneously equal to zero, as this will violate the condition t13 t 21t 32 + t 31t12 t 23 = z −4 − 1. Consider the case t13 = 0. Then the above equation reduces to t 31t12 t 23 = z −4 − 1. From this equation and t 23 t 32 = z −2 , it follows that t 32 = z −2 , t 23 = 1, t 31t12 = z −4 − 1 = ( z −1 − 1)( z −1 + 1)( z −2 + 1). There are four possible realizable sets of values of t 21 and

t 31 satisfying the last equation and t12 t 21 = z −1 ( z −2 − 1). These lead to 4 different ⎡ z−2 z − 2 − 1 0⎤ ⎥ ⎢ −1 realizable transfer matrices for the three-pair: Type 2A: ⎢ z 1 ⎥, z −1 ⎢z − 2 + 1 0⎥ z−2 ⎦ ⎣ − 2 − 1 − 2 − ⎡ ⎡ z z + 1 0⎤ z z 1 − 1 0⎤ ⎥ ⎥ ⎢ ⎢ 1⎥, Type 2C: ⎢ z −1 ( z −1 + 1) 1 ⎥, Type 2B: ⎢ z −1 ( z −1 − 1) z −1 z −1 ⎢( z − 2 + 1)( z −1 − 1) ⎢( z − 2 + 1)( z −1 + 1) 0⎥ 0⎥ z−2 z−2 ⎦ ⎦ ⎣ ⎣ − 2 ⎡ 1 0⎤ z ⎥ ⎢ −1 − 2 Type 2D: ⎢ z ( z − 1) z −1 1 ⎥. A realization of each of the above Type 2 allpass ⎢ z−4 −1 z − 2 0⎥ ⎦ ⎣ structures is obtained by implementing its respective transfer matrix, and then constraining the Y2 and X 2 variables through the multiplier d1 and constraining the Y3 and X 3

variables through the multiplier − d2 resulting in the four structures shown in Figure 8.25 of the text. It can be easily shown that the allpass structures obtained for the case t 31 = 0 are precisely the transpose of the structures of Figure 8.25. 8.37 A cascade of two Type 2D second-order allpass networks is shown below which requires 8 delays. Not for sale

264

w[n _1] x[n] _1

+

_ z 1

_ z 1

a

b

+

w[n]

+

_ z 1

_ z 1

_1

c

+

_ z 1

y[n]

_1

_1

w[n _1]

+

d

+

_1

w[n _2]

w[n _2]

_ z 1

_1

z

z

By delay sharing between adjacent allpass sections we arrive at the following equivalent realizations requiring now 6 delays.

x[n] _1

w[n _ 2]

+

_ z 1

_ z 1

a

b

+

+

_1

w[n _1]

w[n] _1

+

_ z 1

c

_ z 1

+

y[n]

d

+ _1

_ z 1

_ z 1

The minimum number of delays needed to implement a cascade of Type 2D second-order allpass sections is thus 4 + 2( M − 1) = 2( M + 1).

d 2 + d1 z −1 + z −2 in the form of a constrained three-pair as indicated 1 + d1 z −1 + d 2 z − 2 in the solution of Problem 8.36. Comparing the numerator and the denominator of the Type 3 allpass transfer function with N (z ) and D (z ) given in the solution of Problem 8.36 we arrive at t 11= z −2 , t 22 = z −1 , t 22 = z −2 , t 33 = z −2 , t 23 t 32 = z −3 , t 12 t 212 = z −1 ( z −2 − 1),

8.38 We realize A2 ( z ) =

t 13 t 31= z −4 − 1, t 13 t 21t 32 + t 13 t 21t 23 = z −3 ( z −2 − 1) + z −1 ( z −4 − 1). To solve the last four equations to arrive at a set of realizable set of values for the transfer parameters, t12 , t 21 , t13 , t 31 , t 23 , and t 32 , we need to pre-select t 23 and t 32 satisfying the condition t 23t 32 = z −3 , and then determine realizable values for the other transfer parameters. (a) Let t 23 = z −1 , t 32 = z −2 . There are 4 possible realizable sets of values of t12 , t 21 , t13 , and t 31 satisfying the constraints given above resulting in the following transfer matrices of the three-pair:

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265

⎡ z−2 ⎡ z − 2 − 1 z − 2 − 1⎤ z−2 z −1 + 1 z −1 + 1⎤ ⎥ ⎥ ⎢ ⎢ Type 3A: ⎢ z −1 z −1 z −1 ⎥, Type 3B: ⎢ z −1 ( z −1 − 1) z −1 z −1 ⎥, ⎢z − 2 + 1 ⎢( z −1 − 1)( z − 2 + 1) z−2 z−2 ⎥ z−2 z−2 ⎥ ⎦ ⎦ ⎣ ⎣ − 2 − 1 − 1 − 2 ⎡ ⎡ 1 1 ⎤ z z − 1 z − 1⎤ z ⎥ ⎥ ⎢ ⎢ Type 3C: ⎢ z −1 ( z −1 + 1) z −1 z −1 ⎥, Type 3D: ⎢ z −1 ( z − 2 − 1) z −1 z −1 ⎥. ⎢( z −1 + 1)( z − 2 + 1) ⎢ z−4 −1 z−2 z−2 ⎥ z−2 z−2 ⎥ ⎦ ⎦ ⎣ ⎣

(b) t 23 = z −2 , t 32 = z −1. This leads to 4 new realizations which are transpose of the 4 structures developed in Case (a). (c) t 23 = 1, t 32 = z −3 . Here also there are 4 different realizations whose transfer matrices are as given below: ⎡ z−2 z −1 ( z − 2 − 1) z − 2 − 1⎤ ⎥ ⎢ Type 3E: ⎢ 1 z −1 z −1 ⎥, ⎢z − 2 + 1 z−3 z−2 ⎥ ⎦ ⎣ − 2 − 1 − 1 ⎡ z z ( z + 1) z −1 + 1⎤ ⎥ ⎢ 1 ⎥, Type 3F: ⎢ z −1 − 1 z −1 ⎢( z −1 − 1)( z − 2 + 1) z−3 z−2 ⎥ ⎦ ⎣ − 2 − 1 − 1 − 1 ⎡ ⎡ z−2 1 ⎤ z z ( z − 1) z − 1⎤ z −1 ⎥ ⎥ ⎢ ⎢ 1 ⎥, Type 3H: ⎢ z − 2 − 1 z −1 1 ⎥. Type 3G: ⎢ z −1 + 1 z −1 ⎢( z −1 + 1)( z − 2 + 1) ⎢z − 4 − 1 z − 3 z − 2 ⎥ z −3 z−2 ⎥ ⎦ ⎦ ⎣ ⎣ (d) t 23 = z −3 , t 32 = 1. This leads to 4 new realizations which are transpose of the 4 structures developed in Case (b). 8.39 A cascade of two Type 3B second-order allpass networks is shown below which requires 8 delays.

+

+

_1

c

a x[n]

_1

z

+

_1

_1

+

z

w[n]

_1

z

+

_ z 1

+

_1

z

b _1

+

w[n _ 2]

Not for sale

z

d

w[n _1]

_ z 1

y[n] _1

_1

_ z 1

+

266

By delay sharing between adjacent allpass sections we arrive at the following equivalent realizations requiring now 7 delays.

+

+

_1

w[n _1]

a x[n]

_1

z

_1

+

+

z

w[n]

_1

_1

c _ z 1

+

z

z

d

w[n _ 2] _1

_1

+

y[n] _1

_ z 1

b

+

_ z 1

+

The minimum number of delays needed to implement a cascade of Type 3B second-order allpass sections is thus 4 + 3( M − 1) = 3( M + 1). 8.40 From the transfer parameters given in Eq. (8.50d) we arrive at the input-output relations of the two-pair as (1): Y1 = k m X1 + z −1 X 2 , and (2): Y2 = (1 − k m2 ) X1 − k m z −1 X 2 . A threemultiplier realization based on these two equations is shown below: 1 _ km2

+

X1

_k

km Y1

Y2 m

_ z 1

+

X2

8.41 Eq. (2) in the solution of Problem 8.40 can be rewritten as: 2 X − k z −1 X = X − k (k X + z −1 X ). Substituting Eq. (1) of the Y2 = X1 − k m 1 m 2 1 m m 1 2

solution of Problem 8.40 in this equation we then get (3): Y2 = X1 − k m Y1 . A realization of the two-pair based on Eq. (3) and Eq. (1) of the solution of Problem 8.40 results in the lattice structure shown below: X1

Y2 _k

m

km Y1

z–1

X2

From Eqs. (1) and (2) of the solution of Problem 8.40 we arrive at the transfer 2, parameters of the lattice two-pair given by t11 = k m , t12 = z −1 , t 21 = 1 − k m

t 22 = − k m z −1 . The corresponding chain parameters are obtained using Eq. (7.128b)

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and are given by A =

km k z −1 z −1 ,B = m ,C= ,D= . 2 2 2 2 1 − km 1 − km 1 − km 1 − km 1

X1 (z)

W 1 (z) _k

_k

2

1

k1

k2 Y 1(z)

z

–1

z

–1

S 1 (z)

Hence, the input-output relation of the all-pole cascade lattice structure given above can ⎡ 1 k 2 z −1 ⎤ ⎡ 1 k1 z −1 ⎤ ⎡W1 ( z)⎤ ⎡ X ( z )⎤ 1 be expressed as: ⎢ 1 ⎥ = ⎥⎢ ⎥⎢ ⎢ ⎥, from z −1 ⎦⎥ ⎣⎢ k1 z −1 ⎦⎥ ⎣ S1 ( z ) ⎦ ⎣ Y1 ( z ) ⎦ (1 − k 2 )(1 − k 2 ) ⎣⎢ k 2 2

which we obtain H r ( z ) =

1

(1 − k 22 )(1 − k12 ) W1 ( z ) . = X1 ( z ) 1 + k1 (1 + k 2 )z −1 + k 2 z − 2

X 1(z)

W 1 (z) _k

_k

2

1

k1

k2

z–1

Y 1(z)

z–1

S 1 (z)

Likewise, from Section 8.8.1, we arrive at input-output relation of the all-pole cascade ⎡ X ( z )⎤ ⎡ 1 k 2 z −1 ⎤ ⎡ 1 k1 z −1 ⎤ ⎡W1 ( z )⎤ lattice structure given above as: ⎢ 1 ⎥ = ⎢ ⎥⎢ ⎥⎢ ⎥, from z −1 ⎥⎦ ⎢⎣ k1 z −1 ⎥⎦ ⎣ S1 ( z ) ⎦ ⎣ Y1 ( z ) ⎦ ⎢⎣ k 2 W1 ( z) 1 . = X1 ( z) 1 + k1 (1 + k 2 )z −1 + k 2 z − 2 Now a second-order all-pole transfer function can also be expressed in terms of its 1 . Comparing the denominator of with that poles as H ( z) = 1 − 2r cos(ω o )z −1 + r 2 z − 2

which we obtain H s ( z) =

of H s (z) and H r ( z), we observe k 2 = r 2 and k1 (1 + k 2 ) = −2r cos(ωo ). As r ≅ 1, 2r cos(ωo ) k1 = − ≅ − cos(ωo ). At the true resonance peak, we have 1+ r2 1 1 . Likewise, at ω = ωr , H s (e jωr ) = . As H (e ωr ) = (1 − r 2 ) sin(ωo ) (1 − k 2 ) 1 = k12 r → 1, k 2 = r 2 → 1, and (1 − k12 )(1 − k 22 ) = (1 − k12 )(1 − k 2 )(1 + k 2 ) ≅ 2(1 − k12 )(1 − k 2 ).

Hence, at ω = ωr , H r (e jωr ) ≅

2(1 − k12 )(1 − k2 ) (1 − k 2 ) 1 − k12

= 2 1 − k12 = 2 sin(ωr ) . This

indicates that the gain at the peak resonance is approximately independent of the pole

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radius as long as the pole angle is constant and the radius is close to 1. 3.9 + 2.3z −1 + z −2

8.42 (a) H1 ( z) =

. Choose A2 ( z ) =

0.5 + 0.3z −1 + z −2

. Note 1 + 0.3z −1 + 0.5z − 2 1 + 0.3z −1 + 0.5z − 2 k2 = A2 (∞) = d 2 = 0.5 < 1. Using Eq. (8.45) we next determine the coefficient d1' of 0.2 + z −1 the allpass transfer function A1 ( z) arriving at A1 ( z) = . Here, 1 + 0.2 z −1 k1 = A1 (∞ ) = d1' = 0.2 < 1. Therefore, A2 ( z) and hence, H1 ( z) is stable. The feed-forward coefficients are given by α1 = p2 = 1.0, α2 = p1 − α1d 1= 2.0,

α3 = p0 − α1d 2 −α2 d1' = 3.0. Final realization of H1 ( z) is as shown below: X1 _

_

0.5

0.2 0.2

0.5

z

–1

z

–1 3

2

Yo

(b) H 2 ( z) =

2.6 + 0.74 z −1 + 3z −2

. Choose A2 ( z) =

0.4 − 0.42 z −1 + z −2

. Note 1 − 0.42 z −1 + 0.4 z − 2 1 − 0.42 z −1 + 0.4 z − 2 k2 = A2 (∞) = d 2 = 0.4 < 1. Using Eq. (8.45) we next determine the coefficient d1' of − 0.3 + z −1 the allpass transfer function A1 ( z) arriving at A1 ( z ) = . Here, 1 − 0.3z −1 k1 = A1 (∞ ) = d1' = 0.3 < 1. Therefore, A2 ( z) and hence, H 2 ( z) is stable.

The feed-forward coefficients are given by α1 = p2 = 3.0, α2 = p1 − α1d 1= 2.0, α3 = p0 − α1d 2 −α2 d1' = 2.0. Final realization of H 2 ( z) is as shown below: X1 _

0.4

_ 0.3

0.4

0.3

z–1

z 2

3

–1 2

Yo

(c) H 3 ( z) =

− 3 + 5.192 z −1 − 3.56 z −2 + 2 z −3 1 − 0.28 z −1 + 0.056 z − 2 + 0.4 z − 3

Not for sale

. Choose

269

A3 ( z) =

0.4 + 0.056 z −1 − 0.28z −2 + z −3

Note k3 = A3 (∞ ) = d3 = 0.4 < 1. Using 1 − 0.28z −1 + 0.056 z − 2 + 0.4 z − 3 Eq. (8.45) we next determine the coefficient d1' and d 2' of the allpass transfer function 0.2 − 0.36 z −1 + z −2

. Now k2 = A2 (∞) = d 2 = 0.2 < 1. 1 − 0.36 z −1 + 0.2 z − 2 Using Eq. (8.45) we next determine the coefficient d1" of the allpass transfer function

A2 ( z) arriving at A2 ( z) =

− 0.3 + z −1

A1 ( z) arriving at A1 ( z ) =

1 − 0.3z −1

. Here, k1 = A1 (∞ ) = d1' = 0.3 < 1. Therefore,

A3 ( z) and hence, H 3 ( z) is stable. The feed-forward coefficients are given by α1 = p3 = 2.0, α 2 = p2 − α1d 1= −3.0, α3 = p1 − α1d 2 −α2 d1' = 4.0, α 4 = p0 − α1d 3 −α2 d 2' − α3 d1" = −2.0. Final realization of H 3 ( z) is as shown below: X1 _ 0.4

_ 0.2

0.4

0.3 _

0.2 _1

_1

z

2

_

z1

z

_3

0.3

4

_2

Yo

(d) H 4 ( z) =

1.63 + 3.112 z −1 + 0.84 z −2 + 2 z −3

. Choose 1 + 0.42 z −1 + 0.056 z − 2 − 0.3z − 3 − 0.3 + 0.056 z −1 + 0.42 z −2 + z −3 A3 ( z) = Note k3 = A3 (∞) = d3 = 0.3 < 1. Using 1 + 0.42 z −1 + 0.056 z − 2 − 0.3z − 3 Eq. (8.45) we next determine the coefficients d1' and d 2' of the allpass transfer function


0.2 + 0.48z −1 + z −2

. 1 + 0.48z −1 + 0.2 z − 2 Using Eq. (8.45) we next determine the coefficient


0.4 + z −1 1 + 0.4 z −1

Now k2 = A2 (∞) = d 2 = 0.2 < 1. d1" of the allpass transfer function

. Here, k1 = A1 (∞ ) = d1' = 0.4 < 1. Therefore,

A3 ( z) and hence, H 4 ( z) is stable. The feed-forward coefficients are given by α1 = p3 = 2.0, α 2 = p2 − α1d 1= 0, α3 = p1 − α1d 2 −α2 d1' = 3.0, α 4 = p0 − α1d 3 −α2 d 2' − α3 d1" = 1.0. Final realization of H 3 ( z) is as shown on the next page.

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270

X1 _

0.4 _1 _

_1

z1

_1

z

0.6

z

3.2

2.2

Yo

2 + 0.6 z −1 − z −2 + 0.3z −3

(e) H 5 ( z ) =

. Choose 1 − 0.1z −1 + 0.18z − 2 + 0.08z − 3 − 0.08z − 4 − 0.08 + 0.08z −1 + 0.18 z −2 − 0.1z −3 + z −4 A4 ( z) = . Note 1 − 0.1z −1 + 0.18 z − 2 + 0.08 z − 3 − 0.08z − 4 k 4 = A4 (∞) = d 4 = 0.08 < 1. Using Eq. (8.45) we next determine the coefficients

d1' , d 2' and d3' of the allpass transfer function A3 ( z) arriving at

A3 ( z) =

0.0725 + 0.195z −1 − 0.0942 z −2 + z −3

. Now 1 − 0.0942 z −1 + 0.1957z − 2 + 0.0725z − 3 k3 = A3 (∞) = d3' = 0.0725 < 1. Using Eq. (8.45) we next determine the coefficients d1"

and d 2" of the allpass transfer function A2 ( z) arriving at A2 ( z ) =

0.2035 − 0.1090 z −1 + z −2 1 − 0.1090 z −1

+ 0.2035z − 2

. Now k 2 = A2 (∞) = d 2" = 0.2035 < 1. Using Eq.

(8.45) we next determine the coefficient d1' ' ' of the allpass transfer function A1 ( z) arriving at A1 ( z ) =

− 0.009058 + z −1 1 − 0.0905z −1

. Here, k1 = A1 (∞) = d1' ' ' = 0.0905 < 1.

Therefore, A4 ( z) and hence, H 5 ( z ) is stable. The feed-forward coefficients are given by α1 = p 4 = 0, α 2 = p3 − α1d 1= −0.3,

α3 = p2 − α1d 2 −α2 d1' = −1.0283, α 4 = p1 − α1d 3−α 2 d 2' − α 3 d1" = 0.5467, α5 = p0 − α1d 4 −α2 d3' − α3 d 2" − α 4 d1' ' ' = 2.2805. Final realization of H 5 ( z ) is as shown below: X1 _

_ 0.0725

0.08

0.0725 _

0.3

0.0905

0.2035

_ 0.0905

_1

z1 _

0.2035

z _

1.0283

_

z1

0.5467

_

z1

2.2805

Yo

8.43 (a) H1 ( z ) =

0.32 z −1 + 0.66 z −2 − 0.72 z −3 1 − 2.35z −1 + 3.36 z − 2 − 1.575z − 3

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. Choose

271

− 1.575 + 3.36 z −1 − 2.35z −2 + z −3

. Note k3 = A3 (∞) = d3 = 1.575 > 1. 1 − 2.35z −1 + 3.36 z − 2 − 1.575z − 3 Therefore, A3 ( z) and hence, H 3 ( z) is unstable. Using Eq. (8.45) we next determine A3 ( z) =

the coefficients d1' and d 2' of the allpass transfer function A2 ( z) arriving at 3.5105 − 2.6308z −1 + z −2 A2 ( z ) = . Now k2 = A2 (∞) = d 2 = 3.5105 > 1. Using Eq. 1 − 2.6308z −1 + 3.5105z − 2 (8.45) we next determine the coefficient d1" of the allpass transfer function A1 ( z) arriving at A1 ( z ) =

− 0.5833 + z −1 1 − 0.5833z −1

. Here, k1 = A1 (∞ ) = d1" = 0.5833 < 1.

The feed-forward coefficients are given by α1 = p3 = 2.0, α 2 = p2 − α1d 1= 0, α3 = p1 − α1d 2 −α2 d1' = 3.0, α 4 = p0 − α1d 3 −α2 d 2' − α3 d1" = 1.0. Final realization of H 3 ( z) is as shown below: X1 _ 3.5105

1.575 _ 1.575

0.5833 _

3.5105 _1

0.5833

_1

z

_

z1

z

3

2

Yo

(b) H 2 ( z) =

9.3 − 39.3z −1 − 13.025z −2 + 6.552 z −3 + 1.9125z −4

. Choose 1 + 0.55z −1 + 0.095z − 2 − 0.01z − 3 − 0.003z − 4 − 0.003 − 0.01z −1 + 0.095z −2 + 0.55z −3 + z −4 A4 ( z) = . Note 1 + 0.55z −1 + 0.095z − 2 − 0.01z − 3 − 0.003z − 4 k 4 = A4 (∞) = d 4 = 0.003 < 1. Using Eq. (8.45) we next determine the coefficients d1' , d 2' and d3' of the allpass transfer function A3 ( z) arriving at − 0.0084 + 0.0953z −1 + 0.55z −2 + z −3 A3 ( z) = . Now 1 + 0.55z −1 + 0.0953z − 2 − 0.0084 z − 3

k3 = A3 (∞ ) = d3' = 0.0084 < 1. Using Eq. (8.45) we next determine the coefficients d1" and d 2" of the allpass transfer function A2 ( z) arriving at 0.0999 + 0.5508z −1 + z −2 A2 ( z) = . Now k2 = A2 (∞) = d 2 = 0.0999 < 1. Using Eq. 1 + 0.5508z −1 + 0.0999z − 2 (8.45) we next determine the coefficient d1" of the allpass transfer function A1 ( z)

arriving at A1 ( z ) =

0.5088 + z −1 1 + 0.5088z −1

. Here, k1 = A1 (∞ ) = d1" = 0.5088 < 1. As a result,

A3 ( z) , and hence, H 2 ( z) is stable.

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The feed-forward coefficients are given by α1 = p 4 = 1.9125, α 2 = p3 − α1d 1= 5.5001, α 3 = p 2 − α1d 2 −α 2 d1' = −16.2316, α 4 = p1 − α1d 3−α 2 d 2' − α 3 d1" = −30.8644, α 5 = p0 − α1d 4 −α 2 d 3' − α 3 d 3" − α 4 d 4' ' ' = 26.4295. Final realization of H 2 (z ) is as shown below: X1 _

0.003

0.0084

0.003

_ 0.0084

_

0.0999

_

_

z1

_ 30.8644

1.6985 _ z 1

_ z 1 _ 0.7198

+

0.1891

0.1574

+

+

_ z 1

y[n]

+

+ 1.6985

0.2545

26.4295 Yo

. A direct form II realization of

+

1 − 0.1891z −1 + 0.7198 z − 2 − 0.1574 z − 3 H (z ) is shown below:

(b) H ( z ) =

z

_ 16.2316

0.2545(1 + 1.6985 z −1 + 1.6985 z −2 + z −3 )

x[n]

_1

z1

5.5001

1.9125

0.5088 0.5088

_

z1

8.44 (a) H ( z ) =

_

0.0999

0.2545(1 + z −1 )(1 + 0.6985 z −1 + z

−2

)

(1 − 0.2169 z −1 )(1 + 0.0278 z −1 + 0.7258 z H (z ) is shown below:

−2

. A cascade realization of )

0.2545 x[n]

+

+

+

+

y[n]

_ z 1

_ z 1

_

+

0.6985

0.0278

0.2169

+ _ z 1

_ 0.7285

(c) We first form A3 ( z) =

− 0.1574 + 0.71985z −1 − 0.1891z −2 + z −3 1 − 0.1891z −1 + 0.7198z − 2 − 0.1574 z − 3

. Now

k3 = A3 (∞ ) = d3' = 0.1574 < 1. Using Eq. (8.45) we next determine the coefficients d1" and d 2" of the allpass transfer function A2 ( z) arriving at

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273

0.7075 − 0.0777z −1 + z −2

. Now k2 = A2 (∞) = d 2 = 0.7075 < 1. Using Eq. 1 − 0.0777z −1 + 0.7075z − 2 (8.45) we next determine the coefficient d1" of the allpass transfer function A1 ( z) A2 ( z ) =

arriving at A1 ( z ) =

− 0.0455 + z −1

. Here, k1 = A1 (∞ ) = d1" = 0.0455 < 1. 1 − 0.0455z −1 The feed-forward coefficients are given by α1 = p3 = 0.2545, α2 = p2 − α1d 1= 0.4804, α3 = p1 − α1d 2 −α2 d1' = 0.2864, α 4 = p0 − α1d 3 −α2 d 2' − α3 d1" = −0.0323. The Gray-Markel realization of H (z ) is as shown below: X1 _

0.1574 _ 0.1574

0.0455

0.7075

_ 0.0455

0.7075 _1

_1

z

0.4804

0.2545

_1

z

z

_ 0.0323

0.2864

Yo

(d) Using Eqs. (8.132a) and (8.132b) we determine the coefficients of H 2 ( z) : 1.6309 + 0.8456 z −1 + z −2 H 2 ( z) = . Next, using Eqs. (8.132a) and (8.132b) we 1 + 0.0676 z −1 + 0.8529z − 2 − 1.8808 + z −1 . A cascaded lattice determine the coefficients of H1 ( z) : H1 ( z) = 1 + 1.6718z −1 realization of using the method of Problem 8.30 is thus as shown below: X1 _

0.1574

1.6309

0.2545 Y1

0.2545

0.8529

z

–1

z

_

1.6718

_

1.8808

–1

z

–1

A comparison of the hardware requirements of the four realizations is as follows: Direct form: No. of multipliers = 7 and no. of 2-input adders = 6, Cascade form: No. of multipliers = 10 and no. of 2-input adders = 9, Gray-Markel realization: No. of multipliers = 5 and no. of 2-input adders = 6, Cascaded lattice realization: No. of multipliers = 6 and no. of 2-input adders = 6. Hence, the Gray-Markel realization has the smallest hardware requirements. 8.45 (a) A partial-fraction expansion of G (z ) is of the form N /2

N /2 N /2 υi υi υi* d + G( z ) = d + . If we define H ( z ) = + , then we can * − z − λ z λ 2 z − λ i i i i =1 i =1 i =1

∑

∑

∑

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274

write G( z) = H ( z) + H* ( z), where H* ( z) represents the transfer function obtained from H (z ) by conjugating its coefficients. (b) In this case, the partial-fraction expansion of G (z ) is of the form Nr

N /2

N /2

c c ρi υi υi* , where N r and N c are the number G( z) = d + + + * z − ξ z − λ z − λ i i i i =1 i =1 i =1 of real poles ξ i ’s and complex poles λi ’s, respectively, with residues ρi ’s and υi ’s.

∑

∑

∑

We can thus decompose G (z ) as G( z) = H ( z) + H* ( z), where N

N /2

c r ρi / 2 υi d . H (z) = + + z − λ 2 i =1 z − ξ i i i =1

∑

∑

(c) An implementation of real coefficient G (z ) is thus simply a parallel connection of two complex-coefficient filters characterized by transfer functions H (z ) and H* ( z) as indicated in the figure below: x[n]

x[n]

H(z)

real imaginary

H* (z)

=0

However, for a real-valued input x[ n], the output of H (z ) is the complex conjugate of that of H* ( z). As a result, two times the real part of the output of H (z ) is the desired real-valued output sequence y[n] indicating that a single complex-coefficient filter H (z ) is sufficient to realize G (z ) as indicated below: 2

x[n]

H(z)

y[n]

Y ( z) A + jB = , we arrive at the difference equation X ( z ) 1 + (α + j β ) z − 1 representation yre [ n] + j yim [ n] = −(α + jβ )(yre [ n − 1] + j yim [ n − 1]) + Ax[ n] + jBx[ n], which is equivalent to a set of two difference equations involving all real variables and real multiplier coefficients: yre [ n] = −α yre [ n − 1] + β yim [ n − 1] + A x[ n] and yim [ n] = − β yre [ n − 1] − α yim [ n − 1] + B x[ n]. A realization of H (z ) based on the last two equations is shown on next page. To determine the transfer function Yre ( z) / X ( z ), we take the z -transform of the last

8.46 From H ( z) =

two difference equations and arrive at (1 + α z −1 )Yre ( z) − β z −1Yim ( z) = A X ( z) and

β z −1Yre ( z ) + (1 + α z −1 )Yim ( z ) = B X ( z). Solving these two equations we get

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275

Yre ( z ) A + ( Aα + Bβ )z −1 , which is seen to be a second-order transfer = X ( z ) 1 + 2αz −1 + (α 2 + β 2 )z − 2 function. A y re [n]

x[n] _α

–1

z

_β

B

y im [n] _α

z

–1

β

To determine the transfer function Yre ( z) / X ( z ), we take the z -transform of the last two difference equations and arrive at (1 + α z −1 )Yre ( z) − β z −1Yim ( z) = A X ( z) and

β z −1Yre ( z ) + (1 + α z −1 )Yim ( z ) = B X ( z). Solving these two equations we get Yre ( z ) A + ( Aα + Bβ )z −1 , which is seen to be a second-order transfer = X ( z ) 1 + 2αz −1 + (α 2 + β 2 )z − 2 function. 8.47 An m -th order complex allpass function is given by * + α * z −1 + L + α * z − m + 1 + z − m αm m −1 1 . To generate an ( m − 1) –th Am ( z ) = − 1 − 2 1 + α 1 z + α 2 z + L + + α m −1 z − m + 1 + α m z − m * ⎤ ⎡ A (z) − km Pm −1 ( z ) = z⎢ m ⎥. Dm −1 ( z ) ⎢⎣ 1 − k m Am ( z ) ⎦⎥ Substituting the expression for A m (z ) in the above equation we obtain after some −2 + L + α * z − m + 1 + z − m * + α * z −1 + α * algebra Pm −1 ( z ) = z[α m 1 m −1 m−2z

order allpass function, we use the recursion Am −1 ( z) =

− m + 1 + α z − m )] * (1 + α z −1 + α z −2 + L + α − αm 1 2 m −1 z m

* −1 − α m * α1 ) + (α m * − 2 − αm * α 2 ) z − 1 + L + (α1* − α m * α m − 1 ) z − m + 2 + (1 − α m ) z − m + 1 , = (α m 2

and Dm −1 ( z) = 1 + α1 z −1 + α 2 z −2 + L + α m −1 z − m +1 + α m z − m * + α * z −1 + L + α * z − m + 1 + z − m ) − α m (α m 1 m −1 * − 2 ) z − 2 + L + (α m − 1 − α m α1* ) z − m + 1 . * − 1 ) z − 1 + (α 2 − α m α m = (1 − α m ) + (α1 − α m α m Thus, A m −1 ( z) is a complex allpass function of order m − 1 and given by 2

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Am −1 ( z ) =

−1 + L + β * z − m + 2 + z − m + 1 * * βm −1 + β m − 2 z 1 , where 1 2 − − 1 + β1 z + β 2 z + L + β m − 2 z − m + 2 + β m −1 z − m + 1

* α − αm αm − k 1 ≤ k ≤ m − 1. , βk = k 2 1 − αm

To develop a realization of A m (z ) we express A m (z ) in terms of Am −1 ( z ) : k * − z −1 Am −1 ( z ) Y , and compare it with Eq. (8.47) resulting in the Am ( z ) = 1 = m X1 1 + k m z −1 Am −1 ( z ) *, following expressions for the transfer parameters of the two-pair: t11 = k m * )z −1 . As in the case of the realization of a real t 22 = − k m z −1 , and t12 t 21 = (1 − k m k m

allpass function, there are many possible choices for t12 and t 21. We choose * )z −1 , t = 1. The corresponding input-output relations of the two-pair t12 = (1 − k m k m 21 * X + (1 − k * k )z −1 X = k * ( X − k z −1 X ) + z −1 X and are: Y1 = k m 1 m m 2 m 1 m 2 2

Y2 = X1 − k m z −1 X 2 . A realization of A m (z ) based on the above two-pair relations is indicated below: X1 _k

A m (z)

Y2 m

A m _1(z)

* km

Y1

z

X2

–1

−1 ⎞ ⎛ 1 + z −1 ⎞ 1 ⎛ 8 + 8 z −1 ⎞ 1 ⎛ ⎟ = ⎜ 1 + 3 + 5z ⎟ = 1 ( A0 ( z ) + A1 ( z )), ⎟= ⎜ 8.48 (a) H1 ( z ) = 4⎜ ⎜ 5 + 3z −1 ⎟ 2 ⎜ 5 + 3z − 1 ⎟ 2 ⎜ 5 + 3z −1 ⎟⎠ 2 ⎝ ⎠ ⎝ ⎠ ⎝ 3 + 5z −1 where A0 ( z) = 1 and A1 ( z ) = . 5 + 3z − 1 −1 ⎞ ⎛ 5 − 5z − 1 ⎞ 1 ⎛ 5 − 5z −1 ⎞ 1 ⎛ ⎟ = ⎜ 1 − − 2 + 3z ⎟ = 1 ( A0 ( z ) − A1 ( z )), ⎟= ⎜ (b) H 2 ( z ) = ⎜⎜ ⎜ ⎜ −1 ⎟ −1 ⎟ 3 − 2 z −1 ⎟⎠ 2 ⎝ 6 − 4z ⎠ 2 ⎝ 3 − 2z ⎠ 2 ⎝ − 2 + 3z −1 where A0 ( z) = 1 and A1 ( z ) = . 3 − 2 z −1 0.5414(1 − z −1 )(1 − 0.0757z −1 + z −2 ) 1 = [A0 ( z ) − A1 ( z)], where (c) H 3 ( z) = 1 1 2 − − − (1 − 0.1768z )(1 − 0.004 z + 0.9061z ) 2

A0 ( z) =

0.9061 − 0.004 z −1 + z −2 1 − 0.004 z −1 + 0.9061z − 2

(d) H 4 ( z) =

and A1 ( z ) =

− 0.1768 + z −1 1 − 0.1768z −1

0.4547(1 + z −1 )(1 − 0.2859z −1 + z −2 ) (1 + 0.0712 z −1 )(1 − 0.0377z −1 + 0.8482 z − 2 )

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=

.

1 [A0 ( z) − A1( z)] , where 2

277

A0 ( z ) =

0.0712 + z −1 1 + 0.0712 z −1

and A1 ( z) =

0.8482 − 0.0377z −1 + z −2 1 − 0.0377z −1 + 0.8482 z − 2

.

8.49 (a) From the given equation we get y[ 2l] = h[ 0] x[ 2l ] + h[1] x[ 2l − 1] + h[ 2] x[ 2l − 2] + h[3] x[ 2l − 3] + h[ 4] x[ 2l − 4] + h[ 5] x[ 2l − 5]. y[ 2l + 1] = h[ 0] x[ 2l + 1] + h[1] x[ 2l] + h[ 2] x[ 2l − 1] + h[3] x[ 2l − 2] + h[ 4] x[ 2l − 3] + h[ 5] x[ 2l − 4].

Rewriting the above equations in matrix form we arrive at ⎡ y[2l] ⎤ ⎡h[0] 0 ⎤ ⎡ x[2l] ⎤ ⎡h[2] h[1] ⎤ ⎡ x[2l − 2]⎤ ⎢ y[2l + 1]⎥ = ⎢ h[1] h[0]⎥ ⎢ x[2l + 1]⎥ + ⎢ h[3] h[2]⎥ ⎢ x[2l − 1] ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎦⎣ ⎡h[ 4] h[3]⎤ ⎡ x[2l − 4]⎤ ⎡0 h[5]⎤ ⎡ x[2l] ⎤ +⎢ ⎥+⎢ ⎥⎢ ⎥, ⎥⎢ ⎣ h[5] h[ 4]⎦ ⎣ x[2l − 3]⎦ ⎣0 0 ⎦ ⎣ x[2l + 1]⎦ which can be alternately expressed as Yl = H 0 X l + H1X l −1 + H 2 X l − 2 + H 3 X l − 3 ,

⎡ y[2l] ⎤ ⎡ x[2l] ⎤ ⎡h[0] 0 ⎤ ⎡h[2] h[1] ⎤ where Yl = ⎢ , H1 = ⎢ , H0 = ⎢ , Xl = ⎢ ⎥ ⎥ ⎥ ⎥, ⎣ y[2l + 1]⎦ ⎣ x[2l + 1]⎦ ⎣ h[1] h[0]⎦ ⎣ h[3] h[2]⎦ ⎡h[ 4] h[3]⎤ ⎡0 h[5]⎤ H2 = ⎢ , H3 = ⎢ ⎥ ⎥. ⎣ h[5] h[ 4]⎦ ⎣0 0 ⎦ (b) Here we have

y[3l ] = h[ 0] x[3l] + h[1] x[3l − 1] + h[ 2] x[3l − 2] + h[3] x[3l − 3] + h[ 4] x[3l − 4] + h[ 5] x[3l − 5]. y[3l + 1] = h[0]x[3l + 1] + h[1]x[3l] + h[2]x[3l − 1] + h[3]x[3l − 2] + h[ 4]x[3l − 3] + h[5]x[3l − 4], y[3l + 2] = h[0]x[3l + 2] + h[1]x[3l + 1] + h[2]x[3l] + h[3]x[3l − 1] + h[ 4]x[3l − 2] + h[5]x[3l − 3].

Rewriting the above equations in matrix form we arrive at 0 ⎤ ⎡ x[3l] ⎤ ⎡ h[3] h[2] h[1] ⎤ ⎡ x[3l − 3]⎤ ⎡ y[3l] ⎤ ⎡h[0] 0 ⎢ y[3l + 1] ⎥ = ⎢ h[1] h[0] 0 ⎥ ⎢ x[3l + 1] ⎥ + ⎢h[ 4] h[3] h[2]⎥ ⎢ x[3l − 2]⎥ ⎥ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎢ ⎢⎣ y[3l + 2]⎥⎦ ⎢⎣h[2] h[1] h[0]⎥⎦ ⎢⎣ x[3l + 2]⎥⎦ ⎢⎣ h[5] h[ 4] h[3]⎥⎦ ⎢⎣ x[3l − 1] ⎥⎦ ⎡0 h[2] h[1] ⎤ ⎡ x[3l − 6]⎤ + ⎢⎢0 0 h[2]⎥⎥ ⎢⎢ x[3l − 5]⎥⎥, ⎢⎣0 0 0 ⎥⎦ ⎢⎣ x[3l − 4]⎥⎦

which can be alternately expressed as Yl = H 0 X l + H1X l −1 + H 2 X l − 2 , where

0 ⎤ ⎡ h[3] h[2] h[1] ⎤ ⎡h[0] 0 ⎡ x[3l] ⎤ ⎡ y[3l] ⎤ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ Y l = ⎢ y[3l + 1] ⎥, X l = ⎢ x[3l + 1] ⎥, H 0 = ⎢ h[1] h[0] 0 ⎥, H1 = ⎢⎢h[ 4] h[3] h[2]⎥⎥, ⎢⎣ h[5] h[ 4] h[3]⎥⎦ ⎢⎣h[2] h[1] h[0]⎥⎦ ⎢⎣ x[3l + 2]⎥⎦ ⎢⎣ y[3l + 2]⎥⎦

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⎡0 h[2] h[1] ⎤ and H 2 = ⎢⎢0 0 h[2]⎥⎥. ⎢⎣0 0 0 ⎥⎦ (c) Following a procedure similar to that outlined in Parts (a) and (b) above, we can show that here Yl = H 0 X l + H1X l −1 + H 2 X l − 2 , where

⎡ y[ 4l] ⎤ ⎢ y[ 4l + 1] ⎥ ⎥, Yl = ⎢ ⎢ y[ 4l + 2]⎥ ⎢ ⎥ ⎣ y[ 4l + 3]⎦ ⎡h[ 4] h[3] ⎢ h[5] h[ 4] H1 = ⎢ ⎢ 0 h[5] ⎢ 0 ⎣ 0

⎡h[0] ⎡ x[ 4l] ⎤ ⎢ h[1] ⎢ x[ 4l + 1] ⎥ ⎢ ⎥ , H0 = ⎢ Xl = ⎢h[2] ⎢ x[ 4l + 2]⎥ ⎢ ⎢ ⎥ ⎣ h[3] ⎣ x[ 4l + 3]⎦ h[2] h[1] ⎤ ⎡0 0 0 ⎢0 0 0 ⎥ h[3] h[2]⎥ , H2 = ⎢ ⎢0 0 0 h[ 4] h[3]⎥ ⎢ ⎥ h[5] h[ 4]⎦ ⎣0 0 0

0 h[0] h[1] h[2] h[5]⎤ 0 ⎥⎥ . 0 ⎥ ⎥ 0 ⎦

0 0 ⎤ 0 0 ⎥⎥ , h[0] 0 ⎥ ⎥ h[1] h[0]⎦

8.50 (a) d 0 y[2l] + d1 y[2l − 1] + d 2 y[2l − 2] + d3 y[2l − 3] + d 4 y[2l − 4] = p0 x[2l] + p1 x[2l − 1] + p2 x[2l − 2] + p3 x[2l − 3] + p4 x[2l − 4], d 0 y[2l + 1] + d1 y[2l] + d 2 y[2l − 1] + d3 y[2l − 2] + d 4 y[2l − 3] = p0 x[2l + 1] + p1 x[2l] + p2 x[2l − 1] + p3 x[2l − 2] + p 4 x[2l − 3]. Rewriting the above equations in matrix form we arrive at ⎡ d 0 0 ⎤ ⎡ y[2l] ⎤ ⎡ d 2 d1 ⎤ ⎡ y[2l − 2]⎤ ⎡ d 4 d3 ⎤ ⎡ y[2l − 4]⎤ ⎥⎢ ⎢d d ⎥ ⎢ ⎥ + ⎢d ⎥ + ⎢ 0 d ⎥ ⎢ y[2l − 3]⎥ d [ 2 1 ] [ 2 1 ] y y l l + − ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 1 0 3 2 4⎦⎣ ⎣ ⎦ ⎣ ⎦

0 ⎤ ⎡ x[2l] ⎤ ⎡ p2 p1 ⎤ ⎡ x[2l − 2]⎤ ⎡ p 4 ⎡p =⎢ 0 ⎥⎢ ⎥⎢ ⎥+⎢ ⎥+⎢ ⎣ p1 p0 ⎦ ⎣ x[2l + 1]⎦ ⎣ p3 p2 ⎦ ⎣ x[2l − 1] ⎦ ⎣ 0 The above matrix equation can be alternately expressed as

p3 ⎤ ⎡ x[2l − 4]⎤ . p 4 ⎥⎦ ⎢⎣ x[2l − 3]⎥⎦

⎡ y[2l] ⎤ D 0Yl + D1Yl −1 + D 2Yl − 2 = P 0 X l + P1 X l −1 + P 2 X l − 2 , where Yl = ⎢ ⎥, ⎣ y[2l + 1]⎦ d1 ⎤ 0⎤ d3 ⎤ ⎡d ⎡d ⎡d ⎡ x[2l] ⎤ Xl = ⎢ , D0 = ⎢ 0 , D1 = ⎢ 2 , D2 = ⎢ 4 ⎥ ⎥ ⎥, ⎥ ⎣ x[2l + 1]⎦ ⎣ 0 d4 ⎦ ⎣ d1 d 0 ⎦ ⎣ d3 d2 ⎦

⎡p P0 = ⎢ 0 ⎣ p1 ⎡d0 (b) ⎢⎢ d1 ⎢⎣ d 2

0⎤ ⎡p , P1 = ⎢ 2 ⎥ p0 ⎦ ⎣ p3 0 d0 d1

p1 ⎤ ⎡p , and P2 = ⎢ 4 ⎥ p2 ⎦ ⎣0

0 ⎤ ⎡ y[3l] ⎤ ⎡ d3 0 ⎥⎥ ⎢⎢ y[3l + 1] ⎥⎥ + ⎢⎢ d 4 d 0 ⎥⎦ ⎢⎣ y[3l + 2]⎥⎦ ⎢⎣ 0

d2 d3 d4

p3 ⎤ . p4 ⎥⎦

d1 ⎤ ⎡ y[3l] − 3⎤ ⎡0 0 d 4 ⎤ ⎡ y[3l − 6]⎤ d 2 ⎥⎥ ⎢⎢ y[3l − 2]⎥⎥ + ⎢⎢0 0 0 ⎥⎥ ⎢⎢ y[3l − 5]⎥⎥ d3 ⎥⎦ ⎢⎣ y[3l − 1] ⎥⎦ ⎢⎣0 0 0 ⎥⎦ ⎢⎣ y[3l − 4]⎥⎦

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0 ⎤ ⎡ x[3l] ⎤ ⎡ p3 ⎡ p0 0 ⎢ = ⎢ p1 p0 0 ⎥⎥ ⎢⎢ x[3l + 1] ⎥⎥ + ⎢⎢ p 4 ⎢⎣ p2 p1 p0 ⎥⎦ ⎢⎣ x[3l + 2]⎥⎦ ⎢⎣ 0 which can be alternately expressed as D 0 Yl + D1Yl −1 + D 2 Yl − 2

p2 p3 p4

p1 ⎤ ⎡ x[3l] − 3⎤ ⎡0 0 p2 ⎥⎥ ⎢⎢ x[3l − 2]⎥⎥ + ⎢⎢0 0 p3 ⎥⎦ ⎢⎣ x[3l − 1] ⎥⎦ ⎢⎣0 0

p4 ⎤ ⎡ x[3l − 6]⎤ 0 ⎥⎥ ⎢⎢ x[3l − 5]⎥⎥ , 0 ⎥⎦ ⎢⎣ x[3l − 4]⎥⎦

⎡ y[3l] ⎤ = P0 X l + P1X l −1 + P2 X l − 2 , where D 0 = ⎢⎢ y[3l + 1] ⎥⎥, ⎢⎣ y[3l + 2]⎥⎦ 0 0⎤ ⎡ d3 d 2 d1 ⎤ ⎡0 0 d 4 ⎤ ⎥ ⎥ ⎢ d 0 0 ⎥, D1 = ⎢ d 4 d3 d 2 ⎥, D 2 = ⎢⎢0 0 0 ⎥⎥, ⎢⎣ 0 d 4 d3 ⎥⎦ ⎢⎣0 0 0 ⎥⎦ d1 d 0 ⎥⎦

⎡ x[3l] ⎤ ⎡d0 ⎥ ⎢ P 0 = ⎢ x[3l + 1] ⎥, D 0 = ⎢⎢ d1 ⎢⎣ x[3l + 2]⎥⎦ ⎢⎣ d 2 0⎤ ⎡ p0 0 ⎡ p3 ⎢ ⎥ P0 = ⎢ p1 p0 0 ⎥, P1 = ⎢⎢ p 4 ⎢⎣ p2 p1 p0 ⎥⎦ ⎢⎣ 0

p2 p3 p4

p1 ⎤ ⎡0 0 ⎥ p2 ⎥, and P 2 = ⎢⎢0 0 ⎢⎣0 0 p3 ⎥⎦

p4 ⎤ 0 ⎥⎥. 0 ⎥⎦

0 0 ⎤ ⎡ y[ 4l] ⎤ ⎡ d 4 d3 d 2 d1 ⎤ ⎡ y[ 4l − 4]⎤ ⎡d0 0 ⎢d d 0 0 ⎥⎥ ⎢⎢ y[ 4l + 1] ⎥⎥ ⎢⎢ 0 d 4 d3 d 2 ⎥⎥ ⎢⎢ y[ 4l − 3]⎥⎥ 1 0 ⎢ + (c) ⎢ d 2 d1 d 0 0 ⎥ ⎢ y[ 4l + 2]⎥ ⎢ 0 0 d 4 d3 ⎥ ⎢ y[ 4l − 2]⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ 0 0 d 4 ⎦ ⎣ y[ 4l − 1] ⎦ ⎣ d3 d 2 d1 d 0 ⎦ ⎣ y[ 4l + 3]⎦ ⎣ 0 0 0 ⎤ ⎡ x[ 4l] ⎤ ⎡ p 4 p3 p2 p1 ⎤ ⎡ x[ 4l − 4]⎤ ⎡ p0 0 ⎢p p0 0 0 ⎥⎥ ⎢⎢ x[ 4l + 1] ⎥⎥ ⎢⎢ 0 p 4 p3 p2 ⎥⎥ ⎢⎢ x[ 4l − 3]⎥⎥ =⎢ 1 + , which can be ⎢ p2 p1 p0 0 ⎥ ⎢ x[ 4l + 2]⎥ ⎢ 0 0 p 4 p3 ⎥ ⎢ x[ 4l − 2]⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ 0 0 p4 ⎦ ⎣ x[ 4l − 1] ⎦ ⎣ p3 p2 p1 p0 ⎦ ⎣ x[ 4l + 3]⎦ ⎣ 0 alternately written as D 0 Yl + D1Yl −1 + D 2 Yl − 2 = P0 X l + P1X l −1 + P2 X l − 2 , where

⎡d0 ⎡ y[ 4l] ⎤ ⎡ x[ 4l] ⎤ ⎢d ⎢ y[ 4l + 1] ⎥ ⎢ x[ 4l + 1] ⎥ ⎥, X l = ⎢ ⎥, D 0 = ⎢ 1 Yl = ⎢ ⎢d2 ⎢ y[ 4l + 2]⎥ ⎢ x[ 4l + 2]⎥ ⎢ ⎢ ⎥ ⎢ ⎥ ⎣ y[ 4l + 3]⎦ ⎣ x[ 4l + 3]⎦ ⎣ d3 0 ⎡ p0 0 ⎡ d 4 d3 d 2 d1 ⎤ ⎢ ⎢0 d ⎥ p p0 0 4 d3 d 2 ⎥ D1 = ⎢ , P0 = ⎢ 1 ⎢ p2 p1 p0 ⎢0 0 d 4 d3 ⎥ ⎢ ⎢ ⎥ 0 0 d4 ⎦ ⎣0 ⎣ p3 p2 p1

0 d0 d1 d2

0 0 d0 d1

0⎤ 0 ⎥⎥ , 0⎥ ⎥ d0 ⎦

0⎤ ⎡ p4 ⎥ ⎢0 0⎥ , and P 1 = ⎢ ⎢0 0⎥ ⎥ ⎢ p0 ⎦ ⎣0

p3 p4 0 0

p2 p3 p4 0

p1 ⎤ p2 ⎥⎥ . p3 ⎥ ⎥ p4 ⎦

8.51 We first rewrite the second-order block difference equation D 0 Yl + D1Yl −1 + D 2 Yl − 2 = P0 X l + P1X l −1 + P2 X l − 2 , as two separate equations: Wl = P0 X l + P1X l −1 + P2 X l − 2 , and

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Yl = − D 0−1D1Yl −1 − D 0−1D 2 Yl − 2 + D 0−1Wl . As cascade realization of the IIR block

digital filter based on the last two equations is thus as shown below: Wk

Xk

_1

D0

P0

Yk

∆

∆ D1

P1

∆

∆ D2

P2

By interchanging the locations of the two block sections in the above structure we arrive at an equivalent realization as indicated below: _1

Xk

D0

P0

Yk

∆

∆ D1

P1 ∆

∆

P2

D2

Finally, by delay-sharing the above structure reduces to a canonic realization as shown below: Xk

_1

D0

P0

Yk

∆

D1

P1

∆

D2

P2

8.51 By setting α = ± β sin θ in Eq. (8.125), the state-space description of the sine-cosine

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⎡ s [ n + 1] ⎤ ⎡cos ωo generator reduces to ⎢ 1 ⎥=⎢ ⎣ s 2 [ n + 1]⎦ ⎢⎣ m multiplier realization shown below: cos ω o s 2[n+1]

cos ω o + _ sin 2ω o _1

z

8.53 Let −

± sin 2 ωo ⎤ ⎡ s 1[ n] ⎤ ⎥⎢ ⎥, which leads to the threecos ωo ⎥⎦ ⎣ s 2 [ n]⎦

s 2[n]

s 1[n+1]

_1

z

s 1[n]

_ +

− sin 2 ωo cos 2 ωo − 1 β sin ωo α = = cos ωo − 1, and = 1 + cos ωo . Then sin ωo = α β cos ωo + 1 cos ωo + 1

β sin ωo = cos ωo + 1. Substituting these values in Eq. (8.125) we arrive at α cos ωo − 1⎤ ⎡ s 1[ n] ⎤ ⎡ s1[ n + 1] ⎤ ⎡ cos ωo ⎢ s [ n + 1]⎥ = ⎢cos ω + 1 cos ω ⎥ ⎢ s [ n]⎥. These equations can be alternately written o o ⎦⎣ 2 ⎣ 2 ⎦ ⎣ ⎦ as s1[ n + 1] = cos ωo (s1[ n] + s 2 [ n]) − s 2 [ n], and s 2 [ n + 1] = cos ωo (s1[ n] + s 2 [ n]) + s 2 [ n]. A realization based on the last two equations results in a single-multiplier structure as indicated below: −

s 2[n+1]

_1

z

s 2[n]

_1

s 1[n+1]

_

z1

s 1[n]

cos ω o

⎡ ⎡ s1[ n + 1] ⎤ ⎢0 8.54 ⎢ ⎥= ⎣ s 2 [ n + 1]⎦ ⎢0 ⎣

⎡ α(C − cos ωo ) ⎤ C [ 1 ] + s n ⎡ 1 ⎤ ⎢ ⎥ +⎢ β sin ωo ⎥ ⎢⎣ s 2 [ n + 1]⎥⎦ ⎢ β 0 − sin ωo ⎦ ⎣⎢ α

α(1 − C cos ωo ) ⎤ ⎥ ⎡ s1[ n] ⎤ β sin ωo . If ⎥⎢ s 2 [ n]⎥⎦ ⎣ ⎥ cos ωo ⎦⎥

C = 1, then ⎡ s1[ n + 1] ⎤ ⎢ s [ n + 1]⎥ ⎣ 2 ⎦

⎡ α(1 − cos ωo ) ⎤ 1 ⎡ s1[ n + 1] ⎤ ⎢ ⎥ +⎢ β sin ωo ⎥ ⎢⎣ s 2 [ n + 1]⎥⎦ ⎢ β 0 − sin ωo ⎦ ⎢⎣ α β sin ω o . Then the above equation reduces to

⎡ 0 =⎢ ⎢ ⎣0

Choose α =

α(1 − 1 cos ωo ) ⎤ β sin ωo ⎥ ⎡ s1[ n] ⎤. ⎥⎢ ⎥ ⎥ ⎣ s 2 [ n]⎦ cos ωo ⎦⎥

⎡ s1 [ n + 1] ⎤ ⎡0 1 − cos ωo ⎤ ⎡ s1 [ n + 1] ⎤ ⎡ 1 1 − cos ωo ⎤ ⎡ s1 [ n] ⎤ ⎢ s [ n + 1]⎥ = ⎢ ⎥ ⎢ s [ n + 1]⎥ + ⎢ − 1 cos ω ⎥ ⎢ s [ n]⎥. 0 ⎦⎣ 2 o ⎦⎣ 2 ⎣ 2 ⎦ ⎣0 ⎦ ⎣ ⎦ A two-multiplier realization of the above equation is shown below: Not for sale

282

_1

1 _ cos ω o _

_

z1

s 2[n+1] cos ω o

s 2[n]

s 1[n+1]

z1

s 1[n]

To arrive at a one-multiplier realization we observe that the two equations describing the sine-cosine generator are given by s1[n + 1] = (1 − cos ωo ) s2 [n + 1] + s1[n] + (1 − cos ωo ) s2 [n], and s2 [n + 1] = − s1[n] + cosωo ⋅s2 [n]. Substituting the second equation in the first we arrive at an alternate description in the form s1[n + 1] = − cos ωo ⋅ s2 [n + 1] + s2 [n], and s2 [n + 1] = − s1[n] + cosωo ⋅s2 [n]. A realization of these two equations leads to the singlemultiplier structure shown below: _1

s 2[n+1]

_

z1

s 2[n]

cos ω o

s 1[n+1]

_

z1

s 1[n]

_1

_1

z

8.55 From Figure P8.18(a), the input-output relation of the channel is given H12 ( z )⎤ ⎡ X1 ( z ) ⎤ ⎡Y (z) ⎤ ⎡ 1 . Likewise, the input-output relation of the by ⎢ 1 ⎥ = ⎢ 1 ⎥⎦ ⎢⎣ X 2 ( z )⎥⎦ ⎣Y2 ( z )⎦ ⎣ H 21 ( z) channel separation circuit of Figure P8.18(b) is given by 1 − G12 ( z )⎤ ⎡ Y1 ( z) ⎤ ⎡ V1 ( z ) ⎤ ⎡ ⎢V ( z )⎥ = ⎢ − G ( z ) ⎥ ⎢Y ( z)⎥. Hence, the overall system is characterized by 1 21 ⎣ 2 ⎦ ⎣ ⎦⎣ 2 ⎦

H12 ( z)⎤ ⎡ X1 ( z) ⎤ 1 − G12 ( z )⎤ ⎡ 1 ⎡ V1 ( z ) ⎤ ⎡ ⎢V ( z )⎥ = ⎢ − G ( z) ⎥ ⎢ 1 1 ⎥⎦ ⎢⎣ X 2 ( z)⎥⎦ 21 ⎣ 2 ⎦ ⎣ ⎦ ⎣ H 21 ( z) ⎡1 − H 21 ( z)G12 ( z) H12 ( z ) − G12 ( z ) ⎤ ⎡ X1 ( z ) ⎤ =⎢ ⎥⎢ ⎥. ⎣ H 21 ( z ) − G21 ( z) 1 − H12 ( z)G21 ( z )⎦ ⎣ X 2 ( z )⎦ The cross-talk is eliminated if V1 ( z) is a function of either X1 ( z) or X 2 ( z), and similarly, if V2 ( z) is a function of either X1 ( z) or X 2 ( z). From the above equation it follows that if H 12 ( z) = G12 ( z), and H 21( z) = G 21( z), then V1 ( z) = (1 − H 21 ( z)G12 ( z))X1 ( z), and

−1 ( z ), and G ( z ) = H −1 ( z ), V2 ( z) = (1 − H12 ( z)G21 ( z))X 2 ( z). Alternately, if G12 ( z ) = H 21 21 12

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⎛ H ( z )H 21 ( z) − 1 ⎞ ⎛ H ( z )H 21 ( z) − 1 ⎞ ⎟⎟ X 2 ( z), and V2 ( z ) = ⎜⎜ 12 ⎟⎟ X1 ( z). then V1 ( z ) = ⎜⎜ 12 H 21 ( z) H12 ( z) ⎠ ⎠ ⎝ ⎝ M8.1 (a) Using the M-file factor, we factorize H1 ( z) resulting in

H1 ( z) = −0.24(1 − 2.5z −1 )(1 − 0.4 z −1 )(1 + z −1 + z − 2 )(1 + 0.8z −1 + 0.6 z − 2 ). 4 3

6 3

A cascade realization of H1 ( z) based on the above factored form is shown below: _ 0.24

x[n]

+ z

z

z

z

_ 2.5

_1

_1

_1 _ 0.5

y[n]

+

+

+

_1

4/3

0.8

+

+

_1

_1

z

z

6/3

0.6

(b) Using the M-file factor, we factorize H 2 ( z) resulting in H 2 ( z ) = 4(1 − 3.8508 z −1 )(1 + 2.5z −1 )(1 − 1.5403 z −1 )(1 + 0.4 z −1 )(1 − 0.649 z −1 )(1 − 0.2597 z −1 ). A cascade realization of H 2 ( z) based on the above factored form is shown below: 4

x[n]

+ _1

+ _1

z

z _

+

1.5403

z _

0.4

y[n]

_1

z

z _

2.5

3.8508

_1

_1

z

+

+

+

_1

0.649

_

0.2597

(c) Using the M-file factor, we factorize H 3 ( z) resulting in H 3 ( z ) = −0.24(1 + z −1 )(1 − z −1 )(1 − 1.4024z −1 )(1 − 0.71306z −1 )(1 + 1.3488z −1 + z −2 ).

A cascade realization of H 3 ( z) based on the above factored form is shown below: _ 0.24

+

x[n] _1

z

+ _1

+ _1

z

z _1

y[n]

_1

z

z _ 1.4024

+

+ _1 _ 0.71306

1.3488

+ _1

z

(d) Using the M-file factor, we factorize H 4 ( z) resulting in H 4 ( z ) = 4(1 + z −1 )(1 − z −1 )(1 − 4.6735 z −1 )(1 − 0.214 z −1 )(1 + 1.4675 z −1 + z −2 ). A cascade realization of H 4 ( z) based on the above factored form is shown below:

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4

+

x[n] _1

+ _1

z

+ _1

z

z

_1

z

z

_1

y[n]

+

+ _1 _ 0.214

_ 4.6735

1.4675

+ _1

z

M8.2 (a) Using the M-file factor, we factorize the numerator and the denominator of G (z )

resulting in G ( z ) =

0.3901(1 + 1.4891z −1 + z −2 )(1 + 0.1582 z −1 + z −2 ) (1 − 0.06147 z −1 + 0.70426 z − 2 )(1 + 0.5653 z −1 + 0.2228 z − 2 )

.

(b) We can write ⎞ ⎛ ⎞⎛ 1 + 0.1582 z −1 + z − 2 1 + 1.4891z −1 + z − 2 ⎟. ⎜ ⎟⎜ G ( z ) = 0.3901 ⎜ 1 − 0.06147 z −1 + 0.70426 z − 2 ⎟⎜ 1 + 0.5653 z −1 + 0.2228 z − 2 ⎟ ⎠ ⎝ ⎠⎝ cascade realization of G (z ) based on the above decomposition is shown on next page:

A

0.3901

+

+

x[n]

+

_ 0.56527

1.4891

0.1582

+

+

+ _ z 1

_ z 1 _

y[n]

_ z 1

_ z 1

0.06147

+

+

_ 0.2228

0.70426

Alternately, we can write ⎞ ⎛ ⎞⎛ 1 + 0.1582 z −1 + z − 2 1 + 1.4891z −1 + z − 2 ⎟. ⎟⎜ G ( z ) = 0.3901⎜ ⎜ 1 − 0.06147 z −1 + 0.70426 z − 2 ⎟⎜ 1 + 0.5653 z −1 + 0.2228 z − 2 ⎟ ⎠ ⎝ ⎠⎝ A cascade realization of G (z ) based on the above decomposition is shown below: 0.3901 x[n]

+

+

+

0.1582

+

_ 0.56527

0.70426

1.4891

+

+ _ z 1

_ z 1 _

y[n]

_ z 1

_ z 1

0.06147

+

+

_ 0.2228

(c) A partial-fraction expansion of G (z ) in z −1 obtained using the M-file residuez is

given by G ( z ) = 2.4863 +

− 0.3481 − 0.11128 z −1 1 − 0.06147 z −1 + 0.70425 z − 2

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+

− 1.7481 + 0.4094 z −1 1 + 0.5653 z −1 + 0.2228 z − 2

285

.

A Parallel Form I realization of G (z ) based on the above expansion is shown on the left side top of next page. (d) A partial-fraction expansion of G (z ) in z obtained using the M-file residue is given

− 0.1327 z −1 + 0.2451z −1

by G ( z ) = 0.3901 +

+

0.5787 z −1 + 0.3895 z −2

. A 1 − 0.06147 z −1 + 0.70425 z − 2 1 + 0.5653 z −1 + 0.2228 z − 2 Parallel Form II realization of G (z ) based on the above expansion is shown on the right side top of next page. M8.3 (a) Using the M-file factor, we factorize the numerator and the denominator of H (z )

resulting in H ( z ) =

0.3549(1 + 0.5958 z −1 + z −2 )(1 − 0.03169 z −1 + z −2 ) (1 + 0.6209 z −1 + 0.9845 z − 2 )(1 + 0.6313 z −1 + 0.5681z − 2 )

.

2.4863 0.3901

_ 0.3481

x[n]

+

+

+

y[n] x[n]

_ z 1

+

+ _ z 1

0.06147 _1

_ 0.70425

_ 0.11128

+

z

_

_

+

y[n]

_ z 1

0.70425

0.2451

+

_1

_ z 1

z _ 0.5653

_ 0.5653

+

+

1.7481

+

+

_ 0.1327

0.06147

0.4094

0.5787

+

_1

+

_1

z

z

_ 0.2228

_ 0.2228

0.3895

(b) We can write

⎛ ⎞⎛ 1 − 0.03169 z −1 + z − 2 ⎞ 1 + 0.5958 z −1 + z − 2 ⎟. ⎟⎜ H ( z ) = 0.3549⎜ ⎜ 1 + 0.620986 z −1 + 0.9845 z − 2 ⎟⎜ 1 + 0.6313 z −1 + 0.5683 z − 2 ⎟ ⎠ ⎝ ⎠⎝ A cascade realization of H (z ) based on the above decomposition is shown on top of the next page. Alternately, we can write ⎞ ⎛ 1 − 0.03169 z −1 + z − 2 ⎞⎛ 1 + 0.5958 z −1 + z − 2 ⎟. ⎟⎜ H ( z ) = 0.3549⎜ ⎜ 1 + 0.6209 z −1 + 0.9845 z − 2 ⎟⎜ 1 + 0.6313 z −1 + 0.5683 z − 2 ⎟ ⎠ ⎝ ⎠⎝

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286

0.3549

+

+

x[n]

+

+ _ z 1

_ z 1 _

+

y[n]

0.5958

0.6209

+

+

_ 0.03169

_ 0.6313

+ _ z 1

_ z 1 _ 0.9845

_ 0.5683

A cascade realization of H (z ) based on the decomposition given at the bottom of the previous page is shown below: 0.3549

+

+

x[n]

+

+ _ z 1

_ z 1 _

+

_ 0.03169

0.6209

y[n]

+

+

_ 0.6313

0.5958

+ _ z 1

_ z 1 _ 0.9845

_ 0.5683

(c) A partial-fraction expansion of G (z ) in z −1 obtained using the M-file residuez is given by

− 0.0133 + 0.0094 z −1

H ( z ) = 0.6343 +

− 0.2661 − 0.4299 z −1

+ . A Parallel 1 + 0.6209 z −1 + 0.9845 z − 2 1 + 0.6313 z −1 + 0.5683 z − 2 Form I realization of G (z ) based on the above expansion is shown below on the left side. 0.6343

x[n]

_ 0.0133

+

0.3549

+

+

y[n]

x[n]

+

_1

_ z 1

z _ 0.6209

_ 0.6209

0.0094

+

_ z 1 _

_

_ 0.2661

+

0.9845

y[n]

0.0131

_ z 1

z

_ 0.6313

_ 0.4299

+

+

+

_1

_ 0.6313

+

_ z 1

0.9845

+

0.0177

+

_ 0.2619

+

+

_1

_1

z

z _ 0.5683

_ 0.5683

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0.1512

287

(d) A partial-fraction expansion of G (z ) in z obtained using the M-file residue is given

− 0.0177 z −1 + 0.0131z −1

by G ( z ) = 0.3549 +

+

− 0.2619 z −1 + 0.1512 z −2

. A Parallel 1 + 0.6209 z −1 + 0.9845 z − 2 1 + 0.6313 z −1 + 0.5683 z − 2 Form II realization of G (z ) based on the above expansion is shown at the bottom of the previous page on the right side. M8.4

Lattice parameters are 0.1393 0.6973 0.3131

0.1569

Feedforward multipliers are 0.3901 0.4461 0.3212

0.0727

-0.0488

The Gray-Markel tapped cascaded lattice realization of G (z ) is shown on next page: x[n] _ 0.1569

_ 0.3131

0.3131

0.1569 _1

M8.5

0.1393

0.6973 _1

z

0.3901

_ 0.1393

_ 0.6973

0.4461

_1

_1

z

z

z

0.3212

0.0727

_ 0.0488

y[n]

Lattice parameters are 0.3171 0.9949 0.3986

0.5595

Feedforward multipliers are 03549 -0.2442 0.2642

-0.0082

-0.0066

The Gray-Markel tapped cascaded lattice realization of H (z ) is shown below: x[n] _ 0.5595

_ 0.3986

0.3986

0.5595 _1

0.3549

_ 0.9949

_ 0.3171

0.3171

0.9949 _1

z

z

_ 0.2442

0.2642

_1

z

_ 0.0082

_1

z

_ 0.0066

y[n]

M8.6 (a) (b) Program 8_7 generated the following error messages for the FIR transfer functions of Parts (a) and (b): ??? Error using ==> signal\private\levdown At least one of the reflection coefficients is equal to one. The algorithm fails for this case. (c) (d) Program 8_7 generated the following error messages for the FIR transfer functions of Parts (c) and (d): Warning: Divide by zero. Lattice coefficients are Not for sale

288

-1 NaN NaN NaN NaN NaN M8.7 (a) Using the M-file tf2ca we arrive at the parallel allpass decomposition of G (z ) as:

G( z) =

1 2

A1 ( z ) =

[A0 ( z ) + A1 ( z )],

where A0 ( z ) =

0.8041 − 0.8171z −1 + z −2

− 0.2440 + 1.2092 z −1 − 1.1436 z −2 + z −3 1 − 1.1436 z −1 + 1.2092 z − 2 − 0.2440 z − 3

.

1 − 0.8171z −1 + 0.8041z − 2

(b) The power-complementary transfer function is then given by H ( z ) =

=

0.524(1 − 3.0011z −1 + 4.9948 z −2 − 4.9948 z −3 + 3.0011z −4 − z −5

1 − 1.9606 z −1 + 2.9473 z − 2 − 2.1511z − 3 + 1.1713 z − 4 − 0.1961z − 5 (c) jω 2

Magnitude Square

and

1 2

[A0 ( z ) − A1 ( z )]

.

jω 2

|G(e )| +|H(e )|

1 jω 2

0.8

jω 2

|G(e )|

|H(e )|

0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

M8.8 (a) Using the M-file tf2ca we arrive at the parallel allpass decomposition of G (z ) as:

G( z) =

1 2

[A0 (z) − A1 (z)], where

A0 ( z ) =

0.9077 + 0.6205z −1 + z −2

1 + 0.6205z −1 + 0.9077z − 2 0.3325 + 1.2027z −1 + 0.9522 z −2 + z −3 A1 ( z) = . 1 + 0.9522 z −1 + 1.2027z − 2 + 0.3325z − 3

and

(b) The power-complementary transfer function is then given by H ( z) =

=

(1 + 2.3334 z −1 + 3.7759 z −2 + 3.7759 z −3 + 2.3334 z −4 + z −5 1 + 1.5727 z −1 + 2.712 z − 2 + 1.9431z − 3 + 1.2979 z − 4 +0.3018 z − 5

1 2

[A0 (z) + A1 (z)]

.

(c)

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|G(e ω | +|H(e ω |

Magnitude Square

j 2

1

|H(e ω |

|G(e ω |

j 2

0.8

j 2 j 2

0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

M8.9 The transfer function given cannot be realized as a parallel connection of two real allpass sections as the degree difference between the two allpass transfer functions must be 2. M8.10 The MATLAB program for the simulation of the sine-cosine generator of Problem 8.53 is given below: s10 = 0.1; s20 = 0.1; a = 0.8; y1 = zeros(1,50); y2 = y1; for n = 1:50; y1(n) = a*(s10 + s20) - s20; y2(n) = a*(s10 +s20) + s10; s10 = y1(n); s20 = y2(n); end y1max = max(y1);y2max = max(y2); k = 1:50; stem (k-1,y1/y1max); xlabel('Time index n');ylabel('Amplitude'); title('y_1[n]'); pause stem(k-1,y2/y2max); xlabel('Time index n');ylabel('Amplitude'); title('y_2[n]');

The plots generated by the above program for initial conditions s1[-1] s2[-1] = 0.1 are shown below:

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290

y 2[n] 1

0.5

0.5 Amplitude

Amplitude

y 1[n] 1

0 -0.5 -1

0 -0.5

0

10

20 30 Time index n

40

50

-1

0

10

20 30 Time index n

40

50

The outputs are zero for zero initial conditions. Non-zero initial conditions of equal values appear to have no effects on the outputs. However, unequal initial conditions have effects on the amplitudes and phases of the two output sequences. M8.11 The MATLAB program for the simulation of the sine-cosine generator of Problem 8.53 is given below: s10 = 0.1; s20 = 1; a = 0.8; y1 = zeros(1,50); y2 = y1; for n = 1:50; y1(n) = -s20 +a*s10; y2(n) = -a*y1(n) + s10; s10 = y1(n); s20 = y2(n); end y1max = max(y1);y2max = max(y2); k = 1:50; stem (k-1,y1/y1max); xlabel('Time index n');ylabel('Amplitude'); title('y_1[n]'); pause stem(k-1,y2/y2max); xlabel('Time index n');ylabel('Amplitude'); title('y_2[n]');

The plots generated by the above program for initial conditions s1[-1] s2[-1] = 0.1 are shown on top of the next page.: The outputs are zero for zero initial conditions. Non-zero initial conditions of equal values appear to have no effects on the outputs. However, unequal initial conditions have effects on the amplitudes and phases of the two output sequences.

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291

y 1[n]

y 2[n]

1

1 0.5 Amplitude

Amplitude

0.5 0 -0.5 -1

0 -0.5

0

10

20 30 Time index n

40

50

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-1

0

10

20 30 Time index n

40

50

292

Chapter 9 9.1

We obtain the solutions by using Eq. (9.3) and Eq. (9.4). −α p / 20

= 1 − 10 − 0.21 / 20 = 0.0239, δ s = 10 − α s / 20 = 10 − 53 / 20 = 0.0022

−α p / 20

= 1 − 10 − 0.17 / 20 = 0.0194,

(a) δ p = 1 − 10 (b) δ p = 1 − 10

δ s = 10 −α s / 20 = 10 − 78 / 20 = 1.26 × 10 − 4 .

9.2

We obtain the solutions by using Eq. (9.3) and Eq. (9.4). (a) α p = −20 log10 1 − δ p = −20 log10 (1 − 0.02) = 0.1755 dB,

(

)

(

)

α s = −20 log10 (δ s ) = −20 log10 (0.03) = 30.458 dB.

(b) α p = −20 log10 1 − δ p = −20 log10 (1 − 0.055) = 0.4914 dB, α s = −20 log10 (δ s ) = −20 log10 (0.033) = 29.630 dB.

2

9.3

jω jω 2 jω jω 2 jω G( z ) = H 2 ( z ), or equivalently, G(e ) = H (e ). G(e ) = H (e ) = H (e ) .

Let δ p and δ s denote the passband and stopband ripples of H (e jω ) , respectively. Also, let δ p,2 = 2δ p, and δ s,2 denote the passband and stopband ripples of G(e jω ) , respectively. Then δ p,2 = 1 − (1 − δ p ) 2 , and δ s,2 = (δ s ) 2 . For a cascade of M sections, δ p, M = 1 − (1 − δ p ) M , and δ s,M = (δ s ) M . 9.4 HLP (e j ) HHP (e

1 p 1– p

1– p

1

0 – s

– p

)

p

s

s

–

j

š p

s

– –( – p ) –( – ) s

š

0 s

Therefore, the passband edge and the stopband edge of the highpass filter are given by ω p,HP = π − ω p , and ω s,HP = π − ω s , respectively. 9.5

Note that G (z ) is a complex bandpass filter with a passband in the range 0 ≤ ω ≤ π . Its passband edges are at ω p,BP = ωo ± ω p , and stopband edges at ω s, BP = ω o ± ω s . A real coefficient bandpass transfer function can be generated according to

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293

p

G BP ( z ) = H LP (e jωo z ) + H LP (e – jωo z ) which will have a passband in the range 0 ≤ ω ≤ π and another passband in the range – π ≤ ω ≤ 0. However because of the overlap of the two spectra a simple formula for the bandedges cannot be derived. HLP (e

j

)

j

G(e )

1 p 1– p

1

1– p

s

s

–

0 – s

– p

š

0

–

s

p

o

9.6 (a) h p (t ) = ha (t ) ⋅ p(t ) where p(t ) =

∞

n = −∞

h p (t)e stdt

p

o

p

o

s

ha (nT) (t nT) .

ha (t) e st dt and nT)e st dt

h a (nT) (t

h a (nT)e snT . n

g[n]z n

Comparing the above expression with G(z) n

H p (s)

o

n

n

conclude that G(z)

š

o s

∑ δ(t − nT ). Thus, h p (t)

ha (nT). Now, Ha (s)

We also have, g[n] H p (s)

p

s

h a (nT)z n , we n

. 1 ln z T

We can also show that a Fourier series expansion of p(t) is given by 1 e j(2 kt / T) . Therefore, p(t) Tk

h p (t)

1 Tk

e j (2 kt / T) h a (t)

H p (s)

1 Tk

Ha s

G(z)

j

1 Tk

ha (t) e j(2 kt / T) . Hence,

2 kt . As a result, we have T

1 Tk

Ha s

j

2 kt . 1 T s ln z

(7-1)

T

e sT . If we express e oT e j oT . Therefore,

(b) The transformation from the s -plane to z -plane is given by z

s

o

j o , then we can write z

re j

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294

1, for o 1, z 1, for o 1, Or in other words, a point in the left-half s -plane is mapped onto 1, for o 1. a point inside the unit circle in the z -plane, a point in the right-half s -plane is mapped onto a point outside the unit circle in the z -plane, and a point on the j -axis in the s plane is mapped onto a point on the unit circle in the z -plane. As a result, the mapping has the desirable properties enumerated in Section 9.1.3. (c) However, all points in the s -plane defined by s mapped onto a single point in the z -plane as z mapping is illustrated in the figure below

e

j o

o oT

j(

e

o

j

2 k , k T

2 k )T T

j

e

0,,1,2,K , are

oT

e

j

oT

. The

Im z 3š T š T

1

1

Re z

– šT 3š

– T s -plane

z -plane

Note that the strip of width 2π / T in the s -plane for values of s in the range /T / T is mapped into the entire z -plane, and so are the adjacent strips of width 2π / T . The mapping is many-to-one with infinite number of such strips of width 2π / T . It follows from the above figure and also from Eq. (7-1) that if the frequency response 1 j , and there is no aliasing. Ha (j ) 0 for , then G(e ) H a ( j ), for T T T

(d) For z 9.7

ej

e j T . Or equivalently,

T.

∫

Assume ha (t ) is causal. Now, ha (t ) = H a (s )e st ds. Hence,

∫

g[ n] = ha (nT ) = H a ( s ) e snT ds.

G( z ) =

∞

∑

n =0

g[ n]z −n =

∞

∑∫

Therefore,

H a (s )e snT z −n ds = H a (s )

∫

n =0

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∞

∑ z −n e snT ds = ∫ 1 − easT z −1 ds . H (s )

n =0

295

Hence G( z ) =

⎡ H a (s) ⎤ Residues⎢ . sT −1 ⎥ − e z 1 ⎣ ⎦ all poles of H a ( s )

∑

A . The transfer function has a pole at s = −α. Now s+α ⎡ ⎤ A A A G( z) = Residue⎢ . = ⎥= sT − 1 sT − 1 − α T −1 at s = – α ⎣⎢ (s + α)(1 − e z ) ⎦⎥ 1 − e z 1 e z − s= – α

9.8 H a ( s ) =

9.9

(a) H a (s ) =

4(3s + 7) 2

(s + 2)(s + 4s + 5)

=

−4 2−6j 2+6j + + s + 2 (s + 2 + j ) (s + 2 − j )

(− 12 ) ⋅ 12 . 4s − 4 4(s + 2 ) −4 −4 + = + + s + 2 (s + 2 )2 + 12 s + 2 (s + 2 )2 + 12 (s + 2 )2 + 12 Using Eqs. (9.58), (9.60), and (9.62), we get =

G a (z ) =

−4 1− e

− 2T − 1

+

z

(

− 1 − 2T

1 − 2z e

−1 −2 T

z e

+

−4 1 − 0.5488 z −1 =

+

(

z

,

)

1 − 1.0486 z −1 + 0.3012 z −2 +

1 − 0.5488 z −1

(s

sin (T )

4 1 − 0.5243z −1

−4

(b) H b (s ) =

cos(T ) + e

− 4T − 2

cos(T ) + e − 4T z − 2

−1 − 2T

1 − 2z e where T = 0.3. Therefore,

G a (z ) =

)

4 1 − z −1e − 2T cos(T )

+

4 − 1.935z −1 1 − 1.0486 z −1 + 0.3012 z −2

8s 2 + 37s + 56

)

=

0.1622 z −1 1 − 1.0486 z −1 + 0.3012 z − 2 .

2 3 + 0. 5 j s − 0 .5 j + + (s + 4 ) (s + 1 + 3 j ) (s + 1 − 3 j )

+ 2 s + 10 (s + 4 ) 2 6s + 9 2 6(s + 1) 3(3) = + = + . + 2 2 2 s + 4 (s + 1) + 9 s + 4 (s + 1) + 3 (s + 1)2 + 32 Using Eqs. (9.58), (9.60), and (9.62), we get Gb (z ) =

2

2 1 − e − 4T z − 1

+6

1 − 2 z −1e − T cos(3T ) + e − 2T z − 2

+3

where T = 0.3. Therefore,

Gb (z ) =

4 1 − 0.3012 z −1

+

1 − z −1e −T cos(3T )

z −1e −T sin (3T )

1 − 2 z −1e −T cos(3T ) + e − 2T z − 2

(

6 1 − 0.4605z −1

)

1 − 0.9210 z −1 + 0.5489z −2

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+

,

(

3 0.5803z −1

)

1 − 0.9210 z −1 + 0.5489z −2

296

=

4 1 − 0.3012 z −1

(s + 2 )

1 − 0.9210 z −1 + 0.5489 z −2 4s + 1

− 3s + 2

(s 2 + 2s + 5)(s 2 + s + 4) (s 2 + 2s + 5) (s 2 + s + 4)

4(s + 2 ) 2

6 − 1.0221z −1

s 3 + s 2 + 6 s + 14

(c) H c (s ) = =

+

2

+1

+

− 7(1)

(s + 2 )

2

+

2

+1

=

+

− 3(s + 0.5)

(s + 0.5)2 + (

15 / 2

)

2

+

(s + 0.5)2 + (

Using Eqs. (9.58), (9.60), and (9.62), we get Gc (z ) = (4 )

+ (− 3)

1 − e −2T cos(T )z −1

+ (− 7 )

1 − 2e − 2T cos(T )z −1 + e − 4T z − 2

(

)

1 − e −T / 2 cos 15T / 2 z −1

(

)

1 − 2e − 2T cos 15T / 2 z −1 + e − T z − 2

where T = 0.3. Hence, Gc (z ) =

(

4 1 − 0.5243z −1

)

1 − 1.0486 z −1 + 0.3012 z −2

(

− 3 1 − 0.7195z −1

)

1 − 1.4390 z −1 + 0.7408z − 2 9.10 (a) G a (z ) =

3z z − e −1.5

+

4z z − e −1.8

=

−

+

(

15 / 2

1 − 2e − 2T cos(T )z −1 + e − 4T z − 2

+ 7 / 15

(

(

1 − 1.4390 z −1 + 0.7408z − 2

z − e − α1T

+

A2 z

z − e − α 2T

=

.

. 3 4 + . s + 7 .5 s + 9

ze −β T sin (λT )

(b) H b (s ) = Gb (z ) z = 4⎛ 1+ s ⎞ = ⎜ ⎟ ⎝ 1− s ⎠

2

1+ s ⎞⎤ ⎡ 1+ s ⎞⎤ + 4 ⎟ ⎟⎥ + 3⎢4⎛⎜ s 1 − ⎝ ⎠⎦ ⎣ ⎝ 1− s ⎠⎥⎦ 2

1+ s ⎞⎤ ⎡ 1+ s ⎞⎤ + 6 ⎟ ⎟⎥ + 4 ⎢4⎛⎜ ⎝ 1− s ⎠ ⎦ ⎣ ⎝ 1− s ⎠⎥⎦

10 ⎡ 4⎛⎜

⎢⎣

54 ⎡ 4⎛⎜

⎢⎣

)

1 − 2e − 2T cos 15T / 2 z −1 + e − T z − 2

z 2 − 2 ze − β T cos(λT ) + e − 2β T 7.5 Since T = 0.2, λ = 7.5, β = 6, it follows H b (s ) = (s + 6 )2 + 7.52

⎝ 1− s ⎠

)

e −T / 2 sin 15T / 2 z −1

)

0.8538z −1

z 2 − 2 ze −1.2 cos(1.5) + e − 2.4

⎢⎣

.

1 − 1.0486 z −1 + 0.3012 z −2

ze −1.2 sin (1.5)

9.11 (a) H a (s ) = G a (z ) z = 4⎛ 1+ s ⎞ = ⎜ ⎟

)2

1.1353z −1

A1 z

4 ⎡4⎛⎜

)

e −2T sin (T )z −1

Since T = 0.2, α 1 = 7.5, α 2 9, A1 = 3, A2 = 4 , it follows H a (s ) = (b) Gb (z ) =

(

7 / 15 15 / 2

3

=

.

(

16 s 2 + 3s + 4

).

75s 2 + 154 s + 91

2

1+ s ⎞⎤ ⎡ 1+ s ⎞⎤ + 26 ⎡4⎛ 1+ s ⎞⎤ +18 ⎟⎥ + 62 ⎢4⎛⎜ ⎟ ⎢⎣ ⎜⎝ 1− s ⎟⎠⎥⎦ ⎝ 1− s ⎠⎦ ⎣ ⎝ 1− s ⎠⎥⎦

2 ⎛ 3⎡4⎛ 1+ s ⎞⎤ +1 ⎞⎛⎜ 12 ⎡4⎛ 1+ s ⎞⎤ − 4 ⎡4⎛ 1+ s ⎞⎤ + 8 ⎞⎟ ⎜ ⎢ ⎜ 1− s ⎟⎥ ⎟⎜ ⎢ ⎜ 1− s ⎟⎥ ⎢⎣ ⎜⎝ 1− s ⎟⎠⎥⎦ ⎟ ⎠⎦ ⎠⎝ ⎣ ⎝ ⎠⎦ ⎝ ⎣ ⎝ ⎠

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297

,

=

(

− (5 + 3s ) 425s 2 + 846 s + 457

(

4(13s + 11) 27s 2 + 46 s + 23

9.12 For the impulse invariance design:

(

)

).

)(

)

ω p = Ω p T = 2 πFp T = 2π 0.45 × 10 3 0.3 × 10 − 3 = 0.27π For the bilinear transformation method: ⎛ Ω pT ⎞ ⎟ = 2 tan −1 Fp Tπ = 2 tan −1 0.45 × 10 3 ⋅ 0.3 × 10 − 3 ⋅ π = 0.2554 π ω p = 2 tan −1 ⎜⎜ ⎟ ⎝ 2 ⎠ ωp 0.56 π 9.13 For the impulse invariance method: 2 πFp = ⇒ Fp = 1.4 kHz = T 0.2 × 10 −3 For the bilinear transformation method: ⎛ ωp ⎞ 1 1 ⎛ 0.56 π ⎞ ⎟⋅ Fp = tan⎜⎜ tan ⋅ = 1.923 kHz. = ⎟ ⎜ ⎟ ⎝ 2 ⎠ π 0.2 × 10 − 3 ⎝ 2 ⎠ πT

(

)

(

(

)

)

9.14 The passband and the stopband edges of the analog lowpass filter are assumed to Ω p = 0.25π and s 0.55 . The requirements to be satisfied by the analog lowpass filter are thus 20log10 H a ( j 0.25π) ≥ −0.5 dB and

20log10 H a ( j 0.55π) ≤ −15 dB. From α p = 20log10 ( 1 + ε 2 ) = 0.5 we obtain 2 0.1220184543. From 2 we obtain A 2 31.6227766 . From Eq. (4.32), the inverse s 10 log10 (A ) 15

1 A2 − 1 = = 15.841979 and from Eq. (4.31) the inverse k1 ε 1 Ωs = 2.2 . Substituting these values in Eq. (4.35) we transition ratio is given by = k Ωp

discrimination ratio is given by

obtain

N=

log10 (1 / k1 ) log10 (15.841979) = = 3.503885. We choose N = 4. log10 (1 / k ) log10 (2.2) 2N

⎛Ωp ⎞ ⎟ = ε 2 . Substituting the values of From Eq. (4.33) we have ⎜⎜ ⎟ ⎝ Ωc ⎠ Ω c = 1.3007568(Ω p ) = 1.021612.

p,

N, and

2

we get

Using the statement [z,p,k] = buttap(4) we get the poles of the 4-th order Butterworth analog filter with a 3-dB cutoff at 1 rad/s as p1 = −0.3827 + j 0.9239, p2 = −0.3827 − j 0.9239, p3 = −0.9239 + j 0.3827, and p 4 = −0.9239 − j 0.3827. Therefore, Not for sale

298

H an (s ) =

1 1 = . (s − p1 )(s − p 2 )(s − p3 )(s − p 4 ) (s 2 + 0.7654s + 1)(s 2 + 1.8478s + 1)

Next we expand H an (s ) in a partial-fraction expansion using the M-file residue and arrive at H an (s ) =

− 0.9238729 s − 0.7071323

+

0.9238729 s + 1.7071323

. We next s 2 + 0.7654 s + 1 s 2 + 1.8478s + 1 denormalize H an (s ) to move the 3-dB cutoff frequency to c 1.021612 using the M-file s ⎛ ⎞ ⎟ lp2lp resulting in H a ( s ) = H an ⎜ ⎝ 1.021612 ⎠ 0.943847s + 1.78174665 − 0.943847s − 0.738039 + = s 2 + 0.781947948s + 1.0437074244 s 2 + 1.887749436s + 1.0437074244 − 0.943847 s − 0.738039 0.943847s + 1.78174665 = + (s + 0.390974 ) 2 + (0.9438467) 2 (s + 0.94387471) 2 + (0.39090656 ) 2 Making use of the M-file bilinear we finally arrive at G( z) =

− 0.943847 z 2 + 0.68178386 z z 2 − 1.363567724 z + 0.4575139

9.15 The mapping is given by s =

s = σ o + jΩ o , z =

1 T

+

0.943847 z 2 − 0.25640047 z z 2 − 0.77823439 z + 0.1514122

(1 − z −1 ) or equivalently, by z =

.

1 . For 1 − sT 1

1 2 . Therefore, z = . Hence, 1 − σ o T − jΩ o T (1 − σ o T ) 2 + (Ω o T ) 2

z < 1 for o

0. As a result, a stable H a (s ) results in a stable H (z ) after the 1 2 which is equal to 1 only for transformation. However, for σ o = 0, z = 1 + (Ω o T ) 2 Ω o = 0. Hence, only the point Ω o = 0 on the jΩ -axis in the s -plane is mapped onto the point z = 1 on the unit circle. Consequently, this mapping is not useful for the design of digital filters via analog filter transformation.

9.16 For no aliasing T ≤

H1 (z)

π . Figure below shows the magnitude responses of the digital filters Ωc

and H 2 (z) . j

H 2 (e j )

H1(e )

2 1

–š

0

š

Not for sale

–š

0

š

299

(a) The magnitude responses of the digital filters G1 ( z) and G2 ( z) are shown below: G2 (e j )

G1(e j )

3/2 1 –š

0

š

–š

0

š

(b) As can be seen from the above G1 ( z) is a multi-band filter, whereas, G2 ( z) is a highpass filter. 9.17 H a (s ) is causal and stable and H a (s )

s, Now, G( z) = H a (s ) s = 2 ⎛⎜ 1− z −1 ⎞⎟ . Thus,

1,

T ⎜ 1− z −1 ⎟

⎝

⎠

G (z ) is causal and stable. Now,

G ( e jω ) = H a ( s )

= 2 ⎛ 1− e jω ⎞⎟ s= ⎜ T ⎜ 1+ e jω ⎟ ⎝

H a (s ) s = j 2 tan(ω / 2) = H a ( j T

2 T

tan(ω / 2).

⎠

Therefore, G(e jω ) = H a ( j tan(ω / 2) ≤ 1 for all values of . Hence, G (z ) is a BR T 2

function. 9.18 G ( z ) =

1+ α 1 − 2 βz − 1 + z − 2 . For β = cos ωo , the numerator of G (z ) becomes ⋅ 2 1 − β (1 + α ) z −1 + αz − 2 1

2

j

1

j 0 1 z )

1 2cos 0 z z (1 e 0 z )(1 e numerator of G(z N ) is then given by (1 e j N

j

0z

N

which has roots at z

)(1 e

j 0 N z )

e j

0.

The

whose roots are j(2 n

)/ N

0 e , and are given by z e , obtained by solving the equation z N 0 n N 1. Hence G(z ) has N complex conjugate zero-pairs located on the unit 2 n 0 radians, 0 n N 1. circle at angles of N

0

0

/ 2, there are 2N equally spaced zeros on the unit circle starting at

9.19 (a) H ( z ) =

1 [1 + 2

/ 2N.

z −2 D1 ( z −1 ) z −2 D2 ( z −1 ) N ( z) . We can write A4 ( z ) = ⋅ , where A4 ( z )] = D( z ) D1 ( z ) D2 ( z )

D1 ( z ) = 1 − β1 (1 + α1 )z −1 + α1 z −2 and D2 ( z ) = 1 − β2 (1 + α 2 )z −1 + α 2 z −2 . Therefore, N ( z) =

1 2

[D (z)D (z) + z 1

2

−4

]

D1 ( z −1 ) D2 ( z −1 ) . Now,

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z − 4 N ( z −1 ) =

1 2

[z

−4

]

D1 ( z −1 ) D2 ( z −1 ) + D1 ( z ) D2 ( z ) = N ( z ). Hence, N ( z ) is a symmetric

polynomial. It follows then P( z) =

1 2

[(α1 − β1 (1 + α1 )z −1 + z − 2 )(α2 − β2 (1 + α2 )z −1 + z − 2 )

)(

(

+ 1 − β1 (1 + α1 )z −1 + α1 z − 2 1 − β2 (1 + α 2 )z −1 + zα 2 − 2

=

)]

1 + α 1α 2 [1 − (1 + α1 )(1 + α 2 )(β1 + β2 ) z −1 + 2[α1 + α 2 + β1β 2 (1 + α1 )(1 + α 2 )] z − 2 2 1 + α 1α 2 1 + α 1α 2 (1 + α1 )(1 + α 2 )(β1 + β 2 ) − 3 − z + z − 4 ] = a(1 + b1 z −1 + b2 z −2 + b1 z −3 + z −3 ), 1 + α1α 2

(1 + α1 )(1 + α 2 )( β1 + β2 ) , 1 + α1α 2 2[α1 + α2 + β1 β2 (1 + α1 )(1 + α2 )] , b2 = 1 + α1α2 b1 = −

where

(b) a =

(7-a) (7-b)

1 + α 1α 2 . 2

(7-c)

(c) for z = e jω , we can write N (e jω ) = a(1 + b1e − jω + b2 e − j 2ω + b1e − j 3ω + e − j 4ω ) = ae − j 2ω (b2 + 2b1 cos ω + 2 cos 2ω). Now, N (e jω ) = 0 for i = 1, 2. For i = 1, we get b2 + 2b1 cos ω1 + 2 cos 2ω1 = 0, (7-d) for i = 2, we get

b2 + 2b1 cos ω2 + 2 cos 2ω2 = 0,

(7-e)

Solving Eqs. (7-d) and (7-e) we get b1 = −2(cos ω1 + cos ω2 ) b2 = 2(2 cos ω1 cos ω2 + 1). and

(7-f) (7-g)

(1 + α1 )(1 + α 2 )(β1 + β1 ) = 2(cos ω1 + cos ω2 ), (7-h) 1 + α 1α 2 and from Eqs. (7-b) and (7-g) we have 2[α1 + α 2 + β1β 2 (1 + α1 )(1 + α 2 )] = 2(2 cos ω1 cos ω2 + 1). (7-i) 1 + α1α 2 1 − tan( B2 / 2) 1 − tan( B1 / 2) , and after rearrangement we get Substituting α1 = and α 2 = 1 + tan( B2 / 2) 1 + tan( B1 / 2) β + β = (cos ω + cos ω )[1 + tan( B / 2) tan( B / 2)] ∆ θ , (7-j) From Eqs. (7-a) and (7-f) we have

1

2

1

2

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1

2

= 1

301

β1β 2 = [1 + tan( B1 / 2) tan( B2 / 2)] cos ω1 cos ω2 ∆ = θ2 .

and

(7-k)

The above two nonlinear equations can be solved yielding 1

2 1

4 2 and 2

2. 2 1 (d) For the double notch filter with the following specifications: ω1 = 0.2π , ω2 = 0.6π , B1 = 0.2π, and B2 = 0.25π we get the following values for the parameters of the notch filter transfer function: α1 = 0.5095, α 2 = 0.4142, θ1 = 0.5673, θ 2 = −0.1491, β1 = 0.7628, and 1

β 2 = −0.1955 . H (z ) =

1 2

[1 + A4 (z )] :

1

2

Phase, radians

Magnitude

0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

1 0 -1 -2

0

ω/ π

0.2

0.4

0.6

0.8

1

ω/ π

9.20 A zero (pole) of H LP (z) is given by the factor ( z − z k ). After applying the lowpass-tozˆ − α − z k , and hence the new location of the lowpass transformation, this factor becomes 1 − α zˆ zero (pole) is given by the roots of the equation α + zk . For z k = −1, zˆ − α − z k + α z k zˆ = (1 + α z k )zˆ − (α + z k ) = 0 or zˆ k = 1 + αzk a −1 zˆ k = = − 1. 1− a 9.21 The lowpass-to-bandpass transformation is given by z

b a ˆz zˆ 2 where a 1 a zˆ b zˆ 2

2 1

and

1 . A zero (pole) of H LP (z) is given by the factor ( z − z k ). . After applying the 1 b azˆ zˆ 2 zk , and hence, the lowpass-to-bandpass transformation, this factor becomes 1 a zˆ b zˆ 2 b

new location of the zero (pole) of the bandpass transfer function is given by the roots of the

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a(1 z k ) b zk z 1 b zk 1 b zk

equation (1 b zk ) z2 a(1 zk ) z (b zk ) 0, or z 2 a(1 z k ) 2(1 b z k )

solution is given by zˆ k

(

0.3404 1 + z −1

9.22 G LP (z ) =

a(1 z k ) 2(1 b z k )

2

0, whose

b zk . For z k 1 bz k

1, zˆ k

)2

, with ωc = 0.55π. 1 + 0.1842 z −1 + 0.1776 z −2 ˆ c = 0.42 π H LP (z ) for ω 0.55π ⎞ 0.42 π ⎞ 0.55π − 0.42 π ⎞ tan⎛⎜ ⎟ sin⎛⎜ ⎟ − tan⎛⎜ ⎟ 2 ⎝ 2 ⎠= ⎝ 2 ⎠ ⎠ = 0.2030 ⎝ α= 0 . 55 0 . 42 0 . 55 0 . 42 π π π + π ⎞⎟ sin⎛⎜ ⎞⎟ + tan⎛⎜ ⎞⎟ tan⎛⎜ 2 ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ ⎠ H LP (z ) = G LP (z ) z −1 = zˆ −1 − α = 1− αzˆ −1

=

⎛ zˆ −1 − α ⎞ ⎟ 0.3404⎜⎜ 1+ −1 ⎟ ˆ 1 − α z ⎝ ⎠

2

⎛ −1 ⎞ ⎛ zˆ −1 − α ⎞ ⎟⎟ + 0.1776⎜⎜ zˆ − α ⎟⎟ 1 + 0.1842⎜⎜ ⎝ 1− αzˆ −1 ⎠ ⎝ 1− αzˆ −1 ⎠

0.21623 + 0.43245zˆ −1 + 0.21623zˆ −2 0.96993 − 0.2863zˆ −1 + 0.181416zˆ − 2

2

.

Gain Responses 0 GLP(z) Gain in dB

-10 HLP(z) -20 -30 -40 -50

9.23

0

0.2

0.4

0.6

0.8

1

0.55π + 0.47π ⎞ − cos⎛⎜ ⎟ 2 ⎝ ⎠ = 0.0317. α= 0.55π − 0.47π ⎞ cos⎛⎜ ⎟ 2 ⎝ ⎠

H HP (z ) = G LP (z ) z −1 = −⎛⎜ zˆ −1 + α ⎞⎟ = ⎜ 1+ αzˆ −1 ⎟ ⎠ ⎝

0.3192 − 0.6383zˆ −1 + 0.3192 zˆ −2 0.9943 − 0.1097 zˆ −1 + 0.1728zˆ − 2

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.

303

1.

Gain Responses 0

Gain, dB

-10 HHP(z)

-20

GLP(z)

-30 -40 -50

0

0.2

0.4

0.6

0.8

1

ω /π

ˆ c2 − ω ˆ c1 , and α = cos(ω ˆ c ) = 0.2790 9.24 ωc = ω ⎛ zˆ −1 − α ⎞ 0.2790 zˆ −1 − zˆ − 2 −1 − 1 ⎟= z = − zˆ ⎜ . ⎜ 1 − αzˆ −1 ⎟ 1 − 0.2790 zˆ −1 ⎝ ⎠ 0.6918 − 0.1968zˆ −1 − 0.6234 zˆ −2 − 0.1968zˆ −3 + 0.6918zˆ −4 H BP (z ) = . 3.5588 − 2.6824 zˆ −1 + 3.046 zˆ −2 − 1.2546 zˆ −3 + zˆ −4 Gain Responses 0 HBP(z) Gain, dB

-10

GLP(z)

-20 -30 -40 -50

0

0.2

0.4

0.6

0.8

1

ω /π

⎛ 0.52 π − 0.45π ⎞ sin⎜ ⎟ 2 ⎝ ⎠ = 0.1099. ˆ p = 0.45π, and ω p = 0.52π. λ = 9.25 ω ⎛ 0.52π + 0.45π ⎞ sin⎜ ⎟ 2 ⎝ ⎠ 2.873 − 5.622 zˆ −1 + 5.629zˆ −2 − 2.874 zˆ −3 . H HP (z ) = G HP (z ) z = zˆ −1 −0.1099 = −1 − 2 ˆ −3 − 1 ˆ ˆ − z + z + z 9 . 7623 1 . 5359 6 . 711 1−0.1099 zˆ

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Gain Responses 0 HHP(z)

Gain in dB

-10

GHP(z)

-20 -30 -40 -50

0

0.2

0.4

0.6

0.8

1

ω /π

⎛ 1 − z −2 9.26 Eq. (7.79): H BP (z ) = 0.136728736⎜ ⎜ 1 − 0.53353098z −1 + 0.726542528z −2 ⎝ ⎛ 0.5π − 0.4π ⎞ sin⎜ ⎟ 2 ⎝ ⎠ = −0.1584 ˆ 0 = 0.5π, and ω0 = 0.4 π. λ = ω ⎛ 0.5π + 0.4π ⎞ sin ⎜ ⎟ 2 ⎝ ⎠ G BP (z ) = H BP (z ) z −1= zˆ −1 + 0.1584 = 1+ 0.1584

zˆ −1

0.1333 − 0.1333zˆ −2 0.933718 + 0.667122 zˆ −2

⎞ ⎟. ⎟ ⎠

.

Gain Responses 0 GBP(z)

HBP(z)

Gain in dB

-10 -20 -30 -40 -50

0

0.2

0.4

0.6

0.8

1

ω /π

B 1 − tan⎛⎜ ω ⎞⎟ ⎛ 80 ⎞ ⎛ 5 ⎞ ⎝ 2 ⎠ = 0.9244 9.27 ω0 = 2 π⎜ ⎟ = 0.4 π , Bω = 2 π⎜ ⎟ = 0.025π. α = B ⎝ 400 ⎠ ⎝ 400 ⎠ 1 + tan⎛⎜ ω ⎞⎟ ⎝ 2 ⎠ β = cos(ω0 ) = 0.3090.

G (z ) =

=

1 2

(1 + α ) − 2β(1 + α )z −1 + (1 + α )z − 2 1 − β(1 + α )z −1 + αz − 2

0.9622 − 0.59465z 1 − 0.5946 z

−1

−1

+ 0.9622 z

+ 0.9244 z − 2

=

(

0.5 1.9244 − 1.1893 z −1 + 1.9244 z − 2 1 − 0.5946 z

−1

+ 0.9244 z

−2

)

−2

.

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ˆ ω −ω sin⎛⎜ 0 0 ⎞⎟ ⎝ 2 ⎠ = 0.2738. α= ˆ ω +ω sin⎛⎜ 0 0 ⎞⎟ ⎝ 2 ⎠

50 ⎞ ˆ 0 = ⎛⎜ ω ⎟2 π = 0.25π. ⎝ 400 ⎠

H BS (z ) = G BS (z ) z −1 = zˆ −1 −α =

1.0301 − 1.45679zˆ −1 + 1.0301zˆ −2 1.06017 − 1.45674zˆ −1 + zˆ − 2

1− αzˆ −1

.

Gain Responses 0 HBS(z)

Gain in dB

-10

GBS(z)

-20 -30 -40 -50

0

0.2

0.4

0.6

0.8

1

ω /π

ˆ p = 0.45π and ω p = 0.52π. 9.28 ω G HP (z ) =

(

0.2397 1 − 1.5858z −1 + 1.5858z −2 − z −3

1 + 0.3272 z −1 + 0.7459z −2 + 0.179z −3 ⎛ ω p + ωˆ p ⎞ − cos⎜ ⎟ 2 ⎝ ⎠ = −0.0474. α= ˆ ω − ω ⎛ p p⎞ cos⎜ ⎟ ⎝ 2 ⎠ H LP (z ) = G HP (z )

− zˆ −1 − α z −1 = 1+ αzˆ −1

=

)

− 1.03399 − 1.45578zˆ −1 − 1.45578zˆ −2 − 1.03399zˆ −3 − 4.72684 + 2.52265zˆ −1 − 3.7756zˆ − 2 + zˆ −3 Gain Responses

0 HLP(z)

Gain, dB

-10

GHP(z)

-20 -30 -40 -50

0

0.2

0.4

0.6

0.8

1

ω /π

9.29 D = 6.9231 , hence, N = 6. A(z ) =

( )

z −6 D z −1 . D (z )

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Group delay 6

0.75

5.5 Delay, in radians

Magnitude

Magnitude response 0.8

0.7 0.65 0.6 0.55

5 4.5 4 3.5

0

0.2

0.4

0.6

0.8

3

1

0

0.2

0.4

ω/ π

9.30

0.6

0.8

1

ω/ π

From Eq. (2.141) y(nT ) = y((n − 1)T ) +

nT

∫ x(τ)dτ =y((n − 1)T ) + T ⋅ x((n − 1)T ), which

( n −1)T

reduces to y[ n] = y[ n − 1] + T ⋅ x[ n − 1]. Hence, the corresponding transfer function is given by H R ( z ) =

Tz −1

1 − z −1

. From Eq. (2.120) y[ n] = y[ n − 1] +

T (x[ n] + x[ n − 1]) . 2

T 1 + z −1 . From the Hence, the corresponding transfer function is given by H T ( z) = ⋅ 2 1 − z −1 plot given below it can be seen that the magnitude response of H int (z) lies between that of H R ( z) and H T (z) .

Magnitude

1.5

1 HR(z) 0.5 Hint(z)

HT(z) 0 0.2

0.4

0.6

0.8

1

ω/ π

3

1

9.31 H N ( z ) = H R ( z ) + H T ( z ). From the plot given below it can be seen that the magnitude 4 4

response of H N (z) lies between that of H R (z) and H T (z) , and is much closer to that of H int ( z) .

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307

Magnitude

1.5

1 HR(z) 0.5

HN(z) HT(z)

0 0.2

0.4

Hint(z)

0.6

0.8

1

ω/π

3 ⎛ z −1

⎞

1 ⎛ 1+ z −1 ⎞

8(1− z −1 )

1+ 7 z −1

⎟+ ⎜ , ⎟= 9.32 H N ( z ) = ⎜⎜ . Its inverse is given by H ( z ) = 4 ⎝ 1 − z −1 ⎟⎠ 8 ⎜⎝ 1− z −1 ⎟⎠ 8(1 − z −1 ) 1+ 7 z −1 which is unstable as it has a pole at z = −7. A stable transfer function with the same magnitude response is obtained by multiplying H (z ) with an allpass function resulting in H IIR ( z) =

8(1− z −1 ) 1+ 7 z −1 ⋅ 1+ 7 z −1 7 + z −1

=

8(1− z −1 ) 7 + z −1

1+ 7 z −1 7 + z −1

. A plot of the ideal differentiator

H DIF (z) with a frequency response given by Eq. (7.68) and that of the IIR differentiator H IIR (z) is given below. As can be seen the magnitude response of H IIR (z) is very close to that H DIF (z) . 3

HIIR(z)→

Magnitude

2.5 2 HDIF(z)

1.5 1 0.5 0

0

0.2

0.4

0.6

0.8

1

ω/π

M9.1 FT = 100 kHz, α p = 0.4 dB, Fp = 10 kHz, α s = 50 dB, & Fs = 30 kHz , ωp =

2 πF p FT

= 0.628, ωs =

2 πFs = 1.885. FT

⎛ ωp Let T = 2. Ω p = tan⎜⎜ ⎝ 2

⎞ ω ⎟ = 1.376, and Ω s = tan⎛⎜ s ⎞⎟ = 1.376. Therefore, ⎟ ⎝ 2 ⎠ ⎠ ⎛ 1 ⎞ 1 Ωs ⎟ = −0.4. Hence, ε 2 = 0.096. = 4.235. Now, 20 log10 ⎜ = ⎜ 2 ⎟ k Ωp ⎝ 1+ ε ⎠

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308

⎛1⎞ From 20 log10 ⎜ ⎟ = −50 we obtain A 2 = 100,000. Therefore, ⎝ A⎠ 1 k1 = 9.798 × 10 −4 , or = 1020.62. As a result, k1

log10 (1 / k1 ) N= = 4.80 → 5. Next, solving log10 (1 / k )

⎛ Ωs ⎜⎜ ⎝ Ωc

10

⎞ ⎟⎟ ⎠

= A 2 − 1 = 99,999 we get

Ωs

Ωc =

= 0.316Ω s = 0.435 99,9991 / 10 Using the M-file buttap, we determine the normalized analog Butterworth transfer function of 5th order with a 3-dB cutoff frequency at Ω c = 1 , which is: 1 . H an (s ) = 2 (s + 1) s + 0.618s + 1 s 2 + 1.618s + 1 Denormalize H an (s ) to move Ω c to 0.316: 1 s ⎞ H a (s ) = H an ⎛⎜ ⎟= ⎝ 0.435 ⎠ (2.30s + 1)(5.28s 2 + 1.42s + 1)(5.28s 2 + 3.72s + 1)

(

=

)(

)

1 1 + 7.44s + 27.66s 2 + 63.58s 3 + 90.30s 4 + 64.12s 5

.

G (z ) = H a (s ) s = 2 ⎛⎜ 1− z −1 ⎞⎟ T ⎜ 1+ z −1 ⎟ ⎝ ⎠

=

0.0039 + 0.0197z −1 + 0.0394 z −2 + 0.0394z −3 + 0.0197z −4 + 0.0039z −5 1 − 2.3617z −1 + 2.6139z − 2 − 1.5492 z −3 + 0.4865z − 4 − 0.0637z − 5

.

Matlab code is as follows: % Program M9.01 N = 5; [z, p, k] = buttap(N); [num, den] = zp2tf(z, p, k); % s -> s/0.435 den = [64.12 90.30 63.58 27.66 7.44 1]; num = [0 0 0 0 0 1]; % compute z, p, and k [z, p, k] = tf2zp(num, den); % perform bilinear transformation with T = 2; [zd, pd, kd] = bilinear(z, p, k, 1/2); % get the digital transfer function [n2, d2] = zp2tf(zd, pd, kd); % get the frequency response [h, w] = freqz(n2, d2, 512); figure(1); plot(w/pi, 20*log10(abs(h))); grid; axis([0 1 -60 5]); xlabel('\omega/\pi'); ylabel('Gain, dB');

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title('Gain response'); figure(2); plot(w/pi, unwrap(angle(h))); grid; axis([0 1 -8 1]); xlabel('\omega/\pi'); ylabel('Phase, radians'); title('Phase response'); Gain response

Phase response

0

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M9.2 % Problem M9.02 Fp = input('Passband edge frequency in Hz = '); Fs = input('Stopband edge frequency in Hz = '); FT = input('Sampling frequency in Hz = '); Rp = input('Passband ripple in dB = '); Rs = input('Stopband minimum attenuation in dB = '); Wp = 2*Fp/FT; Ws = 2*Fs/FT; [N, Wn] = buttord(Wp, Ws, Rp, Rs) [b, a] = butter(N, Wn); disp('Numerator polynomial'); disp(b) disp('Denominator polynomial'); disp(a) [h, w] = freqz(b, a, 512); plot(w/pi, 20*log10(abs(h))); grid axis([0 1 -60 5]); xlabel('\omega/\pi'); ylabel('Magnitude, dB'); pause plot(w/pi, unwrap(angle(h))); grid axis([0 1 -8 1]); xlabel('\omega/\pi'); ylabel('Phase, radians'); M9.3 % Program #M9.03 close all; clear; clc; % (a.) Ft = 1Hz; [B, A] = besself(5, 0.5);

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[BZ,AZ] = impinvar(B,A,1); [h, w] = freqz(BZ, AZ, 512); [Gd,W] = grpdelay(BZ,AZ,512); figure(1); plot(w/pi, 20*log10(abs(h))); title('Gain response, , Sampling rate = 1 Hz'); xlabel('\omega/\pi'); ylabel('Gain, in dB'); figure(2); plot(W/pi, (Gd)); title('Group delay, Sampling rate = 1 Hz'); xlabel('\omega/\pi'); ylabel('Delay, in samples'); % (b.) Ft = 2Hz; [B, A] = besself(5, 0.5); [BZ,AZ] = impinvar(B,A,2); [h, w] = freqz(BZ, AZ, 512); [Gd,W] = grpdelay(BZ,AZ,512); figure(3); plot(w/pi, 20*log10(abs(h))); title('Gain response, Sampling rate = 2 Hz'); xlabel('\omega/\pi');ylabel('gain response'); figure(4); plot(W/pi, (Gd)); title('Group delay, Sampling rate = 2 Hz'); xlabel('\omega/\pi'); ylabel('Delay, in samples'); (a) FT = 1 Hz Group delay, Sampling rate = 1 Hz 8

0

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Delay, in samples

Gain, in dB

Gain response, Sampling rate = 1 Hz 20

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(b) FT = 2 Hz

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Group delay, Sampling rate = 2 Hz 20

0

15

Delay, in samples

Gain, in dB

Gain response, Sampling rate = 2 Hz 20

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M9.4 ω p = 0.628, ω s = 1.885.

(

0.6

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1

ω/ π

)

(

)

20 log10 G e 0.628 j ≥ −0.4, 20 log10 G e1.885 j ≤ −50.

Impulse invariance method, let T = 1 and assume no aliasing. Then ⎛ 1 ⎞ ⎟ = −0.4 ⇒ ε 2 = 0.096 , 20 log10 ⎜⎜ ⎟ 2 ⎝ 1+ε ⎠ ⎛1⎞ 20 log10 ⎜ ⎟ = −50 ⇒ A 2 = 100,000 . ⎝ A⎠

A2 − 1

1 = k1

= 1020.62 ,

ε2 ⎛1⎞ log10 ⎜ ⎟ ⎝ k1 ⎠ = 6.3 → 7 N= 1 log10 ⎛⎜ ⎞⎟ ⎝k⎠ ⎛ Ωs ⎜⎜ ⎝ Ωc

1 1.885 = = 3 . Hence, k 0.628

10

⎞ ⎟⎟ ⎠

= A 2 − 1 = 99,999 or Ω c =

Ωs

(A − 1) 2

1 / 10

= 0.596 .

Using the M-file butter, we get the analog Butterworth transfer function H a (s ) of 7th order with a 3-dB cutoff frequency at Ω c . H a (s ) is then transformed into a digital transfer function using the M-file impinvar, which yields:

G (z ) =

0.001z −2 + 0.0034 z −3 + 0.0023z −4 + 0.0003z −5 1 − 4.38z −1 + 8.54 z −2 − 9.55z −3 + 6.57z −4 − 2.77z −5 + 0.66 z −6 − 0.07z −7

% Problem M9.04 [B, A] = butter(7, 0.596, 's'); [num, den] = impinvar(B, A, 1);

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.

% get the frequency response [h, w] = freqz(num, den, 512); plot(w/pi, 20*log10(abs(h))); grid; axis([0 1 -60 5]); xlabel('\omega/\pi'); ylabel('Gain, in dB'); title('Gain response'); Gain response 0

Gain, in dB

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M9.5 FT = 100 kHz, Fp = 10 kHz, Fs = 30 kHz, α p = 0.4 dB, α s = 50 dB ωp =

2 πFp FT

2πFs = 0.6 π. FT

= 0.2 π, and . ω s =

(

)

(

)

20 log10 G e 0.628 j ≥ −0.4, 20 log10 G e1.885 j ≤ −50.

Impulse invariance method: Let T = 1 and assume no aliasing. In this case, the specifications for H a (s ) are

same as that for G (z ), i.e., Ω p = 0.628, Ω s = 1.885,

20 log10 H a (0.628 j ) ≥ −0.4, 20 log10 H a (1.885 j ) ≤ −50. ⎛ 1 ⎞ ⎟ = −0.4 ⇒ ε 2 = 0.096 , and Now, 20 log10 ⎜⎜ ⎟ 2 ⎝ 1+ε ⎠ 1 20 log10 ⎛⎜ ⎞⎟ = −50 ⇒ A 2 = 100,000 . Hence, ⎝ A⎠ 1 = k1

A2 − 1 ε

2

= 1020.62 and

1 1.885 = = 3. k 0.628

Order of the Type I Chebyshev filter is N =

cosh −1 (1 / k1 ) cosh −1 (1 / k )

= 4.324 → 5

Bilinear transformation method: Let T = 2. Ω p = 0.325, Ω s = 1.376 . Here,

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1 Ωs = = 4.235 , and k Ωp

313

ε 2 = 0.096, A 2 = 100,000 . Thus,

1 = 1020.62 . k1

Order of the Type I Chebyshev filter is N =

cosh −1 (1 / k1 )

= 3.59 → 4. cosh −1 (1 / k ) Both designs meet the specifications, while the bilinear transformation method meets with a filter of lower order. MATLAB code is as follows:

% Problem #M9.05 % Impulse invariance method [z, p, k] = cheb1ap(5, 0.4); [B, A] = zp2tf(z, p, k); [BT, AT] = lp2lp(B, A, 0.628); [num, den] = impinvar(BT, AT, 1); [h, w] = freqz(num, den, 512); figure(1); plot(w/pi, 20*log10(abs(h))); grid; axis([0 1 -60 5]); xlabel('\omega/\pi'); ylabel('Gain, in dB'); title('Impulse Invariance Method'); figure(2); plot(w/pi, unwrap(angle(h))); grid axis([0 1 -8 1]); xlabel('\omega/\pi'); ylabel('Phase, in radians'); title('Impulse Invariance Method'); % Bilinear transformation method [z, p, k] = cheb1ap(4, 0.4); [B, A] = zp2tf(z, p, k); [BT, AT] = lp2lp(B, A, 0.325); [num, den] = bilinear(BT, AT, 0.5); [h, w] = freqz(num, den, 512); figure(3); plot(w/pi, 20*log10(abs(h))); grid; axis([0 1 -60 5]); xlabel('\omega/\pi'); ylabel('Gain, in dB'); title('Bilinear Transformation Method'); figure(4); plot(w/pi, unwrap(angle(h))); grid axis([0 1 -8 1]); xlabel('\omega/\pi'); ylabel('Phase, in radians'); title('Bilinear Transformation Method');

Impulse Invariance Method:

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Impulse Invariance Method

Impulse Invariance Method

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Gain, in dB

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Bilinear Transformation Method: Bilinear Transformation Method

Bilinear Transformation Method 0

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M9.6 % Problem M9.06 Wp = input('Normalized passband edge = '); Ws = input('Normalized stopband edge = '); Rp = input('Passband ripple in dB = '); Rs = input('Minimum stopband attenuation in dB = '); [N, Wn] = cheb1ord(Wp, Ws, Rp, Rs); [b, a] = cheby1(N, Rp, Wn); [h, omega] = freqz(b,a,256); figure(1); plot(omega/pi, 20*log10(abs(h)));grid; xlabel('\omega/\pi'); ylabel('Gain, in dB'); title('Type I Chebyshev Filter'); axis([0 1 -60 5]); figure(2); plot(omega/pi, unwrap(angle(h))); grid axis([0 1 -8 1]); xlabel('\omega/\pi'); ylabel('Phase, in radians'); title('Type I Chebyshev Filter');

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Type I Chebyshev Filter


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M9.7 % Problem M9.07 Wp = input(' Passband edge in radians = '); Ws = input(' Stopband edge in radians = '); Rp = input('Passband ripple in dB = '); Rs = input('Stopband minimum attenuation in dB = '); [N, Wn] = cheb1ord(Wp,Ws,Rp,Rs, 's'); [B, A] = cheby1(N, 0.5, Wn, 's'); [num, den] = impinvar(B, A, 1); [h, omega] = freqz(num,den,256); figure(1); plot(omega/pi, 20*log10(abs(h)));grid; xlabel('\omega/\pi'); ylabel('Gain, in dB'); title('Type I Chebyshev Filter'); axis([0 1 -60 5]); figure(2); plot(omega/pi, unwrap(angle(h))); grid axis([0 1 -8 1]); xlabel('\omega/\pi'); ylabel('Phase, in radians'); title('Type I Chebyshev Filter'); Type I Chebyshev Filter


0

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Gain, in dB

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M9.8 Impulse invariance method: 2 log10 (4 / k1 ) 1 Ωs 1 , where = = 3.00, = 1020.62. Hence, N ≈ log10 (1 / ρ ) k Ωp k1

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k ' = 1 − k 2 , ρ0 =

1 − k'

(

2 1 + k'

),

ρ = ρ0 + 2 ρ05 + 15 ρ09 + 150 ρ13 0 . In our case, k ' = 0.943, ρ0 = 0.0073, ρ = 0.0073 . Hence, N ≈ 3.38 → 4

Bilinear transformation method: 1 1 Ω = 1020.62. Ω p = 0.325, Ω s = 1.376, = s = 4.235. k Ωp k1

k ' = 0.972, ρ0 = 0.0036 , ρ = 0.0036 . Hence, N ≈ 2.96 → 3 We note that the filter design using the impulse invariance method does not meet the specifications due to aliasing. Increasing N from 4 to 6 will meet the specifications. % Problem M9_08 % Impulse Invariance Method [z, p, k] = ellipap(4, 0.4, 50); [B, A] = zp2tf(z, p, k); [BT, AT] = lp2lp(B, A, 0.628); [num, den] = impinvar(BT, AT, 1); [h, omega] = freqz(num,den,256); figure(1); plot(omega/pi, 20*log10(abs(h)));grid; xlabel('\omega/\pi'); ylabel('Gain, in dB'); title('Impulse Invariance Method'); axis([0 1 -60 5]); figure(2); plot(omega/pi, unwrap(angle(h))); grid axis([0 1 -8 1]); xlabel('\omega/\pi'); ylabel('Phase, in radians'); title('Impulse Invariance Method'); % Bilinear Transformation Method [z, p, k] = ellipap(4, 0.4, 50); [B, A] = zp2tf(z, p, k); [BT, AT] = lp2lp(B, A, 0.325); [num, den] = bilinear(BT, AT, 0.5); [h, omega] = freqz(num,den,256); figure(3); plot(omega/pi, 20*log10(abs(h)));grid; xlabel('\omega/\pi'); ylabel('Gain, in dB'); title('Bilinear Transformation Method'); axis([0 1 -60 5]); figure(4); plot(omega/pi, unwrap(angle(h))); grid axis([0 1 -8 1]); xlabel('\omega/\pi'); ylabel('Phase, in radians'); Not for sale

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title('Bilinear Transformation Method'); Impulse Invariance Method

Impulse Invariance Method 0

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Gain, in dB

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Bilinear Transformation Method

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M9.9 % Program M9_09 Wp = input('Normalized passband edge = '); Ws = input('Normalized stopband edge = '); Rp = input('Passband ripple in dB = '); Rs = input('Stopband ripple in dB = '); [N, Wn] = ellipord(Wp, Ws, Rp, Rs); [b, a] = ellip(N, Rp, Rs, Wn); [h, omega] = freqz(b, a, 256); plot(omega/pi, 20*log10(abs(h)));grid; xlabel('\omega/\pi'); ylabel('Gain, in dB'); title('IIR Elliptic Lowpass Filter'); axis([0 1 -60 5]); IIR Elliptic Lowpass Filter 0

Gain, in dB

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M9.10 FT = 1.5 MHz, Fp = 600 kHz, Fs = 210 kHz.

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. α p = 0.4 dB, α s = 45 dB (a) ω p =

2 πF p FT

= 2.513, ωs =

2πFs = 0.880. FF

ˆ = tan⎛⎜ ωp ⎞⎟ = 3.076, Ω ˆ = tan⎛⎜ ωs ⎞⎟ = 0.471. Ω p s ⎝ 2⎠ ⎝ 2 ⎠ Analog highpass specifications are thus: ˆ = 3.076, Ω ˆ = 0.471, α = 0.4 dB, α = 45 dB Ω p s p s ˆ Ω p (b) Ω p = 1, Ω s = = 6.531 ˆ Ω s % Problem M9.10 close all; clear; clc; [N, Wn] = ellipord(1, 6.531, 0.4, 45, 's'); [B, A] = ellip(N, 0.4, 45, Wn, 's'); [BT, AT] = lp2hp(B, A, 3.076); [num, den] = bilinear(BT, AT, 0.5); [h, omega] = freqz(num, den, 256);

figure(1); plot(omega/pi, 20*log10(abs(h)));grid; xlabel('\omega/\pi'); ylabel('Gain, in dB'); title('Elliptic Highpass Filter'); axis([0 1 -60 5]); Elliptic Highpass Filter 0

Gain, in dB

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ω/ π

(c) Analog lowpass transfer function coefficient are obtained by displaying B and A: H LP (s ) =

0.0556 s 2 + 0.8329 s 3 + 1.3314 s 2 + 1.6434 s + 0.8329

.

Analog highpass transfer function coefficient are obtained by displaying BT and AT:

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H HP (s ) =

s 3 + 0.6315 s s 3 + 6.0693 s 2 + 15.1254 s + 34.9438

.

Digital highpass transfer function coefficient are obtained by displaying num and den: : G HP ( z ) =

0.0286 − 0.0415 z −1 + 0.0415 z −2 − 0.0286 z −3 1 + 1.9407 z −1 + 1.5613 z − 2 + 0.4356 z − 3

.

M9.11 FT = 9kHz, Fp1 = 1.2kHz, Fp2 = 2.2kHz, Fs1 = 650 Hz, Fs 2 = 3kHz,

α p = 0.8dB, αs = 31dB (a) ω p1 =

ωs 2 =

2 πF p1 FT

= 0.8378 , ω p 2 =

2 πF p2 FT

2πFs1 = 0.4538 , and FT

2πFs 2 = 2.094 , with FT

⎛ ω p1 ⎞ ⎛ ω p2 ˆ ˆ ⎜ ⎜ ⎟ , = = Ω tan = 0 . 445 Ω tan p1 p2 ⎜ 2 ⎟ ⎜ 2 ⎝ ⎝ ⎠ ˆ = tan⎛⎜ ωs 2 ⎞⎟ = 1.731 . and Ω s2 ⎝ 2 ⎠ ˆ ˆ B =Ω −Ω = 0.521, ω

= 1.536 , ω s1 =

p2

⎞ ˆ = tan⎛⎜ ωs1 ⎞⎟ = 0.231 , ⎟ = 0.966 , Ω s1 ⎟ ⎝ 2 ⎠ ⎠

p1

ˆ2 =Ω ˆ Ω ˆ ˆ ˆ Ω 0 p1 p2 = 0.430 ≠ Ω s1Ω s 2 = 0.40 ˆ 2 −Ω ˆ2 Ω s1 0 Therefore, we choose to lower the stopband edge to: Ω s = = 3.13 ˆ Ω s1 ⋅ Bω (b) The analog lowpass specifications are thus: Ω p = 1 rad, Ω s = 3.13 rad, α p = 0.8 dB, α s = 31 dB.

% Problem M9.11 [N, Wn] = cheb1ord(1, 3.13, 0.8, 31, 's'); [B, A] = cheby1(N, 0.8, Wn, 's'); [BT, AT] = lp2bp(B, A, sqrt(0.43), 0.521); [num, den] = bilinear(BT, AT, 0.5); [h, omega] = freqz(num, den, 256); plot(omega/pi, 20*log10(abs(h)));grid; xlabel('\omega/\pi'); ylabel('Gain,in dB'); title('Chebyshev I Bandpass Filter'); axis([0 1 -60 5]);

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Chebyshev I Bandpass Filter 0

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(c) Analog lowpass transfer function coefficient are obtained by displaying B and A: 0.5559 H LP (s ) = . s 3 + 1.0719 s 2 + 1.3244 s + 0.5559

Analog bandpass transfer function coefficient are obtained by displaying BT and AT: H BP (s ) =

0.0786 s 3 s 6 + 0.5584 s 5 + 1.6495 s 4 + 0.5589 s 3 + 0.7093 s 2 + 0.1033 s + 0.0795

.

Digital bandpass transfer function coefficient are obtained by displaying den and num: G BP ( z ) =

0.0169 − 0.0506 z −2 + 0.0506 z −4 − 0.0169 z −6 1 − 1.9799 z −1 + 3.3196 z − 2 − 3.1443 z − 3 + 2.6191 z − 4 + 1.1983 z − 5 + 0.476 z − 6

M9.12 FT = 8 kHz, Fp1 = 0.9 kHz, Fp2 = 2.1 kHz, Fs1 = 0.6 kHz, Fs 2 = 3 kHz, α p = 1.5 dB, α s = 30 dB (a) ω p1 =

ωs2 =

2πF p1 FT

= 0.707 , ω p 2 =

2πF p 2 FT

= 1.649 , ω s1 =

2πFs 2 = 2.356 , with FT

ω ω ˆ = tan⎛⎜ p 2 ˆ = tan⎛⎜ p1 ⎞⎟ = 0.369 , Ω Ω p1 p 2 ⎜ 2 ⎟ ⎜ 2 ⎠ ⎝ ⎝ ˆ = tan⎛⎜ ω s 2 ⎞⎟ = 2.414 . and Ω s2 ⎝ 2 ⎠ ˆ −Ω ˆ = 2.174 B =Ω ω

2πFs1 = 0.471 , and FT

s2

⎞ ˆ = tan⎛⎜ ω s1 ⎞⎟ = 0.240 , ⎟⎟ = 1.081 , Ω s1 ⎝ 2 ⎠ ⎠

s1

ˆ2 =Ω ˆ Ω ˆ = 0.399 ≠ Ω ˆ Ω ˆ = 0.579 Ω 0 p1 p2 s1 s2 Therefore, we choose to lower the passband:

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.

ˆ ˆ ˆ = Ω s 2 Ω s1 = 0.536 Ω p1 ˆ Ω p2

The analog bandstop filter specifications are thus: ˆ = 0.240 rad, Ω ˆ = 2.414 rad, ˆ = 0.536 rad, Ω ˆ = 1.081 rad, Ω Ω s1 s2 p1 p2

α p = 1.5 dB, α s = 30 dB. (b) Analog prototype LP filter: ˆ ⋅B Ω ω p1 = 3.995 rad, α p = 1.5 dB, α s = 30 dB. Ω s = 1 rad, Ω p = ˆ 2 −Ω ˆ2 Ω 0 p1 % Problem M9.12 close all; clear; clc; [N, Wn] = ellipord(3.995, 1, 1.5, 30, 's'); [B, A] = ellip(N, 1.5, 30, Wn, 's'); [BT, AT] = lp2bs(B, A, sqrt(0.579), 2.174); [num, den] = bilinear(BT, AT, 0.5); [h, omega] = freqz(num, den, 256); plot(omega/pi, 20*log10(abs(h)));grid; xlabel('\omega/\pi'); ylabel('Gain, in dB'); title('Elliptic Bandstop Filter'); axis([0 1 -60 5]); Elliptic Bandstop Filter 0

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ω/ π

(c) Analog lowpass transfer function coefficient are obtained by displaying B and A: H LP (s ) =

0.0316 s 2 + 12.6706 s 2 + 3.6122 s + 15.059

.

Analog bandstop transfer function coefficients are obtained by displaying BT and AT: H BS (s ) =

0.8414 s 4 + 0.9843 s 2 + 0.2821 s 4 + 0.5215 s 3 + 1.4719 s 2 + 0.3019 s + 0.3352

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322

Digital bandstop transfer function coefficients are obtained by displaying num and den: G BS ( z ) =

0.5806 − 0.6163 z −1 + 1.3145 z −2 − 0.6163 z −3 + 0.5806 z −4 1 − 0.8534 z −1 + 1.3959 z − 2 − 0.6115 z − 3 + 0.5464 z − 4

.

M9.13 % Program M9.13 close all; clear; clc; Wp = 0.7; Ws = 0.5; Rp = 1; Rs = 32; [N,Wn] = cheb1ord(Wp,Ws,Rp,Rs); [b,a] = cheby1(N,Rp,Wn,'high'); [h,omega] = freqz(b,a,256); plot (omega/pi,20*log10(abs(h)));grid; xlabel('\omega/\pi'); ylabel('Gain, in dB'); title('Type I Chebyshev Highpass Filter'); [GdH,w] = grpdelay(b,a,512); plot(w/pi,GdH); grid xlabel('\omega/\pi'); ylabel('Delay, in samples'); title('Original Filter Group Delay'); F = 0.7:0.001:1; g = grpdelay(b,a,F,2); % Equalize the passband Gd = max(g)-g; % Design the allpass delay equalizer [num,den,tau] = iirgrpdelay(2*N, F, [0.7 1], Gd); [GdA,w] = grpdelay(num,den,512); plot(w/pi,GdH+GdA); grid xlabel('\omega/\pi');ylabel('Delay, in samples'); title('Group Delay Equalized Filter'); Group Delay Equalized Filter

Delay, in samples

40 30 20 10 0

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ω/ π

M9.14 % Program M9.14 close all; clear;

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clc; Wp = [0.45 0.65]; Ws = [0.3 0.75]; Rp = 1; Rs = 40; [N,Wn] = buttord(Wp, Ws, Rp, Rs); [b,a] = butter(N,Wn); [h,omega] = freqz(b,a,256); gain = 20*log10(abs(h)); plot (omega/pi,gain);grid; xlabel('\omega/\pi'); ylabel('Gain, in dB'); title('IIR Butterworth Bandpass Filter'); [GdH,w] = grpdelay(b,a,512); plot(w/pi,GdH); grid xlabel('\omega/\pi'); ylabel('Delay, in samples'); title('Original Filter Group Delay'); F = 0.45:0.001:0.65; g = grpdelay(b,a,F,2); % Equalize the passband Gd = max(g)-g; % Design the allpass delay equalizer [num,den,tau] = iirgrpdelay(2*N, F, [0.45 0.65], Gd); [GdA,w] = grpdelay(num,den,512); plot(w/pi,GdH+GdA); grid xlabel('\omega/\pi');ylabel('Delay, in samples'); title('Group Delay Equalized Filter'); Group Delay Equalized Filter

Delay, in samples

60 50 40 30 20 10 0

0

0.2

0.4

0.6

0.8

1

ω/π

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324

Chapter 10 10.1

To compute the filter orders, we use Kaiser’s formula of Eq. (10.3), Bellanger’s formula of Eq. (10.4), and Hermann’s formula of Eq. (10.5). Filter #1:

(

)

− 20 log10 0.0224 ⋅ 0.00012 − 13 = 157.097 ≈ 158 14.6(0.14375π − 0.10625π ) / 2π − 2 log10 (10 ⋅ 0.0224 ⋅ 0.000112 ) Bellanger’s formula - N = − 1 = 162.575 ≈ 163 3(0.14375π − 0.10625π ) / 2π Hermann’s formula 2 D∞ (δ p , δ s ) = 0.005309(log10 0.0224 ) + 0.07114(log10 0.0224 ) − 0.4761 ⋅ log10 0.000112 − Kaiser’s formula - N =

[0.00266(log

[

]

]

0.0224 ) + 0.5941(log10 0.0224 ) + 0.4278 = 2.8326 F (δ p , δ s ) = 11.01217 + 0.51244[log 10 0.0224 − log 10 0.000112 ] = 12.1913

N=

2

10

2.8326 − 12.1913[(0.14375π − 0.10625π ) / 2π ]2 = 150.8434 ≈ 151 (0.14375π − 0.10625π ) / 2π

Filter #2:

(

)

− 20 log10 0.017 ⋅ 0.034 − 13 = 33.186 ≈ 34 14.6(0.2875π − 0.2075π ) / 2π − 2 log10 (10 ⋅ 0.017 ⋅ 0.034 ) Bellanger’s formula - N = − 1 = 36.3012 ≈ 37 3(0.2875π − 0.2075π ) / 2π Hermann’s formula 2 D∞ (δ p , δ s ) = 0.005309(log10 0.017 ) + 0.07114(log10 0.017 ) − 0.4761 ⋅ log10 0.034 − Kaiser’s formula - N =

[0.00266(log

[

]

]

0.017 ) + 0.5941(log10 0.017 ) + 0.4278 = 1.474777 F (δ p , δ s ) = 11.01217 + 0.51244[log 10 0.017 − log 10 0.034] = 10.85791019

N=

2

10

1.474777 − 10.85791019[(0.2875π − 0.2075π ) / 2π ]2 = 36.435 ≈ 37 (0.2875π − 0.2075π ) / 2π

Filter #3:

(

)

− 20 log10 0.0411 ⋅ 0.0137 − 13 = 11.6107 ≈ 12 14.6(0.575π − 0.345π ) / 2π − 2 log10 (10 ⋅ 0.0411 ⋅ 0.0137 ) Bellanger’s formula - N = − 1 = 12.04 ≈ 13 3(0.575π − 0.345π ) / 2π Hermann’s formula Kaiser’s formula - N k =

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324

[

]

D∞ (δ p , δ s ) = 0.005309(log10 0.0411) + 0.07114(log10 0.0411) − 0.4761 ⋅ log10 0.0137 −

[0.00266(log

2

]

0.0411) + 0.5941(log10 0.0411) + 0.4278 = 1.4424 F (δ p , δ s ) = 11.01217 + 0.51244[log 10 0.0411 − log 10 0.0137 ] = 11.256666 2

10

1.4424 − 11.256666[(0.575π − 0.345π ) / 2π ]2 N= = 11.248 ≈ 12 (0.575π − 0.345π ) / 2π 10.2

N = 75 and ω s − ω p = 0.05π and we assume δ s = δ p . (a) Using Kaiser’s formula of Eq. (10.3): ⎛ 75[14.6⋅0.05π / 2 π ]+13 ⎞ ⎜ ⎟ − 20 ⎝ ⎠ = 10

δs = 0.009577 ∴ α s = 40.375 dB. (b) Using Bellanger’s formula of Eq. (10.4): 1/ 2

⎛ − 76⋅3⋅0.05π / 2π ⎞ ⎞ ⎛ ⎜ ⎟⎟ ⎜ 2 ⎝ ⎠ δ s = ⎜ 0.1 ⋅ 10 = 0.0119 ∴ α s = 38.5 dB. ⎟ ⎟ ⎜ ⎠ ⎝ (c) Using Hermann’s formula of Eq. (10.5): 2 F = b1 ∴ D∞ (δ s ) = N (ω s − ω p ) / 2π + b1 (ω s − ω p ) / 2π

[

[

]

]

D∞ (δ s ) = a1 (log10 δ s ) + a 2 (log10 δ s ) + a3 (log10 δ s ) −

[a (log 4

10

2

δ s )2 + a5 (log10 δ s ) + a6

]

D∞ (δ s ) = a1 (log10 δ s ) + a 2 (log10 δ s ) + a3 (log10 δ s ) − 3

2

a 4 (log10 δ s ) − a5 (log10 δ s ) − a 6 2

D∞ (δ s ) = a1 (log10 δ s ) + (a 2 − a 4 )(log10 δ s ) + (a3 − a5 )(log10 δ s ) − a 6 Let x = (log10 δ s ) , and thus 3

2

D∞ (δ s ) = 0.005309 x 3 + 0.06848 x 2 − 1.0702 x − 0.4278 = 1.875697 Solving for x gives us three possible solutions: x = −21.3787 , x = −1.94654, x = 10.4263 The most reasonable solution is the second. Therefore, δ s = 0.0113 ∴ α s = 38.93

10.3

N = 75 and ω s − ω p = 0.05π = ∆ω

α s = 2.285(∆ω )N + 8 = 34.9 dB. 10.4

The ideal L -band digital filter H ML (z) has a frequency response given by H ML (e jω ) = Ak , for ωk −1 ≤ ω ≤ ωk , 1 ≤ k ≤ L, and can be considered as sum of L ideal k k 0 bandpass filters with cutoff frequencies at ωc1 = ωk −1 and ωc 2 = ωk , where ωc1 = 0 L and ωc 2 = π. Now from Eq. (10.47) the impulse response of an ideal bandpass filter is

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325

sin(ωc 2 n) sin(ωc1n) . Therefore , − πn πn sin(ωk n) sin(ωk −1n) k − [ n] = h BP . Hence, πn πn

given by h BP [ n] =

h ML [ n] =

L

∑

k [ n] = h BP

k =1

L

⎛ sin(ωk n) sin(ωk −1n) ⎞ − ⎟ πn πn ⎠

∑ Ak ⎜⎝

k =1

L −1 sin(ωk n) L −1 ⎛ sin(ωk n) sin(ωk −1n) ⎞ ⎛ sin(ω1n) sin(0n) ⎞ = A1 ⎜ − − − Ak Ak ⎜ ⎟ ⎟+ πn ⎠ k = 2 πn π n π n ⎝ πn ⎝ ⎠ k =2

∑

∑

⎛ sin(ω L n) sin(ω L −1 n) ⎞ + AL ⎜ − ⎟ πn πn ⎝ ⎠ sin(ω1n) L −1 sin(ωk n) L −1 sin(ωk −1n) sin(ωL −1n) = A1 + − − AL Ak Ak πn πn πn πn k =2 k =2

∑

=

∑

L −1

∑

L sin(ωk n) sin(ωk −1n) − . Ak πn π n k =2

∑

Ak

k =1

sin(ωL n) Since ωL = π, sin(ωL n) = 0. We add a term AL to the first sum in the above πn expression and change the index range of the second sum, resulting in h ML [ n] =

L

∑

Ak

k =1

sin(ωk n) L −1 sin(ωk n) − Ak + 1 . πn π n k =1

∑

Finally, since AL +1 = 0, we can add a term AL +1 L

sin(ωL n) to the second sum. This leads to πn

L sin(ωk n) L sin(ωk n) sin(ωk n) h ML [ n] = Ak Ak +1 − = ( Ak − Ak + 1 ) . π n π n π n k =1 k =1 k =1

∑

10.5

∑

∑

⎧ j, − π < ω < 0, H HT (e jω ) = ⎨ Therefore, ⎩− j, 0 < ω < π. 1 hHT [ n] = 2π

1 = 2π

0

∫ je

−π

jωn

0

∫ H HT (e

jω

)e

j ωn

−π

1 dω − 2π

π

1 dω + H HT (e jω )e jωn dω 2π

∫ 0

π

∫ je 0

1 For n = 0, h HT [0] = 2π

0

∫

−π

jωn

2 2 sin 2 (πn / 2) (1 − cos(πn)) = dω = if n ≠ 0. πn 2 πn

1 jdω − 2π

π

∫ jdω = 0. 0

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326

0, if n = 0, ⎧ ⎪ 2 Hence, h HT [n] = ⎨ 2sin (πn / 2) , if n ≠ 0. ⎪⎩ πn Since hHT [ n] = −hHT [ − n], and the length of the truncated impulse response is odd, it is a Type 3 linear-phase FIR filter.

Magnitude

1 0.8 0.6 0.4

N=11 N=21 N=61

0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

From the frequency response plots given above, we observe the presence of ripples at the bandedges due to the Gibbs phenomenon caused by the truncation of the impulse response. 10.6

∞

H {x[ n]} =

∑ hHT [n − k ]x[k ] . Hence,

k = −∞

⎧⎪ jX (e jω ), − π < ω < 0, F {H {x[ n]}} = H HT (e jω ) X (e jω ) = ⎨ ⎪⎩− jX (e jω ), 0 < ω < π. ⎧⎪ j 4 X (e jω ), − π < ω < 0, jω = Y ( e ) = X (e jω ). (a) Let y[ n] = H {H {H {H {x[n]}}}}. Hence, ⎨ 4 ω j ⎪⎩(– j ) X (e ), 0 < ω < π , Therefore, y[ n] = x[ n]. (b) Define g[ n] = H {x[ n]}, and h * [ n] = x[ n]. Then But from the Parseval's' relation in Table 3.4, Therefore,

∞

∞

∞

∞

∑ l = –∞ H {x[l]}x[l] = ∑ l = – ∞ g[l]h * [l]. π

∑ l = –∞ g[l]h * [l] = 2π ∫– π G(e jω )G(e jω )dω. 1

π

∑ l = – ∞ H {x[l]}x[l] = 2π ∫− π H HT (e jω )X (e jω )X (e – jω )dω where 1

⎧ j, − π < ω < 0, H HT (e jω ) = ⎨ Since the integrand H HT (e jω ) X (e jω ) X (e – jω ) is an odd − < < , 0 . j ω π ⎩ function of ω,

π

∫− π H HT (e

jω

) X (e jω ) X (e – jω )dω = 0. As a result,

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∞

∑ l = –∞ H {x[l]}x[l] = 0.

327

10.7

N

∑ hLP [n] z − n .

H LP ( z ) =

Its frequency response H LP (e jω ) is shown in Figure (a)

n=0

π 2

j (ω − )

jω

π 2

j (ω + )

) + H LP (e below. A plot of the frequency response H HT (e ) = H LP (e is shown in Figure (b) below. It is evident from this figure that H HT (e jω ) is the frequency response of an ideal Hilbert transformer. Therefore, we have

)

HLP (e jω ) 1

π _ __ 2

_π

π __ 2

0

π

ω

(a) HHT (e jω )

π __

HLP(e j(ω + 2 ) )

_π

HLP(e j(ω

1

π _ __ 2

π _ __ ) 2 )

π __ 2

0

π

ω

(b)

H HT (e

= =

N

jω

∑ hLP

) = H LP

π j (ω + ) 2 )+ (e

π − jn (ω + ) 2 [ n] e

n=0 N

N

+

H LP (e

∑ hLP

π 2

j (ω − )

)

π − jn(ω − ) 2 [ n] e

n=0

∑ 2 hLP [n]e − jnω cos

n=0

N

=

(

∑ hLP [n]e − jnω e − jnπ / 2 + e jnπ / 2

n=0

)

( ). Now, for n odd, cos(nω / 2) = 0 and hence, we can drop all nπ 2

odd terms in the above expression. Let N = 2 M with N even and let r = 2n. Then, we can rewrite the above equation as H HT (e jω ) =

M

∑ 2hLP [2r ] cos(rπ)e − j 2rω .

The

r =0

corresponding transfer function of the Hilbert transformer is therefore given by

H HT ( z ) =

10.8

M

∑ 2hLP [2n] cos(nπ) z − 2n

n =0

=

M

∑ 2(−1) n hLP [2n]z − 2n .

n=0

H DIF (e jω )= jω. Hence, j π j ⎛⎜ ωe jωn e jωn 1 π jωn jωn h DIF [ n] = j ωe dω= dω = ωe + 2π − π 2π − π 2π ⎜⎝ jn n2 cos(πn) sin(πn) cos(πn) h DIF [ n] = , if n ≠ 0. − = n n πn 2

∫

∫

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⎞ ⎟ ⎟ ⎠

π −π

. Therefore,

328

1 For n = 0 , h DIF [0] = 2π

π

∫ jωdω = 0.

−π

n = 0, ⎧⎪ 0, n cos( ) π h n [ ] = ⎨ Hence, DIF , | n | > 0. Since hDIF [ n] = −hDIF [ − n], the truncated impulse ⎪⎩ n response is a Type 3 linear-phase FIR filter. The magnitude responses of the above differentiator for several values of M are given below: 4 M=10 M=20 M=30

Magnitude

3

2

1

0

0

0.2

0.4

0.6

0.8

1

ω/π

10.9

N = 2 M + 1.

ω ⎧ for n = M , 1− c , ⎪ π ⎪ ˆh [ n] = ⎪⎨− sin (ωc (n − m)) , if n ≠ M , 0 ≤ n < N , HP π (n − m) ⎪ ⎪ otherwise. ⎪⎩ 0,

Now, Hˆ HP ( z ) + Hˆ LP ( z ) =

=

∞

∑

hˆ HP [ n]z − n +

n = −∞ N −1

∞

∑

hˆ LP [ n]z − n =

n = −∞

N −1

N −1

n=0

n=0

∑

hˆ HP [ n]z − n +

∑ hˆLP [n]z −n

∑ (hˆHP [n] + hˆLP [n])z −n .

n=0

⎧0, 0 ≤ n ≤ N–1, n ≠ M , But hˆ HP [ n] + hˆ LP [ n] = hˆ HP [ n] + hˆ LP [ n] = ⎨ , n=M . ⎩1, Hence, Hˆ HP ( z ) + Hˆ LP ( z ) = z – M , i.e. the two filters are delay-complementary.

⎧ ω , ω < ωc , jω 10.10 H LLP (e ) = ⎨ Therefore, ⎩ 0, otherwise.

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329

ωc ⎞ ⎛ 0 ⎟ 1 ⎜ j ωn j ωn hLLP [ n] = dω + ωe dω ⎟ − ωe ⎜ 2π ⎜ ⎟ 0 ⎠ ⎝ − ωc ω ⎞ 0 ⎛ ⎡ ωe jωn e jωn ⎤ c ⎟ 1 ⎜ ⎡ ωe jωn e jωn ⎤ +⎢ + = + ⎥ ⎟ ⎥ ⎜ −⎢ 2 2π ⎜ ⎢⎣ jn jn n 2 ⎥⎦ n ⎥⎦ 0 ⎟ ⎢ –ωc ⎣ ⎠ ⎝ jω n − jω n cos(ωc n) − 1 1 ⎛⎜ ωc e c − ωc e c e jωc n + e − jωc n − 2 ⎞⎟ ωc = sin(ωc n) + . + = 2 2 ⎟ πn 2π ⎜ jn π n n ⎝ ⎠

∫

∫

⎧ω, ω < ωc , jω Hence, 10.11 H BLDIF (e ) = ⎨ ⎩ 0, otherwise. ωc ⎞ ⎛ ωc ⎟ 1 1 ⎜ ⎡ ωe jωn e jωn ⎤ jωn + h BLDIF [ n] = dω = ωe ⎥ ⎜⎢ ⎟ 2 2π 2π ⎜ ⎣⎢ jn ⎥ n ⎦ − ωc −ωc ⎟⎠ ⎝

∫

1 ⎛⎜ ωc e = 2π ⎜ ⎝

jωc n

+ ωc e − jωc n e jωc n − e − jωc n + jn n2

⎞ ⎟ = – j ωc cos(ω n) + j 1 sin(ω n). c c ⎟ πn πn 2 ⎠

10.12 The frequency response of a causal ideal notch filter can thus be expressed as ( ( H notch (e jω ) = H notch (ω) e jθ (ω) where H notch (ω) is the amplitude response which can

( ( ⎧ 1, 0 ≤ ω ≤ ωo , be expressed as H notch (ω) = ⎨ It follows then that H notch (ω) is ⎩− 1, ωo < ω < π. ( related to the amplitude response H LP (ω) of the ideal lowpass filter with a cutoff at ωo ( ( Hence, the impulse response of the ideal through H notch (ω) = ±[ 2 H LP (ω) − 1]. notch filter is given by hnotch [ n] = ±[2hLP [ n] − δ[ n]], where sin(ωo n) , − ∞ < n < ∞. The magnitude responses of a length 41 notch filter πn with a notch frequency at ωo = 0.4π and its associated length-41 lowpass filter are shown below. hLP [ n] =

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Amplitude

1.5

1

0.5

0

o

0.2

0

1

0.8

0.6

0.4

= 0.4š

/

1 2π

10.13 Φ R =

( )2 dω , where H t (e jω ) = ∑ ht [n]e − jωn . n=−M

∫ ( ) −π π

M

H t e jω − H d e jω

∞

∑ ht [n] − hd [n]

Using Parseval’s relation, we can write Φ R =

2

n = −∞ M

ht [n] − hd [n] +

∑

=

2

n=−M

Now, Φ Haan =

− M −1

∞

n = −∞

n = M +1

∑ hd2 [n] + ∑ hd2 [n] .

∞

∑ hd [n]⋅ w Hann [n] − hd [n]

2

n = −∞

− M −1 ∞ ⎛1 1 ⎛ 2πn ⎞ ⎞ 2 = hd [n]⎜⎜ + cos⎜ hd [n] + hd2 [n] ⎟ ⎟⎟ − hd [n] + ⎝ 2M + 1⎠ ⎠ ⎝2 2 n=−M n = −∞ n = M +1 Hence, Φ Excess = Φ R − Φ Haan 2

M

∑ M

=

∑ hd [n]⋅ w R [n] − hd [n]

n=−M

=−

2

∑

10.14 Φ R =

∞

∑

M

⎛1 1 ⎛ 2πn ⎞ ⎞ − hd [n] ⋅ ⎜⎜ + cos⎜ ⎟ ⎟⎟ − hd [n] + 2 1 M 2 2 ⎝ ⎠⎠ ⎝ n= −M

∑

hd [n] ⎛ 2πn ⎞ hd [n] cos⎜ ⎟− 2 2 1 2 M + ⎝ ⎠ n= −M M

∑

2

∑ ht [n] − hd [n] and Φ Hamm = 2

n = −∞

2

2

=

1 (1 + 2 M ) cos⎛⎜ 2πM ⎞⎟ − 1 . 2 ⎝ 2M + 1⎠

∞

∑ hd [n]⋅ w Hamm [n] − hd [n]

2

.

n = −∞ M

⎛ ⎞ ⎛ 2πn ⎞ Therefore, Φ Excess = Φ R − Φ Hamm = − hd [n]⎜⎜ 0.46 cos⎜ ⎟ − 0.46 ⎟⎟ ⎝ 2M + 1 ⎠ ⎝ ⎠ n= −M

∑

M

⎛ ⎛ 2πn ⎞ ⎞ =− 0.46hd [n]⎜⎜ cos⎜ ⎟ − 1⎟⎟ + M 2 1 ⎝ ⎠ ⎠ ⎝ n=−M

∑

2

2

2

⎛ 2πM ⎞ = 0.46(2 M + 1) cos⎜ ⎟ −1 . ⎝ 2M + 1 ⎠

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331

10.15 (a) ω p = 0.47π , ω s = 0.59π , δ p = 0.001 , δ s = 0.007 , ∆ω = 0.12π ,

α s = −20 log10 δ s = 43.1 dB From Table 10.2, we see that for fixed-window functions, we can achieve the minimum stopband attenuation by using Hann, Hamming, or Blackman windows. Hann will have the lowest filter length. 3.11π N Haan = ⎡2 M + 1⎤ = 53 since M = = 25.917. 0.12π Lowpass filter using Hann window 20 0

Magnitude Response

-20 -40 -60 -80 -100 -120 -140

0

0.1

0.2

0.3

0.4

0.5 ω/π

0.6

0.7

0.8

0.9

1

(b) ω p = 0.61π , ω s = 0.78π , δ p = 0.001 , δ s = 0.002 , ∆ω = 0.17π ,

α s = −20 log10 δ s = 54 dB From Table 10.2, we see that for fixed-window functions, we can achieve the minimum stopband attenuation by using either Hamming, or Blackman windows. Hamming will have the lowest filter length. 3.32π N Hamm = ⎡2 M + 1⎤ = 41 since M = = 19.53 . 0.17π

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Lowpass filter using Hamming window 20

Magnitude Response

0

-20

-40

-60

-80

-100

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

ω/π

10.16 ω p1 = 0.45π , ω p 2 = 0.65π , ω s1 = 0.3π , ω s 2 = 0.8π , ∆ω1 = ∆ω 2 = 0.15π ,

δ p = 0.01 , δ s1 = 0.008 , δ s 2 = 0.05

α s1 = −20 log10 δ s1 = 42 dB, α s 2 = −20 log10 δ s 2 = 26 dB From Table 10.2, we see that the Hann window will have minimum length and meet the minimum stopband attenuation. 3.11π M= = 21. Therefore, N = 43 . 0.15π

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333

Bandpass filter using Hann window 20 0

Magnitude Response

-20 -40 -60 -80 -100 -120 -140

0

0.1

0.2

0.3

0.4

0.5 ω/π

0.6

0.7

0.8

0.9

1

10.17 ω p1 = 0.3π , ω p 2 = 0.8π , ω s1 = 0.45π , ω s 2 = 0.65π , ∆ω1 = ∆ω 2 = 0.15π ,

δ p1 = 0.05 , δ p 2 = 0.009 , δ s = 0.02 ,

α s = −20 log10 δ s = 34 dB From Table 10.2, we see that the Hann window will have minimum length and meet the minimum stopband attenuation.

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334

Bandstop filter using Hann window 20

0

Magnitude Response

-20

-40

-60

-80

-100

-120

0

0.1

0.2

0.3

0.4

0.5 ω/π

0.6

0.7

0.8

0.9

1

10.18 Consider another filter with a frequency response G(e jω ) given by

0, 0 ≤ ω ≤ ωp , ⎧ ⎪ π (ω − ω p ) ⎞ ⎪ − π sin⎛⎜ ⎟, ω p < ω ≤ ωs , ⎟ ⎪ 2∆ω ⎜⎝ ∆ω jω ⎠ G (e ) = ⎨ + ( ) π ω ω ⎛ p ⎞ ⎪ −π ⎜ ⎟, − ωs ≤ ω ≤ −ω p , sin ⎪ 2∆ω ⎜ ⎟ ∆ ω ⎝ ⎠ ⎪ 0 , elsewhere. ⎩ jω Clearly G(e ) =

dH (e jω ) . Now, dω

⎛ π (ω − ω p ) ⎞ ⎛ π (ω − ω p ) ⎞ ⎛ ⎟ ⎟ ⎜ ωs j⎜ ωs − j⎜ ⎟ j n ⎜ ⎟ jωn ⎜ π ω ω ∆ ∆ −π ⎜ ⎠ e ω dω ⎝ ⎠e ⎝ 1 e d ω e = − j ω j ωn g[ n] = G (e )e dω 8π∆ωj ⎜ 2π ωp ⎜⎜ ω p −π ⎝

∫

∫

+

∫

−ω p

⎛ π ( ω+ ω p ) ⎞ ⎛ π ( ω+ ω p ) ⎞ ⎞ ⎟ ⎟ −ω p − j ⎜ ⎟ j⎜ ⎟ ⎜ ⎟ ⎜ ∆ω ⎠ jωn ∆ω ⎠ jωn ⎝ ⎝ e e dω − e e dω ⎟

− ωs

− ωs

∫

Not for sale

∫

⎟ ⎟ ⎠

335

π ⎞⎤ π ⎞ π ⎞ ⎤ ⎛ ⎛ ⎛ ⎛ ⎡ jω ⎛⎜ n − π ⎞⎟ ⎡ jω p ⎜ n + jω p ⎜ n − ⎟) jπω p ⎟ ⎜ jπω p ⎢ jωs ⎜ n + ⎟ s ⎢ ⎥ ∆ ∆ ∆ ω ∆ ω⎠ ⎥ ω ω ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ − − e −e −1 ⎜ −e ⎥ ⎥ − e ∆ω ⎢ e e ∆ω ⎢ = ⎜ ⎥ ⎢ ⎥ ⎢ π ⎞ π ⎞ j8∆ω ⎛ ⎛ ⎜ j⎜ n + j⎜ n − ⎟ ⎟ ⎥ ⎢ ⎥ ⎢ ⎜ ∆ω ⎠ ∆ω ⎠ ⎝ ⎝ ⎦ ⎣ ⎦ ⎣ ⎝ π ⎞⎤ π ⎞ ⎤⎞ ⎛ ⎛ ⎡ − jω ⎛⎜ n + π ⎞⎟ ⎡ − jω ⎛⎜ n − π ⎞⎟ − jω s ⎜ n − − jω s ⎜ n + jπω p jπω p ⎟ ⎟ ⎟ p p ⎢ ⎢ ⎝ ∆ω ⎠ ⎥ ⎟ ⎝ ∆ω ⎠ − e ⎝ ∆ω ⎠ ⎥ ⎝ ∆ω ⎠ − e − e e ⎥ − e ∆ω ⎢ ⎥ + e ∆ω ⎢ ⎢ ⎥ ⎢ ⎥⎟ π π ⎛ ⎞ ⎛ ⎞ + − j n j n ⎜ ⎟ ⎜ ⎟ ⎢ ⎥ ⎢ ⎥ ⎟⎟ ∆ ω ∆ ω ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ ⎣ ⎦⎠ ⎛ ⎞ ⎛ ⎜ ⎟ ⎜ − 1 ⎜ − sin(ωs n) − sin(ω p n) − sin(ωs n) − sin(ω p n) ⎟ = sin(ωs n) + sin(ω p n) ⎜ − 2π / ∆ω = − ⎜ ⎟ 4∆ωj π ⎞ π ⎞ 4∆ωj ⎜ ⎛ ⎛ π2 2 ⎜ ⎟ ⎜ ⎜n + ⎟ ⎜n − ⎟ ⎜n − ∆ω ⎠ ∆ω ⎠ ⎝ ⎝ ⎠ ⎝ ∆ω 2 ⎝ ⎞ sin(ωc n) cos(∆ωn / 2) ⎛⎜ 1 ⎟. = ⎜ π 1 − (∆ω / π ) 2 n 2 ⎟ ∆ωj ⎝ ⎠ ⎛ cos(∆ωn / 2) ⎞⎛ sin(ωc n) ⎞ j ⎜ ⎟⎜ h n = [ ] ⎟. Now, h[ n] = g[ n]. Therefore, ⎜ 1 − (∆ω / π ) 2 n 2 ⎟⎝ πn ⎠ n ⎝ ⎠

(

(

)

)

⎡ ⎛ 4πn ⎞ ⎤ ⎛ 2πn ⎞ ⎟ ⎥ w R [ n] ⎟ + γ cos⎜ 10.19 wGC [ n] = ⎢α + β cos⎜ ⎝ 2 M + 1 ⎠⎦ ⎝ 2M + 1 ⎠ ⎣ ⎡ ⎛ 2 πn ⎞ ⎞ ⎛ 4 πn ⎞ ⎞ ⎤ ⎛ j ⎛⎜ 2πn ⎞⎟ ⎛ j ⎛⎜ 4πn ⎞⎟ − j⎜ − j⎜ ⎟⎟ ⎟ ⎟⎥ ⎜ ⎜ ⎢ = ⎢ α + 2 β ⎜ e ⎝ 2 M + 1 ⎠ + e ⎝ 2 M + 1 ⎠ ⎟ + 2 γ⎜ e ⎝ 2 M + 1 ⎠ + e ⎝ 2 M + 1 ⎠ ⎟ ⎥ w R [ n ] ⎜ ⎟ ⎜ ⎟⎥ ⎢ ⎝ ⎠ ⎝ ⎠⎦ ⎣ ⎛ j ⎛⎜ ω − 2 π ⎞⎟ ⎞ ⎛ j ⎛⎜ ω + 2 π ⎞⎟ ⎞ ⎜ ⎟ ⎜ ⎟ Hence, ΨGC (e jω ) = αΨR (e jω ) + 2 βΨR ⎜ e ⎝ 2 M + 1 ⎠ ⎟ + 2 βΨR ⎜ e ⎝ 2 M + 1 ⎠ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ j ⎛⎜ ω − 4π ⎞⎟ ⎞ ⎛ j ⎛⎜ ω + 4π ⎞⎟ ⎞ ⎜ ⎟ ⎜ ⎟ 2 M +1 ⎠ 2 M +1 ⎠ ⎝ + 2 γΨR ⎜ e ⎟ + 2 γΨR ⎜ e ⎝ ⎟. ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

For the Hann window :

ΨHaan (e

jω

) = 0.5ΨR (e

= 0.5,

jω

= 0.5 and

= 0. Hence,

⎛ j ⎛⎜ ω − 2π ⎞⎟ ⎞ ⎛ j ⎛⎜ ω + 2 π ⎞⎟ ⎞ ⎜ ⎟ ⎜ ⎟ 2 + 1 2 M +1 ⎠ M ⎠ + 2 βΨ e ⎝ ) + ΨR ⎜ e ⎝ ⎟ ⎟ R⎜ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

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336

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

⎛ (2 M + 1)ω ⎞ sin⎛⎜ (2 M + 1)⎛ ω − π ⎞ ⎞⎟ sin⎛⎜ (2 M + 1)⎛ ω + π ⎞ ⎞⎟ ⎟⎟ ⎜ ⎟⎟ ⎜ sin⎜ ⎟ ⎜ ⎜ 2 ⎝ 2 2M + 1 ⎠ ⎠ ⎝ 2 2M + 1 ⎠ ⎠ ⎝ ⎝ ⎠ ⎝ . + + = 0.5 sin(ω / 2) π ⎞ π ⎞ ⎛ω ⎛ω sin⎜ − sin⎜ + ⎟ ⎟ ⎝ 2 2M + 1 ⎠ ⎝ 2 2M + 1 ⎠ For the Hamming window,

= 0.54,

= 0.46, and

= 0. Hence,

⎛ j ⎛⎜ ω − 2 π ⎞⎟ ⎞ ⎛ j ⎛⎜ ω + 2π ⎞⎟ ⎞ ⎜ ⎟ ⎜ ⎟ ΨHam min g (e jω ) = 0.54ΨR (e jω ) + 0.92ΨR ⎜ e ⎝ 2 M + 1 ⎠ ⎟ + 0.92ΨR ⎜ e ⎝ 2 M + 1 ⎠ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎛ π ⎞⎞ π ⎞⎞ ⎛ω ⎛ω ⎛ (2 M + 1)ω ⎞ sin⎜⎜ (2 M + 1)⎜ + sin⎜⎜ (2 M + 1)⎜ − ⎟ ⎟⎟ ⎟ ⎟⎟ sin⎜ ⎟ 2 ⎝ 2 2M + 1⎠ ⎠ ⎝ 2 2M + 1 ⎠ ⎠ ⎝ ⎝ ⎠ ⎝ . + 0.92 + 0.92 = 0.54 F sin(ω / 2) π ⎞ π ⎞ ⎛ω ⎛ω sin⎜ + sin⎜ − ⎟ ⎟ ⎝ 2 2M + 1 ⎠ ⎝ 2 2M + 1⎠ or the Blackmann window = 0.42, = 0.5 and = 0.08 ⎛ j ⎛⎜ ω − 2π ⎞⎟ ⎞ ⎛ j ⎛⎜ ω + 2 π ⎞⎟ ⎞ ⎜ ⎟ ⎜ ⎟ jω jω 2 + 1 2 M +1 ⎠ M ⎠ +Ψ e ⎝ ΨBlackman (e ) = 0.42ΨR (e ) + ΨR ⎜ e ⎝ ⎟ ⎟ R⎜ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ j ⎛⎜ ω− 4π ⎞⎟ ⎞ ⎛ j ⎛⎜ ω− 4π ⎞⎟ ⎞ ⎟ ⎜ ⎟ ⎜ + 0.16ΨR ⎜ e ⎝ 2 M +1 ⎠ ⎟ + 0.16ΨR ⎜ e ⎝ 2 M +1 ⎠ ⎟ ⎟ ⎜ ⎟ ⎜ ⎠ ⎝ ⎠ ⎝ ⎛ (2 M + 1)ω ⎞ sin⎛⎜ (2 M + 1)⎛ ω − π ⎞ ⎞⎟ sin⎛⎜ (2 M + 1)⎛ ω + π ⎞ ⎞⎟ ⎟⎟ ⎜ ⎟⎟ ⎜ sin⎜ ⎟ ⎜ ⎜ 2 ⎝ 2 2M + 1 ⎠⎠ ⎝ 2 2M + 1 ⎠ ⎠ ⎝ ⎝ ⎠ ⎝ + + = 0.42 sin(ω / 2) π ⎞ π ⎞ ⎛ω ⎛ω sin⎜ − sin⎜ + ⎟ ⎟ ⎝ 2 2M + 1 ⎠ ⎝ 2 2M + 1 ⎠ ⎛ ⎛ 2π ⎞ ⎞ 2π ⎞ ⎞ ⎛ω ⎛ω sin⎜⎜ (2 M + 1)⎜ − sin⎜⎜ (2 M + 1)⎜ − ⎟ ⎟⎟ ⎟ ⎟⎟ ⎝ 2 2M + 1⎠ ⎠ ⎝ 2 2M + 1⎠ ⎠ ⎝ ⎝ + 0.16 + 0.16 . 2π ⎞ 2π ⎞ ⎛ω ⎛ω sin⎜ − sin⎜ − ⎟ ⎟ ⎝ 2 2M + 1 ⎠ ⎝ 2 2M + 1⎠ N

10.20 (a) H ( z ) = z − D ≈ ∑ h[n ]z − n = h[0] + h[1]z −1 + h[2]z − 2 + L + h[N ]z − N n =0

We see that if xˆ a (t ) =

N2

N2

⎛ t − tl ⎜⎜ l = − N1 ⎝ t k − t l

∑ Pk (t )x[n + k ], then Pk (t ) = ∏

k = − N1

⎞ ⎟⎟ for − N1 ≤ k ≤ N 2 . ⎠

l≠k

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337

Here, we have Hˆ (z ) =

N

∑ h[n]z −n and the solution follows if

n=0

Pk (t ) = h[n] , N1 = 0 , N 2 = N , k = n , t = D , t l = k , and t k = n Therefore, we have h[n] =

N

D−k

∏ n−k

for 0 ≤ n ≤ N .

k =0

k≠n

(b) N = 21, D = 90/13, L = 22. H (z ) ≈

21

∑

h[n]z − n , where h[n] =

n=0

21

90 / 13 − k . n − k k =0

∏

k≠n

% Problem #10.20 D = 90/13; N = 21; for n = 0:N, for k = 0:N, if n ~= k, tmp(n+1,k+1) = (D-k)/(n-k); else tmp(n+1,k+1) = 1; end end end h = prod(tmp'); [Gd,W] = grpdelay(h,1,512); [H, w] = freqz(h,1,512); figure(1); plot(W/pi, Gd); xlabel('\omega/\pi'); ylabel('Group Delay'); title('Group delay of z^-^D'); grid; figure(2); plot(w/pi, (abs(H))); xlabel('\omega/\pi'); ylabel('Magnitude'); title('Magnitude response of z^-^D'); grid;

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338

Group delay of z

Magnitude response of z

-D

-D

1.5

8

Magnitude

Group Delay

1.4 7.5

7

1.3 1.2 1.1 1

6.5

0.9 0

0.2

0.4

0.6

0.8

1

0

0.2

0.4

10.21 (a) x[ n] = s[ n] +

0.6

0.8

1

ω/π

ω/π

M

∑ Ak sin(kωo n +φ k ) = s[n] + r[n], where s[n] is the desired signal and

k =0

r[ n] =

M

∑ Ak sin(kωo n +φ k ) is the harmonic interference with fundamental

k =0

frequency ωo . Now, r[ n − D] =

M

∑ Ak sin[kωo (n − D) +φ k ]

k =0

=

M

∑ Ak sin(kωo n +φ k −2πk ) = r[n].

k =0

(b) y[ n] = x[ n] − x[ n − D] = s[ n] + r[ n] − s[ n − D] − r[ n − D] = s[ n] + r[ n] − s[ n − D] − r[ n]

= s[ n] − s[ n − D]. Hence, y[ n] does not contain any harmonic disturbances. (c) H c ( z ) =

(1 − cos( Dω)) + j sin( Dω) . 1 − e − jDω = . Thus, H c (e jω ) = −D −D − D − jDω 1 − ρ D cos( Dω) + jρ D sin( Dω) 1− ρ e 1− ρ z 1 − z −D

Then, H c (e jω ) =

(

2(1 − cos( Dω)) D

1 − 2 ρ cos( Dω) + ρ ρ = 0.99 is shown below:

2D

Not for sale

)

. A plot of H c (e jω ) for ωo = 0.22π and

339

Amplitude

1 0.8 0.6 0.4 0.2 0 -1

-0.5

0

0.5

1

ω/π

(d) y[n]

x[n] z–D –

+

P( z) 1 − N ( z) , with D = 2π / 0.18π , N = 18, and ρ = 0.98 . = Q( z) 1 − 0.98 D N ( z) % Problem #10.23 close all; clear; clc; D = 2/.18; N = 18; rho = 0.98; for n = 0:N, for k = 0:N, if n ~= k, tmp(n+1,k+1) = (D-k)/(n-k); else tmp(n+1,k+1) = 1; end end end h = prod(tmp'); [H,w] = freqz(h,1,1024); Hc = (1-H)./(1-(rho^D)*H); x = sqrt((2*(1-cos(D*w)))./(12*(rho^D)*cos(D*w)+rho^(2*D))); %plot(w/pi, abs(Hc)); plot(w/pi, x); grid; xlabel('\omega/\pi'); ylabel('Magnitude'); title(‘Comb filter using FIR fractional delay');

10.22 H c ( z) =

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340

Comb filter using FIR fractional delay

Magnitude

1 0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

P( z) D( z) − z −11 D( z −1 ) = , with D = 2π / 0.18π = 11.11 and N = 11. Q( z) D( z) − 0.98 90 / 13 z −11 D( z −1 ) % Problem #10.23 close all; clear; clc; D = 2/0.18; rho = 0.98; N = floor(D); for k = 1:N, for n = 0:N, p(n+1) = (D-N+n)/(D-N+k+n); end d(k) = ((-1)^k)*nchoosek(N,k)*prod(p); end [H,w] = freqz(fliplr(d)-d, fliplr(d)-(rho^D).*d , 512); plot(w/pi, abs(H));grid; xlabel('\omega/\pi'); ylabel('Magnitude'); title(Comb filter using allpass IIR fractional delay');

10.23 H c ( z) =

Comb filter using allpass IIR fractional delay

Magnitude

1 0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

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341

−ω p ≤ ω ≤ ω p , 1, ⎧ ⎪ ⎛ ⎪1 − ⎜ ω − ω p ⎞⎟, ω < ω ≤ ω , p s ⎪⎪ ⎜ ωs − ω p ⎟ jω ⎝ ⎠ = ( ) ⎨ 10.24 H LP e ⎪ ⎛⎜ ω + ω p ⎞⎟ ⎪1 + ⎜ ω − ω ⎟, −ωs < ω ≤ ω p , p⎠ ⎪ ⎝ s ⎪⎩ 0, elsewhere.

Now, for n ≠ 0, hLP [ n] =

1 2π

π

∫ H LP (e

jω

)e jωn dω

−π

−ωp ⎡ ωp ωs ω − ω p ⎞ jωn ω + ωp ⎛ ⎛ 1 ⎢ jωn ⎜1 − ⎟e ⎜1 + = + + e d ω d ω ⎜ ⎜ 2π ⎢ ∆ω ⎟⎠ ∆ω − ωs ⎝ ωp ⎝ ⎢⎣ − ω p −ω p ⎡ ωs ⎤ ωs ω − ω p jωn ω + ω p jωn ⎥ 1 ⎢ jωn = e dω − e dω + e dω⎥ 2π ⎢ ∆ω ∆ω − ωs ωp ⎢⎣ − ωs ⎥⎦

∫

∫

∫

∫

∫

⎤ ⎞ jωn ⎥ ⎟e dω⎥ ⎟ ⎠ ⎥⎦

∫

ωs – ωp ⎫ ⎧ jωn jωn ⎪⎪ e jωn ⎤⎥ e jωn ⎤⎥ 1 ⎪⎪ 2sin(ωs n) 1 ⎡⎢ (ω − ω p )e 1 ⎡⎢ (ω + ω p )e + + = − + ⎬ ⎨ 2 ⎥ 2 ⎥ jn jn 2π ⎪ ∆ω ⎢ ∆ω ⎢ πn n n ⎦ – ωs ⎪⎪ ⎣ ⎦ ωp ⎣ ⎪⎩ ⎭ jω n − jω p n ⎧ ⎫ − ∆ωe − jωs n e 1 ⎪ 2sin(ωs n) − e − jωs n ⎤ ⎪ 1 ⎡ ∆ωe jωs n e jωs n − e p ⎢ ⎥ = − + + − ⎨ ⎬ 2 2 2π ⎪ jn πn ∆ω ⎢ jn ⎥⎪ n n ⎣ ⎦⎭ ⎩

=

1 ⎧⎪ 2sin(ωs n) 2 ⎡ ∆ω sin(ωs n) cos(ωs n) − cos(ω p n) ⎤ ⎫⎪ − + ⎥⎬ ⎢ ⎨ 2π ⎪⎩ πn ∆ω ⎣⎢ n n2 ⎦⎥ ⎪⎭

1 ⎧ cos((ωc − ∆ω / 2)n) cos((ωc + ∆ω / 2)n) ⎫ 1 ⎧⎪ cos(ω p n) cos(ωs n) ⎫⎪ = − − ⎨ ⎬ ⎨ ⎬ ∆ω ⎪⎩ πn 2 πn 2 πn 2 ⎭ πn 2 ⎪⎭ ∆ω ⎩ 2 sin(∆ωn / 2) sin(ωc n) = . ∆ωn πn

=

1 Next, for n = 0, h LP [0] = 2π =

(

)

π

∫ H LP (e

jω

) dω =

−π

1 (area under the curve) 2π

ω 1 2 ωs + ω p = c. π 2π 2

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342

ωc ⎧ , if n = 0, ⎪ π h n [ ] = ⎨ 2sin(∆ωn / 2) sin(ω n) Hence, LP c ⎪ , if n ≠ 0. ∆ωn πn ⎩ An alternate approach to solving this problem is as follows. Consider the frequency response ⎧ 0, − ω p ≤ ω ≤ ω p , ⎪ 1 jω d H LP (e ) ⎪⎪ – ∆ω , ω p < ω < ωs , G ( e jω ) = =⎨ Its inverse DTFT is given by dω ⎪ 1 , − ωs < ω < − ω p , ⎪ ∆ω ⎪⎩ 0, elsewhere. 1 g[ n] = 2π

π

∫ G (e

jω

)e

jωn

–π

1 dω = 2π

– ωp

∫

– ωs

1 jωn 1 e dω − ∆ω 2π

ωs

∫

ωp

1 jωn e dω ∆ω

−ωp ωs ⎞ ⎛ ⎟ e jωn 1 ⎜ e jωn 1 ⎜ ⎟ = = − cos(ω p n) − cos(ωs n) . jn π 2 ∆ω ⎜ jn j n ∆ π ω ⎟ ⎜ − ωs ω p ⎟⎠ ⎝ 1 j cos(ω p n) − cos(ωs n) Thus, hLP [ n] = g[ n] = 2 n πn ∆ω ⎛⎛ 1 ⎡ ⎛⎛ ∆ω ⎞ ⎞ ∆ω ⎞ ⎞ ⎤ cos⎜⎜ ⎜ ωc − = ⎟n ⎟⎥ ⎟ n ⎟⎟ − cos⎜⎜ ⎜ ωs + ⎢ 2 ⎠ ⎠ 2 ⎠ ⎟⎠ ⎦ ⎝⎝ πn 2 ∆ω ⎣ ⎝ ⎝

(

(

=

)

)

2sin(∆ωn / 2) sin(ωc n) ⋅ , for n ≠ 0. πn ∆ωn

For n = 0, hLP [ n] =

ωc . π

10.25 Consider the case when the transition region is approximated by a second order spline. In this case the ideal frequency response can be constructed by convolving an ideal, notransition-band frequency response with a triangular pulse of width ∆ω = ωs − ω p ,

which in turn can be obtained by convolving two rectangular pulses of width ∆ω / 2 . In the time domain this implies that the impulse response of a filter with transition band approximated by a second order spline is given by the product of the impulse response of an ideal low pass filter with no transition region and square of the impulse response of a rectangular pulse. Now,

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343

H LP (ideal ) [ n] =

sin(ωc n) sin(∆ωn / 4) and H rec [ n] = . Hence, πn ∆ωn / 4

H LP [ n] = H LP(ideal) [ n](H rec [ n])2 .

Thus for a lowpass filter with a transition region approximated by a second order spline ωc ⎧ if n = 0, , ⎪⎪ π hLP [ n] = ⎨ 2 ⎪⎛⎜ sin(∆ωn / 4) ⎞⎟ sin(ωc n) , otherwise. ⎪⎩⎝ ∆ωn / 4 ⎠ πn Similarly the frequency response of a lowpass filter with the transition region specified by a P -th order spline can be obtained by convolving in the frequency domain an ideal filter with no transition region with P rectangular pulses of width ∆ω / P. Hence, H LP [ n] = H LP(ideal) [ n](H rec [ n]) P , where the rectangular pulse is of width ∆ω / P. Thus ωc ⎧ , if n = 0 , ⎪⎪ π hLP [ n] = ⎨ P ⎪⎛⎜ sin(∆ωn / 2 P ) ⎞⎟ sin(ωc n) , otherwise. πn ⎩⎪⎝ ∆ωn / 2 P ⎠ ( ( 10.26 From Step 2, we have G (e jω ) = G (ω)e − jNω = δ s( F ) e − jNω + F (ω)e − jNω . ( The amplitude response G (ω) has been obtained by raising the amplitude response ( F (ω) by δ s( F ) and hence, the filter G (z ) has double zeros in the stopband.

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344

We may factorize as follows: G ( z ) = z − N H m ( z ) H m ( z −1 ), where H m (z ) is a real-coefficient ( minimum-phase FIR lowpass filter with half the degree of the original H (z ). Since, G (ω) ≥ 0, ( the amplitude response H m (ω) of the minimum-phase filter H m (z ) does not oscillate about ( unity in the passband. Since the original frequency response was raised by δ s( F ) , H m (ω) must be normalized by a factor 1 + δ s . Therefore,

( For H m (ω) , we can see δs(F ) =

1 + δ p + δs δp 2δ s and δ (pF ) = −1 = 1+ −1. 1 + δs 1 + δs 1 + δs

10.27 (a) N = 1 and hence, x a (t ) = α 0 + α1t. Without any loss of generality, for L = 5. we first fit the data set {x[ k ]}, − 5 ≤ k ≤ 5, by the polynomial x a (t ) = α 0 + α1t with a minimum mean-square error at t = −5, − 4,K , − 1, 0,1,K ,5, , and then replace x[ 0] with a new value x[ 0 ] = x ( 0 ) = α 0 .

Now, the mean-square error is given by ε(α0 , α1 ) =

5

∑ (x[k ] − α0 − α1k )2 . We set

k = −5

5 5 ∂ ε(α 0 , α1 ) ∂ ε(α 0 , α1 ) = 0 and = 0 which yields 11α 0 + α1 ∑ k = ∑ x[ k ], and ∂α 0 ∂α1 k = −5 k = −5 5

5

5

k = −5

k = −5

k = −5

α0 ∑ k + α1 ∑ k 2 = ∑ k x[ k ]. From the first equation we get x[0] = α 0 =

5

∑

1 x[ k ]. In the general case we thus 11 k = −5

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345

5

∑

1 have x[ n] = α0 = x[ k ] which is a moving average filter of length 11. 11 k = −5 (b) N = 2, and hence, x a (t ) = α 0 + α1t + α 2 t 2 . Here, we fit the data set {x[ k ]}, − 5 ≤ k ≤ 5, by the polynomial x a (t ) = α 0 + α1t + α 2 t 2 with a minimum mean-square

error at t = −5, − 4,K , − 1, 0,1,K ,5, and then replace x[ 0] with a new value x[0] = x a (0) = α 0 . The mean-square error is now given by 5

∑ (x[k ] − α0 − α1k − α2 k 2 ) .

ε (α 0 , α1 , α 2 ) =

2

We set

k = −5

∂ ε(α 0 , α1 , α 2 ) = 0, ∂α 0

∂ ε(α 0 , α1 , α 2 ) ∂ ε(α 0 , α1 , α 2 ) = 0, and = 0, which yields ∂α1 ∂α 2 5

∑

11α 0 + 110α 2 =

5

x[ k ], 110α1 =

k = −5

∑

k x[ k ], 110α 0 + 1958α 2 =

k = −5

5

∑ k 2 x[k ].

From the

k = −5

first and the third equations we then get 5

1958

∑

x[ k ] − 110

k = −5

5

∑ k 2 x[k ]

k = −5 (1958 × 11) − (110) 2

α0 =

5

∑(

)

1 = 89 − 5k 2 x[ k ]. 429 k = −5

Hence, here we replace x[n] with a new value x[ n] = α0 which is a weighted combination of the original data set {x[ k ]}, − 5 ≤ k ≤ 5 : 5

∑(

)

1 x[ n ] = 89 − 5k 2 x[ n − k ] 429 k = −5 =

1 (−36 x[ n + 5] + 9 x[ n + 4] + 44 x[ n + 3] + 69 x[ n + 2] + 84 x[ n + 1] + 89 x[ n] 429 +84 x[ n − 1] + 69x[ n − 2] + 44 x[ n − 3] + 9 x[ n − 4] − 36 x[ n − 5]).

(c) The impulse response of the FIR filter of Part (a) is given by 1 h1[ n] = {1 1 1 1 1 1 1 1 1 1 1}, 11

whereas, the impulse response of the FIR filter of Part (b) is given by 1 h1[ n] = {− 36 9 44 69 84 89 84 69 44 9 − 36}. 429 The corresponding frequency responses are given by H1

( e jω ) =

5

5

∑(

∑

)

1 1 e − jωk , and H 2 (e jω ) = 89 − 5k 2 e − jωk . 11 k = −5 429 k = −5

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A plot of the magnitude responses of these two filters are shown below from which it can be seen that the filter of Part (b) has a wider passband and thus provides smoothing over a larger frequency range than the filter of Part (a). 1 H1(z) H2(z)

Magnitude

0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

10.28 y[ n] =

1 {−3 x[ n − 7] − 6 x[ n − 6] − 5 x[n − 5] + 3 x[ n − 4] + 21x[ n − 3] + 46 x[ n − 2] 320 + 67 x[ n − 1] + 74 x[ n] + 67 x[ n + 1] + 46 x[ n + 2] + 21x[ n + 3] + 3 x[ n + 4]

+ 5 x[ n + 5] − 6 x[ n − 6] − 3 x[ n + 7]}.

{

Y ( e jω ) 1 = − 3e − j 7ω − 6e − j 6ω − 5e − j 5ω + 3e − j 4ω + 21e − j 3ω X (e jω ) 320 46e − j 2ω +67e − jω + 74 +67e jω + 46e j 2ω + 21e j 3ω + 3e j 4ω − 5e j 5ω − 6e j 6ω − 3e j 7ω . 1 {74 + 67cos ω + 46cos(2ω) + 21cos(3ω) + 3cos(4ω)– 5cos(5ω) – 6cos(6ω) – 3cos(7ω)} = 160 – 5cos(5ω) – 6cos(6ω) – 3cos(7ω)}.

Hence, H 3 (e jω ) =

}

The magnitude response of the above FIR filter H 3 ( z) is plotted below (solid line) along with that of the FIR filter H 2 ( z) of Part (b) of Problem 10.27 (dashed line). Note that both filters have roughly the same passband but the Spencer's filter has very large attenuation in the stopband and hence it provides better smoothing than the filter of Part (b). 1 H3(z) H2(z)

Magnitude

0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

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10.29 (a) L = 3. P ( x ) = α1 x + α 2 x 2 + α 3 x 3 . Now P (0) = 0 is satisfied by the way P (x ) has been defined. Also to ensure P (1) = 1 we require α1 + α 2 + α3 = 1 . Choose m = 1 and

n = 1. Since

dP ( x ) = 0, hence α1 + 2α 2 x + 3α3 x 2 = 0, implying α1 = 0 . Also x =0 dx x = 0

dP( x ) = 0, hence α1 + 2α 2 + 3α3 = 0 . Thus solving the three equations: dx x =1 α1 + α 2 + α3 = 1 , α1 = 0 ,and α1 + 2α 2 + 3α3 = 0

since

we arrive at

1

0,

2

3 , and

–2. Therefore, P ( x ) = 3 x 2 − 2 x 3 .

3

(b) L = 4. Hence, P ( x ) = α1 x + α 2 x 2 + α 3 x 3 + α 4 x 4 . Choose m = 2 and n = 1 (alternatively one can choose m = 1 and n = 2 for better stopband performance ). Then, P (1) = 1 ⇒ α1 + α 2 + α3 + α 4 = 1. Also,

dP( x ) = 0 ⇒ α1 + 2α2 x + 3α3 x 2 + 4α 4 x 3 x = 0 = 0, dx x = 0 d 2 P( x ) = 0 ⇒ 2α2 + 6α3 x + 12α 4 x 2 x = 0 = 0, 2 dx x=0 dP( x ) = 0 ⇒ α1 + 2α2 + 3α3 + 4α 4 = 0. dx x =1 Solving the above four simultaneous equations we get

1

0,

2

0,

3

4, and

4

– 3.

Therefore, P ( x ) = 4 x 3 − 3 x 4 . (c) L = 5. Hence P ( x ) = α1 x + α 2 x 2 + α3 x 3 + α 4 x 4 + α 5 x 5 . Choose m = 2 and n = 2 . Following a procedure similar to that in parts (a) and (b) we get 1 0, 2 0, 3 10, 4 –15, and 5 6. M ( 10.30 From Eq. (7.102) we have H (ω) = c[ k ] sin(ωk ). Now

∑

k =1 M

M ( H (π − ω) = c[ k ] sin((π − ω)k ) = −

∑

k =1

∑

c[ k ] sin(ωk ) cos(π k ) =

k =1

M

∑ c[k ](−1) k +1 (ωk ).

k =1

M M ( ( Thus, H (ω) = H (π − ω) implies c[ k ] sin(ωk ) = c[ k ] (−1) k + 1 sin(ωk ), or

∑

k =1

∑

k =1

M

equivalently, k = 2, 4, 6,K.

∑ (1 − (−1) k +1 )c[k ] sin(ωk ) = 0, which in turn implies that c[k ] = 0 for

k =1

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1 2

But from Eq. (7.103) we have c[ k ] = 2 h[ M − k ], 1 ≤ k ≤ M , or, h[ k ] = c[ M − k ]. For k even, 1 2

i.e., k = 2 R, h[2 R ] = c[ M − 2 R] = 0 if M is even. 10.31 (a) H [ k ] = H (e jωk ) = H (e j 2πk / M ), 0 ≤ k ≤ M − 1. Thus, h[ n] =

M −1

∑

1 − kn , H [ k ] WM M k =0

where WM = e − j 2πk / M . Now, H ( z) =

M −1

∑

M −1 M −1 M −1 ⎛ M −1 ⎞ 1 − kn z − n = 1 − kn z − n ⎟. H [ k ] WM H[ k ]⎜ WM ⎜ ⎟ M n=0 k =0 M k =0 ⎝ n=0 ⎠

∑ ∑

h[ n] z − n =

n=0 M −1

We can write

=

∞

∑

∑

− kn z − n WM n=0

∞

∑

− kn z − n WM n=0

=

− kn z − n − W − kM z − M WM M

n=0

∑

∞

∑

−

∑

∞

∑ WM− kn z − n

n= M

(

− kn z − n = 1 − z − M WM

n=0

∞

)∑ WM− kn z − n = 1 −1 W− z− k z −1 . −M

n=0

M

M −1

H[ k ] 1 − z− M Therefore, H ( z) = . − k z −1 M − W 1 M k =0

∑

(b) 1 1 z 1

x[n]

1 z M

H[0] M H[1] M

1 1 z 1e j2 / M

1 1 z

y[n]

H[M 1] M

1e j2 (M 1)/ M

M −1 M −1 ⎞ ⎛ M −1 1− z−M H[ k ] 1 − kn z − n ⎟. H[ k ] ⎜ WM = (c) Note H ( z) = − k −1 ⎟ ⎜ M M k =0 k = 0 1 − WM z ⎠ ⎝ n=0

∑

∑

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∑

349

⎛ M −1 ⎞ − kn e − jωn ⎟. For H[ k ] ⎜ WM ⎜ ⎟ M k =0 ⎝ n=0 ⎠ M −1 M − 1 ⎞ ⎛ 1 − kn e − 2πln / M ⎟ H[ k ] ⎜ WM ω = j 2πl / M , we then get from the above H (e j 2πl / M ) = ⎟ ⎜M k =0 ⎠ ⎝ n=0 M −1 M − 1 M − 1 M − 1 ⎞ ⎞ ⎛ 1 ⎛ 1 − ( k − l)n ⎟ − knW ln ⎟ = ⎜ H[ k ] ⎜ WM H [ k ] W . Using the identity of Eq. = M ⎟ M ⎟ ⎜M ⎜M k =0 ⎠ k =0 ⎠ ⎝ n=0 ⎝ n=0

1 On the unit circle the above reduces to H (e jω ) =

M −1

∑

∑

∑

∑

∑

∑

∑

∑

M −1 if l = k, 1 − ( k − l )n ⎧1, =⎨ WM (5.23) of text we observe that M n=0 ⎩0, otherwise. Hence, H (e j 2 πl / M ) = H [ l].

∑

10.32 (a) For Type 1 FIR filter,

H ( e jω ) =

e

−j

ω( M − 1) 2

H (e jω ) . Since in the frequency

2πk , M 0 ≤ k ≤ M − 1, therefore H [ k ] = H (e j 2πk / M ) = H d (e j 2πk / M ) e j 2πk ( M −1) / M ,

sampling approach we sample the DTFT H (e jω ) at M points given by ω =

0 ≤ k ≤ M − 1. Since the filter is of Type 1, M − 1 is even, thus, e j 2πk ( M −1) / 2 = 1. . Moreover, h[n] being real, H (e jω ) = H * (e jω ). Thus, H (e jω ) = e jω( M −1) / 2 H (e jω ) ,

π ≤ ω < 2π. Hence, ⎧ H (e j 2 πk / M ) e − j 2 πk ( M −1) / 2 M , d ⎪ H[ k ] = ⎨ ⎪ H d (e j 2πk / M ) e − j 2 π ( M − k )( M −1) / 2 M ⎩

k = 0,1, 2,K, k=

M −1 , 2

M +1 M + 3 , , K , M − 1. 2 2

(b) For the Type 2 FIR filter M −1 ⎧ H (e j 2πk / M ) e − j 2πk ( M −1) / 2 M , , k = 0,1, 2,K, d ⎪ 2 ⎪ M H[ k ] = ⎨ 0, k= , 2 ⎪ M j k M j M k M M π − π − − 2 / 2 ( )( 1 ) / 2 )e , k = + 1,K, M − 1. ⎪ H d (e 2 ⎩

10.33 (a) ω p = 0.55π = 1.72788. The frequency spacing between 2 consecutive DFT samples

is given by ω = 2π

2π 19

5 and 19

= 0.3307. The desired passband edge is between the frequency samples at ω = 2π

6 . 19

Therefore, the 19-point DFT is given by

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⎧e − j (2 π / 19) 9k , k = 0,1,K,5,13,14,K,18, H[k ] = ⎨ 0, k = 6,K,12. ⎩ A 19-point IDFT of the above DFT samples yields the impulse response coefficients given below in ascending powers of z −1 :

Columns 1 through 10 -0.0037 -0.0022 0.0224 -0.0316 -0.1128 0.2888

-0.0211 0.6316

Columns 11 through 19 0.2888 -0.1128 -0.0316 0.0224 -0.0022 -0.0037

0.0674

-0.0231

-0.0231

0.0674

-0.0211

(b)

Magnitude

1 0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

% Problem #10.33 close all; clear; clc; ind = 1; for k = 0:5, H(ind) = exp(-i*2*pi*9*k/19); ind = ind + 1; end for k = 6:12, H(ind) = 0; ind = ind + 1; end for k = 13:18, H(ind) = exp(-i*2*pi*9*k/19); ind = ind + 1; end h = ifft(H); figure(1); stem(real(h)); [FF, w] = freqz(h, 1, 512);

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figure(2); plot(w/pi, abs(FF)); axis([0 1 0 1.2]);grid; ylabel('Gain, in dB'); xlabel('\omega/\pi'); 10.34 (a) ω p = 0.35π = 1.09956. The frequency spacing between 2 consecutiveDFT samples

is given by

2π 39

= 0.1611. The desired passband edge is between the frequency samples

6 7 and ω = 2π . Therefore, the 39-point DFT is given by 39 39 ⎧e − j (2 π / 39)19 k , k = 0,1,K,6, 32, 33,K,38,

at ω = 2π

H[k ] = ⎨ 0, k = 7,K,31. ⎩ A 39-point IDFT of the above DFT samples yields the impulse response coefficients given below in ascending powers of z −1 :

Columns 1 through 10 0.0006 0.0031 0.0017 0.0128 0.0146 -0.0034

-0.0054 -0.0237

Columns 11 through 20 -0.0192 0.0134 0.0405 -0.0249 0.1222 0.2870 Columns 21 through 30 0.2870 0.1222 -0.0249 0.0405 0.0134 -0.0192

-0.0091

0.0227 0.3590 -0.0753 -0.0237

Columns 31 through 39 -0.0034 0.0146 0.0128 0.0017 0.0031 0.0006

-0.0362

-0.0362

-0.0010

-0.0091

-0.0010

-0.0753

0.0227

-0.0054

(b)

Magnitude

1 0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

% Problem #10.34 close all; clear; clc; ind = 1; for k = 0:6,

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H(ind) = exp(-i*2*pi*19*k/39); ind = ind + 1; end for k = 7:31, H(ind) = 0; ind = ind + 1; end for k = 32:38, H(ind) = exp(-i*2*pi*19*k/39); ind = ind + 1; end h = ifft(H); figure(1); stem(real(h)); [FF, w] = freqz(h, 1, 512); figure(2); plot(w/pi, abs(FF)); axis([0 1 0 1.2]); grid; xlabel('\omega/\pi'); ylabel('Gain, dB'); 10.35 By expressing cos(ωn) = Tn (cos ω), where Tn (x ) Tn(x) is the n -th order Chebyshev polynomial in x, we first rewrite Eq. (10.48) in the form: M M ( H (ω) = a[ n] cos(ωn) = α n cos n (ω).

∑

∑

n=0

n=0

Therefore, we can rewrite Eq. (10.70) repeated below for convenience P(ωi )[H (ωi ) − D(ωi )] = (−1)i ε, 1 ≤ i ≤ M + 2, in a matrix form as ⎡1 ⎤ ⎡ α 0 ⎤ ⎡ D(ω1 ) ⎤ cos(ω1 ) cos M (ω1 ) 1 / P (ω1 ) L ⎢ ⎥⎢ ⎥ ⎥ ⎢ L − 1 / P (ω2 ) cos(ω2 ) cos M (ω2 ) ⎢1 ⎥ ⎢ α1 ⎥ ⎢ D(ω2 ) ⎥ ⎢M ⎥⎢ M ⎥ = ⎢ ⎥. M O M M ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ (−1) M / P (ω M + 1 ) ⎥ ⎢α M ⎥ ⎢ D(ω M + 1 ) ⎥ ⎢1 cos(ω M + 1 ) L cos M (ω M + 1 ) ⎢1 cos(ω ⎥⎢ M M + 1 / P (ω ⎥ ⎥ ⎢ M + 2 ) L cos (ω M + 2 ) ( −1) M + 2 )⎦ ⎣ ε ⎦ ⎣ D(ω M + 2 )⎦ ⎣ Note that the coefficients {αi } are different from the coefficients {a[i ]} of Eq. (10.70). To determine the expression of 1 1 ∆ = det M 1 1

we use Cramer's rule arriving at

, where

L cos M (ω1 ) 1 / P (ω1 ) M L − 1 / P (ω2 ) cos(ω2 ) cos (ω2 ) , and M O M M cos(ω M +1 ) L cos M (ω M +1 ) (−1) M / P (ω M + 1 ) cos(ω M + 2 ) L cos M (ω M + 2 ) (−1) M + 1 / P (ω M + 2 ) cos(ω1 )

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L D(ω1 ) cos M (ω1 ) L D(ω2 ) cos(ω2 ) cos M (ω2 ) .. M O M M cos(ω M + 1 ) L cos M (ω M + 1 ) D(ω M + 1 ) cos(ω M + 2 ) L cos M (ω M + 2 ) D(ω M + 2 ) cos(ω1 )

1 1 ∆ε = M 1 1

Expanding both determinants using the last column we get ∆ ε =

M +2

∑ bi D(ωi +1 ) and i =1

M +2

(−1) i −1

i =1

P(ωi )

∑ bi

∆=

1 1 M bi = det 1 1

, where cos(ω1 ) cos(ω2 )

cos 2 (ω1 ) cos 2 (ω2 )

L L

cos M (ω1 ) cos M (ω2 )

M cos(ωi −1 ) cos(ωi + 1 )

M cos 2 (ωi −1 ) cos 2 (ωi + 1 )

O L L

cos M (ωi −1 ) cos M (ωi + 1 )

M .

M M M O M 1 cos(ω M + 2 ) cos 2 (ω M + 2 ) L cos M (ω M + 2 )

The above matrix is seen to be a Vandermonde matrix and is determinant is given by

∏ (cos ωk − cos ωl ).

bi =

k ≠ l, k > l k, l ≠ i

ci =

bi Define ci = . It can be shown by induction that M +2 br

∏

r =1 r ≠i

M +2

∏ cos ωi − cos ωn . Therefore, 1

n =1 n≠i

M +2

∑ bi D(ωi )

c D(ω1 ) + c2 D(ω2 ) + L + c M + 2 D(ω M + 2 ) = 1 . M +2 c1 c2 c M + 2 (−1) M + 1 (−1) i − +L+ bi P(ω1 ) P(ω2 ) P (ω M + 2 ) P ( ω ) i i =1

ε = i =1

∑

10.36

⎧ 1 ⎪ W (ω) = ⎨ δ p ⎪δ ⎩ s

ω ∈ passband, ω ∈ stopband.

⎧ 1 0 ≤ ω ≤ 0.45π, W (ω) = ⎨ ⎩4.5 0.55π ≤ ω ≤ π.

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⎧ 1 ⎪ 10.37 W (ω) = ⎨ δ p ⎪δ ⎩ s

ω ∈ passband,

⎧ 1 0 ≤ ω ≤ 0.55π, W (ω) = ⎨ ⎩3.4 0.7π ≤ ω ≤ π.

ω ∈ stopband.

⎧ δp ⎪ ⎪⎪ δs1 10.38 W (ω) = ⎨ 1 ⎪ δp ⎪ ⎪⎩ δs 2

0 ≤ ω ≤ ωs1 ,

0 ≤ ω ≤ 0.44π, ⎧1.43 ⎪ ω p1 ≤ ω ≤ ω p2 , = ⎨ 1 0.55π ≤ ω ≤ 0.7π, ⎪ 5 0.82π ≤ ω ≤ π. ⎩ ωs 2 ≤ ω ≤ π ,

10.39 It follows from Eq. (10.22) that the impulse response of an ideal Hilbert transformer is an antisymmetric sequence. If we truncate it to a finite number of terms between − M ≤ n ≤ M the impulse response is of length (2 M + 1) which is odd. Hence the FIR Hilbert transformer obtained by truncation and satisfying Eq. (10.90) cannot be satisfied by a Type 4 FIR filter.

10.40 (a) X ( z) =

N −1

∑ x[n] z − n .

Thus,

n=0

( ( X ( z ) = X ( z ) z −1 = − α + z −1 = ( 1− αz −1

( P( z ) = ( D( z ) =

N −1

⎛ − α + z( −1 ⎞ P ( z( ) ⎟ x[ n]⎜⎜ ( −1 ⎟ = D( z( ) , where 1 α z − ⎠ ⎝ n=0

∑

N −1

∑ p[n] z( −1 = ∑ x[n](1 − αz( −1 ) N −1− n (−α + z( −1 ) n ,

and

n=0 N −1

∑ d[n]z( − n = (1 − αz( −1 ) N −1.

n=0

( ( P( z ) ( (b) X[ k ] = X ( z ) ( j 2 πk / N = ( z =e D( z )

( z = e j 2 πk / N

( ( P[ k ] ( = ( , where P[ k ] = P ( z ) z( = e j 2 πk / N is D[ k ]

( ( the N -point DFT of the sequence p[n] and D[ k ] = D( z ) z( = e j 2 πk / N is the N -point DFT

of the sequence d[n]. (c) Let P = [ p[0]

p[1] L p[ N − 1]]T and X = [x[0] x[1] L x[ N − 1]]T . Without

( any loss of generality, assume N = 4 in which case P( z ) =

( ) ( + (− 3αx[ 0] + (1 + 2α 2 ) x[1] − α (2 + α 2 ) x[ 2] + 3α 2 x[3])z −1 ( + (3α 2 x[ 0] − α(2 + α 2 ) x[1] + (1 + 2α 2 ) x[ 2] − 3α x[3])z − 2 =

3

∑ p[n] z( − n

n=0

x[ 0] − α x[1] + α 2 x[ 2] − α 3 x[3]

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(

)

( ( + − α 3 x[ 0] + α 2 x[1] − α x[ 2] + α x[3] z − 3 . Equating like powers of z −1 we can write P = Q ⋅ X, where P = [ p[0]

⎡ 1 ⎢ − 3α Q=⎢ ⎢ 3α 2 ⎢ 3 ⎢⎣ − α

p[1] p[2] p[3]]T , X = [x[0] x[1] x[2] x[3]]T and

α2 −α − α3 ⎤ ⎥ − α(2 + α 2 ) 3α 2 ⎥ 1 + 2α 2 . − α (2 + α 2 ) − 3α ⎥ 1 + 2α 2 ⎥ α2 − α2 1 ⎥⎦

It can be seen that the elements q r , s , 0 ≤ r , s ≤ 3, q r,s , of the 4 4 matrix Q can be determined as follows: (i) The first row is given by q 0, s = (− α ) s , 3! (−α ) r , and r! (3 − r )! (iii) the remaining elements can be obtained using the recurrence relation q r , s = qr −1, s −1 − αq r , s −1 + αq r −1, s . In the general case, we only change the computation of the elements of the first column ( N − 1)! (−α )r . using the relation qr ,0 = N −1C r (− α ) r = r! ( N − 1 − r )! (ii) The first column is given by qr ,0 = 3C r (−α ) r =

M10.1 The impulse response coefficients of the truncated FIR highpass filter with cutoff frequency at 0.4π can be generated using the following MATLAB statements: n = -M:M; num = -0.4*sinc(0.4*n); num(M+1) = 0.6;

The magnitude responses of the truncated FIR highpass filter for two values of M are shown below:

Magnitude

1 0.8 0.6 0.4 M=5 M=20

0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

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M10.2 The impulse response coefficients of the truncated FIR bandpass filter with cutoff frequencies at 0.7π and 0.3π can be generated using the following MATLAB statements: n = -M:M; num = 0.7*sinc(0.7*n) - 0.3*sinc(0.3*n);

The magnitude responses of the truncated FIR bandpass filter for two values of M are shown below:

Magnitude

1 0.8 0.6 0.4 M=5 M=20

0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

M10.3 The impulse response coefficients of the truncated Hilbert transformer can be generated using the following MATLAB statements: n = 1:M; c = 2*sin(pi*n/2).*sin(pi*n/2);b = c./(pi*n); num = [-fliplr(b) 0 b];

The magnitude responses of the truncated Hilbert transformer for two values of M are shown below:

Magnitude

1 0.8 0.6 0.4 M=5 M=20

0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

M10.4 % Problem #M10.4 % Cascade of 2 boxcar filters of length 4 K = 2;N = 4;b = firgauss(K,N); figure(1); stem(b);xlabel(‘n’);ylabel(‘h[n]’); title('Cascade of 2 boxcar filters of length 4');

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% Cascade of 4 boxcar filters of length 4 K = 4;N = 4;b = firgauss(K,N); figure(2); stem(b);xlabel(‘n’);ylabel(‘h[n]’); title('Cascade of 4 boxcar filters of length 4'); % Cascade of 2 boxcar filters of length 12 K = 2;N = 12;b = firgauss(K,N); figure(3); stem(b);xlabel(‘n’);ylabel(‘h[n]’); title('Cascade of 2 boxcar filters of length 12'); % Cascade of 4 boxcar filters of length 12 K = 4;N = 12;b = firgauss(K,N); figure(4); stem(b);xlabel(‘n’);ylabel(‘h[n]’); title('Cascade of 4 boxcar filters of length 12'); Cascade of 2 boxcar filters of length 4

Cascade of 4 boxcar filters of length 4

4

50 40

3 h[n]

h[n]

30 2 1 0

20 10

1

2

3

4 n

5

6

0

7

0

5

10

1000

8

800

6

600

4

400

2

200

0

5

10

15

Cascade of 4 boxcar filters of length 12 1200

h[n]

h[n]

Cascade of 2 boxcar filters of length 12 12

0

10 n

15

20

25

0

0

n

10

20

30

40

50

n

We can see that by increasing either K or N, the approximation to a Gaussian function gets better. It is noted that increasing the number of boxcar filters K to a number greater than 3 greatly affects the Gaussian shape of the impulse response. M10.5 % Problem #M10.05 N = 36;fc = 0.2*pi; M = N/2; n = -M:1:M;t = fc*n; lp = fc*sinc(t);

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b = 2*[lp(1:M) (lp(M+1) - 0.5) lp((M+2):N+1)]; bw = b.*hamming(N+1)'; [h2, w] = freqz(bw, 1, 512); plot(w/pi, abs(h2));axis([0 1 0 1.2]); xlabel('\omega/\pi');ylabel('Magnitude'); title(['\omega_c = ', num2str(fc), ', N = ', num2str(N)]); ωc = 0.62832, N = 36

Magnitude

1 0.8 0.6 0.4 0.2 0

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M10.6 D( x) = 3.2 x 2 + 4.05 x − 5.5. Its approximation over the range − 3 ≤ x ≤ 2 is given by A( x) = a0 + a1 x. We want to minimize the peak value of the absolute error, i.e.,

minimize max 3.2 x 2 + 4.05 x − 5.5 − a0 − a1 x . Since there are 3 unknowns, a0 , a1 , − 3≤ x ≤ 2

and ε, we need 3 extremal points on x , which we arbitrarily choose as x1 = −3, x2 = 0, and x3 = 2. We then solve the 3 linear equations: a0 + a1 x − ( −1) l ε = D ( xl ), l = 1, 2, 3. ⎡1 − 3 1 ⎤ ⎡a0 ⎤ ⎡11.15⎤ This leads to ⎢1 0 − 1⎥ ⎢ a1 ⎥ = ⎢ − 5.5 ⎥, whose solution yields a0 = 4.1, a1 = 0.85, 1 ⎥⎦ ⎢ ε ⎥ ⎢⎣ 15.4 ⎥⎦ ⎢⎣1 2 ⎣ ⎦ and ε = 9.6. A plot of the corresponding error E1 ( x) = 3.2 x 2 + 3.2 x − 9.6 is shown below in Figure (c). Result of second guess

10

10

5

5 Error

Error

Result of first guess

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(c)

(d)

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After looking at E1 (x) , we move the second extremal point x2 to the location where E1 (x) is a minimum. The next extrtemal points are therefore given by x1 = −3, x2 = −0.5, and x3 = 2. The new values of the unknowns are obtained by solving 1 ⎤ ⎡a0 ⎤ ⎡ 11.15 ⎤ ⎡1 − 3 ⎢1 − 0.5 − 1⎥ ⎢ a1 ⎥ = ⎢− 6.725⎥, which yields a0 = 3.7, a1 = 0.85, and ε = 10. A plot 2 1 ⎥⎦ ⎢ ε ⎥ ⎢⎣ 15.4 ⎥⎦ ⎢⎣1 ⎣ ⎦ of the corresponding error E 2 ( x) = 3.2 x 2 + 3.2 x − 9.2 is shown on the previous page in Figure (d). % x d D A C

Problem #M10.06 = [-3 0 2]; = 3.2.*x.^2 + 4.05.*x-5.5; = d'; = [1 -3 1;1 0 -1;1 2 1]; = inv(A)*D;

y = -3:0.05:2; E = 3.2.*y.^2 + 4.05.*y - 5.5 - C(1) - C(2).*y; % Results of first guess figure(1); plot(y,E); axis([-3 2 -12 12]); xlabel('x'); ylabel('Error'); title('Result of first guess'); hold on; plot([-3 -3], [E(1) E(1)], 'o'); plot([2 2], [E(end) E(end)], 'o'); plot([0 0], [E(61) E(61)], 'o'); hold off; x d D A C

= = = = =

[-3 -0.5 2]; 3.2.*x.^2 + 4.05.*x-5.5; d'; [1 -3 1;1 -0.5 -1;1 2 1]; inv(A)*D;

y = -3:0.05:2; E = 3.2.*y.^2 + 4.05.*y - 5.5 - C(1) - C(2).*y; % Results of second guess figure(2); plot(y,E); axis([-3 2 -12 12]); xlabel('x'); ylabel('Error');

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title('Result of second guess'); hold on; plot([-3 -3], [E(1) E(1)], 'o'); plot([2 2], [E(end) E(end)], 'o'); plot([-0.5 -0.5], [E(51) E(51)], 'o'); hold off; M10.7 D(x ) = −5 x 3 − 0.2 x 2 + 8 x + 5.5 . Its approximation over the range − 2 ≤ x ≤ 2 is given by A( x) = a0 + a1 x + a2 x 2 . We want to minimize the peak value of the absolute error, i.e.,

minimize max − 5 x 3 − 0.2 x 2 + 8 x + 5.5 − a0 − a1 x − a2 x 2 . Since there are 4 −2≤ x≤2

unknowns, a0 , a1 , a2 , and ε, we need 4 extremal points on x , which we arbitrarily choose as x1 = −2, x2 = −1, x3 = 1, and x4 = 2. We then solve the 4 linear equations: a0 + a1 x + a 2 x 2 − ( −1) l ε = D ( xl ), l = 1, 2, 3, 4. This leads to ⎡1 − 2 4 1 ⎤ ⎡a0 ⎤ ⎡ 28.7 ⎤ ⎢1 − 1 1 − 1⎥ ⎢ a1 ⎥ = ⎢ 2.3 ⎥, whose solution yields a = 5.5, a = −7, a = −0.2, 0 1 2 ⎢1 1 1 1 ⎥ ⎢a2 ⎥ ⎢ 8.3 ⎥ ⎢ ⎥ − 1 2 4 1 19 3 − . ⎣⎢ ⎦⎥ ⎣ ε ⎦ ⎣⎢ ⎦⎥

and ε = 10. A plot of the corresponding error E1 ( x) = −5 x 3 + 15 x is shown below in Figure (e). We observe that these values maximize the error (ε = 10). Result of first guess 10

Error

5 0 -5 -10 -2

% x d D A C

-1

0 x

1

2

Problem #M10.07 = [-2 -1 1 2]; = -5.*x.^3 - 0.2.*x.^2 + 8.*x + 5.5; = d'; = [1 -2 4 1;1 -1 1 -1;1 1 1 1;1 2 4 -1]; = inv(A)*D;

y = -2:0.05:2; E = -5.*y.^3 - 0.2.*y.^2 + 8.*y + 5.5 - C(1) - C(2).*y C(3).*y.^2; % Results of first guess figure(1);

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plot(y,E); axis([-2 2 -12 12]); xlabel('x'); ylabel('Error'); title('Result of first guess'); hold on; plot([-2 -2], [E(1) E(1)], 'o'); plot([2 2], [E(end) E(end)], 'o'); plot([-1 -1], [E(21) E(21)], 'o'); plot([1 1], [E(1) E(1)], 'o'); hold off; 8π 12π 10π , ωs = , ωT = 18 18 18 % Problem #M10.8 wp = 4*(2*pi)/18; ws = 6*(2*pi)/18; wc = (wp + ws)/2; dw = ws - wp; % Hamming M = ceil(3.32*pi/dw);N = 2*M+1;n = -M:M; num = (6/18)*sinc(6*n/18); wh = hamming(N)';b = num.*wh;

M10.8 ω p =

figure(1); k=0:2*M:stem(k,b); title('Impulse Response Coefficients'); xlabel('Time index n'); ylabel('Amplitude'); figure(2); [h, w] = freqz(b,1,512); plot(w/pi, 20*log10(abs(h))); grid; xlabel('\omega/\pi'); ylabel('Gain, in dB'); title('Lowpass filter designed using Hamming window'); axis([0 1 -80 10]); % Hann M = ceil(3.11*pi/dw);N = 2*M+1;n = -M:M; num = (6/18)*sinc(6*n/18); wh = hann(N)';b = num.*wh; figure(3); k=0:2*M:stem(k,b); title('Impulse Response Coefficients'); xlabel('Time index n'); ylabel('Amplitude'); figure(4); [h, w] = freqz(b,1,512); plot(w/pi, 20*log10(abs(h)));grid; xlabel('\omega/\pi');ylabel('Gain, in dB');

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title('Lowpass filter designed using Hann window'); axis([0 1 -80 10]); % Blackman M = ceil(5.56*pi/dw);N = 2*M+1;n = -M:M; num = (6/18)*sinc(6*n/18); wh = blackman(N)';b = num.*wh; figure(5); k=0:2*M:stem(k,b); title('Impulse Response Coefficients'); xlabel('Time index n'); ylabel('Amplitude'); figure(6); [h, w] = freqz(b,1,512); plot(w/pi, 20*log10(abs(h)));grid; xlabel('\omega/\pi');ylabel('Gain, in dB'); title('Lowpass filter designed using Blackman window'); axis([0 1 -80 10]);

Lowpass filter design using Hamming window: N = 31 Lowpass filter designed using Hamming window

Impulse Response Coefficients 0.4

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Lowpass filter design using Hann window: N = 29 Impulse Response Coefficients

Lowpass filter designed using Hann window

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Lowpass filter design using Blackman window: N = 53

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Lowpass filter designed using Blackman window

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Comments: The Hann window method results in using the lowest filter order. All filters meet the requirements of the specifications. M10.9 α s = 42 , β = 0.5842(42 − 21)0.4 + 0.07886(42 − 21) = 3.631 using Eq. (10.41). 42 − 8 N= using Eq. (10.42). ⎛ 2π ⎞ 2.285⎜ ⎟ ⎝ 18 ⎠ N = 42.627 ≥ 43 and we choose 44 since N must be even. M = 22. % Problem #M10.9 beta = 3.631;N = 44;n = -N/2:N/2; num = (6/18)*sinc(6*n/18); wh = kaiser(N+1,beta)';b = num.*wh; figure(1); stem(b); title('Impulse Response Coefficients'); xlabel('Time index n');ylabel('Amplitude') figure(2); [h, w] = freqz(b,1,512); plot(w/pi, 20*log10(abs(h)));grid; xlabel('\omega/\pi');ylabel('Gain, in dB'); title('Lowpass filter designed using Kaiser window'); axis([0 1 -80 10]); Impulse Response Coefficients

Lowpass filter designed using Kaiser window

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M10.10 ω p = 0.4π , ω s = 0.6π , α s = 42 dB, ω c = 0.5π , ∆ω = 0.2π We will use the Hann window since it meets the requirements and has the lowest order from Table 10.2. 3.11π M= = 15.55 → 16 ⇒ N = 32. ∆ω % Problem M10.10 n = -16:16; lp = 0.5*sinc(0.5*n);wh = hanning(33); b = lp.*wh'; figure(1); k=0:2*n;stem(k,b); title('Impulse Response Coefficients'); xlabel('Time index n');ylabel(‘Amplitude'); figure(2); [h, w] = freqz(b,1,512); plot(w/pi, 20*log10(abs(h)));grid; xlabel('\omega/\pi');ylabel('Gain, in dB'); title('Lowpass filter designed using Hann window'); axis([0 1 -80 10]); Impulse Response Coefficients

Lowpass filter designed using Hann window 0

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M10.11 We use the same specifications from Problem M10.10, but we use the Dolph-Chebyshev 2.056(42 ) − 16.4 = 48.7 . We use N = 50, which is a much higher order window. N = 2.285(0.2π ) than in Problem M10.10. % Problem M10.11 n = -25:25; lp = 0.5*sinc(0.5*n);wh = chebwin(51,42); b = lp.*wh'; figure(1); stem(b); title('Impulse Response Coefficients'); xlabel('Time index n');ylabel('Amplitude'); figure(2); [h, w] = freqz(b,1,512);

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plot(w/pi, 20*log10(abs(h)));grid; xlabel('\omega/\pi');ylabel('Magnitude'); title('Filter designed using Dolph-Chebyshev window'); axis([0 1 -80 10]); Impulse Response Coefficients

Filter designed using Dolph-Chebyshev window

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M10.12 n = -16:16; b = fir1(32, 0.5, hanning(33)); figure(1); stem(b); title('Impulse Response Coefficients'); xlabel('Time index n');ylabel('Amplitude'); figure(2); [h, w] = freqz(b,1,512); plot(w/pi, 20*log10(abs(h))); grid; xlabel('\omega/\pi');ylabel('Magnitude'); title(' Lowpass filter designed using Hann window'); axis([0 1 -80 10]); Impulse Response Coefficients

Lowpass filter designed using Hann window

0.6 0 Magnitude

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M10.13 N = 35; for k = 1:N+1, w(k) = 2*pi*(k-1)/(N+1); if(w(k) >= 0.45*pi & w(k) <= 1.45*pi) H(k) = 1; else H(k) = 0; end if (w(k) <= pi) phase(k) = i*exp(-i*w(k)*N/2); else phase(k) = -i*exp(i*(2*pi-w(k))*N/2);

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end end H = H.*phase; f = ifft(H); [FF, w] = freqz(f, 1, 512); k = 0:N; figure(1); stem(k, real(f)); xlabel('Time index n');ylabel('Amplitude'); figure(2); plot(w/pi,20*log10(abs(FF)));grid xlabel('\omega/\pi');ylabel('Gain, dB'); axis([0 1 -50 5]); 0.5

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M10.14 N = 45; L = N + 1; for k = 1:L, w = 2*pi*(k-1)/L; if (w >= 0.5*pi & w <= 0.7*pi) H(k) = i*exp(-i*w*N/2); elseif (w >= 1.3*pi & w <= 1.5*pi) H(k) = i*exp(i*(2*pi-w)*N/2); else H(k) = 0; end end f = ifft(H); [FF, w] = freqz(f, 1, 512); k = 0:N; figure(1); stem(k, real(f)); xlabel('Time index, n');ylabel('h[n]'); figure(2); plot(w/pi, 20*log10(abs(FF)));grid; ylabel('Gain, dB');xlabel('\omega/\pi'); axis([0 1 -50 5]);

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0.2 0 -10 Gain, dB

h[n]

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M10.15 ind = 1; for k = 0:6, H(ind) = exp(-i*2*pi*19*k/39); ind = ind + 1; end k = 7; H(ind) = 0.5*exp(-i*2*pi*19*k/39); ind = ind + 1; for k = 8:30, H(ind) = 0; ind = ind + 1; end k = 31; H(ind) = 0.5*exp(-i*2*pi*19*k/39); ind = ind + 1; for k = 32:38, H(ind) = exp(-i*2*pi*19*k/39); ind = ind + 1; end h = ifft(H); [FF, w] = freqz(h, 1, 512); plot(w/pi, 20*log10(abs(FF)));grid; xlabel('\omega/\pi');ylabel('Gain, dB'); axis([0 1 -50 5]); 0

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M10.16 ind = 1;

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for k = 0:6, H(ind) = exp(-i*2*pi*19*k/39); ind = ind + 1; end k = 7; H(ind) = (2/3)*exp(-i*2*pi*19*k/39); ind = ind + 1; k = 8; H(ind) = (1/3)*exp(-i*2*pi*19*k/39); ind = ind + 1; for k = 9:29, H(ind) = 0; ind = ind + 1; end k = 30; H(ind) = (1/3)*exp(-i*2*pi*19*k/39); ind = ind + 1; k = 31; H(ind) = (2/3)*exp(-i*2*pi*19*k/39); ind = ind + 1; for k = 32:38, H(ind) = exp(-i*2*pi*19*k/39); ind = ind + 1; end h = ifft(H); [FF, w] = freqz(h, 1, 512); plot(w/pi, 20*log10(abs(FF)));grid; xlabel('\omega/\pi');ylabel('Gain, dB'); axis([0 1 -50 5]); 0

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8π 12π 10π 4π , ωs = , ωc = , ∆ω = 18 18 18 18 wp = 4*(2*pi)/18; ws = 6*(2*pi)/18; wc = (wp + ws)/2; dw = ws - wp; % Hamming M = ceil(3.32*pi/dw); N = 2*M; b = fir1(N, ws/(2*pi));

M10.17 ω p =

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[H, w] = freqz(b,1,512); figure(1); stem(b); title('Impulse Response Coefficients'); xlabel('Time index n');ylabel('h[n]'); figure(2); plot(w/pi, 20*log10(abs(H)));grid; xlabel('\omega/\pi');ylabel('Gain, dB'); title('Lowpass filter designed using Hamming window'); axis([0 1 -80 10]); % Hann M = ceil(3.11*pi/dw); N = 2*M; b = fir1(N, ws/(2*pi), hanning(N+1)); [H, w] = freqz(b,1,512); figure(3); stem(b); title('Impulse Response Coefficients'); xlabel('Time index n');ylabel('h[n]'); figure(4); plot(w/pi, 20*log10(abs(H)));grid; xlabel('\omega/\pi');ylabel('Gain, dB'); title('Lowpass filter designed using Hann window'); axis([0 1 -80 10]); % Blackman M = ceil(5.56*pi/dw); N = 2*M; b = fir1(N, ws/(2*pi), blackman(N+1)); [H, w] = freqz(b,1,512); figure(5); stem(b); title('Impulse Response Coefficients'); xlabel('Time index n');ylabel('h[n]'); figure(6); plot(w/pi, 20*log10(abs(H))); grid; xlabel('\omega/\pi');ylabel('Gain, dB'); title('Lowpass filter designed using Blackman window'); axis([0 1 -80 10]); % Kaiser beta = 3.631; N = 44; b = fir1(N, ws/(2*pi), kaiser(N+1, beta)); [H, w] = freqz(b,1,512); figure(7); stem(b); title('Impulse Response Coefficients'); Not for sale

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xlabel('Time index n');ylabel('h[n]'); figure(8); plot(w/pi, 20*log10(abs(H)));grid; xlabel('\omega/\pi');ylabel('Gain, dB'); title('Lowpass filter designed using Kaiser window'); axis([0 1 -80 10]); (a) Hamming window using fir1 Impulse Response Coefficients

Lowpass filter designed using Hamming window

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(b) Hann window using fir1 Lowpass filter designed using Hann window


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(c) Blackman window using fir1 Impulse Response Coefficients

Lowpass filter designed using Blackman window

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Lowpass filter designed using Kaiser window

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M10.18 ω s = 0.4π , ω p = 0.55π , α p = 0.1 dB, α s = 42 dB, ∆ω = 0.15π , ωc = 0.475π 3.32 π (a) Hamming: use Eq. (10.33): M = = 22.133 → 23 ∴ N = 2 M = 46 0.15π 3.11π (b) Hann: M = = 20.733 → 21 ∴ N = 2 M = 42 0.15π 5.56 π (c) Blackman: M = = 37.067 → 38 ∴ N = 2 M = 76 0.15π (d) Kaiser: δ s = 10 −α s / 20 = 0.00794 % Hamming N = 46; b = fir1(N, 0.475, 'high'); [H,w] = freqz(b,1,512); figure(1); stem(b); title('Impulse Response Coefficients'); xlabel('Time index n');ylabel('h[n]'); figure(2); plot(w/pi, 20*log10(abs(H)));grid; xlabel('\omega/\pi');ylabel('Gain, dB'); title('Highpass filter designed using Hamming window'); axis([0 1 -80 10]); % Hann N = 42; b = fir1(N, 0.475, 'high', hanning(N+1)); [H,w] = freqz(b,1,512); figure(3); stem(b); title('Impulse Response Coefficients'); xlabel('Time index n');ylabel('h[n]'); figure(4); plot(w/pi, 20*log10(abs(H)));grid; xlabel('\omega/\pi');ylabel('Gain, dB'); title('Highpass filter designed using Hann window'); axis([0 1 -80 10]); Not for sale

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% Blackman N = 76; b = fir1(N, 0.475, 'high', blackman(N+1)); [H,w] = freqz(b,1,512); figure(5); stem(b); title('Impulse Response Coefficients'); xlabel('Time index n');ylabel('h[n]'); figure(6); plot(w/pi, 20*log10(abs(H)));grid; xlabel('\omega/\pi');ylabel('Gain, dB'); title('Highpass filter designed using Blackman window'); axis([0 1 -80 10]); % Kaiser ds = 0.00794; [N,Wn,beta,type] = kaiserord([0.4 0.55],[1 0],[ds ds]); b = fir1(N, 0.475,'high',kaiser(N+1,beta)); [H,w] = freqz(b,1,512); figure(7); stem(b); title('Impulse Response Coefficients'); xlabel('Time index n');ylabel('h[n]'); figure(8); plot(w/pi, 20*log10(abs(H)));grid; xlabel('\omega/\pi');ylabel('Gain, dB'); title('Highpass filter designed using Kaiser window'); axis([0 1 -80 10]); (a) Hamming window using fir1 Impulse Response Coefficients

Highpass filter designed using Hamming window

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Highpass filter designed using Hann window

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(c) Blackman window using fir1 Highpass filter designed using Blackman window


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(d) Kaiser window using fir 1 Impulse Response Coefficients

Highpass filter designed using Kaiser window

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M10.19 ω p1 = 0.65π , ω p2 = 0.85π , ωs1 = 0.55π , ωs 2 = 0.75π , α p = 0.2 dB, α s = 42 dB ∆ω1 = ω p1 − ωs1 = 0.1π , ∆ω2 = ω p2 − ωs 2 = 0.1π = ∆ω (a) Hamming window: M = (b) Hann: M =

3.32π = 33.2 → 34 ∴ N = 2 M = 68 0.1π

3.11π = 31.1 → 32 ∴ N = 2 M = 64 0.1π

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(c) Blackman: M =

5.56π = 55.6 → 56 ∴ N = 2 M = 112 0.1π − α / 20

(d) Kaiser: δs = 10 − α s / 20 = 0.00794 , δ p = 10 p = 0.97724 % Problem #M10.19 % Hamming N = 68; b = fir1(N, [0.6 0.8]); [H, w] = freqz(b,1,512); figure(1); stem(b); title('Impulse Response Coefficients'); xlabel('Time index n');ylabel('h[n]'); figure(2); plot(w/pi, 20*log10(abs(H)));grid; xlabel('\omega/\pi');ylabel('Gain, dB'); title('Bandpass filter designed using Hamming window'); axis([0 1 -80 10]); % Hann N = 64; b = fir1(N, [0.6 0.8], hanning(N+1)); [H, w] = freqz(b,1,512); figure(3); stem(b); title('Impulse Response Coefficients'); xlabel('Time index n');ylabel('h[n]'); figure(4); plot(w/pi, 20*log10(abs(H)));grid; xlabel('\omega/\pi');ylabel('Gain, dB'); title('Bandpass filter designed using Hann window'); axis([0 1 -80 10]); % Blackman N = 112; b = fir1(N, [0.6 0.8], blackman(N+1)); [H, w] = freqz(b,1,512); figure(5); stem(b); title('Impulse Response Coefficients'); xlabel('Time index n');ylabel('h[n]'); figure(6); plot(w/pi, 20*log10(abs(H)));grid; xlabel('\omega/\pi');ylabel('Gain, dB'); title('Bandpass filter designed using Blackman window'); axis([0 1 -80 10]); % Kaiser [N, Wn, beta, type] = kaiserord([0.6 0.8], [1 0], [0.97724 0.00794]);

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b = fir1(2*N, [0.6 0.8], kaiser(2*N+1, beta)); [H, w] = freqz(b,1,512); figure(7); stem(b); title('Impulse Response Coefficients'); xlabel('Time index n');ylabel('h[n]'); figure(8); plot(w/pi, 20*log10(abs(H)));grid; xlabel('\omega/\pi');ylabel('Gain, dB'); title('Bandpass filter designed using Kaiser window'); axis([0 1 -80 10]);

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2π ⋅ 15 = 0.4286 π 70 % Problem #M10.20 N = 31; d1 = fir1(N-1,0.4286);d2 = fir1(N-1,0.4286,'high'); [h1,w] = freqz(d1,1,512);h2 = freqz(d2,1,w); plot(w/pi,20*log10(abs(h1)),'-r',w/pi,20*log10(abs(h2)),'-b');grid; xlabel('\omega/\pi');ylabel('Gain, dB'); title('Crossover pair'); axis([0 1 -80 10]);

M10.20 ωc =

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M10.21 ωc1 = 0.2494 , ωc 2 = 0.5442 % Problem #M10.21 N = 32; d1 = fir1(N, 0.2494, hanning(33)); d2 = fir1(N, 0.5442, 'high', hanning(33)); d3 = -d1-d2; d3(17) = 1-d1(17)-d2(17); [h1,w] = freqz(d1,1,512);h2 = freqz(d2,1,w);h3 = freqz(d3,1,w); g1 = 20*log10(abs(h1))); g2 = 20*log10(abs(h2))); g3 = 20*log10(abs(h3))); plot(w/pi, g1,’—-b’,w/pi,g2,’-.g’,w/pi,g3,’-r’);grid; xlabel('\omega/\pi');ylabel('Gain, dB'); title('Crossover triple'); axis([0 1 -80 10]); Crossover triple

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M10.22 % Problem #M10.22 fpts = [0 0.35 0.4 0.7 0.72 1]; mval = [0.2 0.2 1 1 0.6 0.6]; b = fir2(70, fpts, mval); [H,w] = freqz(b,1,512); figure(1); plot(w/pi,abs(H));grid; xlabel('\omega/\pi');ylabel('Magnitude'); title('Multilevel FIR filter'); axis([0 1 0 1.2]); Multilevel FIR filter

Magnitude

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M10.23 From Problem 10.36, ω p = 0.45π , ω s = 0.6π , δ p = 0.2043 , δ s = 0.0454 and we assume that

0.45π ⋅ FT = 0.45 and Fs = 0.6. 2π α p = −20 log10 1 − δ p = 1.985 dB, α s = −20 log10 (δ s ) = 26.858 dB. After obtaining the length N using ‘remezord’, the specifications of the filter were not met. We increased N to 11 to meet the specifications. % Program #M10.23 Ft = 2;Fp = 0.45;Fs = 0.6; ds = 0.0454;dp = 0.2043; F = [Fp Fs];A = [1 0];DEV = [dp ds]; [N,Fo,Ao,W] = remezord(F,A,DEV,Ft); b = remez(N,Fo,Ao,W); [H,w] = freqz(b,1,512); figure(1); plot(w/pi, 20*log10(abs(H))); xlabel('\omega/\pi');ylabel('Gain, dB');title('N = 9'); %axis([0 0.45 -3 3]); N = 11; b = remez(N,Fo, Ao, W); [H,w] = freqz(b,1,512); figure(2); plot(w/pi, 20*log10(abs(H))); xlabel('\omega/\pi');ylabel(Gain, dB);title('N = 11');

FT = 2. Therefore, F p =

(

)

Using remezord, we get N = 9. The corresponding gain response is shown in Figure (e) below: N = 11 20

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(f)

However, specifications are not met in the passband with this filter, so we increase N to 11. The corresponding gain response is shown in Figure (f) above. The specifications are now met. Not for sale

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M10.24 From problem 10.37, ω p = 0.7π , ω s = 0.55π , δ p = 0.03808 , δ s = 0.0112 and we assume that

0.7π ⋅ FT = 0.7 and Fs = 0.55. 2π α p = −20 log10 1 − δ p = 0.3372 dB, α s = −20 log10 (δ s ) = 39.016 dB. Using ‘remezord’, we get an estimate of N = 20. However, using this order, the specifications are not met in the stopband, so we need to increase N up to 23 to meet the specifications. % Program #M10.24 Ft = 2;Fp = 0.7;Fs = 0.55; ds = 0.0112;dp = 0.03808; F = [Fs Fp];A = [0 1];DEV = [ds dp]; [N,Fo,Ao,W] = remezord(F,A,DEV,Ft); b = remez(N,Fo,Ao,W); [H,w] = freqz(b,1,512); figure(1); plot(w/pi, 20*log10(abs(H))); xlabel('\omega/\pi');ylabel('Gain, dB');title('N = 20'); N = 23; b = remez(N,Fo,Ao,W); [H,w] = freqz(b,1,512); figure(2); plot(w/pi,20*log10(abs(H))); xlabel('\omega/\pi');ylabel('Gain,dB ');title('N = 23');

FT = 2. Therefore, F p =

(

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Note: As odd order symmetric FIR filters must have a gain of zero at the Nyquist frequency. The order has been increased by one by remez. M10.25 From Problem 10.38, ω p1 = 0.55π , ω p 2 = 0.7π , ω s1 = 0.44 π , ω s 2 = 0.82π , δ p = 0.01 , δ s1 = 0.007 ,

(

)

δ s 2 = 0.002 . α p = −20 log10 1 − δ p = 0.087 dB,

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α s1 = −20 log10 (δ s1 ) = 43 dB, α s 2 = −20 log10 (δ s 2 ) = 54 dB. % Program #M10.25 Ft = 2;Fp1 = 0.55;Fp2 = 0.7;Fs1 = 0.44;Fs2 = 0.82; ds1 = 0.007;ds2 = 0.002;dp = 0.01; F = [Fs1 Fp1 Fp2 Fs2];A = [0 1 0];DEV = [ds1 dp ds2]; [N,Fo,Ao,W] = remezord(F,A,DEV,Ft); b = remez(N,Fo,Ao,W); [H, w] = freqz(b, 1, 512); figure(1); plot(w/pi,20*log10(abs(H)));grid; xlabel('\omega/\pi');ylabel('Gain, dB');title('N = 39'); axis([0 1 -80 10]); N = 41; b = remez(N, Fo, Ao, W); [H, w] = freqz(b, 1, 512); figure(2); plot(w/pi, 20*log10(abs(H))); xlabel('\omega/\pi');ylabel('Gain, dB');title('N = 41'); axis([0 1 -80 10]);

Using remezord, we estimate the filter length to be N = 39. However, the minimum stopband attenuation specifications are not met in both stopbands, so we increase N to 41 N = 41

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. M10.26 % Program #M10.26 b = remez(29, [0 1], [0 pi], 'differentiator'); [H, w] = freqz(b, 1, 512); plot(w/pi,abs(H));grid xlabel('\omega/\pi');ylabel('Magnitude'); axis([0 1 0 pi]);

The magnitude response of the differentiator is given below:

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M10.27 % Program #M10.27 f = [0.02 0.05 0.07 0.95 0.97 1]; m = [0 0 1 1 0 0]; wt = [1 60 1]; b = remez(30, f, m, wt, 'hilbert'); [H,w] = freqz(b,1,512); plot(w/pi,abs(H));grid xlabel('\omega/\pi');ylabel('Magnitude'); axis([0 1 0 1.2]);

The magnitude response of the Hilbert transformer is shown below:

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M10.28 ω p = 0.35π , ω s = 0.5π , R p = 1 dB, and Rs = 28 dB. % Program #M10_28 % Design of a minimum-phase lowpass FIR filter Wp = 0.35; Ws = 0.5; Rp = 1; Rs = 28; % Desired ripple values of minimum-phase filter dp = 1- 10^(-Rp/20); ds = 10^(-Rs/20); % Compute ripple values of prototype linear-phase filter Ds = (ds*ds)/(2 - ds*ds); Dp = (1 + Ds)*((dp + 1)*(dp + 1) - 1); % Estimate filter order [N,fpts,mag,wt] = remezord([Wp Ws], [1 0], [Dp Ds]); % Design the prototype linear-phase filter H(z) [b,err,res] = remez(N,fpts,mag,wt); K = N/2; b1 = b(1:K);

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% Design the linear-phase filter G(z) lenerr= res.error(length(res.error)); c = [b1 (b(K+1) + lenerr)) fliplr(b1)]/(1+Ds); zplane(c);title('Zeros of G(z)');pause c1 = c(K+1:N+1); [y, ssp, iter] = minphase(c1); zplane(y);title(‘Zeros of the minimum-phase filter’); pause [hh,w] = freqz(y,1, 512); % Plot the gain response of the minimum-phase filter plot(w/pi, 20*log10(abs(hh)));grid xlabel('\omega/\pi');ylabel('Gain, dB'); Zeros of G(z)

Zeros of the minimum-phase filter 1 Imaginary Part

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M10.29 % Program #M10.29 c = [2.4 6.76 26.15 68.43 186.83 326.51 565.53 678.95 805.24 678.95 565.53 326.51 186.83 68.43 26.15 6.76 2.4]; h = firminphase(c)

The coefficients of the minimum phase spectral factor are: h = 7.6730 8.5329 18.0722 12.5696 12.6822 4.8388 2.0784 0.5332 0.3128 M10.30 % Program #M10.30 [h,g]=ifir(6,'low',[.1 .15],[.001 .001]); [hh,w]=freqz(h,1,1024); hg=freqz(g,1,1024); h = hh.*hg; % Compounded response Fg = 20*log10(abs(hh)); Ig = 20*log10(abs(hg)); plot(w/pi,Fg,'-r',w/pi,Ig,'--b'); grid; axis([0 1 -90 5]);

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legend('F(z^6)','I(z)'); xlabel('\omega/\pi');title('Gain responses, in dB'); pause; plot(w/pi,20*log10(abs(h))); grid; axis([0 1 -90 5]); xlabel('\omega/\pi');title('Gain response, in dB'); gtext('H_{IFIR}(z)'); Gain response, in dB

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M10.31 % Program #M10.31 [h,g]=ifir(6,'high',[.9 .95],[.002 .004]); [hh,w]=freqz(h,1,1024); hg=freqz(g,1,1024); h = hh.*hg; % Compounded response Fg = 20*log10(abs(hh)); Ig = 20*log10(abs(hg)); plot(w/pi,Fg,'-r',w/pi,Ig,'--b'); grid; axis([0 1 -90 5]); legend('F(z^6)','I(z)'); xlabel('\omega/\pi');title('Gain responses, in dB'); pause; plot(w/pi,20*log10(abs(h))); grid; axis([0 1 -90 5]); xlabel('\omega/\pi');title('Gain response, in dB'); gtext('H_{IFIR}(z)'); Gain responses, in dB

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M10.32 % Problem #M10.32 InpN = ceil(2*pi/(0.15*pi)); % Plotting function % plots the result of using an equalizer of length InpN %function [N] = plotfunc(InpN); % Creating filters Wfilt = ones(1, InpN); Efilt = remezfunc(InpN, Wfilt); % Plot running sum filter response figure(1); Wfilt = Wfilt/sum(Wfilt); [hh, w] = freqz(Wfilt, 1, 512); plot(w/pi, 20*log10(abs(hh))); axis([0 1 -50 5]);grid; xlabel('\omega/\pi');ylabel('Gain, dB'); title('Prefilter H(z)'); % Plot equalizer filter response figure(2); [hw, w] = freqz(Efilt, 1, 512); plot(w/pi, 20*log10(abs(hw))); axis([0 1 -50 5]);grid; xlabel('\omega/\pi');ylabel('Gain, dB'); title('Equalizer F(z)'); % Plot cascade filter response figure(3); Cfilt = conv(Wfilt, Efilt); [hc, w] = freqz(Cfilt, 1, 512); plot(w/pi, 20*log10(abs(hc))); axis([0 1 -80 5]);grid; xlabel('\omega/\pi');ylabel('Gain, dB'); title('Cascade filter H(z)F(z)'); % Remez function using 1/P(z) as desired amplitude % and P(z) as weighting function [N] = remezfunc(Nin, Wfilt); % Nin : number of tuples in the remez equalizer filter % Wfilt : the prefilter a = [0:0.001:0.999]; % The accuracy of the computation w = a.*pi;wp = 0.05*pi;ws = 0.15*pi; i = 1;n = 1; for t = 1:(length(a)/2),

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if w(2*t) < wp pas(i) = w(2*t - 1); pas(i+1) = w(2*t); i = i+2; end if w(2*t-1) > ws sto(n) = w(2*t - 1); sto(n+1) = w(2*t); n = n+2; end end w = cat(2, pas, sto); bi = length(w)/2; for t1 = 1:bi, bw(t1) = (w(2*t1) + w(2*t1-1))/2; W(t1) = Weight(bw(t1), Wfilt, ws); end W = W/max(W); for t2 = 1:length(w), G(t2) = Hdr(w(t2), Wfilt, wp); end G = G/max(G); N = remez(Nin, w/pi, G, W); % Weighting function function[Wout] = Weight(w, Wfilt, ws); K = 22.8; L = length(Wfilt); Wtemp = 0; Wsum = 0; for k = 1:L, Wtemp = Wfilt(k)*exp((k-1)*i*w); Wsum = Wsum + Wtemp; end Wout = abs(Wsum); if w > ws, Wout = K*max(Wout); end % Desired function function [Wout] = Hdr(w, Wfilt, ws); if w <= ws, L = length(Wfilt); Wtemp = 0; Wsum = 0; for k = 1:L, Wtemp = Wfilt(k)*exp(i*(k-1)*w); Wsum = Wsum + Wtemp; end Wsum = abs(Wsum); Wout = 1/Wsum; Not for sale

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else Wout = 0; end Equalizer F(z) 0

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Chapter 11 11.1 Analysis yields w1[ n] = β1{x[ n − 1] + α 2 w2 [ n − 1]}, w2 [ n] = w1[ n] + w3 [ n], w3 [ n] = β1{α 2 w2 [ n − 1] + α 2 w 4 [ n − 1]}, w 4 [ n] = w3 [ n] + w5 [ n], w5 [ n] = β3 {α 2 w 4 [ n − 1] + α3 w5 [ n − 1]}, y[ n] = w1[ n] + α 0 x[ n]. In matrix form the above equations can be written as 0 0 0 ⎤ ⎡ w1[n − 1] ⎤ ⎡ β1 x[n − 1]⎤ ⎡ w1[n] ⎤ ⎡0 0 0 0 0 0 ⎤ ⎡ w1[n] ⎤ ⎡0 α1β1 0 ⎥ ⎢ w [n]⎥ ⎢1 0 1 0 0 0 ⎥ ⎢ w [n]⎥ ⎢0 0 0 0 0 0 0 ⎥⎥ ⎢⎢ w2 [n − 1]⎥⎥ ⎢⎢ ⎥ ⎢ 2 ⎥ ⎢ ⎥⎢ 2 ⎥ ⎢ ⎥ ⎢ w3[n]⎥ ⎢0 0 0 0 0 0 ⎥ ⎢ w3 [n]⎥ ⎢0 α 2 β 2 0 α 2 β 2 0 0 0 ⎥ ⎢ w3 [n − 1]⎥ ⎢ ⎥ ⎥⎢ ⎥+⎢ ⎢ ⎥=⎢ ⎥⎢ ⎥+⎢ 0 0 0 0 0 0 ⎥ ⎢ w4 [n − 1]⎥ ⎢ ⎥ ⎢ w4 [n]⎥ ⎢0 0 1 0 1 0 ⎥ ⎢ w4 [n]⎥ ⎢0 ⎥ ⎢ w5 [n]⎥ ⎢0 0 0 0 0 0 ⎥ ⎢ w5 [n]⎥ ⎢0 0 0 0 α 2 β 3 α 3 β3 0 ⎥ ⎢ w5 [n − 1]⎥ ⎢ ⎥ ⎥⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ 0 0 0 0 0 ⎦ ⎣ y[n − 1] ⎦ ⎢⎣ α 0 x[n] ⎥⎦ ⎣ y[n] ⎦ ⎣1 0 0 0 0 0 ⎦ ⎣ y[n] ⎦ ⎣0 ⎡0 0 0 0 0 0 ⎤ ⎢1 0 1 0 0 0 ⎥ Here the F matrix is given by: F = ⎢⎢0 0 0 0 0 0⎥⎥. 0 0 1 0 1 0 ⎢0 0 0 0 0 0 ⎥ ⎢⎣1 0 0 0 0 0⎥⎦ Since the F matrix contains nonzero entries above the main diagonal, the above set of equations are not computable. 11.2 A computable set of equations are given by: w1[ n] = β1{x[ n − 1] + α1w2 [ n − 1]}, w3[ n] = β1{α2 w2 [ n − 1] + α2 w4 [ n − 1]},

w5[ n] = β3 {α2 w4 [ n − 1] + α3w5[ n − 1]}, w2 [ n] = w1[ n] + w3[ n], y[ n] = w1[ n] + α0 x[ n] , w4 [ n] = w3[ n] + w5[ n] . In matrix form the above equations can be written as 0 0 0 ⎤ ⎡ w1[n − 1] ⎤ ⎡ β1 x[n − 1]⎤ ⎡ w1[n] ⎤ ⎡0 0 0 0 0 0 ⎤ ⎡ w1[n] ⎤ ⎡0 α1β1 0 ⎢ w [ n]⎥ ⎢ 0 0 0 0 0 0 ⎥ ⎢ w [ n]⎥ ⎢ 0 α β 0 α β ⎥ 0 0 ⎥⎥ ⎢⎢ w3 [n − 1]⎥⎥ ⎢⎢ 0 2 2 2 2 ⎢ 3 ⎥ ⎢ ⎥ ⎥⎢ 3 ⎥ ⎢ ⎢ w5 [n]⎥ ⎢0 0 0 0 0 0 ⎥ ⎢ w5 [n]⎥ ⎢0 ⎥ 0 0 α 2 β 3 α 3 β 3 0 ⎥ ⎢ w5 [n − 1]⎥ ⎢ 0 ⎥+⎢ ⎢ ⎥=⎢ ⎥+⎢ ⎥⎢ ⎥ ⎥⎢ 0 0 0 0 0 ⎥ ⎢ w2 [n − 1]⎥ ⎢ 0 ⎢ w2 [n]⎥ ⎢1 0 1 0 0 0 ⎥ ⎢ w2 [n]⎥ ⎢0 ⎥ ⎢ ⎥ ⎢ y[n] ⎥ ⎢1 0 0 0 0 0 ⎥ ⎢ y[n] ⎥ ⎢0 ⎥ ⎢ 0 0 0 0 0 y[n − 1] α 0 x[n] ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎥⎢ 0 0 0 0 0 ⎦ ⎢⎣ w4 [n − 1]⎥⎦ ⎣ 0 ⎢⎣ w4 [n]⎥⎦ ⎣0 0 1 0 1 0 ⎦ ⎢⎣ w4 [n]⎥⎦ ⎣0 ⎦

⎡0 ⎢0 Here the F matrix is given by: F = ⎢⎢0 1 ⎢1 ⎢⎣0

0 0 0 0 0 0

0 0 0 1 0 1

0 0 0 0 0 0

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0 0 0 0 0 1

0⎤ 0⎥ 0 ⎥. 0⎥ 0⎥ 0⎥⎦

388

Since the F matrix has no nonzero entries above the main diagonal, the new set of equations are computable. 11.3 Analysis yields w1[ n] = β1 {α1w2 [ n − 1] + α 2 w 4 [ n − 1] + α3 w6 [ n − 1]}, w2 [ n] = w1[ n] + x[ n], w3 [ n] = β2 {α2 w 4 [ n − 1] + α3 w6 [ n − 1]}, w 4 [ n] = w2 [ n] + w3 [ n], w 5 [ n] = β 3 {α3 w6 [ n − 1]}, w6 [ n] = w4 [ n] + w5[ n], y[ n] = α1w2 [ n] + α2 w 4 [ n] + α3 w6 [ n] + α 0 x[ n]. In matrix form the above set of equations are given by ⎡ w1[n] ⎤ ⎡0 0 ⎢ w [ n ]⎥ ⎢ 1 0 ⎢ 2 ⎥ ⎢ ⎢ w3 [n]⎥ ⎢0 0 ⎢ ⎥ ⎢ ⎢ w4 [n]⎥ = ⎢0 1 ⎢ w5 [n]⎥ ⎢0 0 ⎢ ⎥ ⎢ ⎢ w6 [n]⎥ ⎢0 0 ⎢ y[n] ⎥ ⎢0 a ⎣ ⎦ ⎣ 1

0 0

0 0

0 0

0 0

0 1 0

0 0 0

0 0 0

0 0 0

0 1 0 a2

0 ⎤ ⎡ w1[n] ⎤ ⎡0 β1α1 0 ⎥⎥ ⎢ w2 [ n]⎥ ⎢0 0 ⎢ ⎥ ⎢ 0 ⎥ ⎢ w3 [ n]⎥ ⎢0 0 ⎥⎢ ⎥ ⎢ 0 ⎥ ⎢ w4 [ n]⎥ + ⎢0 0 ⎥ ⎢ ⎥ ⎢ 0 w5 [n] 0 0 ⎥⎢ ⎥ ⎢ 0 ⎥ ⎢ w6 [ n]⎥ ⎢0 0 0 ⎥⎦ ⎢⎣ y[ n] ⎥⎦ ⎢⎣0 0

1 0 0 a3

⎡0 0 ⎢1 0 ⎢0 0 Here the F matrix is given by: F = ⎢0 1 ⎢0 0 ⎢0 0 ⎢⎣0 α1

β1α 2

0

β1α 3

0

0 0 0 β 2α 2

0 0 0 β 2α 3

0 0 0

0 0 0

0 0 0 β 3α 3 0 0

0

0

0

0

0 0 0 0 0 0 0 0 0 0 1 0 0 α3

0⎤ 0⎥ 0⎥ 0 ⎥. 0⎥ 0⎥ 0 ⎥⎦

0 0 0 0 0 0 1 0 0 0 0 1 0 α2

0 ⎤ ⎡ w1[n − 1] ⎤ ⎡ 0 ⎤ 0 ⎥⎥ ⎢⎢ w2 [n − 1]⎥⎥ ⎢⎢ x[ n] ⎥⎥ 0 ⎥ ⎢ w3 [n − 1]⎥ ⎢ 0 ⎥ ⎥ ⎥⎢ ⎥ ⎢ 0 ⎥ ⎢ w4 [n − 1]⎥ + ⎢ 0 ⎥ 0 ⎥ ⎢ w5 [n − 1]⎥ ⎢ 0 ⎥ ⎥ ⎥⎢ ⎥ ⎢ 0 ⎥ ⎢ w6 [ n − 1]⎥ ⎢ 0 ⎥ 0 ⎥⎦ ⎢⎣ y[ n − 1] ⎥⎦ ⎢⎣α 0 x[n]⎥⎦

Since the diagonal of the F matrix has all zeros, and no nonzero entries above the main diagonal, the new set of equations are computable. 11.4 α0 _1

x[n]

β1 z

_1

β2 α 1 z

1 β1 α 1 z

w1[n]

_1

β2 α 2 z

1

w2 [n]

β3 α 2 z

1 _1

_1

1

w5 [n]

w4 [n]

w3 [n]

y[n]

_1

1

β3 α 3 z

Reduced signal flow graph obtained by removing the branches going out of the input node and the delay branches is: 1 w1[n]

1 1

1 w2 [n]

w3 [n]

w4 [n]

w5 [n]

y[n]

1

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The node sets of the precedence graph are as follows: {N 1 } = {w 1 [n], w 3 [n], w 5 [ n] }, {N 2 } = {w 2 [ n], w 4 [ n], wy[ n] }. N1

N2

w1[n]

w2 [n]

w3 [n]

w4 [n]

w5 [n]

y[n]

In the above precedence graph, N 1 contains only outgoing nodes and N 2 contains only incoming branches. The structure has no delay free loops. A valid computational algorithm is:

w1[ n] = β1{x[ n − 1] + α1w2 [ n − 1]}, w3[ n] = β1{α2 w2 [ n − 1] + α2 w4 [ n − 1]}, w5[ n] = β3 {α2 w4 [ n − 1] + α3w5[ n − 1]}, w2 [ n] = w1[ n] + w3[ n], y[ n] = w1[ n] + α0 x[ n] , w4 [ n] = w3[ n] + w5[ n] .

11.5 Reduced signal flow graph is: 1

1 w1[n]

1

1

w2 [n]

w3 [n]

α3

1

w4 [n]

w5 [n]

α2

y[n]

w6 [n]

α1

The node sets of the precedence graph are as follows: {N 1 } = {w1[ n], w3 [ n], w5 [ n]}, {N 2 } = {w2 [ n]}, {N 3 } = {w4 [ n]}, {N 4 } = {w6 [ n]},{N 5 } = {y[ n]}. N1

w1[n] w3 [n]

N2

N3

N4

N5

w2 [n]

w5 [n]

y[n]

w4 [n]

w6 [n]

A valid computational algorithm is:

w1[ n] = β1{α1w2 [ n − 1] + α2 w 4 [ n − 1] + α3 w6 [ n − 1]}, w3 [ n] = β2 {α2 w 4 [ n − 1] + α3 w6 [ n − 1]}, w 5 [ n] = β 3 {α3 w6 [ n − 1]},

w2 [ n] = w1[ n] + x[ n], w4 [ n] = w2 [ n] + w3 [ n], w6 [ n] = w4 [ n] + w5[ n], y[ n] = α1w2 [ n] + α2 w 4 [ n] + α3 w6 [ n] + α 0 x[ n]. 11.6 (a) Analysis yields w1[ n] = w2 [ n] − k1w1[ n − 1], w2 [ n] = w3 [ n] − k 2 s 2 [ n − 1], w3 [ n] = x[ n] − k3 s3 [ n − 1], s 2 [ n] = k1w1[ n] + w1[ n − 1], s3 [ n] = k 2 w2 [ n] + s 2 [ n − 1], y[ n] = k3 w3 [ n] + s3 [ n − 1]. In matrix form we have

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⎡ w1 [ n ] ⎤ ⎡ 0 ⎤ ⎡ 0 1 0 0 0 0 ⎤ ⎡ w 1 [ n ] ⎤ ⎢ w 2 [ n]⎥ ⎢ 0 ⎥ ⎢ 0 0 1 0 0 0⎥ ⎢ w 2 [ n]⎥ ⎢ w [ n]⎥ ⎢ x[ n]⎥ ⎢ 0 0 0 0 0 0⎥ ⎢ w [ n]⎥ 3 ⎢ 3 ⎥=⎢ 0 0 0 0 0 ⎥ ⎢ s [ n] ⎥ ⎥ + ⎢k 2 ⎥ ⎢ ⎢ s 2 [ n] ⎥ ⎢ 0 ⎥ ⎢ 1 ⎥ ⎢ s 3 [ n] ⎥ ⎢ 0 ⎥ ⎢ 0 k 2 0 0 0 0 ⎥ ⎢ s 3 [ n] ⎥ 0 0⎦ ⎢ y[ n] ⎥ ⎢⎣ y[ n] ⎥⎦ ⎣ 0 ⎦ 1 3 4044 ⎣ 044044k2 ⎦ 43 ⎣ F 0 0 0 ⎤ ⎡ w1 [ n − 1] ⎤ ⎡ − k1 0 0 ⎢ 0 0 0 − k2 0 0 ⎥ ⎢ w 2 [ n − 1]⎥ ⎥ ⎢ ⎢ 0 − k3 0 ⎥ ⎢ w 3 [ n − 1]⎥ . 0 0 0 +⎢ ⎥ s [ n − 1] 0 0 0 0 0⎥ ⎢ 2 ⎥ ⎢ 1 s n − [ 1 ] 0 0 0 1 0 0 ⎥ ⎢ ⎢ ⎥ 3 0 0 0 0 1 0 ⎢ ⎣1444442444443⎦ ⎣ y[ n − 1] ⎥⎦ G As the diagonal elements of F-matrix are all zeros, there are no delay-free loops. However, the above set of equations are not computable as there are non-zero elements above the diagonal of F. (b) The reduced signal flow-graph representation of Figure P11.3 is shown below:

From the above flow-graph we observe that the set composed of nodes with only outgoing branches is N 1 = {w 3 [ n]}. The set of nodes with incoming branches from N 1 and outgoing branches is N 2 = {w2 [ n]}. The set of nodes with incoming branches from N 1 and N 2 and outgoing branches is N 3 = {w1[ n]}. Finally, the set of nodes with only incoming branches from N 1 , N 2 and N 3 , and no outgoing branches N 4 = {s 2 [ n], s3 [ n], y[ n]}. Therefore, one possible ordered set of equations that is

⎡ w3 [ n]⎤ ⎡ 0 0 0 0 0 0⎤ ⎡ w3 [ n]⎤ ⎢ w2 [ n]⎥ ⎢ 1 0 0 0 0 0⎥ ⎢ w2 [ n]⎥ ⎢ w [ n] ⎥ ⎢ 0 1 0 0 0 0 ⎥ ⎢ w [ n] ⎥ computable is given by ⎢ 1 ⎥ = ⎢ 0 0 k 0 0 0⎥ ⎢ 1 ⎥ 1 s [ n] ⎢ s 2 [ n] ⎥ ⎢ ⎥⎢ 2 ⎥ k 0 0 0 0 0 2 ⎢ s1[ n] ⎥ ⎢ ⎥ ⎢ s1[ n] ⎥ k3 0 0 0 0 0⎦ ⎢ y[ n] ⎥ ⎢⎣ y[ n] ⎥⎦ ⎣1 ⎦ 4444244443 ⎣ F − k3 0⎤ ⎡ w3 [ n − 1]⎤ 0 0 ⎡0 0 ⎢0 0 − k2 0 0 0⎥ ⎢ w2 [ n − 1]⎥ ⎢ ⎥ ⎢0 0 − k 0 0 0⎥ ⎢ w1[ n − 1] ⎥ . 3 +⎢ ⎥ s [ n − 1] 1 0 0 0⎥ ⎢ 2 ⎥ ⎢0 0 − s n [ 1 ] 0 0 0 1 0 0 ⎢ ⎥ ⎥ 1 ⎢ 0 0 1 0⎥⎦ ⎢⎣ y[ n − 1] ⎥⎦ ⎢⎣0 0 1444442444443 G Note: All elements on the diagonal and above diagonal of F are zeros.

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p + p1 z −1 + p2 z −2 11.7 H ( z ) = 0 . From Eq. (11.18) we get, 1 − 3 z − 1 + 1 .5 z − 2 0 0 ⎤ ⎡ 1 ⎤ ⎡ 3.2 ⎤ ⎡ p0 ⎤ ⎡3.2 ⎢ p ⎥ = ⎢5.6 3.2 ⎥ ⎢ − 3⎥ = ⎢ − 4.0⎥. Hence, H ( z ) = 3.2 − 4 z −1 − 5z −2 . 0 1 ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ 1 − 3z −1 + 1.5z − 2 ⎢⎣ p2 ⎥⎦ ⎢⎣7.0 5.6 3.2⎥⎦ ⎢⎣1.5 ⎥⎦ ⎢⎣ − 5.0 ⎥⎦

11.8 From Eqn (11.15),

⎡ d1 ⎤ ⎡4 ⎢ d ⎥ = − ⎢8 ⎣ ⎣ 2⎦ ⎡ p0 ⎤ ⎡ 2 ⎢ p ⎥ = ⎢− 2 ⎢ 1⎥ ⎢ ⎢⎣ p2 ⎥⎦ ⎢⎣ 4

− 2⎤ 4 ⎥⎦

0 2 −2

11.9 From Eqn. (11.15)

⎡ d 1 ⎤ ⎡− 8 ⎢ d ⎥ = ⎢ 12 ⎢ 2⎥ ⎢ ⎢⎣ d 3 ⎥⎦ ⎢⎣ 16 ⎡ p0 ⎤ ⎡ 2 ⎢p ⎥ ⎢ ⎢ 1 ⎥ = ⎢− 4 ⎢ p2 ⎥ ⎢ 8 ⎢ ⎥ ⎢ ⎣ p3 ⎦ ⎣ − 8 H ( z) =

0 0⎤ ⎡ p0 ⎤ ⎡ 2 ⎢ p ⎥ ⎢ −2 2 0 ⎥ 1 ⎢ 1⎥ ⎢ ⎥⎡ ⎤ ⎢ p2 ⎥ = ⎢ 4 −2 2 ⎥ ⎢ d1 ⎥ and from Eqn (11.20), ⎢ ⎥ ⎢ ⎥⎢ ⎥ − 0 8 4 2 ⎢ ⎥ ⎢ ⎥ ⎣⎢ d 2 ⎦⎥ ⎢⎣ 0 ⎥⎦ ⎢⎣ −12 8 4 ⎥⎦ −1

⎡ 8 ⎤ ⎡ − 0.25⎤ ⎢ − 12⎥ = ⎢ 3.5 ⎥. . Finally, from Eqn (11.21), ⎣ ⎦ ⎣ ⎦ 0⎤ ⎡ 1 ⎤ ⎡ 2 ⎤ 2 − 2.5z −1 + 11.5z −2 0⎥⎥ ⎢⎢ − 0.25⎥⎥ = ⎢⎢ − 2.5⎥⎥. Hence, H ( z) = . 1 − 0.25z −1 + 3.5z − 2 2⎥⎦ ⎢⎣ 3.5 ⎥⎦ ⎢⎣ 11.5 ⎥⎦

⎡ p0 ⎤ ⎡ 2 ⎢ ⎥ ⎢ ⎢ p1 ⎥ ⎢ −4 ⎢ p2 ⎥ ⎢ 8 ⎢ ⎥ ⎢ ⎢ p3 ⎥ = ⎢ −8 ⎢ 0 ⎥ ⎢12 ⎢ ⎥ ⎢ ⎢ 0 ⎥ ⎢16 ⎢ 0 ⎥ ⎢ −8 ⎣ ⎦ ⎣

8 −8 12

− 4⎤ 8 ⎥⎥ − 8 ⎥⎦

0 2 −4 8

0 0 2 −4

−1

0 0 0⎤ 2 0 0 ⎥⎥ ⎡ ⎤ −4 2 0 ⎥ ⎢⎢ 1 ⎥⎥ ⎥ 8 −4 2 ⎥ ⎢ d1 ⎥ . Next, from Eqn. (11.20), ⎢ ⎥ −8 8 −4 ⎥ ⎢ d 2 ⎥ ⎥ 12 −8 8 ⎥ ⎢⎣ d3 ⎥⎦ 16 12 −8⎥⎦

⎡ 12 ⎤ ⎡ 0.4 ⎤ ⎢ 16 ⎥ = ⎢ − 4.8⎥. Finally, from Eqn. (11.21), ⎥ ⎢ ⎥ ⎢ ⎢⎣ − 8⎥⎦ ⎢⎣ − 7.4⎥⎦ 0⎤ ⎡ 1 ⎤ ⎡ 2.0 ⎤ 0⎥⎥ ⎢⎢ 0.4 ⎥⎥ ⎢⎢ − 3.2 ⎥⎥ = . Hence, 0⎥ ⎢ − 4.8⎥ ⎢ − 3.2 ⎥ ⎥⎢ ⎥ ⎢ ⎥ 2⎦ ⎣ − 7.4 ⎦ ⎣ − 0.4⎦

2 − 3.2 z −1 − 3.2 z −2 − 0.4 z −3 1 + 0.4 z −1 − 4.8z − 2 − 7.4 z − 3

.

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11.10

⎡ p0 ⎤ ⎡ 2 0 0 ⎢ ⎥ ⎢ ⎢ p1 ⎥ = ⎢ −4 2 0 ⎢ p2 ⎥ ⎢ 4 −4 2 ⎢ ⎥ ⎢ ⎣ p3 ⎦ ⎣ −6 −6 −4

0⎤ ⎡ 1 ⎤ ⎡ 2 ⎤ 0 ⎥⎥ ⎢⎢ −0.6 ⎥⎥ ⎢⎢ −5.2⎥⎥ = . Hence, P ( z ) = 2 − 5.2 z −1 + 6.8 z −2 − 5.6 z −3 . 0 ⎥ ⎢ 0.2 ⎥ ⎢ 6.8 ⎥ ⎥⎢ ⎥ ⎢ ⎥ 2 ⎦ ⎣ 1.8 ⎦ ⎣ −5.6⎦

⎡ p0 ⎤ ⎡ 2 0 0 0 ⎢ ⎥ ⎢ ⎢ p1 ⎥ ⎢ −2 2 0 0 11.11 ⎢ p2 ⎥ = ⎢ 4 −2 2 0 ⎢ ⎥ ⎢ ⎢ p3 ⎥ ⎢ 6 4 −2 2 ⎢⎣ p4 ⎥⎦ ⎢⎣ −8 6 4 −2

0⎤ ⎡ 1 ⎤ ⎡ 2 ⎤ 0 ⎥⎥ ⎢⎢ 2 3 ⎥⎥ ⎢⎢ − 2 3 ⎥⎥ 0⎥ ⎢ 0 ⎥ = ⎢ 2 2 3 ⎥ . ⎥⎢ ⎥ ⎢ ⎥ 0 ⎥ ⎢ −2 ⎥ ⎢ 4 2 3 ⎥ 2 ⎥⎦ ⎢⎣ 5 3 ⎥⎦ ⎢⎣ −3 1 3 ⎥⎦

{ pi } = {2, − 2 3 , 2 2 3 , 4 2 3 , −3 13}

11.12 The k –th sample of an N –point DFT is given by X[ k ] =

N −1

∑ n = 0 x[n]WNnk .

Thus, the

computation of X [k ] requires N complex multiplications and N − 1 complex additions. Now, each complex multiplication, in turn, requires 4 real multiplications and 2 real additions. Likewise, each complex addition requires 2 real additions. As a result, the N complex multiplications needed to compute X [k ] require a total of 4 N real multiplications and a total of 2 N − 2 real additions. Therefore, each sample of the N –point DFT involves 4 N real multiplications and 4 N − 2 real additions. Hence, the computation of all DFT samples thus requires 4N 2 real multiplications and (4 N − 2) N real additions. 11.13 Let the two complex numbers be α = a + j b and β = c + j d. Thus, αβ = ( a + j b)(c + j d ) = ( ac − bd ) + j ( ad + bc ), which requires 4 real multiplications and 2 real additions. Consider the product ( a + b)(c + d ), ac, and bd, which require 3 real multiplications and 2 real additions. The imaginary part of αβ can be formed from ( a + b)(c + d ) − ac − bd = ad + bc, which now requires 2 real additions. Likewise, the real part of αβ can be formed from ac − bd requiring an additional real additions. Hence, the complex multiplication can be computed using 3 real multiplications and 5 real additions. 11.14 Recall, WN = e − j 2π / N = c + j s, where c = cos( 2π / N ) and s = − sin( 2π / N ). Thus, c 2 + s 2 = 1. Now, Ψ r +1 = W N ⋅Ψ r = (c + j s )(Re{Ψ r } + j Im{Ψ r })

= (c ⋅ Re{Ψr } − s ⋅ Im{Ψr }) + j (c ⋅ Im{Ψr } + s ⋅ Re{Ψr }). Thus, Re{Ψr +1} = c ⋅ Re{Ψr } − s ⋅ Im{Ψr } and Im{Ψr +1} = c ⋅ Im{Ψr } + s ⋅ Re{Ψr }. Figure P11.4 with internal node label is shown below. Its analysis yields U

Re{ψr } c _1 ____ s

Re{ψr+1} s

Im{ψ r }

c _1 ____ s Im{ψr+1}

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c −1 Im{Ψr }, (2): Im{Ψr +1} = Im{Ψr } + sU , and s c −1 (3): Re{Ψr + 1} = U + Im{Ψr + 1}. Substituting Eqs. (1) in Eq. (2) we get s ⎛ c −1 ⎞ (4): Im{Ψr + 1} = Im{Ψr } + s⎜ Im{Ψr } + Re{Ψr }⎟ = c ⋅ Im{Ψr } + s ⋅ Re{Ψr }. Next, ⎝ s ⎠ substituting Eqs. (1) and (4) in Eq. (3) we get (5): c −1 (5): Re{Ψr + 1} = (c ⋅ Im{Ψr } + s ⋅ Re{Ψr }) + c − 1 Im{Ψr } + Re{Ψr } s s 2 2 c −1 −s = Im{Ψr } + c ⋅ Re{Ψr } = ⋅ Im{Ψr } + c ⋅ Re{Ψr } = c ⋅ Re{Ψr } − s ⋅ Im c ⋅ Re{Ψr }. s s It thus follows that the structure of Figure P11.4 implements the multiplication of a complex signal with the twiddle factor using only 3 real multiplications.

(1): U = Re{Ψr } +

. In the case of multiplication by complex twiddle factor W N−1 = c − js, we have Ψ r +1 = W N−1 ⋅Ψ r = (c − j s )(Re{Ψ r } + j Im{Ψ r })

= (c ⋅ Re{Ψr } + s ⋅ Im{Ψr }) + j (c ⋅ Im{Ψr } − s ⋅ Re{Ψr }). Thus, here Re{Ψr +1} = c ⋅ Re{Ψr } + s ⋅ Im{Ψr } and Im{Ψr +1} = c ⋅ Im{Ψr } − s ⋅ Re{Ψr }. It follows then that the real and imaginary parts are simply obtained by reversing the sign of s of the real and imaginary parts derived in the case of multiplication by W N . As a result, the corresponding structure is obtained by cascading an inverter to the real multipliers in Figure P11.4 as indicated below: Re{ψr }

Im{ψ r }

U _1 c _1 ____ s

s

_1

_1

c _1 ____ s

Re{ψr+1}

Im{ψr+1}

kFT , where N = # of bins, and FT is the sampling N ⎢ f N⎥ frequency. Inverting we have k ( f ) = ⎢ ⎥. Therefore, the absolute difference from one of ⎣ FT ⎦ the given four tones (150 Hz, 375 Hz, 620 Hz, and 850 Hz) to the center of its bin is given by F ⎢ f N⎥ dist ( N , f ) = f − T ⎢ ⎥ . It follows from this equation that the distance goes to zero if N ⎣ FT ⎦

11.15 The center frequency bin k : f c (k ) =

FT ⎢ f N ⎥ fN is an integer. ⎢ ⎥ = f or N ⎣ FT ⎦ FT

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fN The total distance is reduced to zero if i is an integer for i = 1,K , 4. The minimum value FT of N for which this true is 500. However, the total distance can be small, but nonzero, for significantly smaller values of N . 11.16 H k ( z ) =

1 1 − WN− k z −1

where V ( z ) =

. Hence, Y ( z ) =

1 1 − WN− k z −1

{

= 1,

WN−1 ,

1 − WN− k z −1

=

1 + z −N / 2 1 − WN− k z −1

. Or, in other words, y[ n] = v[ n] + v[ n −

Consider k = 1 : Then V ( z ) = v[ n] = WN− n µ[ n] =

X ( z)

1 1 − WN−1 z −1

= V ( z ) + z − N / 2V ( z ), N ]. 2

. This implies,

N ⎧ ⎫ − ( + 1) ⎪ ⎪ − 1 − 2 − N / 2 2 , WN , L⎬ ⎨1, WN , WN , L, WN ⎪ ⎪ ⎩ ⎭

WN− 2 ,

son on. Thus, v[ n −

L, − 1, N ]= 2

{0,

− WN−1 ,

N − ( + 1) − / 2 N L , since WN = −1, WN 2 = −WN−1 , and

}

}

0, 0, L, 1, WN−1 , WN− 2 , L . Hence,

N ⎧ ⎫ − ( − 1) ⎪ ⎪ − 1 − 2 2 y[ n] = v[ n] + v[ n − , 0, 0, 0, L⎬. ⎨1, WN , WN , L, WN ⎪ ⎪ ⎩ ⎭ 1 N Now, consider k = : V ( z ) = . This implies, 2 1 − WN− N / 2 z −1 N ]= 2

v[ n] = WN− ( N / 2 )n µ[ n] =

N ] = 0, 0, 0, L, 2 N N N − ⋅ − N / 2 WN 2 2 = (−1) , WN 2

Thus, v[ n − Now,

{

N N N N N ⎧ ⎫ − − ⋅ − ( + 1) ⋅ ⎪ ⎪ N − 2 2 2 2 2 , WN , L⎬. ⎨1, WN , WN , L, WN ⎪ ⎪ ⎩ ⎭

}

0, 1, WN− N / 2 , WN− N , L .

N N − ( + 1) N − 2 = (−1) N / 2 (−1), etc. Hence, = −1, WN = 1, WN 2

⎧ ⎫ N +1 ⎪ ⎪ N / 2 y[ n] = ⎨1, − 1, 1, − 1, L, 1 + (−1) , 1 + (−1) 2 , L, L⎬. Therefore, if N / 2 N n= N / 2 ⎪ ⎪ n = +1 2 ⎩ ⎭ ⎫ ⎧ ⎪ ⎪ is even, y[ n] = ⎨1, − 1, 1, − 1, L, − 2, 2, 2, L⎬, and if N / 2 is odd, n = N / 2 n = N +1 ⎪ ⎪ 2 ⎭ ⎩

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⎫ ⎧ ⎪ ⎪ y[ n] = ⎨1, − 1, 1, − 1, L, 0, 0, 0, L⎬, n = N / 2 n = N +1 ⎪ ⎪ 2 ⎭ ⎩ 11.17

11.18

11.19 X[ k ] =

N −1

∑ x[n]WNnk

n=0

=

N −1 r1

∑

n=0

nkr1

x[ nr1 ]WN

+

N −1 r1

∑

n=0

nkr1 + k

x[ nr1 + 1]WN

+K+

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N −1 r1

∑

n=0

nkr1 + (r1 −1)k

x[ nr1 + r 1−1]WN

396

⎛ N −1 ⎞ ⎟ r1 −1⎜ r1 ⎜ nkr ⎟ = x[nr 1+i ]W N 1 ⎟ W Nki . ⎜ ⎟ i =0 ⎜ n=0 ⎜ ( N / r1 ) − point DFT ⎟ ⎝ ⎠ Thus, if the ( N / r1 ) -point DFT has been calculated, we need at the first stage an additional (r1 − 1) multiplications to compute one sample of the N –point DFT X [k ] and, as a result, additional (r1 − 1) N multiplications are required to compute all N samples of the N –point DFT. Decomposing it further, it follows that additional (r2 − 1) N multiplications are needed at the second stage, and so on. Therefore, the total number of multiply (add) operations ν ri − ν ⎞⎟ N . = (r1 − 1) N + (r2 − 1) N + K + (rν − 1) N = ⎛⎜ ⎠ ⎝ i =1

∑∑

∑

11.20 An examination of the flow-graph of the 8-point DIT FFT algorithm shown in Figure 11.24 reveals that in the first stage all twiddle factors are WN0 = 1 and in the second stage the

twiddle factors are either WN0 = 1 or W NN / 4 = j. Hence, there are no twiddle factors with nonunity magnitudes in the first two stages. In all succeeding stages, one of the twiddle factors is WN0 = 1 and another one is W NN / 4 = j. The number of twiddle factors with nonzero magnitudes in the i -th stage, i ≥ 3, is 2 i −1 − 2. Hence, the total number of twiddle factors with nonzero magnitudes in an N-point radix-2 FFT algorithm is ⎛ ν ⎞ R (ν) = ⎜⎜ 2 i −1 ⎟⎟ − 2(ν − 2), ν ≥ 3, where N = 2 ν. ⎜ ⎟ ⎝ i =3 ⎠

∑

11.21 Direct computation of M samples of an N -point DFT requires M 2 multiplications,

whereas, the Radix-2 FFT algorithm requires

N 2

log 2 N multiplications. In order for a N-

point radix-2 FFT algorithm to be computationally more efficient than a direct computation ⎢ N ⎥ of M samples of an N -point DFT, the following inequality must hold: M > ⎢ log 2 N ⎥. ⎣ 2 ⎦ a) N = 512, M = 49, b) N = 1024, M = 72, c) N = 2048, M = 107 11.22 X ( z) = X 0 ( z 3 ) + z −1 X1 ( z 3 ) + z −2 X 2 ( z 3 ). Thus, the N –point DFT can be expressed as X[ k ] = X 0 [〈 k 〉 N / 3 ] + WNk X1[〈 k 〉 N / 3 ] + WN2 k X 2 [〈 k 〉 N / 3 ]. Hence, the structural

interpretation of the first stage of the radix-3 DFT is as indicated below:

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x[n]

3 z

+

X[k]

x 1[n] N_ -point X 1[< k >N/3 ] WNk + 3 3 DFT

z

3

11.23 X[ k ] =

x 0[n] N _ -point X 0 [< k >N/3 ] 3 DFT

2k x 2[n] N_ X 2[< k >N/3 ] WN -point 3 DFT

∑ x[n]W9nk = (x[0]W90⋅ k + x[3]W93k + x[6]W96k ) 8

n=0

(

) (

+ x[1]W9k + x[ 4]W94 k + x[7]W97k + x[2]W92 k + x[5]W95k + x[8]W98k

)

( ) ( ) + (x[2]W90 ⋅ k + x[5]W93k + x[8]W96 k )W92 k = G0 [〈 k 〉 3 ] + G1[〈 k 〉 3 ]W9k + G2 [〈 k 〉 3 ]W92 k , where

= x[0]W90 ⋅ k + x[3]W93k + x[6]W96 k + x[1]W90 ⋅ k + x[ 4]W93k + x[7]W96 k W9k

G0 [〈 k 〉 3 ] = x[0]W30 ⋅ k + x[3]W3k + x[6]W32 k , G1[〈 k 〉 3 ] = x[1]W30 ⋅ k + x[ 4]W3k + x[7]W32 k , G2 [〈 k 〉 3 ] = x[2]W30 ⋅ k + x[5]W3k + x[8]W32 k , are 3-point DFTs. A flow-graph representation of this radix-3 DIT FFT computation scheme is shown below,

where the twiddle factors for computing the DFT samples are indicated below for a typical DFT sample:

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In the above diagram, the 3-point DFT computation is carried out as indicated below:

11.24 X[ k ] =

∑ x[n]W15nk = (x[0]W150⋅ k + x[3]W153k + x[6]W156k + x[9]W159k + x[12]W1512k ) 14

n=0 k + x[ 4]W 4 k + x[ 7]W 7k + x[10]W 10 k + x[13]W 13k + x[1]W15 15 15 15 15 2 k 5 k 8 k 11 k 14 k + x[2]W15 + x[5]W15 + x[8]W15 + x[11]W15 + x[14]W15 k + G [ 〈 k 〉 ]W 2 k , where = G0 [〈 k 〉 5 ] + G1[〈 k 〉 5 ]W15 2 5 15 0 ⋅ k k G0 [〈 k 〉 5 ] = x[0]W5 + x[3]W5 + x[6]W52 k + x[9]W53k + x[12]W54 k , G1[〈 k 〉 5 ] = x[1]W50 ⋅ k + x[ 4]W5k + x[7]W52 k + x[10]W53k + x[13]W54 k , and G2 [〈 k 〉 5 ] = x[2]W50 ⋅ k + x[5]W5k + x[8]W52 k + x[11]W53k + x[14]W54 k .

( (

) )

A flow-graph representation of this mixed-radix DIT FFT computation scheme is shown below:

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11.25

11.26 Now, by definition, q[ n] = Im{X [ n]} + j Re{X [ n]}. Its N -point DFT is given by

Q[ k ] =

N −1

∑ q[n]WNnk .

Thus,

n=0

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(1): Re{Q[ k ]} = (2): Im{Q[ k ]} =

N −1

⎡

⎛ 2πmk ⎞ ⎛ 2πmk ⎞ ⎤ ⎟ + Re{X[ n] sin⎜ ⎟⎥ , N ⎠ ⎝ N ⎠⎦

∑ ⎢⎣Im{X[n] cos⎜⎝

m=0 N −1

⎡

⎛ 2πmk ⎞ ⎛ 2πmk ⎞ ⎤ ⎟ + Re{X[ n] cos⎜ ⎟⎥ . N ⎠ ⎝ N ⎠⎦

∑ ⎢⎣− Im{X[n] sin⎜⎝

m=0

N −1

∑

1 X[ m]WN− mk . Hence, From the definition of the inverse DFT we observe x[ k ] = N m=0 (3): Re{x[ k ]} =

N −1

⎡ 1 ⎛ 2πmk ⎞ ⎛ 2πmk ⎞ ⎤ Re{X[ m]} cos⎜ ⎟ − Im{X[ m]}sin⎜ ⎟ ⎥, ⎢ N m=0 ⎣ ⎝ N ⎠ ⎝ N ⎠⎦

∑

N −1 ⎡ 1 ⎛ 2πmk ⎞ ⎛ 2πmk ⎞ ⎤ Im{X[ m]} cos⎜ ⎟ + Re{X[ m]}sin⎜ ⎟ ⎥. ⎢ N m=0 ⎣ ⎝ N ⎠ ⎝ N ⎠⎦ 1 Comparing Eqs. (2) and (3) we get Re{x[ n]} = ⋅Im{Q[ k ]} k = n , and comparing Eqs. (1) N 1 and (4) we get Im{x[ n]} = ⋅Re{Q[ k ]} k = n . N

∑

(4): Im{x[ k ]} =

if n = 0, ⎧ X[0], Therefore, 11.27 r[ n] = X[〈− n〉 N ] = ⎨ − X N n n ≠ 0. [ ], if ⎩ R[ k ] =

N −1

∑

r[ n]WNnk n=0 N −1

= X [ 0] +

= r[0] +

x[ n] =

∑

r[ n]WNnk n =1 N −1

= X [ 0] +

N −1

∑ X[ N − n]WNnk

n =1 N −1 X[ n]WN− nk = X[ n]WN− nk n=0 n =1

∑ X[n]WN( N − n)k = X[0] + ∑

n =1 1 N

N −1

∑

= N ⋅ x[ k ]. Thus,

⋅R[ k ] k = n .

11.28 Let y[n] denote the result of convolving a length-L sequence x[n] with a length-N sequence h[n]. The length of y[n] is then L + N – 1. Here L = 16 and N = 9, hence length of y[n] is 24.

Method #1: Direct linear convolution - For a length-L sequence x[n] and a length-N sequence h[n], # of real mult. = 2

N

9

n =1

n=1

∑ n + N (L − N − 1) = 2∑ n + 9(16 − 9 − 1) = 135 .

Method # 2: Linear convolution via circular convolution - Since y[n] is of length 24, to get the correct result we need to pad both sequences with zeros to increase their lengths to 24 before carrying out the circular convolution. # of real mult. = 24 ×24 = 576.

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Method #3: Linear convolution via radix-2 FFT - The process involves computing the 16point FFT G[k ] of the length-16 complex sequence g[ n] = x[ n] + j he [ n], where he [n] is obtained by zero-padding h[n] to length 16. Then, the 16-point DFTs, X [k ] and H e [k ], of x[n] and he [n] , respectively, are recovered from G[k ] . Finally, the IDFT of the product Y [ k ] = X[ k ] ⋅ H e [ k ] yields y[n]. Now, the first stage of the 16-point radix-2 FFT requires 0 complex multiplications, the second stage requires 0 complex multiplications, the third stage requires 4 complex multiplications, and the last stage requires 6 complex multiplications, resulting in a total of 10 complex multiplications. # of complex mult. to implement G[k ] = 10 # of complex mult. to recover X [k ] and H e [k ] from G[k ] = 0 # of complex mult. to form Y [k ] = 16 # of complex mult. to form the IDFT of Y [k ] = 10 Hence, the total number of complex mult. = 36 A direct implementation of a complex multiplication requires 4 real multiplications resulting in a total of 4× 36 = 144 real multiplications for Method #3. However, if a complex multiply can be implemented using 3 real multiplies (see Problem 11.13), in which case Method #3 requires a total of 3× 36 = 108 real multiplications. 11.29 Let y[n] denote the result of convolving a length-L sequence x[n] with a length-N sequence h[n]. The length of y[n] is then L + N – 1. Here, L = 16 and N = 10, hence length of y[n] is 25.

Method #1: Direct linear convolution - For a length-L sequence x[n] and a length-N sequence h[n], # of real mult. = 2

N

10

n=1

n=1

∑ n + N (L − N − 1) = 2∑ n + 10(16 − 10 − 1) = 160 .

Method # 2: Linear convolution via circular convolution - Since y[n] is of length 24, to get the correct result we need to pad both sequences with zeros to increase their lengths to 24 before carrying out the circular convolution. # of real mult. = 25 ×25 = 625. Method #3: Linear convolution via radix-2 FFT - The process involves computing the 16point FFT G[k ] of the length-16 complex sequence g[ n] = x[ n] + j he [ n], where he [n] is obtained by zero-padding h[n] to length 16. . Then, the 16-point DFTs, X [k ] and H e [k ], of x[n] and he [n] , respectively, are recovered from G[k ] . Finally, the IDFT of the product Y [ k ] = X[ k ] ⋅ H e [ k ] yields y[n]. Now, the first stage of the 16-point radix-2 FFT requires 0 complex multiplications, the second stage requires 0 complex multiplications, the third stage requires 4 complex

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multiplications, and the last stage requires 6 complex multiplications resulting in a total of 10 complex multiplications. # of complex mult. to implement G[k ] = 10 # of complex mult. to recover X [k ] and H e [k ] from G[k ] = 0 # of complex mult. to form Y [k ] = 16 # of complex mult. to form the IDFT of Y [k ] = 10 Hence, the total number of complex mult. = 36 A direct implementation of a complex multiplication requires 4 real multiplications resulting in a total of 4× 36 = 144 real multiplications for Method #3. However, if a complex multiply can be implemented using 3 real multiplies (see Problem 11.13), in which case Method #3 requires a total of 3× 36 = 108 real multiplications. 11.30 (a) Since the impulse response of the filter is of length 72, the transform length N should be greater than 72. If L denotes the number of input samples used for convolution, then L = N – 71. So for every L samples of the input sequence, an N -point DFT is computed and multiplied with an N -point DFT of the impulse response sequence h[n] (which needs to be computed only once), and finally an N -point inverse of the product sequence is evaluated. Hence, the total number R M of complex multiplications required (assuming N is a powerN 2048 ⎤ ( N log 2 N + N ) + log 2 N . of-2) is given by R M = ⎡ 2 ⎢ N − 71 ⎥

It should be noted that in developing the above expression, multiplications due to twiddle factors of values ±1 and ± j have not been excluded. The values of R M for different values of N are as follows: 1) For N = 128, R M = 37312, 2) For N = 256, R M = 28672, 3) For N = 512, R M = 27904, 4) For N = 1024, R M = 38912. Hence, N = 512 is the appropriate choice for the transform length requiring 27904 complex multiplications or equivalently, 27904 × 3 = 83712 real multiplications. Since the first stage of the FFT calculation process requires only multiplications by ±1, the total number of complex multiplications for N = 128 is actually N N 2048 ⎤ RM = ⎡ ( N log 2 N + N ) + log 2 N − = 27648, or equivalently, 27648× 3 = 82944 ⎢ N − 71 ⎥ 2 2 real multiplications. (b) For direct convolution, # of real multiplications = N

2

∑

n=1

n + N ( L − N − 1) = 2

72

∑ n + 72(2048 − 72 − 1) = 147456 .

n =1

11.31 (a)

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⎡1 ⎢ ⎢0 ⎢0 ⎢ ⎢0 V8 = ⎢ ⎢1 ⎢0 ⎢ ⎢0 ⎢0 ⎣ ⎡1 ⎢ ⎢1 ⎢0 ⎢ ⎢0 V2 = ⎢ ⎢0 ⎢0 ⎢ ⎢0 ⎢0 ⎣

0 0 0 W80 1 0 0

0

0

0

W81

0

0 1 0

0

0

W82

0 0 1

0

0

0

0 0 0

W84

0

0

1 0 0

0

W85

0

0 1 0

0

0

W86

0 0 1

0

0

0

0 ⎤ ⎥ 0 ⎥ 0 ⎥ ⎥ W83 ⎥ ⎥, 0 ⎥ 0 ⎥ ⎥ 0 ⎥ W87 ⎥⎦

W80

0

0

0

0

0

W84

0

0

0

0

0

0

1 W80

0

0

0

0

W84

0

0

0 0 0

1

0

0

0

1

0

0

0

1

W80 W84

0

0

0

0

0

1

0

0

0

0

0

1

⎡1 ⎢ ⎢0 ⎢1 ⎢ ⎢0 V4 = ⎢ ⎢0 ⎢0 ⎢ ⎢0 ⎢0 ⎣

0 ⎤ ⎥ 0 ⎥ 0 ⎥ ⎥ 0 ⎥ E= ⎥, 0 ⎥ 0 ⎥ ⎥ W80 ⎥ W84 ⎥⎦

⎡1 ⎢0 ⎢0 ⎢0 ⎢0 ⎢0 ⎢0 ⎢⎣0

0 W80

0

0 0

0

1

0

W82

0 0

0

0

W80

0

0 0

0

1

0

W82

0 0

0

0

0

0

1 0 W80

0

0

0

0 1

0

0

0

1 0 W80

0

0

0

0 1

0 0 0 0 1 0 0 0

0 0 1 0 0 0 0 0

0 0 0 0 0 0 1 0

0 1 0 0 0 0 0 0

0 0 0 0 0 1 0 0

0 0 0 1 0 0 0 0

0 0

0 ⎤ ⎥ 0 ⎥ 0 ⎥ ⎥ 0 ⎥ ⎥, 0 ⎥ W82 ⎥ ⎥ 0 ⎥ W82 ⎥⎦

0⎤ 0⎥ 0⎥ 0 ⎥. 0⎥ 0⎥ 0⎥ 1⎥⎦

As can be seen from the above, multiplication by each matrix Vk , k = 1, 2, 3, requires at most 8 complex multiplications. (b) The transpose of the matrices given in Part (a) are as follows: ⎡ 1 0 0 0 1 0 0 0 ⎤ ⎡ 1 0 1 0 0 0 0 0 ⎤ ⎢ 0 1 0 0 0 1 0 0 ⎥ ⎢ 0 1 0 1 0 0 0 0 ⎥ ⎢ 0 0 1 0 0 0 1 0 ⎥ ⎢W 0 0 W 0 0 0 0 0 0 ⎥ 8 ⎢ 0 0 0 1 0 0 0 1 ⎥ ⎢ 8 ⎥ 2 0 W2 0 W 0 0 0 0 ⎥ 0 0 ⎥ ⎢W ⎢ t t 8 8 0 0 0 0 0 0 W V8 = ⎢ 8 8 ⎥, V4 = ⎢ 0 0 0 0 1 0 1 0 ⎥, 1 1 ⎢ 0 W8 0 0 0 W8 0 0 ⎥ ⎢ 0 0 0 0 0 1 0 1 ⎥ ⎢ 0 0 W2 0 0 0 W2 0 ⎥ ⎢ 0 0 0 0 W0 0 W0 0 ⎥ 8 8 8 8 ⎥ ⎢ ⎢ ⎥ 2 2⎥ ⎢ 0 0 0 W3 0 0 0 W3⎥ ⎢ W W 0 0 0 0 0 0 8 8 ⎦ ⎣ 8 8 ⎦ ⎣ ⎡ 1 1 0 0 0 0 0 0 ⎤ ⎢W 0 W 4 0 0 0 0 0 0 ⎥ 8 ⎢ 8 ⎥ 0 0 1 1 0 0 0 0 ⎥ ⎢ 0 0 W80 W84 0 0 0 0 ⎥ E t = E = V2t = ⎢⎢ , 0 0 0 0 1 1 0 0 ⎥ ⎢ 0 0 0 0 W0 W4 0 0 ⎥ 8 8 ⎢ ⎥ ⎢ 0 0 0 0 0 0 1 1 ⎥ ⎢⎣ 0 0 0 0 0 0 W80 W84 ⎥⎦

⎡1 ⎢0 ⎢0 ⎢0 ⎢0 ⎢0 ⎢0 ⎢⎣0

0 0 0 0 1 0 0 0

0 0 1 0 0 0 0 0

0 0 0 0 0 0 1 0

0 1 0 0 0 0 0 0

0 0 0 0 0 1 0 0

0 0 0 1 0 0 0 0

0⎤ 0⎥ 0⎥ 0 ⎥. 0⎥ 0⎥ 0⎥ 1⎥⎦

It is easy to show that the flow-graph representation of D8 = E t V2t V4t V8t is precisely the 8point DIF FFT algorithm of Figure 11.28.

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11.32 X [2l] =

n+

N 2

N −1

∑

n=0

x[ n]WN2 ln =

N −1 2

N −1

n=0

n= N / 2

∑

∑ x[n]WN2ln , 0 ≤ l ≤ 2 − 1.

x[ n]WN2ln +

N

Replacing n by

in the right-most sum we get

X [ 2l ] =

N −1 2

N −1 2

n=0

n=0

N −1 2

∑ x[n]WN2ln + ∑ x[n + 2 ]WN2ln = ∑ ⎛⎜⎝ x[n] + x[n + 2 ]⎞⎟⎠WNln/ 2 , 0 ≤ l ≤ 2 − 1. N

N

N

n=0

Likewise,

X[ 4l + 1] =

N −1 4

∑

n=0

where 0 ≤ l ≤

x[ n]WN( 4l + 1)n

N 4

+

∑

+

n=

x[ n]WN( 4l + 1)n

3N 4

N 4

+

N 2

N −1

∑ x[n]WN(4l +1)n ,

n=

3N 4

N 2

in the third

in the fourth sum, we get

N −1 4

N −1 4

n=0 N −1 4

n=0

∑ x[n]WN4lnWNn + ∑ x[n + 4 ]WN4lnWNn WNlN WNN / 4 N

N −1 4

∑ x[n + 2 ]WN4lnWNn WN2lN WNN / 2 + ∑ x[n + N

N −1 4

n=0 WNN / 2

N

N

n=0

N −1 4

N −1 4

N −1 4

n=0

n=0

3N ]WN4lnWNn WN3lN WN3 N / 4 . 4

= −1, and WN3 N / 4 = + j. Therefore,

∑ ⎧⎨⎩⎛⎜⎝ x[n] − x[n + 2 ]⎞⎟⎠ − j⎛⎜⎝ x[n + 4 ] − x[n +

Similarly, X [ 4l + 3] =

N 3 N ⎞ ⎫ n ln ]⎟ ⎬W W , 0 ≤ l ≤ − 1. 4 ⎠⎭ N N / 4 4

∑ x[n]WN(4l + 3)n + ∑ x[n + 4 ]WN(4l + 3)nWN(4l + 3)N / 4 N

N −1 4

∑ x[n + 2 ]WN(4l + 3)nWN(4l + 3)N / 2 + ∑ x[n + N

n=0

=

x[ n]WN( 4l + 1)n

in the second sum, n by n +

Now, WNlN = WN2 lN = WN3lN = 1, WNN / 4 = − j,

+

∑

n=

n=0

X [ 4l + 1] =

+

3N −1 4

N 4

− 1. Replacing n by n +

sum, and n by n +

X [ 4l + 1] =

N −1 2

n=0

N −1 4

N −1 4

n=0

n=0

3N ]WN( 4l + 3)nWN( 4l + 3)3 N / 4 4

∑ x[n]WN4lnWN3n + ∑ x[n + 4 ]WN4lnWN3nWNlN WN3N / 4 N

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+

N −1 4

N −1 4

∑ x[n + 2 ]WN4lnWN3nWN2lN WN6 N / 4 + ∑ x[n + 4 ]WN4lnWN3nWN3lN WN9 N / 4 N

n=0

=

N −1 4

n=0

∑ ⎧⎨⎩⎛⎜⎝ x[n] − x[n + 2 ]⎞⎟⎠ + j⎛⎜⎝ x[n + 4 ] − x[n +

n=0

N

N

N

N 3 N ⎞ ⎫ 3 n ln ]⎟ ⎬W WN / 4 , 0 ≤ l ≤ − 1. 4 4 ⎠⎭ N

The butterfly here is as shown below which is seen to require two complex multiplications.

11.33 From the flow-graph of the 8-point split-radix FFT algorithm given below it can be seen that the total number of complex multiplications required is 2. On the other hand, the total number of complex multiplications required for a standard DIF FFT algorithm is also 2.

11.34 If multiplications by ± j, ± 1 are ignored, the flow-graph shown below requires 8 complex multiplications = 24 real multiplications. A radix-2 DIF 16-point FFT algorithm, on the other hand requires 10 complex multiplications = 30 real multiplications.

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0 ≤ n1 ≤ 3, 11.35 (a) N = 12. Choose N1 = 4 and N 2 = 3. Thus, n = n1 + 4 n2 , ⎧⎨ and 0 ⎩ ≤ n 2 ≤ 2, ⎧ 0 ≤ k1 ≤ 3, The corresponding index mappings are indicated below: k = 3k1 + k 2 , ⎨ ⎩0 ≤ k 2 ≤ 2. k n 1 3 2 1 3 2 k2 1 0 n2 1 0 X[0] X[3] X[6] X[9] x[0] x[1] x[2] x[3] 0 0 X[1] X[4] X[7] X[10] x[4] x[5] x[6] x[7] 1 1 2 X[2] X[5] X[8] X[11] 2 x[8] x[9] x[10] x[11]

0 ≤ n1 ≤ 2, N = 15. Choose N1 = 3 and N 2 = 5. Thus, n = n1 + 3n2 , ⎧⎨ and ⎩0 ≤ n2 ≤ 4, ⎧ 0 ≤ k1 ≤ 2, The corresponding index mappings are indicated below: k = 5k1 + k 2 , ⎨ ⎩0 ≤ k 2 ≤ 4.

(b)

n n2 1 0 x[0] 0 x[3] 1 x[6] 2 3 x[9] 4

1 x[1] x[4] x[7]

2 x[2] x[5] x[8]

x[10] x[11]

x[12] x[13] x[14]

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k k2 1 0 X[0] 0 X[1] 1 X[2] 2 3 X[3] 4

1

2

X[5] X[10] X[6] X[11] X[7] X[12] X[8] X[13]

X[4] X[9] X[14]

407

0 ≤ n1 ≤ 6, (c) N = 21. Choose N1 = 7 and N 2 = 3. Thus, n = n1 + 7n2 , ⎧⎨ and ⎩0 ≤ n2 ≤ 2, ⎧ 0 ≤ k1 ≤ 6, The corresponding index mappings are indicated below: k = 3k1 + k 2 , ⎨ ⎩0 ≤ k 2 ≤ 2. n n2 1 0 x[0] 0 1 x[7] 2 x[14]

1 x[1] x[8]

2

3

4

5

x[2] x[3] x[4] x[5] x[9] x[10] x[11] x[12] x[15] x[16] x[17] x[18] x[19]

6 x[6] x[13] x[20]

k 4 5 6 1 3 2 k2 1 0 X[18] X[3] X[15] X[6] X[9] X[12] X[0] 0 1 X[1] X[4] X[7] X[10] X[13] X[16] X[19] 2 X[2] X[5] X[8] X[11] X[14] X[17] X[20]

0 ≤ n1 ≤ 6, (d) N = 35. Choose N1 = 7 and N 2 = 5. Thus, n = n1 + 7n2 , ⎧⎨ and 0 ⎩ ≤ n 2 ≤ 4, ⎧ 0 ≤ k1 ≤ 6, The corresponding index mappings are indicated below: k = 5k1 + k 2 , ⎨ ⎩0 ≤ k 2 ≤ 4. n 1 2 n2 1 0 x[0] x[1] x[2] 0 x[7] x[8] x[9] 1 2 x[14] x[15] x[16] 3 x[21] x[22] x[23] 4

3

4

5

x[3] x[4] x[5] x[10] x[11] x[12] x[17] x[18] x[19]

x[13] x[20]

x[24] x[25] x[26]

x[27]

x[28] x[29] x[30] x[31] x[32] x[33]

x[34]

k 4 5 1 3 2 k2 1 0 X[0] X[5] X[10] X[15] X[20] X[25] 0 1 X[1] X[6] X[11] X[16] X[21] X[26] 2 X[2] X[7] X[12] X[17] X[22] X[27] 3 X[3] X[8] X[13] X[18] X[23] X[28] 4

6 x[6]

6 X[30] X[31] X[32] X[33]

X[4] X[9] X[14] X[19] X[24] X[29] X[34]

11.36 (a) N = 12. Choose N1 = 4 and N 2 = 3. A = 3, B = 4, C = 3〈3 −1 〉 4 = 9, ⎧ 0 ≤ n1 ≤ 3, D = 4〈 4 −1 〉 3 = 4. Thus, n = 〈3n1 + 4 n2 〉 12 , ⎨ k = 〈 9k1 + 4 k 2 〉 12 , ⎩0 ≤ n2 ≤ 2, The corresponding index mappings are indicated below: n 1 3 2 n2 1 0 x[0] x[3] x[6] x[9] 0 x[4] x[7] x[10] x[1] 1 2 x[8] x[11] x[2] x[5]

⎧ 0 ≤ k1 ≤ 3, ⎨0 ≤ k ≤ 2. 2 ⎩

k 1 3 2 k2 1 0 X[0] X[9] X[6] X[3] 0 X[4] X[1] X[10] X[7] 1 2 X[8] X[5] X[2] X[11]

(b) N = 15. Choose N1 = 3 and N 2 = 5. A = 5, B = 3, C = 5〈5 −1 〉 3 = 10, D = 3〈3 −1 〉 5 = 6.

Not for sale

408

0 ≤ n1 ≤ 2, Thus, n = 〈 5n1 + 3n2 〉 15 , ⎧⎨ k = 〈10 k1 + 6 k 2 〉 15 , 0 ⎩ ≤ n 2 ≤ 4, The corresponding index mappings are indicated below: n n2 1 0 x[0] 0 x[3] 1 x[6] 2 3 x[9] 4

1 2 x[5] x[10] x[8] x[13] x[11] x[1] x[14] x[4]

x[12] x[2]

x[7]

⎧ 0 ≤ k1 ≤ 2, ⎨0 ≤ k ≤ 4. 2 ⎩

k 1 2 k2 1 0 X[0] X[10] X[5] 0 X[6] X[1] X[11] 1 X[12] X[7] X[2] 2 3 X[3] X[13] X[8] 4 X[9] X[4] X[14]

(c) N = 21. Choose N1 = 7 and N 2 = 3. A = 3, B = 7, C = 3〈3 −1 〉 4 = 3 × 5 = 15, ⎧ 0 ≤ n1 ≤ 6, D = 7〈 7 −1 〉 3 = 7 × 1 = 7. Thus, n = 〈3n1 + 7n2 〉 21 , ⎨ ⎩0 ≤ n2 ≤ 2, ⎧ 0 ≤ k1 ≤ 6, The corresponding index mappings are indicated k = 〈15k1 + 7k 2 〉 21 , ⎨ ⎩0 ≤ k 2 ≤ 2. below: n 4 5 6 1 3 2 n2 1 0 x[18] x[0] x[3] x[6] x[9] x[12] x[15] 0 1 x[7] x[10] x[13] x[16] x[19] x[1] x[4] 2 x[14] x[17] x[20] x[2] x[5] x[8] x[11]

k 4 5 6 1 3 2 k2 1 0 X[0] X[15] X[9] X[3] X[18] X[12] X[6] 0 1 X[7] X[1] X[16] X[10] X[4] X[19] X[13] 2 X[14] X[8] X[2] X[17] X[11] X[5] X[20]

(d) N = 35. Choose N1 = 7 and N 2 = 5. A = 5, B = 7, C = 5〈 5 −1 〉 7 = 5 × 3 = 15, ⎧ 0 ≤ n1 ≤ 6, D = 7〈7 −1 〉 5 = 7 × 3 = 21. Thus, n = 〈 5n1 + 7n2 〉 35 , ⎨ ⎩0 ≤ n2 ≤ 4, ⎧ 0 ≤ k1 ≤ 6, The corresponding index mappings are indicated k = 〈15k1 + 21k 2 〉 35 , ⎨ ⎩0 ≤ k 2 ≤ 4. below: n 1 3 2 n2 1 0 x[0] x[5] x[10] x[15] 0 x[7] x[12] x[17] x[22] 1 2 x[14] x[19] x[24] x[29] 3 x[21] x[26] x[31] x[1] 4

4

5

x[20] x[25] x[27] x[32] x[34] x[4]

6 x[30] x[2] x[9]

x[6] x[11]

x[16]

x[28] x[33] x[3] x[8] x[13] x[18]

x[23]

k 4 5 6 1 3 2 k2 1 0 X[0] X[15] X[30] X[10] X[25] X[5] X[20] 0 X[21] X[1] X[16] X[31] X[11] X[26] X[6] 1 2 X[7] X[22] X[2] X[17] X[32] X[12] X[27] 3 X[28] X[8] X[23] X[3] X[18] X[33] X[13] 4

X[14] X[29] X[9] X[24] X[4] X[19] X[34]

Not for sale

409

11.37 N = 12. Choose N1 = 4 and N 2 = 3. A = 3, B = 4, C = 3〈3 −1 〉 4 = 9, D = 4〈 4 −1 〉 3 = 4. 0 ≤ n1 ≤ 3, ⎧ 0 ≤ k1 ≤ 3, Thus, n = 〈3n1 + 4 n2 〉 12 , ⎧⎨ k = 〈 9k1 + 4 k 2 〉 12 , ⎨ ⎩0 ≤ n2 ≤ 2, ⎩0 ≤ k 2 ≤ 2. The corresponding index mappings are indicated below: n 1 3 2 n2 1 0 x[0] x[3] x[6] x[9] 0 x[4] x[7] x[10] x[1] 1 2 x[8] x[11] x[2] x[5]

k 1 3 2 k2 1 0 X[0] X[9] X[6] X[3] 0 X[4] X[1] X[10] X[7] 1 2 X[8] X[5] X[2] X[11]

0 ≤ n1 ≤ 3, Alternately, k = 〈 9n1 + 4 n2 〉 12 , ⎧⎨ k = 〈3k1 + 4 k 2 〉 12 , ⎩0 ≤ n2 ≤ 2, The corresponding index mappings are indicated below: n 1 3 2 n2 1 0 x[0] x[9] x[6] x[3] 0 x[4] x[1] x[10] x[7] 1 2 x[8] x[5] x[2] x[11]

⎧ 0 ≤ k1 ≤ 3, ⎨0 ≤ k ≤ 2. 2 ⎩

k 1 3 2 k2 1 0 X[0] X[3] X[6] X[9] 0 X[4] X[7] X[10] X[1] 1 2 X[8] X[11] X[2] X[5]

Hence, X [2 k ] = Y [2 k ], and X [2k + 1] = Y [〈6 + (2k + 1)〉12 ], k = 0,1,K,5. 11.38 (a) N = 6. Choose N1 = 2 and N 2 = 3. A = 3, B = 2, C = 3〈3 −1 〉 2 = 3, D = 2〈 2 −1 〉 3 = 4. 0 ≤ n1 ≤ 1, ⎧ 0 ≤ k1 ≤ 1, Thus, n = 〈3n1 + 2 n2 〉 6 , ⎧⎨ k = 〈3k1 + 4 k 2 〉 6 , ⎨ ⎩0 ≤ n2 ≤ 2, ⎩0 ≤ k 2 ≤ 2. The corresponding index mappings are indicated below: n 1 n2 1 0 x[0] x[3] 0 x[2] x[5] 1 2 x[4] x[1] x[0] x[3] x[2] x[5] x[4] x[1]

n 1 k2 1 0 G[0,0] G[1,0] 0 G[0,1] G[1,1] 1 2 G[0,2] G[1,2] G[0,0]

2-pt DFT G[1,0]

k 1 k2 1 0 X[0] X[3] 0 X[4] X[1] 1 2 X[2] X[5] X[0]

3-pt DFT

X[4] X[2]

G[0,1]

2-pt DFT G[1,1] G[0,2]

2-pt DFT G[1,2]

Not for sale

X[3]

3-pt DFT

X[1] X[5]

410

(b) N = 10. Choose N1 = 2 and N 2 = 5. A = 5, B = 2, C = 5〈5 −1 〉 2 = 5, D = 2〈 2 −1 〉 5 = 6. 0 ≤ n1 ≤ 4, ⎧ 0 ≤ k1 ≤ 4, Thus, n = 〈 5n1 + 2 n2 〉 10 , ⎧⎨ k = 〈 5k1 + 6 k 2 〉 10 , ⎨ n ≤ ≤ , 0 2 2 ⎩ ⎩0 ≤ k 2 ≤ 2. The corresponding index mappings are indicated below: n 1 n2 1 0 x[0] x[5] 0 x[2] x[7] 1 x[4] x[9] 2 x[6] x[1] 3 x[8] x[3] 4 x[0] x[5] x[2] x[7]

x[4] x[9] x[6]

x[1] x[8]

x[3]

n 1 k2 1 0 G[0,0] G[1,0] 0 G[0,1] G[1,1] 1 G[0,2] G[1,2] 2 G[0,3] G[1,3] 3 G[1,4] G[0,4] 4

k 1 k2 1 0 X[0] X[5] 0 X[6] X[1] 1 2 X[2] X[7] 3 X[8] X[3] 4 X[4] X[9]

G[0,0]

X[0]

2-pt DFT G[1,0] G[0,1]

2-pt DFT G[1,1]

X[6]

5-pt DFT

X[2] X[8]

X[4]

G[0,2]

2-pt DFT G[1,2] X[5]

G[0,3]

2-pt DFT G[1,3] G[0,4]

2-pt DFT G[1,4]

X[1]

5-pt DFT

X[7]

X[3] X[9]

(c) N = 12. Choose N1 = 3 and N 2 = 4. A = 4, B = 3, C = 4〈 4 −1 〉 3 = 4, D = 3〈3 −1 〉 4 = 3. 0 ≤ n1 ≤ 2, ⎧ 0 ≤ k1 ≤ 2, Thus, n = 〈 4 n1 + 3n2 〉 12 , ⎧⎨ k = 〈 4 k1 + 3k 2 〉 12 , ⎨ n 0 ≤ ≤ 3 , 2 ⎩ ⎩0 ≤ k 2 ≤ 3. The corresponding index mappings are indicated below: n n2 1 0 x[0] 0 x[3] 1 2 x[6] 3 x[9]

1 2 x[4] x[8] x[7] x[11] x[10] x[2] x[1] x[5]

k 1 2 n2 1 0 G[0,0] G[1,0] G[2,0] 0 G[0,1] G[1,1] G[2,1] 1 2 G[0,2] G[1,2] G[2,2] 3 G[0,3] G[1,3] G[2,3]

Not for sale

k k2 1 0 X[0] 0 X[3] 1 2 X[6] 3 X[9]

1

2

X[4] X[7] X[10] X[1]

X[8] X[11] X[2] X[5]

411

G[0,0]

x[0] x[4] x[8]

3-pt DFT G[2,0]

x[11]

G[2,2]

x[9]

G[0,3]

x[1] x[5]

4-pt DFT

3-pt G[1,2]

x[2]

X[6]

X[4]

G[0,2]

DFT

X[3]

X[9]

3-pt G[1,1] DFT G[2,1]

x[6] x[10]

4-pt DFT

G[0,1]

x[3] x[7]

X[0]

G[1,0]

X[7] X[10]

X[1] X[8]

4-pt DFT

G[1,3]

3-pt DFT G[2,3]

X[11]

X[2] X[5]

(d) N = 15. Choose N1 = 5 and N 2 = 3. A = 3, B = 5, C = 3〈3 −1 〉 5 = 10, ⎧ 0 ≤ n1 ≤ 4, ⎧0 ≤ k1 ≤ 4, D = 5〈 5 −1 〉 3 = 6. Thus, n = 〈 5n1 + 3n2 〉 10 , ⎨ k = 〈10 k1 + 6 k 2 〉 10 , ⎨ n 0 ≤ ≤ 2 , 2 ⎩ ⎩0 ≤ k 2 ≤ 2. The corresponding index mappings are indicated below: n n 1 2 3 2 4 1 3 k2 1 0 n2 1 0 G[0,0] G[1,0] G[2,0] G[3,0] 0 x[0] x[3] x[6] x[9] x[12] 0 1 x[5] x[8] x[11] x[14] x[2] 1 G[0,1] G[1,1] G[2,1] G[3,1] x[1] x[10] x[13] x[4] G[0,2] G[1,2] G[2,2] G[3,2] x[7] 2 2

k k2 1 0 X[0] 0 1 X[10] X[5] 2

1

2

3

4

X[6] X[12] X[3] X[9] X[1] X[7] X[13] X[4] X[11] X[2] X[8] X[14]

Not for sale

412

G[0,0]

x[0]

G[0,1]

x[3]

5-pt G[0,2] DFT G[0,3]

x[6] x[9] x[12]

G[0,4]

G[1,0] G[1,1]

x[5] x[8]

5-pt G[1,2] DFT G[1,3]

x[11] x[14] x[2]

X[0]

3-pt DFT

X[5]

3-pt DFT

x[10] x[13]

5-pt G[2,2] DFT G[2,3]

x[1] x[4]

G[2,4]

x[7]

X[6] X[1] X[11] X[12]

3-pt DFT

X[7] X[2]

G[1,4] G[2,0] G[2,1]

X[10]

X[3]

3-pt DFT

X[13]

X[8]

3-pt DFT

X[9] X[4] X[14]

11.39 Note that 1536 = 256 × 6. Now an N -point DFT , with N divisible by 6, can be computed

as follows: X[ k ] =

N −1

∑ x[n]WNnk = X 0 [〈 k〉 N / 6 ] + WNk X1[〈 k〉 N / 6 ] + WN2k X 2 [〈 k〉 N / 6 ]

n=0 k 3 + WN X 3 [〈 k 〉 N / 6 ] + WN4 k X 4 [〈 k 〉 N / 6 ] + WN5k X 5 [〈 k 〉 N / 6 ], where N −1 6 X l [〈 k 〉 N / 6 ] = x[6r + l]WNrk/ 6 , 0 ≤ l ≤ 5. For N = 1536, we thus get r =0 k X [〈 k 〉 2k 3k X[ k ] = X 0 [〈 k 〉 256 ] + W1536 1 256 ] + W1536 X 2 [ 〈 k 〉 256 ] + W1536 X 3 [ 〈 k 〉 256 ] 5111 rk , 4 5 k k + W1536 X 4 [〈 k 〉 256 ] + W1536 X 5 [〈 k 〉 256 ], where X l [〈 k 〉 256 ] = x[6r + l]W256 r =0

∑

∑

x000[n] 256-point X000 [ k 256 ] 6 DFT

x[n] z

6 z z

x [n] 256-point X [ k 256 ] 010 6 010 DFT

6 z 6 z

x001[n] 256-point X001 [ k 256 ] DFT

x011[n] 256-point X [ 011 k 256 ] DFT x100[n]

] X [ 256-point 100 k 256 DFT

x101[n] 256-point X101 [ k 256 ] 6 DFT

Not for sale

0 ≤ l ≤ 5.

X[k] k W768 k W768

k W1536

k W768 k W768

413

Now an N-point FFT algorithm requires complex additions. Hence, an multiplications and

N 6

N 6

N 2

log 2 N complex multiplications and N log 2 N

-point FFT algorithm requires

N log 2 ( N / 6) 12

complex

log 2 ( N / 6) complex additions. In addition, we need 5N complex

multiplications and 5N complex additions to compute the N-point DFT X[k]. Hence, for N = 1536, the evaluation of X [k ] using six 256-point FFT modules requires

= 128 × log 2 (256) + 5 × 1536 = 8704 complex multiplications and

N 6

N log 2 ( N / 6) + 5N 12

log 2 ( N / 6) + 5N

= 256 × log 2 (256) + 5 × 1536 = 9728 complex additions. It should be noted that a direct computation of the 3072-point DFT would require 6431296 complex multiplications and 2357760 complex additions. 11.40 (a) # of zero-valued samples to be added is 512 – 498 = 14. (b) Direct computation of a 512-point DFT of a length-498 sequence requires (498)2 = 248004 complex multiplications and 497 ×498 = 247506 complex additions. (c) A 512-point Cooley-Tukey type FFT algorithm requires 256 × log 2 (512) = 2304 complex multiplications and 512 × log 2 (512) = 46308 complex additions. 11.41 z l = α l . Hence, AoVo− l e jθ o e − jlφ o = α l . Since α is real, we have Ao = 1, Vo = 1 / α, θ o = 0, φ o = 0. 11.42 (a) Y ( z ) = H ( z ) X ( z ) or y[ 0] + y[1]z −1 + y[ 2]z −2 = (h[ 0] + h[1]z −1 )( x[ 0] + x[1]z −1 ). Now, Y ( z 0 ) = Y (−1) = y[0] − y[1] + y[2] = H (−1) X (−1) = (h[0] − h[1])( x[0] − x[1]),

Y ( z1 ) = Y (∞) = y[0] = H (∞) X (∞ ) = h[0]x[0], Y ( z 2 ) = Y (1) = y[0] + y[1] + y[2] = H (1) X (1) = {h[0] + h[1])( x[0] + x{1}). From Eqs. (6.114) and (6.115), we arrive at I ( z) I (z) I ( z) Y (z0 ) + 1 Y ( z1 ) + 2 Y ( z 2 ), where I 0( z ) = (1 − z1z −1 )(1 − z 2 z −1 ), Y ( z) = 0 I 0 (z0 ) I1 ( z1 ) I 2 (z2 )

I 1( z ) = (1 − z 0 z −1 )(1 − z 2 z −1 ), I 2( z ) = (1 − z 0 z −1 )(1 − z1z −1 ). Therefore, I 0 ( z) (1 − z1 z −1 )(1 − z 2 z −1 ) 1 = = − z −1 (1 − z −1 ), 2 I 0 ( z 0 ) (1 − z1 z 0−1 )(1 − z 2 z 0−1 ) I1 ( z) (1 − z 0 z −1 )(1 − z 2 z −1 ) = = (1 − z − 2 ), and − 1 − 1 I1 ( z1 ) (1 − z 0 z1 )(1 − z 2 z1 ) I 2 ( z) (1 − z 0 z −1 )(1 − z1 z −1 ) 1 −1 = = z (1 + z −1 ). Hence, 1 1 − − I 2 ( z 0 ) (1 − z 0 z 2 )(1 − z1 z 2 ) 2

Not for sale

414

Y ( z ) = − z −1 (1 − z −1 )Y ( z 0 ) + (1 − z − 2 )Y ( z1 ) + z −1 (1 + z −1 )Y ( z 2 ) 1 2

1 2

1 1 1 1 = Y ( z1 ) + ⎛⎜ − Y ( z 0 ) + Y ( z 2 ) ⎞⎟ z −1 + ⎛⎜ Y ( z 0 ) − Y ( z1 ) + Y ( z 2 ) ⎞⎟ z − 2 2 2 ⎝ 2 ⎠ ⎝2 ⎠ 1 1 = h[0]x[0] + ⎛⎜ − (h[0] − h[1])( x[0] − x[1]) + (h[0] + h[1])( x[0] + x[1]) ⎞⎟ z −1 2 ⎝ 2 ⎠ 1 1 + ⎛⎜ (h[0] − h[1])( x[0] − x[1]) − h[0]x[0] + (h[0] + h[1])( x[0] + x[1]) ⎞⎟ z − 2 2 ⎝2 ⎠ − 1 − 2 = h[ 0] x[ 0] + (h[ 0] x[1] + h[1] x[ 0])z + h[1] x[1]z .

Ignoring the multiplications by

1 , 2

computation of the coefficients of Y ( z ) require the values

of Y ( z 0 ), Y ( z1 ), and Y ( z2 ) , which can be evaluated using only 3 multiplications. (b) Y ( z ) = H ( z ) X ( z ) or

(

)(

)

y[ 0] + y[1]z −1 + y[2]z − 2 + y[3]z − 3 + y[ 4]z − 4 = h[ 0] + h[1]z −1 + h[ 2]z − 2 x[ 0] + x[1]z −1 + x[ 2]z − 2 . Now, 1 2

Y ( z 0 ) = Y (− ) = (h[0] − 2h[1] + 4h[2])( x[0] − 2 x[1] + 4 x[2]),

Y ( z1 ) = Y (−1) = (h[0] − h[1] + h[2])( x[0] − x[1] + x[2]), Y ( z2 ) = Y (∞) = h[0]x[0], Y ( z3 ) = Y (1) = (h[0] + h[1] + h[2])( x[0] + x[1] + x[2]), 1 2

Y ( z 4 ) = Y ( ) = (h[0] + 2h[1] + 4h[2])( x[0] + 2 x[1] + 4 x[2]). From Eqs. (6.114) and (6.115), we arrive at I ( z) I ( z) I ( z) I (z) I ( z) Y (z) = 0 Y (z0 ) + 1 Y ( z1 ) + 2 Y (z2 ) + 3 Y ( z3 ) + 4 Y ( z 4 ), where I 0 (z0 ) I1 ( z1 ) I 2 (z2 ) I 3 ( z3 ) I 4 (z 4 )

I 0 ( z) = (1 − z1z −1 )(1 − z 2 z −1 )(1 − z3 z −1 )(1 − z 4 z −1 ), I 1( z) = (1 − z0 z −1 )(1 − z 2 z −1 )(1 − z3 z −1 )(1 − z 4 z −1 ), I 2 ( z) = (1 − z0 z −1 )(1 − z1z −1 )(1 − z3 z −1 )(1 − z 4 z −1 ), I 3( z) = (1 − z 0 z −1 )(1 − z1z −1 )(1 − z 2 z −1 )(1 − z 4 z −1 ), I 4( z) = (1 − z 0 z −1 )(1 − z1z −1 )(1 − z 2 z −1 )(1 − z3 z −1 ). Therefore, I 0 ( z) (1 − z1z −1 )(1 − z2 z −1 )(1 − z3 z −1 )(1 − z4 z −1 ) 1 −1 1 = = z (1 − z −1 )(1 − z − 2 ), − 1 − − − 1 1 1 2 I 0 ( z0 ) (1 − z1z0 )(1 − z2 z0 )(1 − z3 z0 )(1 − z4 z0 ) 12

I1 ( z) (1 − z 0 z −1 )(1 − z 2 z −1 )(1 − z3 z −1 )(1 − z 4 z −1 ) 1 2 = = − z −1 (1 − z −1 )(1 − z − 2 ), 4 3 I1 ( z1 ) (1 − z 0 z1−1 )(1 − z 2 z1−1 )(1 − z3 z1−1 )(1 − z 4 z1−1 ) I 2 ( z) (1 − z 0 z −1 )(1 − z1 z −1 )(1 − z3 z −1 )(1 − z 4 z −1 ) 1 = = (1 − z − 2 )(1 − z − 2 ), 4 I 2 ( z 2 ) (1 − z 0 z 2−1 )(1 − z1 z 2−1 )(1 − z3 z 2−1 )(1 − z 4 z 2−1 ) I 3 ( z) (1 − z 0 z −1 )(1 − z1 z −1 )(1 − z 2 z −1 )(1 − z 4 z −1 ) 2 −1 1 = = z (1 + z −1 )(1 − z − 2 ), 1 1 1 1 − − − − 4 3 I 3 ( z3 ) (1 − z 0 z3 )(1 − z1 z3 )(1 − z 2 z3 )(1 − z 4 z3 ) Not for sale

415

(1 − z 0 z −1 )(1 − z1 z −1 )(1 − z 2 z −1 )(1 − z3 z −1 ) I 4 ( z) 1 1 = = − z −1 (1 + z −1 )(1 − z − 2 ). Hence, 1 1 1 1 − − − − 2 12 I 4 ( z 4 ) (1 − z 0 z 4 )(1 − z1 z 4 )(1 − z 2 z 4 )(1 − z3 z 4 ) Y ( z) =

1 −1 (z 12

− z − 2 − z − 3 + z − 4 )Y ( z 0 ) − ( z −1 − z − 2 − z − 3 + z − 4 )Y ( z1 ) 1 2

+ (1 −

1 4 5 −2 z 4

1 4

2 3

+ z − 4 )Y ( z 2 ) + ( z −1 + z − 2 − 1 4

2 3

1 4 1 −3 z 4

− z − 4 )Y ( z3 ) 1 4

1 1 −1 1 −1 + z − z − 3 − z − 4 )Y ( z 4 ) (z 2 2 12 2 2 1 Y ( z1 ) + Y ( z3 ) − Y ( z 4 ) ⎞⎟ z −1 3 3 12 ⎠

−

1 = Y ( z 2 ) + ⎛⎜ Y ( z 0 ) − 12 ⎝ 1 2 5 2 1 + ⎛⎜ − Y ( z 0 ) + Y ( z1 ) − Y ( z 2 ) + Y ( z3 ) − Y ( z 4 ) ⎞⎟ z − 2 3 4 3 24 ⎝ 24 ⎠ 1 1 1 1 + ⎛⎜ − Y ( z 0 ) + Y ( z1 ) − Y ( z3 ) + Y ( z 4 ) ⎞⎟ z − 3 12 6 6 12 ⎝ ⎠ 1 1 1 1 1 + ⎛⎜ Y ( z 0 ) − Y ( z1 ) + Y ( z 2 ) − Y ( z3 ) + Y ( z 4 ) ⎞⎟ z − 4 . 6 4 6 24 ⎝ 24 ⎠ Substituting the expressions for Y ( z 0 ), Y ( z1 ), Y ( z 2 ), Y ( z 3 ), and Y ( z 4 ), in the above equation, we then arrive at the expressions for the coefficients {y[ n]} in terms of the coefficients {h[ n]} and {x[ n]}. Thus, y[0] = Y ( z 2 ) = h[0]x[0], 1 2 2 1 Y ( z 0 ) − Y ( z1 ) + Y ( z3 ) − Y ( z 4 ) = h[0]x[1] + h[1]x[0], 12 3 3 12 1 2 5 2 1 y[2] = − Y ( z 0 ) + Y ( z1 ) − Y ( z 2 ) + Y ( z3 ) − Y ( z 4 ) = h[0]x[2] + h[1]x[1] + h[2]x[0], 24 3 4 3 24 1 1 1 1 y[3] = − Y ( z 0 ) + Y ( z1 ) − Y ( z3 ) + Y ( z 4 ) = h[1]x[2] + h[2]x[1], 12 6 6 12 1 1 1 1 1 y[ 4] = Y ( z 0 ) − Y ( z1 ) + Y ( z 2 ) − Y ( z3 ) + Y ( z 4 ) = h[2]x[2]. 24 6 4 6 24 1 2 5 1 1 1 Hence, ignoring the multiplications by , , , , , and , computation of the 12 3 4 4 6 24

y[1] =

coefficients of Y ( z ) require the values of Y ( z 0 ), Y ( z1 ), Y ( z 2 ), Y ( z3 ), and Y ( z 4 ), which can be evaluated using only 5 multiplications.

(

)(

11.43 Y ( z ) = H ( z ) X ( z ) or y[ 0] + y[1]z − 1 + y[2]z − 2 = h[ 0] + h[1]z − 1 x[ 0] + x[1]z −1 h[ 0] x[0] + (h[ 0] x[1] + h[1] x[ 0]) z −1

)

+ h[ 2] x[[ 2]z −2 .

Hence, y[ 0] = h[ 0] x[ 0], = y[1] = h[ 0] x[1] + h[1] x[ 0], y[ 2] = h[1] x[1]. Now, (h[ 0] + h[1])( x[ 0] + x[1]) − h[ 0] x[ 0] − h[1] x[1] = h[ 0] x[1] + h[1] x[0] = y[1]. As a result, evaluation of H ( z ) X ( z ) requires the computation of 3 products, h[ 0] x[ 0], h[1] x[1] , and (h[ 0] + h[1])( x[ 0] + x[1]) . In addition, it requires 4 additions, h[0] + h[1], x[ 0] + x[1], and (h[ 0] + h[1])( x[ 0] + x[1]) − h[ 0] x[ 0] − h[1] x[1] . 11.43 Let the two length- N sequences be denoted by {h[ n]} and {x[ n]}. Denote the sequence

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generated by the linear convolution of h[n] and x[n] as y[n]. Let H ( z) =

X ( z) =

N −1

∑ n = 0 x[n]z − n .

and

Rewrite H ( z ) and X ( z ) in the form H ( z) = H 0 ( z) + z − N / 2 H1 ( z), and

X ( z) = X 0 ( z) + z − N / 2 X1 ( z), where H 0 ( z ) = ( N / 2 ) −1

N −1

∑ n = 0 h[n]z − n

( N / 2) −1

∑n=0

h[ n]z − n ,

( N / 2) −1

∑ n = 0 h[n + 2 ]z − n , X 0 (z) = ∑ n = 0 x[n]z − n , ( N / 2) −1 N X1 ( z ) = ∑ x[ n + ]z − n . Therefore, we can write n=0 2

H1 ( z ) =

N

(

)(

)

Y ( z) = H 0 ( z ) + z − N / 2 H1 ( z) X 0 ( z ) + z − N / 2 X1 ( z ) = H 0 ( z ) X 0 ( z) + z − N / 2 (H 0 ( z) X1 ( z) + H1 ( z) X 0 ( z)) + z − N H1 ( z) X1 ( z) = Y0 ( z) + z − N / 2Y1 ( z ) + z − N Y2 ( z ), where Y0 ( z) = H 0 ( z) X 0 ( z), Y1 ( z ) = H 0 ( z) X1 ( z ) + H1 ( z) X 0 ( z ), and Y2 ( z) = H1 ( z) X1 ( z). Note that Y0 ( z) and Y1 ( z) are 2

N each, and hence, require ⎛⎜ ⎞⎟ multiplications each. ⎝2⎠ Now, we can write Y1 ( z ) = (H 0 ( z ) + H1 ( z ))(X 0 ( z ) + X1 ( z)) − Y0 ( z) − Y1 ( z ). Since,

products of two polynomials of degree

N 2

(H0 ( z) + H1( z))(X 0 ( z) + X1( z)) is a product of two polynomials of degree

N 2

each, it can be

2

N computed using ⎛⎜ ⎞⎟ multiplications. As a result, Y ( z ) = H ( z ) X ( z ) can be computed using ⎝2⎠

2

N N is even, and 3⎛⎜ ⎞⎟ multiplications instead of N 2 multiplications. If N is a power-of-2, 2 ⎝2⎠ the same procedure can be applied to compute Y0 ( z), Y1 ( z), and Y2 ( z), reducing further the number of multiplications. This process can be continued until the sequences being convolved are of length 1 each.

Let R ( N ) denote the total number of multiplications required to compute the linear convolution of two N –length sequences. Then, in the method outlined above, we have R ( N ) = 3 ⋅ R ( N / 2) with R (1) = 1. A solution of this equation is given by R ( N ) = 3 log 2 N .

11.45 The dynamic range of a signed B –bit integer η is given by − (2 ( B −1) − 1) ≤ η < (2 ( B −1) − 1)

which for B = 32 is given by − (2 31 − 1) ≤ η < (2 31 − 1). (a) For E = 6 and M = 25, the value of a 32-bit floating-point number is given by η = ( −1) s 2 E − 31 ( M ). Hence, the value of the largest number is ≈ 2 32 , and the value of the

smallest number is ≈ −2 32. The dynamic range is therefore ≈ 2 × 2 32. (b) For E = 7 and M = 24, the value of a 32-bit floating-point number is given by η = ( −1) s 2 E − 63 ( M ). Hence, the value of the largest number is ≈ 2 64 , and the value of the

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smallest number is ≈ −2 64. The dynamic range is therefore ≈ 2 × 2 64. For E = 8 and M = 23, the value of a 32-bit floating-point number is given by η = ( −1) s 2 E −127 ( M ). Hence, the value of the largest number is ≈ 2128 , and the value of the

(c)

smallest number is ≈ −2128. The dynamic range is therefore ≈ 2 × 2128. Hence, the dynamic range in a floating-point representation is much larger than that in a fixedpoint representation with the same wordlength. 11.46 A 32-bit floating-point number in the IEEE Format has E = 8 and M = 23. Also, the exponent E is coded in a biased form as E − 127 with certain conventions for special cases such as E = 0, 255, and M = 0 (See page 637 of text).

Now, a positive 32-bit floating-point number represented in the “normalized” form have an exponent in the range 0 < E < 255, and is of the form η = ( −1) s 2 E −127 (1∆ M ). Hence, the smallest positive number that can be represented will have E = 1, and M = 0 0 K 0 0, 1442443 22 bits

and has therefore a value given by 2 −126 ≅ 1.18 × 10 −38. For the largest positive number, E = 254, and M = 1 1 K 1 1. Thus, here η = (−1) 0 2127 (1∆ 1 1 K 1 1) ≈ 2127 × 2 1442443 1442443 22 bits

22 bits

3.4 × 10 38.

≅ Note: For representing numbers less than 2 −126 , IEEE format uses the “de-normalized” form where E = 0, and η = ( −1) s 2 −126 (0 ∆ M ). In this case, the smallest positive number that can be represented is given by η = (−1) 0 2126 (0 ∆ 0 0 K 0 1) ≈ 2 −149 1442443 22 bits

≅ 1.4013 × 10 −45. 11.47 For a two’s-complement binary fraction s ∆ a −1a − 2 K a − b , the decimal equivalent for

s = 0 is simply

∑ i =1 a − i 2 − i. For s = 1, the decimal equivalent is given by b

b b b ⎛ b ⎞ − i − b − − − − i b b i ⎜ ⎟ (1 − a − i )2 + 2 − =− 2 + a − i 2 − 2 = −(1 − 2 ) + a− i 2 − i − 2 − b ⎜ ⎟ i =1 i =1 i =1 ⎝ i =1 ⎠

∑

∑

= −1 + −s+

∑

∑

b

∑ a− i 2 − i.

i =1 b

Hence, the decimal equivalent of s ∆ a −1a − 2 K a − b is given by

∑ a − i 2 − i. i =1

11.48 For a ones’-complement binary fraction s ∆ a −1a − 2 K a − b , the decimal equivalent for

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s = 0 is simply

∑ i =1 a − i 2 − i. For s = 1, the decimal equivalent is given by b

b b b ⎛ b ⎞ − i − − − i b i ⎜ ⎟ (1 − a − i )2 − =− 2 + a − i 2 = −(1 − 2 ) + a− i 2 − i ⎜ ⎟ i =1 i =1 i =1 ⎝ i =1 ⎠

∑

= −1 +

∑

∑

∑

b

∑ a− i 2 − i. i =1

− s(1 − 2 − b ) +

Hence, the decimal equivalent of s ∆ a −1a − 2 K a − b is given by

b

∑ a − i 2 − i. i =1

11.49 η1 = 0.7812510 = 0 ∆11001, η2 = 0.6875010 = 0 ∆10110, η3 = 0.5312510 = 1∆ 01111. η1 + η2 = 0 ∆11001 + 0 ∆10110 = 1∆ 01111. Dropping thee overflow bit and adding (η1 + η2 ) to η3 : (η1 + η2 ) + n3 = 0 ∆ 01111 + 1∆ 01111 = 0 ∆11110 = 0.937510 , where we have dropped the carry bit in the MSB location. Note that the final sum is correct inspite of the overflow. ⎛ e jωˆ + e − jωˆ ⎞ ⎟, β ⎜⎜ ⎟ 2 ⎝ ⎠ − 1 − 1 ⎛ zˆ + zˆ ⎞ z+z ⎟. Now, H ( z ) let which by analytic continuation can be expressed as = α + β ⎜⎜ ⎟ 2 ⎝ 2 ⎠ be a Type 1 linear-phase FIR transfer function of degree 2M. As indicated in Eq. (8.128),

11.49 The transformation cos ω = α + β cos ωˆ is equivalent to

H ( z ) can be expressed as H ( z) = z − M

given by H (e jω ) = e − jMω

∑

e jω + e − jω =α+ 2

⎛ z + z −1 ⎞ M ⎟ a[ n]⎜⎜ n=0 2 ⎟ ⎝

(

⎠

n

with a frequency response

∑ n = 0 a[n](cos ω) n , with H (ω) = ∑ n = 0 a[n](cos ω) n M

M

denoting

the amplitude function or the zero-phase frequency response. The amplitude function or the zero-phase frequency response of the transformed filter obtained by applying the mapping ( M a[ n](α + β cos ωˆ ) n . Or, equivalently, cos ω = α + β cos ωˆ is therefore given by H (ωˆ ) =

∑n=0

the transfer function of the transformed filter is given by

H ( zˆ ) = z − M

∑

⎛ M zˆ + zˆ −1 ⎞⎟ a[ n]⎜⎜ α + β n=0 2 ⎟⎠ ⎝

n

A convenient way to realize H ( zˆ ) is to consider the

realization of the parent transfer function H ( z ) in the form of a Taylor structure as outlined in Problem 8.17 which is obtained by expressing H ( z ) in the form

H ( z) =

∑

n −2 ⎞ ⎛ + 1 M z ⎟ . Similarly, the transformed filter can be realized by a[ n]z − M + n ⎜⎜ n=0 2 ⎟

⎝

⎠

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1 + z −2 1 + zˆ −2 in the Taylor structure by the block αzˆ −1 + β . 2 2 Now, for a lowpass-to-lowpass transformation we can impose the condition ( ( H (ωˆ ) = H (ω) . This condition is met if α + β = 1 and 0 ≤ α < 1. In this case, the

replacing each block

ωˆ = 0

ω=0

transformation reduces to cos ω = α + (1 − α ) cos ωˆ . From the plot of the mapping given below, it follows that as α is varied between 0 and 1, ωˆ c < ωc . 1

cos ω

0.5 0 -0.5 -1 -1

-0.5

0 ^ cos ω

0.5

1

On the other hand if ωˆ c > ωc . is desired along with a lowpass-to-lowpass transformation, we ( ( = H (ω) . This condition is met if β = 1 + α and can impose the condition H (ωˆ ) ωˆ = π

ω=π

− 1 < α ≤ 1. The corresponding transformation is now given by cos ω = α + (1 + α ) cos ωˆ . From the plot of the mapping given below, it follows that as α is varied between 0 and 1, ωˆ c > ωc . 1

cos ω

0.5 0 -0.5 -1 -1

11.51 (a) X ( z ) =

N −1

∑

x[n]z

n=0

−n

-0.5

0 ^ cos ω

0.5

1

n ( N −1 ⎛ − α + z( −1 ⎞ ( ( −1 ⎟ = P( z() , . X ( z ) = X ( z ) z −1 = − α + z = ∑ x[n]⎜ ⎜ 1 − αz( −1 ⎟ D( z ) ( n=0 ⎠ ⎝ 1− αz −1

N −1 ( N −1 ( ( ( where P( z ) = ∑ p[n]z − n = ∑ x[n](1 − α z −1 ) N −1− n (−α + z −1 ) n , and n=0

n=0

( D( z ) =

N −1

(

(

∑ d [n]z − n = (1 − αz −1 ) N −1.

n=0

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( ( ( ( P( z ) P[k ] ( ( = ( , where P[k ] = P( z ) z( = e j 2 πk / N is (b) X [k ] = X ( z ) z( = e j 2 πk / N = ( D( z ) z( = e j 2 πk / N D[k ] ( ( the N –point DFT of the sequence p[n] and D[k ] = D( z ) z( = e j 2 πk / N is the N –point DFT of the sequence d [n]. (c) Let P = [ p[0] p[1] L p[ N − 1]]t and X = [x[0] x[1] L x[ N − 1]]t . Without any loss of generality, assume N = 4 in which case

(

)

3 ( ( P( z ) = ∑ p[n]z − n = x[0] − αx[1] + α 2 x[2] − α 3 x[3]

( ( (

n=0

)

( + − 3αx[0] + (1 + 2α 2 ) x[1] − α(2 + α 2 ) x[2] + 3α 2 x[3] z −1 ( + 3α 2 x[0] − α(2 + α 2 ) x[1] − (1 + α 2 ) x[2] − 3αx[3] z −2 ( ( + − α 3 x[0] + α 2 x[1] − αx[2] + αx[3] z −3 . Equating like powers of z −1 we can write ⎡ −α α2 − α 3 ⎤ ⎡ x[0]⎤ ⎡ p[0]⎤ ⎢ 1 ⎥ 2 2 ⎢ p[1] ⎥ ⎢ − 3α − α(2 + α ) 3α 2 ⎥ ⎢ x[1] ⎥ 1 + 2α P = Q ⋅ X or ⎢ . It can be shown = p[2]⎥ ⎢ 2 − α(2 + α 2 ) − 3α ⎥ ⎢⎢ x[2]⎥⎥ 1 + 2α 2 ⎢ p[3]⎥ ⎢ 3α ⎥ x[3]⎦ 3 ⎦ ⎣ α2 −α 1 ⎦⎣ ⎣− α that the elements q r , s , 0 ≤ r , s ≤ 3, of the 4 × 4 matrix Q can be determined as follows: (i)

)

)

The first row is given by q0, s = (−α) s ,

3! (−α ) r , and r!(3 − r )! (iii) The remaining elements can be obtained using the recurrence relation q r , s = q r −1, s −1 − α q r , s −1 + α q r −1, s .

(ii)

The first column is given by q r ,0 = 3C r (−α ) r =

In the general case, we only change the computation of the elements of the first column ( N − 1)! ( −α ) r . using the relation q r ,0 = N −1C r (−α) r = r!( N − 1 − r )!

A 0 ( z) =

0.3038 + z −1

[

]

1 A ( z) − A1 ( z) , 2 0 0.639 + 0.4012 z −1 + z −2

11.52 The highpass transfer function can be expressed as H ( z ) =

where

and A1 ( z) = . The tunable highpass 1 + 0.3038z −1 1 + 0.4012 z −1 + 0.639z − 2 transfer function is obtained by applying the lowpass-to-lowpass transformation of Eq. sin[(0.6π − ωˆ p ) / 2] (11.115) where the tuning parameter is given by α = . The tunable sin[(0.6π + ωˆ p ) / 2] highpass transfer function is then given by H ( z, α ) =

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1 2

[Aˆ 0 (z) − Aˆ1 (z)], where from Eqn.

421

[ 0.3038 − 0.9077α ] + z −1 and from Eqn. (11.123) (11.119) Aˆ 0 ( z ) = 1 + [ 0.3038 − 0.9077α ]z −1 [ 0.639 − 0.1448α ] + [0.4012 + 3.439α ]z −1 + z −2 . Aˆ 1( z ) = 1 + +[ 0.4012 + 3.439α ]z −1 + [ 0.639 − 0.1448α ]z − 2 11.53

[

]

1 A ( z) + A1 ( z) , 2 0 0.3554 − 0.4191 z −1 + z −2

The lowpass transfer function can be expressed as H ( z) = − 0.1584 + z −1

where

. The tunable bandpass and A1 ( z ) = 1 − 0.4191 z −1 + 0.3554 z − 2 1 − 0.1584 z −1 transfer function is obtained by applying the lowpass-to-bandpass transformation of Eq. (11.124) where the tuning parameter is given by β = cos ωo with ωo denoting the center frequency of the bandpass filter. The tunable bandpass transfer function is then given by ⎛ β + z −1 ⎞ ⎟ − 0.1584 − z −1 ⎜⎜ −1 ⎟ 1 + β z 1 ˆ ⎝ ⎠ and H ( z, α ) = A 0 ( z) + Aˆ 1 ( z) , where Aˆ 0 ( z ) = 2 ⎛ β + z −1 ⎞ ⎟⎟ 1 + 0.1584 z −1 ⎜⎜ ⎝ 1+ βz −1 ⎠

A 0 (z) =

[

Aˆ 1 ( z ) =

]

⎛ β + z −1 ⎞ ⎛ β + z −1 ⎞ ⎟⎟ + z − 2 ⎜⎜ ⎟ 0.3554 + 0.4191 z −1 ⎜⎜ −1 −1 ⎟ + z + z 1 1 β β ⎝ ⎠ ⎝ ⎠

2

⎛ ⎞ ⎛ ⎞ ⎟⎟ + 0.3554 z − 2 ⎜⎜ ⎟⎟ 1 + 0.4191 z −1 ⎜⎜ − 1 − 1 z z + β + β 1 1 ⎝ ⎠ ⎝ ⎠ β + z −1

β + z −1

M11.1 (a) Numerator coefficients = [0.0528 0.0797 0.1295 Denominator coefficients = [1.0000 -1.8107 2.4947

2

.

0.1295

0.0797

0.0528]

-1.8801

0.9537

-0.2336]

Impulse response samples 0.6

Amplitude

0.4 0.2 0 -0.2

0

5


(b) Numerator coefficients = [0.0084 -0.0335 0.0502 Denominator coefficients =

-0.0335

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30

0.0084]

422

[1.0000

2.3741

2.7057

1.5917

0.4103]

Impulse response samples 0.4

Amplitude

0.2 0 -0.2 -0.4

0

5


25

30

(c) Numerator coefficients = [0.0003 0 -0.0019 0 0.0057 0 -0.0095 0 0.0095 0 -0.0057 0 0.0019 0 -0.0003] Denominator coefficients = [1.0000 1.7451 4.9282 6.1195 9.8134 9.2245 10.4323 7.5154 6.4091 3.4595 2.2601 0.8470 0.4167 0.0856 0.0299] Impulse response samples 0.4

Amplitude

0.2 0 -0.2 -0.4

0

5


25

30

M11.2 The modified MATLAB program is given below: n = 0:60; w = input('Normalized angular frequency vector = '); num = input('Numerator coefficients = '); den = input('Denominator coefficients = '); x1 = cos(w(1)*pi*n); x2 = cos(w(2)*pi*n); x = x1+x2; % Generate the output sequence by filtering the input y = filter(num,den,x); % Plot the input and the output sequences figure(1) stem(n,x); xlabel('Time index n'); ylabel('Amplitude'); title('Input sequence'); figure(2) stem(n,y); xlabel('Time index n'); ylabel('Amplitude');

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title('Output sequence');

The plots generated by the above program for the filter of Example 9.14 for an input composed of a sum of two sinusoidal sequences of angular frequencies, 0.3π and 0.6π, are given below: Output sequence 1

1

0.5 Amplitude

Amplitude

Input sequence 2

0 -1 -2

0 -0.5

0

10


50

-1

60

0

10


50

60

The blocking of the high-frequency signal by the lowpass filter can be demonstrated by replacing the statement stem(n,x); with stem(n,x2); and the statement y=filter(num,den,x); with y=filter(num,den,x2);. The plots of the highfrequency input signal and the corresponding output are shown below: Output sequence 0.3

0.5

0.2 Amplitude

Amplitude

Input sequence 1

0 -0.5 -1

0.1 0 -0.1

0

10


50

-0.2

60

0

10


50

60

M11.3 The plots generated by the program of Exercise M11.2 for the filter of Example 9.15 for an input composed of a sum of two sinusoidal sequences of angular frequencies, 0.3π and 0.6π, are given below: Output sequence 0.4

1

0.2 Amplitude

Amplitude

Input sequence 2

0 -1 -2

0 -0.2 -0.4

0

10


50

60

-0.6

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10


50

60

424

The blocking of the low-frequency signal by the highpass filter can be demonstrated by replacing the statement stem(n,x); with stem(n,x1); and the statement y=filter(num,den,x); with y=filter(num,den,x1);. The plots of the lowfrequency input signal and the corresponding output are shown below: Output sequence 0.2

0.5

0.1 Amplitude

Amplitude

Input sequence 1

0 -0.5 -1

0 -0.1

0

10


50

60

-0.2

0

10


50

60

M11.4 % The factors for the transfer function of % the lowpass filter of Example 9.14 are % num1 = [0.0528 0.0528 0]; % den1 = [1.0000 -0.4909 0]; % num2 = [1.0000 0.6040 1.0000]; % den2 = [1.0000 -0.7624 0.5390]; % num3 = [1.0000 -0.0949 1.0000]; % den3 = [1.0000 -0.5574 0.8828]; N = input(‘Total number of sections = ‘); for k = 1:N; num(k,:) = input('Numerator factor = '); den(k,:) = input('Denominator factor = '); end n = 0:60; w = input('Normalized angular frequency vector = '); x1 = cos*w(1)*pi*n); x2 = cos*w(2)*pi*n); x = x1 + x2; % Plot the input sequence figure(1) stem(n,x); xlabel('Time index n'); ylabel('Amplitude'); title('Input sequence'); si = [0 0]; for k = 1:N; y(k,:) = filter(num(k,:),den(k,:),x,si); x = y(k,:); end % Plot the output sequence figure(2) stem(n,y(N,:)); xlabel('Time index n'); ylabel('Amplitude'); title('Output sequence');

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The plots generated by the above program for the filter of Example 9.14 for an input composed of a sum of two sinusoidal sequences of angular frequencies, 0.3π and 0.6π, are given below: Output sequence

Input sequence

1

2

0.5 Amplitude

Amplitude

1 0

-0.5

-1 -2

M11.5 % % % % % %

0

-1 0

10


50

60

0

10


50

60

The factors for the transfer function of the highpass filter of Example 9.15 are num1 = [0.0084 -0.0167 0.0084]; den1 = [1.0000 1.3101 0.5151]; num2 = [1.0000 -2.0000 1.0000]; den2 = [1.0000 1.0640 0.7966];

The plots generated by the program of Exercise M11.4 for the filter of Example 9.15 for an input composed of a sum of two sinusoidal sequences of angular frequencies, 0.3π and 0.6π, are given below: Input sequence

Output sequence

2

0.4 0.2 Amplitude

Amplitude

1 0 -1 -2

0 -0.2 -0.4

0

10


50

60

-0.6

0

10


50

60

M11.6 To apply the function direct2 to filter a sum of two sinusoidal sequences, we replace the statement y = filter(num,den,x,si); in the MATLAB program given in the solution of Exercise M11.2 with the statement y = direct2(num,den,x,si);. The plots generated by running the modified program for the data given in this problem are given below:

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Output sequence 1

1

0.5 Amplitude

Amplitude

Input sequence 2

0

-0.5

-1 -2

0

0

10


50

60

-1

0

10


50

60

M11.7 The MATLAB program that can be used to compute all DFT samples using the function goertzel and the function fft is as follows: clear N = input('Desired DFT length = '); x = input('Input sequence = '); for j = 1:N Y(j)=goertzel(x,j); end disp('DFT samples computed using goertzel are ') ; disp(Y) disp('DFT samples computed using fft are ') ; X = fft(x); disp(X); Results obtained for two input sequences of lengths 8 and 12, respectively, are given below: Desired DFT length = 8 Input sequence = [1 2 3 4 4 3 2 1] DFT samples computed using goertzel are Columns 1 through 4 20.0000 -5.8284-2.4142i -0.0000-0.0000i -0.1716-0.4142i Columns 5 through 8 0-0.0000i -0.1716+0.4142i 0.0000-0.0000i -5.8284+2.4142i DFT samples computed using fft are Columns 1 through 4 20.0000 -5.8284-2.4142i 0 Columns 5 through 8 0 -0.1716+0.4142i 0 Desired DFT length = 12 Input sequence = [1 2 4 8 10 12 7

-0.1716-0.4142i -5.828 +2.4142i 3 -4 5 0

DFT samples computed using goertzel are Columns 1 through 4 54.0000 -13.0622-21.0885i 1.5000+19.9186i Columns 5 through 8

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-4.0000-2.0000i

427

4.5000+2.5981i -0.9378-10.0885i

-0.9378+10.0885i

Columns 9 through 12 4.5000-2.5981i -4.0000+2.0000i 13.0622 +21.0885i

-18.0000+0.0000i

1.5000-19.9186i

DFT samples computed using fft are 54.0000 -13.0622-21.0885i 1.5000+19.9186i Columns 5 through 8 4.5000+2.5981i -0.9378-10.0885i

-0.9378+10.0885i

Columns 9 through 12 4.5000-2.5981i -4.0000+2.0000i 13.0622 +21.0885i

-

-4.0000-2.0000i

-18.0000+0.0000i

1.5000-19.9186i

-

M11.8 The MATLAB program that can be used to verify the plots of Figure 11.43 is given below: %Program_11_8.m [z,p,k] = ellip(5,0.5,40,0.4); a = conv([1 -p(1)],[1 -p(2)]);b =[1 -p(5)]; c = conv([1 -p(3)],[1 -p(4)]); w = 0:pi/255:pi; alpha = 0; an1 = a(2) + (a(2)*a(2) - 2*(1 + a(3)))*alpha; an2 = a(3) + (a(3) -1)*a(2)*alpha; g = b(2) - (1 - b(2)*b(2))*alpha; cn1 = c(2) + (c(2)*c(2) - 2*(1 + c(3)))*alpha; cn2 = c(3) + (c(3) -1)*c(2)*alpha; a = [1 an1 an2];b = [1 g]; c = [1 cn1 cn2]; h1 = freqz(fliplr(a),a,w); h2 = freqz(fliplr(b),b,w); h3 = freqz(fliplr(c),c,w); ha = 0.5*(h1.*h2 + h3);ma = 20*log10(abs(ha)); alpha = 0.1; an1 = a(2) + (a(2)*a(2) - 2*(1 + a(3)))*alpha; an2 = a(3) + (a(3) -1)*a(2)*alpha; g = b(2) - (1 - b(2)*b(2))*alpha; cn1 = c(2) + (c(2)*c(2) - 2*(1 + c(3)))*alpha; cn2 = c(3) + (c(3) -1)*c(2)*alpha; a = [1 an1 an2];b = [1 g]; c = [1 cn1 cn2]; h1 = freqz(fliplr(a),a,w); h2 = freqz(fliplr(b),b,w); h3 = freqz(fliplr(c),c,w); hb = 0.5*(h1.*h2 + h3);mb = 20*log10(abs(hb)); alpha = -0.25; an1 = a(2) + (a(2)*a(2) - 2*(1 + a(3)))*alpha; an2 = a(3) + (a(3) -1)*a(2)*alpha; g = b(2) - (1 - b(2)*b(2))*alpha; cn1 = c(2) + (c(2)*c(2) - 2*(1 + c(3)))*alpha; cn2 = c(3) + (c(3) -1)*c(2)*alpha;

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a = [1 an1 an2];b = [1 g]; c = [1 cn1 cn2]; h1 = freqz(fliplr(a),a,w); h2 = freqz(fliplr(b),b,w); h3 = freqz(fliplr(c),c,w); hc = 0.5*(h1.*h2 + h3);mc = 20*log10(abs(hc)); plot(w/pi,ma,'r-',w/pi,mb,'b--',w/pi,mc,'g-.');axis([0 1 -80 5]); xlabel('Normalized frequency');ylabel('Gain, dB'); legend('b--','\alpha = 0.1 ','r-','\alpha = 0 ','g-.','\alpha = -0.25 ');

M11.9 The MATLAB program that can be used to verify the plots of Figure 11.44 is given below: %Program_11_9.m w = 0:pi/255:pi; wc2 = 0.31*pi; f = [0 0.36 0.46 1];m = [1 1 0 0]; b1 = remez(50, f, m); h1 = freqz(b1,1,w); m1 = 20*log10(abs(h1)); n = -25:-1; c = b1(1:25)./sin(0.41*pi*n); d = c.*sin(wc2*n);q = (b1(26)*wc2)/(0.4*pi); b2 = [d q fliplr(d)]; h2 = freqz(b2,1,w); m2 = 20*log10(abs(h2)); wc3 = 0.51*pi; d = c.*sin(wc3*n);q = (b1(26)*wc3)/(0.4*pi); b3 = [d q fliplr(d)]; h3 = freqz(b3,1,w); m3 = 20*log10(abs(h3)); plot(w/pi,m1,'r-',w/pi,m2,'b--',w/pi,m3,'g-.'); axis([0 1 -80 5]); xlabel('Normalized frequency');ylabel('Gain, dB'); legend('b--','\omega_c = 0.3\pi','r-','\omega_c = 0.41\pi','g.','\omega_c = 0.51\pi') M11.10 The MATLAB program to evaluate Eqs. (11.155) is given below: %Program_11_10.m x = 0:0.001:0.5; y = 3.140625*x + 0.0202636*x.^2 - 5.325196*x.^3 + 0.5446778*x.^4 + 1.800293*x.^5; x1 = pi*x; z = sin(x1); plot(x,y); xlabel('Normalized angle, radians');ylabel('Amplitude'); title('Approximate sine values');grid;axis([0 0.5 0 1]); pause plot(x,y-z); xlabel('Normalized angle, radians');ylabel('Amplitude'); title('Error of approximation');grid;

The plots generated by the above program are as indicated below:

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-5

Approximate sine values

3 Amplitude

Amplitude

0.8 0.6 0.4 0.2 0

Error of approximation

x 10

4

1

2 1 0 -1

0

0.1 0.2 0.3 0.4 Normalized angle, radians

-2

0.5

0

0.1 0.2 0.3 0.4 Normalized angle, radians

0.5

M11.11 The MATLAB program to evaluate Eqs. (11.156) and (11.135) is given below: %Program_11_11.m k = 1; for x = 0:.01:1 op1 = 0.318253*x+0.00331*x^2-0.130908*x^3+0.068524*x^40.009159*x^5; op2 = 0.999866*x-0.3302995*x^3+0.180141*x^50.085133*x^7+0.0208351*x^9; arctan1(k) = op1*180/pi; arctan2(k) = 180*op2/pi; actual(k) = atan(x)*180/pi; k = k+1; end subplot(211) x = 0:.01:1; plot(x,arctan2); ylabel('Angle, degrees');xlabel('Tangent Values'); subplot(212) plot(x,actual-arctan2,'--'); ylabel('Tangent Values');xlabel('error, radians');

The plots generated by the above program are as indicated below: 1

40

0.5

Tangent Values

Angle, degrees

-3

50

30 20 10 0

0

0.2

0.4 0.6 Tangent Values

0.8

1

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x 10

0 -0.5 -1

0

0.2

0.4 0.6 error, radians

0.8

1

430

Chapter 12 12.1 Two’s Complement Truncation - Assume x > 0. The relative error et is given by

∑

∑

b

∑

β

β

a 2−i − a 2−i − a 2−i Q( x ) − x Q( M ) − M i =1 − i i =1 − i i = b +1 − i = = = et = . Now will x M M M be a minimum if all a − i ‘s are 1 and will be a maximum if all a − i ‘s are 0 for b + 1 ≤ i ≤ β. δ 0 Thus, − ≤ et ≤ . Since, 0.5 ≤ M < 1 , hence − 2δ ≤ et ≤ 0. M M Next, consider x < 0. Here, the relative error et is given by

∑

∑

b

β

−i −i Q( x ) − x Q( M ) − M − 1 + i = 1 a − i 2 + 1 − i = 1 a − i 2 et = = = . As before, x M M δ 0 − ≤ et ≤ . In this case, − 1 ≤ M ≤ −0.5, and hence, 0 ≤ et ≤ 2δ. M M

Ones’ Complement Truncation - Assume x > 0. The relative error et is given by Q( x ) − x Q( M ) − M et = = = x M

∑ i =1 a − i 2 − i − ∑ i = 1 a − i 2 − i = − ∑ i = b + 1 a − i 2 − i . β

b

β

M

∑

M

∑

b

∑

β

β

a 2−i − a 2−i − a 2−i Q( x ) − x Q( M ) − M i =1 − i i =1 − i i = b +1 − i et = . Now will = = = x M M M be a minimum if all a − i ‘s are 1 and will be a maximum if all a − i ‘s are 0 for b + 1 ≤ i ≤ β. δ 0 Thus, − ≤ et ≤ . Since, 0.5 ≤ M ≤ 1 , hence − 2δ ≤ et ≤ 0. M M Next, consider x < 0. Here, the relative error et is given by −b Q( x ) − x Q( M ) − M − (1 − 2 ) + = = et = x M

=

(2 − b − 2 − β ) +

∑ i = b +1 a − i 2 − i .

∑ i =1 a − i 2 − i + (1 − 2 − β ) − ∑ i =1 a− i 2 − i β

b

M

β

M

Now will be a minimum if all a − i ‘s are 1 and will be a

maximum if all a − i ‘s are 0 for b + 1 ≤ i ≤ β. In this case,

− 0.5 ≤ M < −1 , hence, − 2δ ≤ et ≤ 0.

0 δ ≤ et ≤ . Since, M M

Sign-Magnitude Truncation - Assume x > 0. The relative error et is given by

∑

b

∑ i =1 a − i 2 − i = ∑ i = b + 1 a − i 2 − i .

−i Q( M ) − M − i = 1 a − i 2 + et = = M M − 0.5 ≤ M < −1 , hence − 2δ ≤ et < 0.

β

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β

M

Since,

431

Rounding - ε = Q( M ) − M . Hence, − − δ ≤ et ≤ δ.

δ δ ≤ε≤ . Since, − 1 < M ≤ −0.5, hence, 2M 2M

12.2 The denominator is given by D( z ) = z − α. Here, the pole is at z = re jθ = α, where r = α ∂r and θ = 0. The components of the pole sensitivities are therefore given by = 1 and ∂α ∂θ = 0. ∂α 12.3 The denominator is given by D( z ) = z 2 − β (1 + α ) z + α = ( z − r e jθ )( z − r e − jθ ) = z 2 − 2r cos θ z + r 2 . Comparing we get 2r cos θ = β (1 + α ) and r 2 = α. Taking the partials of both sides of the last two equations we get 1 0 ⎤ ∆α 0 0 ⎤ ⎡∆α ⎤ = ⎡ 2r cos ⎡ ⎤ ⎤ ⎡ ∆r ⎤ = ⎡ 1 ⎡ 2r θ ⎢ 1 + r 2 ⎥ ⎢⎣∆β ⎥⎦ , or ⎢⎣2 cos θ − 2r sin θ ⎥⎦ ⎢⎣∆θ ⎥⎦ ⎢⎣ β 1 + α ⎥⎦ ⎢⎣∆β ⎥⎦ ⎢ ⎥⎦ ⎣ 1+ r 2 1 ⎡ ⎤ 0 −1 ⎡ 1 0 ⎤ ⎢ ⎥ ⎡∆α ⎤ 2 r r r α ∆ ∆ 2 0 ⎤ ⎡ ⎤ ⎡ ⎤=⎡ 2r cos θ . 2 ⎥ 1 + r 2 ⎥ ⎢⎣∆β ⎥⎦ = ⎢ (1− r 2 ) cos θ + r 1 ⎢⎣∆β ⎥⎦ ⎢⎣∆θ ⎥⎦ ⎢⎣2 cos θ − 2r sin θ ⎥⎦ ⎢⎢ 2 ⎥ − ⎢ ⎥ ⎣ 1+ r ⎦ 2r sin θ ⎦ ⎣ 2r 2 (1+ r 2 ) sin θ Hence, the components of the pole sensitivities are given by 1 ∂r (1 − r 2 ) cos θ ∂r ∂θ 1+ r2 ∂θ , and = , = 0, = . =− ∂ α 2r ∂ β ∂ α 2r 2 (1 + r 2 ) sin θ 2r sin θ ∂β

12.4 The digital filter structure of Figure P12.1 with internal variables labeled is shown below. Analysis yields X(z)

W(z)

U(z)

_ z 1

c

V(z)

d

_ z 1

c

Y(z)

d

_1

W ( z ) = X ( z ) − Y ( z ), U ( z ) = W ( z ) + c d z −1U ( z ) ⇒ U ( z ) =

W ( z) 1− cd z

−1

=

X ( z) − Y ( z) 1 − c d z −1

, and

zY ( z ) . Also, V ( z ) = c z −1U ( z ) + d Y ( z ). Thus, c − 1 − 1 Y ( z ){1 − c d z − c d z (1 − − c d z −1 ) + c 2 z −2 } = c 2 z −2 X ( z ). Hence, c z −1V ( z ) = Y ( z ) ⇒ V ( z ) =

H (z) =

Y ( z) c2 . = X ( z) z 2 − 2c d z + c 2 (1 + d 2 )

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Let c 2 (1 + d 2 ) = r 2 and c d = r cos θ . Solving we get cos θ =

d 1+ d2

. Thus,

d = cot θ and c = r sin θ. Hence, ∆d = − cosec 2 θ (∆θ ) and ∆c = sin θ (∆θ ) + r cos θ (∆θ ). Or, r cos θ ⎤ ⎡ ∆r ⎤ ⎡ ∆c ⎤ = ⎡sin θ ⎡ ∆r ⎤ = − sin θ ⎡ − cosec 2 θ − r cos θ ⎤ ⎡ ∆c ⎤ ⇒ 2 ⎢⎣ ⎢⎣∆d ⎥⎦ ⎢⎣ 0 ⎢⎣∆θ ⎥⎦ − cosec θ ⎥⎦ ⎢⎣∆θ ⎥⎦ 0 sin θ ⎥⎦ ⎢⎣∆d ⎥⎦ ⎡ 1 − r cos θ ⋅ cos θ ⎤ ⎡ ∆c ⎤ ⎥ . = ⎢ sin θ ⎢⎣∆d ⎥⎦ 2 ⎥ ⎢⎣ 0 − sin θ ⎦ The digital filter structure of Figure 12.54 with internal variables labeled is shown below. U(z)

X(z)

V(z)

c

c W(z) _ z 1

d

_ z 1

Y(z)

_d

Analysis yields U ( z) = X ( z ) − dY ( z ), V ( z) = U ( z) + c z −1V ( z ) ⇒ V ( z) = W ( z ) = c d z −1V ( z ) + c z −1W ( z ), or W ( z) = Y ( z ) = c z −1W ( z ) or,

Y ( z) c z −1

=

c d z −1 1 − c z −1

c d z −1 (X ( z) − dY ( z)) (1 − c z −1 ) 2

V (z) =

c d z −1 (1 − c z −1 ) 2

U ( z) 1 − c z −1

, and

U ( z ). As a result,

. Thus,

Y ( z ){(1 − c z −1 ) 2 + c 2 d z −2 } = c 2 d z −2 X ( z ). Hence, H ( z) =

Y (z) c2d = . X ( z) z 2 − 2c z + c 2 (1 + d 2 )

We let c = r cosθ , and c 2 (1 + d 2 ) = r 2 . Therefore, 1 + d 2 = sec 2 θ ⇒ d = tan θ. Taking partials we then get ∆c = cos θ (∆r ) − r sin θ (∆θ ) and ∆d = sec 2 θ (∆θ ). Thus, ⎡ ∆c ⎤ = ⎡cos θ ⎢⎣∆d ⎥⎦ ⎢⎣ 0

⎡ 1 − r sin θ ⎤ ⎡ ∆r ⎤ ∆ r ⎡ ⎤ = ⎢ cos θ ⇒ ⎢⎣∆θ ⎥⎦ ⎢ sec 2 θ ⎥⎦ ⎢⎣∆θ ⎥⎦ ⎣ 0

r sin⋅ cos θ ⎤ ⎡ ∆c ⎤ ⎥ . ⎢ ∆d ⎥ cos 2 θ ⎥⎦ ⎣ ⎦

12.5 The digital filter structure of Figure P12.2 with internal variables labeled is shown on the next page. Analysis yields X ( z ) + 2 z −1S ( z ) − bz −1S ( z ) − Y ( z ) − cY ( z ) = S ( z ) and Y ( z ) = z −2 S ( z ).

Thus, H ( z) =

1 Y (z) z −2 . = = X ( z) 1 + (b − 2)z −1 + (1 + c)z − 2 z 2 + (b − 2)z + (1 + c)

Let 1 + c = r 2 and 2 − b = 2r cos θ. Taking partials, we get ∆c = 2r (∆r ) and − ∆b = 2 cos θ (∆r ) − 2r sin θ (∆θ ). Therefore, we have

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S(z) _1 z

X(z)

_ z 1

Y(z)

2 _

b _1 _c

⎡ 1 0 ⎤ ⎥ ⎡ ∆c ⎤ 0 ⎤ ⎡ ∆r ⎤ or, ⎡ ∆r ⎤ = ⎢ 2r ⎡ ∆c ⎤ = ⎡ 2r 1 ⎥ ⎢⎣ ∆b ⎥⎦. ⎢⎣∆θ ⎥⎦ ⎢ 1 ⎢⎣∆b ⎥⎦ ⎢⎣2 cos θ − 2r sin θ ⎥⎦ ⎢⎣∆θ ⎥⎦ − ⎢⎣ 2r 2 tan θ 2r sin θ ⎥⎦ From Eq. (12.19) the pole sensitivities for Figure 12.11 are given by θ sin θ ⎤ ⎡∆α ⎤ ⎡ ∆r ⎤ = ⎡ cos 1 1 . ⎢ cos θ ⎥⎥ ⎢⎣∆β ⎥⎦ ⎢⎣∆θ ⎥⎦ ⎢ − sin θ r ⎣ r ⎦ 12.6 (a) For direct form implementation B( z) = ( z − z1 )( z − z 2 )( z − z3 ), where

z1 = r1e jθ1 , z 2 = r2 e jθ 2 , and z 3 = r3 e jθ 3 . Thus, B( z) = ( z 2 − 2r1 cos θ1 z + r12 )( z − r3 ) = ( z 2 + 0.5625 z + 0.7386 )( z − 0.0606 ). This implies,

2r1 cos θ1 = −0.5625, r12 = 0.7386, r3 = 0.0606, and θ3 = π. Thus, r1 = 0.7386 = 0.8594 and cos θ1 =

− 0.5625

2 0.7386

= −0.3273. Now,

1 1 = = B( z) ( z 2 + 0.5625z + 0.7386)( z − 0.0606) − 0.6440 + j 0.2711 1.2811 − 0.6440 − j 0.2711 + + . z + 0.2813 + j 0.8121 z + 0.2813 − j 0.8121 z + 0.0606

[ [

] ]

P1 = cos θ1 r1 r12 cos θ1 = [− 0.3273 0.8594 − 0.2417], Q1 = sin θ1 0 r12 sin θ1 = [− 0.9449 0 0.6979], R1 = −0.6440, and X1 = 0.2711. Likewise, P3 = cos θ3 r3 r32 cos θ3 = [− 1 0.0606 − 0.0037],

[

[

]

]

Q 3 = sin θ3 0 r32 sin θ3 = [0 0 0], R3 = 1.2881, and X 3 = 0. Thus, ∆r 1= (− R1P1 + X1Q1 ) ⋅ ∆B = − 0.4669 ∆b0 + 0.5535 ∆b1 + 0.0335 ∆b 2 , 1 ∆θ1 = − ( X1P1 + R1Q1 ) ⋅ ∆B = 0.8113∆b0 − 0.2711∆b1 − 0.4467∆b2 , r1 ∆r 3 = (− R3 P3 + X 3Q 3 ) ⋅ ∆B = 1.2881∆b0 − 0.0781∆b1 + 0.0048∆b2 , and 1 ∆θ3 = − ( X 3 P3 + R3Q 3 ) ⋅ ∆B = 0. r3

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Cascade form: B( z) = ( z 2 + c1z + c0 )(z + d 0 ) = B1 ( z) B2 ( z ), where B1 ( z ) = z 2 + c1 z + c 0 = z 2 + 0.5625 z + 0.7386 = ( z − r1e jθ1 )( z − r1e − jθ1 ) = z 2 − 2r cos θ1 z + r12

(b)

and B2 ( z ) = z + d 0 = z − 0.0606 = z −r 3 e jθ 3 . Comparing we get 2r 1cos θ1 = 0.5625, r12 = 0.7386, r3 = 0.0606, θ3 = π. Solving the first two equations we get r1 = 0.7386 = 0.8594 and cos θ1 =

− 0.5625

= −0.3273. 2 0.7386 1 j 0.6157 − j 0.6157 Now, . Hence, R1 = 0 and = + B( z1 ) z + 0.2813 − j 0.8121 z + 0.2813 + j 0.8121 X1 = −0.6157. P1 = [cos θ1 r1 ] = [ − 0.3273 0.8594], Q1 = [ − sin θ1 0] = [ − 0.9449 0]. 1 1 Next, . Hence, R3 = 1 and X 3 = 0. Here, P3 = cos θ3 = −1, and = B2 ( z ) z + 0.0606 Q 3 = − sinθ 3 = 0. Thus, ∆r = ( − R P + X Q ) ⋅ [ ∆ c ∆c ] t = X Q ⋅ [ ∆c ∆c ]t = 0.5818 ∆c , 1

1 1

1 1

0

1 1

0

1

0

1 ⋅ X1P1 ⋅ [ ∆c 0 ∆c1 ]t = −0.23458 ∆c0 + 0.6157∆c1 , r1 1 1 ∆r3 = (− R3 P3 + X 3Q 3 ) ⋅ ∆d 0 = − ∆d 0 , ∆θ3 = − ( X 3 P3 + R3 Q 3 ) ⋅ ∆d 0 = − R3 Q 3 ⋅∆d 0 = 0 r3 r3 . 12.7 In terms of the transfer parameters, the input-output relation of the two-pair of Figure P12.3 is given by Y1 ( z) = t11 X1 ( z) + t12 X 2 ( z) and Y2 ( z) = t 21 X1 ( z) + t 22 X 2 ( z). The constraining equation of the two-pair is given by X 2 ( z) = αY2 ( z). From Eq. (7.135b) we get Y ( z) αt t t12 t 21 ∂ H ( z) t12 t 21 (1 − αt 22 ) + t 22 α t12 t 21 H ( z) = 1 = t11 + 12 21 . Hence, = . = ∂α 1 − α t 22 X1 ( z ) (1 − αt 22 ) 2 (1 − αt 22 ) 2 ∆θ1 = −

1 ( X1P1 + R1Q1 ) ⋅ [ ∆c 0 r1

1

∆c1 ]t = −

Substituting X 2 ( z) = αY2 ( z) in Y2 ( z) = t 21 X1 ( z) + t 22 X 2 ( z), we get the expression for the Y ( z) t 21 . = scaling transfer function Fα ( z ) = 2 X1 ( z) 1 − α t 22 Next, from the structure given below we observe that the noise transfer function is given by Y ( z) . From the structure we also observe that X 2 ( z) = αY2 ( z) + U ( z). Gα ( z) = 1 U ( z) X ( z) = 0 1

Substituting this in the transfer relations we arrive at, with X1 ( z) = 0, Y1 ( z) = t12 X 2 ( z) = t12 (α X 2 ( z) + U ( z)) and Y2 ( z) = t 22 X 2 ( z) = t 22 (αY2 ( z) + U ( z)). From these two equations we obtain, after some algebra, t12 αt t Y1 ( z) = 12 22 U ( z ) + t12U ( z ) = U ( z). Hence, the noise transfer function is given by 1 − α t 22 1 − α t 22

t 21 t ∂ H (z) Y (z) t12 = ⋅ 12 = Fα ( z)Gα ( z). . Therefore, = Gα ( z) = 1 ∂α 1 − αt 22 1 − αt 22 U ( z ) X ( z ) = 0 1 − α t 22 1

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N /2 ⎡ 1 ⎛ ⎞⎤ ⎜1 + 4 cos 2 (ωn) ⎟ ⎥ =⎢ ⎟⎥ ⎢ 2 N + 1 ⎜⎝ n =1 ⎠⎦ ⎣ maximum value of cos 2 (ωn) is 1 and the minimum value is 0, hence,

⎡ 1 ⎛ sin[( N + 1)ω] ⎞ ⎤ 12.8 (a) WN (ω) = ⎢ ⎜N + ⎟⎟ ⎥ ⎜ sin(ω) ⎠⎦ ⎣ 2N + 1 ⎝

⎡ 1 ⎛ 4 N ⎞⎤ max{WN (ω)} = ⎢ ⎟ ⎜1 + 2 ⎠ ⎥⎦ ⎣ 2N + 1 ⎝ 0 < WN (ω) ≤ 1. N / 2 ⎞⎤ ⎡ 1 ⎛ ⎜ 1 + 4 1⎟ ⎥ (b) WN (0) = ⎢ ⎟ ⎢ 2 N + 1 ⎜⎝ m =1 ⎠ ⎥⎦ ⎣

∑

1/ 2

1/ 2

1/ 2

1/ 2

∑

⎡ 1 ⎤ = 1 and min{WN (ω)} = ⎢ ⎥ ⎣ 2 N + 1⎦

1/ 2

N / 2 ⎞⎤ ⎡ 1 ⎛ ⎜ 1 + 4 1⎟ ⎥ = 1. WN (π ) = ⎢ ⎟ ⎢ 2 N + 1 ⎜⎝ m =1 ⎠ ⎥⎦ ⎣

∑

. Since the

> 0. Therefore,

1/ 2

= 1.

1/ 2

1 sin(( N + 1)ω) ⎤ N ⎡ (c) lim WN (π ) = ⎢ lim . Since + lim ⋅ ⎥ sin(ω) N →∞ ⎣N →∞ 2N + 1 N →∞ 2N + 1 ⎦ 1 1 sin(( N + 1)ω) sin(( N + 1)ω) ≤ 1, lim ⋅ = 0. Hence, lim WN (π ) = . sin(ω) N →∞ N →∞ 2N + 1 2

12.9

From Eq. (12.75), SNR A / D = 6.0206 b + 16.81 − 20 log10 ( K ) dB. For a constant value of K, an increase in b by 1 bit, increases the SNR A / D by 6.0206 dB and an increase in b by 2 bits, increases the SNR A / D by 12.0412 dB. If K = 4 and b = 7, then SNR A / D = 6.0206 × 7 + 16.81 − 20 log10 (4) = 46.9122 dB. Therefore, for K = 4 and b = 9, SNR A / D = 46.91 + 12.04 = 58.95 dB; for K = 4 and b = 11, SNR A / D = 58.95 + 12.04 = 70.99 dB; for K = 4 and b = 13, SNR A / D = 70.99 + 12.04 = 83.03 dB; for K = 4 and b = 15, SNR A / D = 83.03 + 12.04 = 95.08 dB. If K = 6 and b = 7, then SNR A / D = 6.0206 × 7 + 16.81 − 20 log10 (6) = 43.3902 dB. Therefore, for K = 6 and b = 9, SNR A / D = 43.39 + 12.04 = 55.43 dB; for K = 6 and b = 11, SNR A / D = 55.43 + 12.04 = 67.47 dB; for K = 6 and b = 13, SNR A / D = 67.47 + 12.04 = 79.51 dB; for K = 6 and b = 15, SNR A / D = 79.51 + 12.04 = 91.56 dB. If K = 8 and b = 7, then SNR A / D = 6.0206 × 7 + 16.81 − 20 log10 (8) = 40.8922 dB. Therefore, for K = 8 and b = 9, SNR A / D = 40.89 + 12.04 = 52.93 dB; for K = 8 and b = 11, SNR A / D = 52.93 + 12.04 = 64.97 dB; for K = 8 and b = 13, SNR A / D = 64.97 + 12.04 = 77.01 dB; for K = 8 and b = 15, SNR A / D = 77.01 + 12.04 = 891.05 dB.

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N N Ak Ak ak z + 1 N , 12.10 (a) H ( z) = = Ck − Ck ak = C k Fk ( z ) + B, where C k = 2 z + a z + a − 1 a k k k =1 k =1 k =1 k =1 k N

∑

∑

∑

∑

N a z +1 , and B = − C k a k . Fk ( z ) = k z + ak k =1

∑

N ⎤ ⎤⎡ N 1 1 ⎡ 1 1 − − ⎢ C k Fk ( z ) + B⎥ ⎢ C k Fk ( z −1 ) + B⎥ z −1dz H ( z )H ( z )z dz = = 2πj 2πj ⎢ k =1 ⎥⎦ ⎥⎦ ⎢⎣ k =1 C C⎣

(b)

∫ ∑

∫

σ o2

N

∑

N

∑ ∫

∑ ∫

1 = [ B 2 z −1dz + B C k Fk (z)z −1dz + B C l Fl (z −1 )z −1dz 2πj k =1 l =1

∫

C

+

C N N

C

∑ ∑ C k Cl ∫ Fk (z)Fl (z)z −1dz] .

k = 1l = 1

C

1 1 1 B 2 z −1dz = B 2 , Fk ( z)z −1dz = a k , Fl ( z −1 )z −1dz = a l , and 2πj 2πj 2πj

Now,

∫

∫

∫

C

C

C

1 − a k2 − a l2 + a k a l 1 − 1 . Therefore, Fk ( z)Fl ( z)z dz = 2πj 1 −a k a l

∫

C

N 1 − a k2 − a l2 + a k a l = + 2 B Ck ak + . Since, C k a k = − B, we have − a a 1 k l k = 1l = 1 k =1 k =1 N N 1 − a2 − a2 + a a k l l k σ o2 = − B 2 + C k Cl 1 − a a k l k = 1l = 1 N N

N

σ o2

∑

B2

∑∑

∑

∑∑

1 − a k2 − a l2 + a k a l =− C k C l a k al + Ck Cl 1 − a k al k = 1l = 1 k = 1l = 1 N N

N N

∑∑

∑∑

N N ⎡1 − a 2 − a 2 + a k al ⎤ ⎤ Ak A l ⎡1 − a k2 − a l2 + a k a l l k ⎢ − ak a⎥ = ⋅ − ak a⎥ C k Cl ⎢ 1 − ak al 1 − ak al ⎢⎣ ⎥⎦ k =1l =11 − a k2 1 − a l2 ⎢⎣ ⎥⎦ k = 1l = 1 N N Ak A l ⎡ (1 − a k2 )(1 − a l2 ) ⎤ N N A k A l ⎢ ⎥= = ⋅ . (A) 2 2 1 − a k al ⎥⎦ k =1l =11 − a k a l k =1l =1 1 − a k 1 − a l ⎢⎣ N A2 N −1 N N −1 N A k Al A k Al k + + Further, σ o2 = 2 1 − a k a l l =1 k = l + 1 1 − a k a l k = 11 − a k =1 l = k + 1

=

N N

∑∑

∑∑

∑∑

∑∑

∑

=

N

N −1

Ak2

∑ ∑

∑ ∑

k N

∑ 1 − a2 + 2 ∑ ∑

k =1

k

A k Al . − a a 1 k l k =1 l = k + 1

(c) See Eq. (A).

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12.11 (a) H1 ( z) =

− 16.1 18.9 ( z − 2)( z + 4) + = 1+ . Making use of Eq. (12.87) and Table z − 0.6 z − 0.2 ( z − 0.2)( z − 0.6)

(−16.1) 2

2 × (−16.1) × (18.9) 1 − (0.6) × (0.2) 1 − (0.6) 2 1 − (0.2) 2 = 1 + 405.0156 + 372.0937 − 691.5682 = 86.5411 .

12.4, we get σ12, n = 1 +

+

(18.9) 2

+

Output of Program 12_4.m is 86.5412. (b) H 2 ( z) =

2(3z − 1)(2 z 2 − 4 z + 1) (2 z − 1)(z + 0.5)( z 2 + 0.8z + 0.5)

=

25 − 0.2174 − 18.7826 z − 21.2174 . + + z + 0.5 z − 0.5 z 2 + 0.8z + 0.5

Making use of Eq. (12.87) and Table 12.4, we get σ 22, n =

+ + +

2 × 25 × (−0.2174) + 1 + 0.25

(25) 2

1 − (0.5) 2 2 × 25 × [ −18.7826 + (−21.2174) × (−0.5)]

+

(0.2174) 2 1 − (0.5) 2

1 + 0.8 × (−0.5) + 0.5 × (−0.5) 2 2 × (−0.2174) × [ −18.7826 + (−21.2174) × 0.5] 1 + 0.8 × 0.5 + 0.5 × (0.5) 2

[(−18.7826) 2 + (−21.2174) 2 ] × (1 − 0.5) 2 − 2 × (−18.7826) × (−21.2174) × (1 − 0.5) × 0.8

[1 − (0.5) 2 ]2 + 2 × (0.5) × (0.8) 2 − [1 + (0.5) 2 ] × (0.8) 2 = 833.3333 + 0.0630 − 8.6960 − 563.7172 + 8.3799 + 704.1211 = 973.4841. (25) 2 (0.2174) 2 2(25)(−0.2174) 2(25)(−18.7826 + (−21.2174) *(−0.5)) σ 2,2 n = + + + 1 − (0.5) 2 1 − (0.5) 2 1 + 0.25 1 + 0.8(−0.5) + 0.5(−0.5) 2 2(−0.2174)(−18.7826 + (−21.2174) *(0.5)) + 1 + 0.8(0.5) + 0.5(0.5) 2 [(−18.7826) 2 + (−21.2174) 2 ](1 − 0.5) 2 − 2(−18.7826)(−21.2174)(1 − 0.5)(0.8) (1 − 0.52 ) 2 + 2(0.5)(0.8) 2 − (1 + 0.52 )(0.8) 2 = 833.3333 + 0.0630 − 8.6960 − 563.7172 − 8.3799 − 704.1211 = 973.4841 Output of Program 12_4.m is 973.4831. +

( z + 1) 2 1.4 z + 0.6 (c) H 3 ( z) = . Making use of Eq. (12.87) and Table 12.4, = 1+ z 2 + 0.6 z + 0.4 z 2 + 0.6 z + 0.4 [(1.4) 2 + (0.6) 2 ][1 − (0.4) 2 ] − 2 × 1.4 × 0.6 × 0.6(1 − 0.4) we get σ 32, n = 1 + [1 − (0.42) 2 ] + 2 × 0.4 × (0.6) 2 − [1 + (0.4) 2 ](0.6) 2 = 1 + 2.3333 = 3.3333. Output of Program 12_4.m is 3.3333. (d) H 4 ( z) =

− 0.825 4.1667 − 4.4917 ( z − 2.5)( z + 1.25)( z + 1) = 1+ + + . z + 0.8 z + 0.5 z − 0.4 ( z − 0.4)( z _ 0.8)( z + 0.5)

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Making use of Eq. (12.87) and Table 12.4, we get (0.825) 2 (4.1667) 2 (4.4917) 2 2 × 0.825 × 4.1667 2 × 0.825 × 4.4917 σ 42, n = 1 + + + + + 1 − 0.4 1 + 0.32 1 − (0.8) 2 1 − (0.5) 2 1 − (0.4) 2 2 × 4.4917 × 4.1667 + = 1 + 1.8906 + 23.1485 + 24.0183 − 11.4584 + 5.6146 − 31.1926 1 + 0.25 = 13.021. Output of Program 12_4.m is 13.0208. 12.12 H ( z ) = A +

B C . + z+ β z+γ

σ 2n = A 2 +

B2

1−β

2

+

C2

1− γ

2

+

2 BC . For β = 0.6, γ = −0.8, 1 − βγ

A = 3, B = −4, C = 2, we get

σ 2n = 3 2 +

42 2

+

22 2

+

2 × ( −4 ) × 2 = 34.3003. 1 − ( − 0 .6 ) × ( 0 .8 )

1 − ( 0 .6 ) 1 − ( 0 .8 ) Output of Program 12_4.m is 34.3003.

12.13 (a) Quantization of products before addition Direct Form II - The noise model is shown below. e 2[n] p

0

x[n]

y[n]

_1

z _d 1

e 1[n]

p

1

e 3[n]

The noise transfer function from the noise source e1[ n] to the filter output is

p + p1 z −1 p 0 z + p1 p −p d G1 ( z ) = 0 = = p0 + 1 0 1 . − 1 z + d1 z + d1 1 + d 1z The corresponding normalized noise variance at the output is σ12 = p02 +

( p1 − p0 d1 ) 2 . 1 − d12

The noise transfer function from the noise sources e2 [ n] and e3 [ n] to the filter output is G2 ( z) = G3 ( z) = 1. The corresponding normalized noise variance at the output is

σ 22 = σ 32 = 1. Hence the total noise variance at the output is σ o2 = σ12 + σ 22 + σ 32 ( p − p0 d1 ) 2 = 2 + p02 + 1 . 1 − d12

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Direct Form IIt - The noise model is shown below. e 2[n] p

x[n]

0

y[n] _

z1

p

_d

1

1

e 3[n]

e 1[n]

E ( z ) z −1 + E 2 ( z ) + E 3 ( z ) z −1 Analyzing the structure, we have that Y ( z ) = 1 . 1 + d 1z −1 The noise transfer function from e1[ n] to the filter output is Y ( z) z −1 1 . The corresponding normalized noise = = E1 ( z ) E ( z ) = E ( z ) = 0 1 + d1 z −1 z + d1 2 3 1 variance at the output is σ12 = . 1 − d12 G1 ( z) =

The noise transfer function from e2 [ n] to the filter output is d 1 z Y ( z) G2 ( z ) = = = = 1 − 1 . The corresponding z + d1 E1 ( z ) E ( z ) = E ( z ) = 0 1 + d1 z −1 z + d1 1

3

normalized noise variance at the output is σ 22 = 1 +

d12 1 − d12

=

1 1 − d12

.

The noise transfer function from e3 [ n] to the filter output is

Y (z) z −1 1 . The corresponding normalized noise = = − 1 E 3 ( z) E ( z ) = E ( z ) = 0 1 + d1 z z + d1 1 2 1 variance at the output is σ 32 = . 1 − d12 G3 ( z ) =

Hence the total noise variance at the output is σ o2 = σ12 + σ 22 + σ 32 =

3 1 − d12

.

(b) Quantization of products after addition

( p − p0 d1 ) 2 Direct Form II - σ o2 = σ12 + σ 22 + σ 32 = 2 + p02 + 1 . 1 − d12 Direct Form IIt - σ o2 = σ12 + σ 22 + σ 32 =

3 1 − d12

.

12.14 (a) Quantization of products before addition

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Cascade Structure #1: G( z ) =

(1 + 0.4 z −1 )(1 − 0.3z −1 ) (1 + 0.2 z −1 )(1 − 0.6 z −1 )

. The noise model of this structure is as

below x[n]

+

+

+

+ _ z 1

_ z 1

_

+

0.2

0.4

+

+

_ 0.3

0.6

+

e 4[n]

e 2[n] e 3[n]

e 1[n]

y[n]

The noise transfer function from the noise source e1[ n] to the filter output is ( z + 0.4)( z − 0.3) 0.375 0.125 G1 ( z ) = . = 1+ + z − 0.6 z + 0.2 ( z + 0.2)( z − 0.6) The corresponding normalized noise variance at the output is (0.375) 2 (0.125) 2 20.375 × 0.125 σ12, n = 1 + + + = 1.3197. 1 − (0.6) 2 1 − (−0.2) 2 1 − 0.6 × (−0.2) Output of Program 12_4.m is 1.3197. The noise transfer function from the noise sources e2 [ n] and e3 [ n] to the filter output is z − 0.3 0.3 . = 1+ G2 ( z ) = z − 0.6 z − 0.6 The normalized noise variance at the output due to each of these noise sources is (0.3) 2 σ 22, n = 1 + = 1.1406. 1 − (0.6) 2 Output of Program 12_4.m is 1.1406. The noise transfer function from the noise source e4 [ n] to the filter output is G 4 ( z) = 1. The corresponding normalized noise variance at the output is σ 42, n = 1. Hence the total normalized noise variance at the output is σ n2 = σ12, n + 2σ 22, n + σ 42, n = 4.6009. (1 − 0.3z −1 )(1 + 0.4 z −1 ) . The noise model of this structure is as Cascade Structure #2: G( z ) = (1 + 0.2 z −1 )(1 − 0.6 z −1 ) below: x[n]

+

+ _ z 1

_

+

0.2

e 1[n]

+

+

y[n]

_ z 1

_ 0.3

+

+

0.6

e 2[n] e 3[n]

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0.4

+ e 4[n]

441

σ12, n = 1.3197 as in Structure#1.

The noise transfer function from the noise sources e2 [ n] and e3 [ n] to the filter output is (1) 2 z + 0.4 1 = 2.5625. . Hence, σ 22, n = 1 + = 1+ G2 ( z ) = z − 0.6 z − 0.6 1 − (0.6) 2 Output of Program 12_4.m is 2.5625. The noise transfer function from the noise source e4 [ n] to the filter output is G 4 ( z) = 1. The corresponding normalized noise variance at the output is σ 42, n = 1. Hence the total normalized noise variance at the output is σ n2 = σ12, n + 2σ 22, n + σ 42, n = 7.4447. Cascade Structure #3: G( z ) =

(1 − 0.3z −1 )(1 + 0.4 z −1 ) (1 − 0.6 z −1 )(1 + 0.2 z −1 )

. The noise model of this structure is as

below: x[n]

+

+

+

+ _ z 1

_ z 1

+

_ 0.3

0.6

_

+

+

0.4

0.2

+ e 4[n]

e 2[n] e 3[n]

e 1[n]

y[n]


The noise transfer function from the noise sources e2 [ n] and e3 [ n] to the filter output is (0.2) 2 z + 0.4 0.2 = 1.0417. . Hence, σ 22, n = 1 + = 1+ G2 ( z ) = z + 0.2 z + 0.2 1 − (0.2) 2 Output of Program 12_4.m is 1.0417. The noise transfer function from the noise source e4 [ n] to the filter output is G 4 ( z) = 1. The corresponding normalized noise variance at the output is σ 42, n = 1. Hence the total normalized noise variance at the output is σ n2 = σ12, n + 2σ 22, n + σ 42, n = 4.4031. (1 + 0.4 z −1 )(1 − 0.3z −1 ) . The noise model of this structure is as Cascade Structure #4: G( z ) = (1 − 0.6 z −1 )(1 + 0.2 z −1 ) below: x[n]

+

+

+

+ _ z 1

_ z 1

+

0.6

e 1[n]

y[n]

0.4

_

+

+

0.2

e 2[n] e 3[n]

_

0.3

+

e 4[n]


The noise transfer function from the noise sources e2 [ n] and e3 [ n] to the filter output is

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(−0.5) 2 z − 0.3 − 0.5 = 1.2604. . Hence, σ 22, n = 1 + = 1+ z + 0.2 z + 0.2 1 − (0.2) 2 Output of Program 12_4.m is 1.2604. The noise transfer function from the noise source e4 [ n] to the filter output is G 4 ( z) = 1. G2 ( z ) =

The corresponding normalized noise variance at the output is σ 42, n = 1. Hence the total normalized noise variance at the output is σ n2 = σ12, n + 2σ 22, n + σ 42, n = 4.8405. Hence, the Cascade Structure #3 has the smallest round off noise variance. (b) Quantization of products after addition. From the results of Part (a) Cascade Structure #1: σ n2 = σ12, n + σ 22, n + σ 42, n = 3.4603.

Cascade Structure #2: σ n2 = σ12, n + σ 22, n + σ 42, n = 4.8822. Cascade Structure #3: σ n2 = σ12, n + σ 22, n + σ 42, n = 3.614. Cascade Structure #4: σ n2 = σ12, n + σ 22, n + σ 42, n = 3.5801. Hence, the Cascade Structure #1 has the smallest round off noise variance. 12.15 (a) Quantization of products before addition. − 0.625 (1 + 0.4 z −1 )(1 − 0.3z −1 ) 0.625 = 1+ + . Parallel Form #1: G( z ) = − 1 − 1 − 1 (1 + 0.2 z )(1 − 0.6 z ) 1 − 0.6 z 1 + 0.2 z −1 The noise model of this structure is shown below:

0.625

e 2[n]

+

+

x[n]

y[n]

_ z 1

e 1[n] 0.6

+

0.6

e 4[n]

+ _ z 1

e 3[n]

+

_ 0.2

The noise transfer function from the noise sources e1[n] and e2 [n] to the filter output is z 0.6 G2 ( z ) = . The corresponding normalized noise variance at the output is = 1+ z − 0.6 z − 0.6

σ 22, n = 1 +

(0.6) 2 1 − (0.6) 2

= 1.5625.

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The noise transfer function from the noise sources e3 [n] and e4 [n] to the filter output is − 0.2 z = 1+ . The corresponding normalized noise variance at the output is G4 ( z) = z + 0.2 z + 0.2

σ 22, n = 1 +

(−0.2) 2 1 − (0.2) 2

= 1.0417.

Hence the total noise variance at the output is σ n2 = 2σ 22, n + 2σ 42, n = 5.2804.

0.375 0.125 0.375z −1 0.125z −1 . + = 1+ + z − 0.6 z + 0.2 1 − 0.6 z −1 1 + 0.2 z −1 The noise model of this structure is shown below: Parallel Form #2: G( z) = 1 +

0.375

e 2[n] _ z 1

+

x[n]

e 1[n] 0.125

+

y[n]

0.6

e 4[n]

+

e 3[n]

+

+

_ z 1

_ 0.2

The noise transfer function from the noise sources e1[n] and e2 [ n] to the filter output is

G2 ( z) =

z −1 1 − 0.6 z −1

=

1 . The corresponding normalized noise variance at the output is z − 0.6

(1) 2

σ 22, n =

= 1.5625. 1 − (0.6) 2 The noise transfer function from the noise sources e3 [n] and e4 [n] to the filter output is G4 (z) =

σ 22, n =

z −1 1 + 0.2 z −1

(1) 2 1 − (0.2) 2

=

1 . The corresponding normalized noise variance at the output is z + 0.2

= 1.0417.

Hence the total noise variance at the output is σ n2 = 2σ 22, n + 2σ 42, n = 5.2804. Therefore, both parallel forms have identical round off noise variances. (b) Quantization of products after addition. From the results of Part (a)

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Parallel Form #1: σ n2 = σ 22, n + σ 42, n = 2.6402. Parallel Form #2: σ n2 = σ 22, n + σ 42, n = 2.6402. Hence, both parallel forms have the same round off noise variances.

2 − 0.4 z −1 − 0.6 z −2

12.16 (a) Direct Form II realization H ( z) =

1 + 0.2 z −1 − 0.15z − 2

is shown below along with the

noise model. x[n]

2

+ _

+

_ z 1

+

+ e1 [n]

_ z 1

y[n]

e3 [n] _ 0.4

0.2

+

+

+

e4 [n] _ 0.6

0.15

+

+

e2 [n]

e5 [n]

The noise transfer function from e1[ n] and e2 [ n] to filter output is

G1 ( z ) = H ( z) =

2 − 0.4 z −1 − 0.6 z −2 1 + 0.2 z −1

=2+

− 0.15z − 2

− 0.125 − 0.675 . The corresponding normalized + z + 0.5 z − 0.3

noise variance at the output is σ12, n = (2) 2 +

(−0.125) 2

+

(−0.675) 2

1 − (−0.5) 2 1 − (0.3) 2 Output of Program 12_4.m is 4.6683.

+

2 × (−0.125) × (−0.675) = 4.6683. 1 − 0.3 × (−0.5)

The noise transfer function from the noise sources e3 [ n] , e4 [ n] to the filter output is G3 ( z ) = 1. The corresponding normalized noise variance at the output is σ 32, n = 1. Hence the total noise variance at the output is σ n2 = 2σ12, n + 3σ 32, n = 12.1898. (b) Cascade Form: H ( z ) =

2(1 − 0.6568z −1 )(1 + 0.4568z −1 )

. There are more than 2 possible (1 − 0.3z −1 )(1 + 0.5z −1 ) cascade realizations. We consider here only two structures. ⎛ 1 − 0.6568 z −1 ⎞⎛ 1 + 0.4568 z −1 ⎞ ⎟. The noise model of this ⎟⎜ Cascade Form #1 - H ( z ) = 2⎜⎜ − 1 ⎟ ⎜ 1 + 0 .5 z − 1 ⎟ 1 0 . 3 z − ⎠ ⎠⎝ ⎝ structure is on the next page.

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2

+

x[n]

+

+ _ z 1

e 1[n]

+

0.3

_

+

+

y[n]

_ z 1

0.6568

+

+

_ 0.5

0.4568

+

e 3[n] e 4[n]

e 2[n]

e 5[n]

The noise transfer function from e1[ n] and e2 [ n] to filter output is 1 − 0.2 z −1 − 0.3z −2 − 0.0625 − 0.3375 G1 ( z ) = . The corresponding normalized noise = 1+ + z + 0.5 z − 0.3 1 + 0.2 z −1 − 0.15z − 2 variance at the output is (−0.0625) 2 (−0.3375) 2 (−0.0625) × (−0.3375) σ12, n = (1) 2 + + + = 1.1671. 1 − 0.3 × (−0.5) 1 − (−0.5) 2 1 − (0.3) 2 Output of Program 12_4.m is 1.1671. The noise transfer function from e3 [ n] and e4 [ n] to filter output is

G3 ( z ) =

1 + 0.4568z −1 1 + 0.5z −1

=

− 0.0432 z + 0.4568 . The corresponding normalized noise = 1+ z + 0.5 z + 0.5

variance at the output is σ 32, n = (1) 2 +

(−0.0432) 2 1 − (0.5) 2

= 1.0025.

Output of Program 12_4.m is 1.0025. The noise transfer function from the noise source e5 [ n] to the filter output is G5 ( z ) = 1. The corresponding normalized noise variance at the output is σ 32, n = 1. Hence the total noise variance at the output is σ n2 = 2σ12, n + 2σ 32, n + σ 52, n = 5.3392 ⎛ 1 + 0.4568 z −1 ⎞⎛ 1 − 0.6568 z −1 ⎞ ⎟. The noise model of this ⎟⎜ Cascade Form #2 - H ( z ) = 2⎜⎜ − 1 ⎟ ⎜ 1 + 0 .5 z − 1 ⎟ 1 0 . 3 z − ⎠ ⎠⎝ ⎝ structure is shown below: 2 x[n]

+

+

+

+

+

_ z 1

e 1[n]

+

0.3

e 2[n]

_ z 1

0.4568

+

+

_ 0.5

e 3[n] e 4[n]

_

y[n]

0.6568

+ e 5[n]

The noise transfer function from e1[ n] and e2 [ n] to filter output is same as for the Cascade Structure #1. Hence, the corresponding normalized noise variance at the output is σ12, n = 1.1671 .

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The noise transfer function from e3 [ n] and e4 [ n] to filter output is

G3 ( z ) =

1 − 0.6568z −1 1 + 0.5z −1

=

− 1.1568 z + 0.6568 . The corresponding normalized noise = 1+ z + 0.5 z + 0.5

variance at the output is σ 32, n = (1) 2 +

(−1.1568) 2 1 − (0.5) 2

= 2.7842.

Output of Program 12_4.m is 2.7842. The noise transfer function from the noise source e5 [ n] to the filter output is G5 ( z ) = 1. The corresponding normalized noise variance at the output is σ 52, n = 1. Hence the total noise variance at the output is σ n2 = 2σ12, n + 2σ 32, n + σ 52, n = 6.1184. (c) Parallel Form I realization - H ( z) = 4 +

0.25 1 + 0.5z −1

+

− 2.25

1 − 0.3z −1

. The noise model of this

structure is shown below: 4

0.25

e 2[n]

+ e 1[n]

+

+

x[n]

y[n]

_ z 1

e 3[n] _ 2.25

+

_ 0.5

e 4[n]

+ _ z 1

e 5[n]

+

0.3

The noise transfer function from e2 [ n] and e3 [ n] to filter output is 1 − 0.5 . The corresponding normalized noise variance at the output is = 1+ G2 ( z ) = z + 0.5 1 + 0.5z −1

σ 22, n = (1) 2 +

(−0.5) 2

=

4 . 3

1 − (0.5) 2 The noise transfer function from e4 [ n] and e5 [ n] to filter output is 1 0.3 . The corresponding normalized noise variance at the output is = 1+ G4 (z) = z − 0.3 1 − 0.3z −1 σ 42, n = (1) 2 +

(0.3) 2

= 1.0989. 1 − (−0.3) 2 The noise transfer function from the noise source e1[ n] to the filter output is G1 ( z) = 1.

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The corresponding normalized noise variance at the output is σ12, n = 1. Hence the total noise variance at the output is σ n2 = σ12, n + 2σ 22, n + σ 42, n = 5.8645. Parallel Form II realization. - H ( z) = 2 +

− 0.125 − 0.675 + . The noise model of this structure z + 0.5 z − 0.3

is shown below: 2

_ 0.125

e 2[n]

e 3[n] _ 0.675

+

+

y[n]

_ 0.5

e 4[n]

+

e 5[n]

e 1[n] _ z 1

+

x[n]

+

+

_ z 1

0.3

1 . The z + 0.5 1 4 = . corresponding normalized noise variance at the output is σ 22, n = 2 3 1 − (0.5) 1 . The The noise transfer function from e4 [ n] and e5 [ n] to filter output is G 4 ( z ) = z − 0.3 1 = 1.0989. corresponding normalized noise variance at the output is σ 42, n = 1 − (−0.3) 2 The noise transfer function from the noise source e1[ n] to the filter output is G1 ( z) = 1.

The noise transfer function from e2 [ n] and e3 [ n] to filter output is G2 ( z) =

The corresponding normalized noise variance at the output is σ12, n = 1. Hence the total noise variance at the output is σ n2 = σ12, n + 2σ 22, n + σ 42, n = 5.8645. 12.17 H ( z) =

2 − 0.4 z −1 − 0.6 z −2

. Using Program 8_5.m, the Gray-Markel realization of H(z) is 1 + 0.2 z −1 − 0.15z − 2 found to be as shown on top of the next page where d 2 = 0.2353, d1' = −0.15, α1 = −0.6, α 2 = −0.28, α3 = 1.9759.

(a) For quantization of products before addition, the noise model is shown on the next page:

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e1 [n]

e2 [n]

x[n] _d

_ d' 1

2

d 1'

d2 _1

_

z1

z

e4 [n]

α1

e3 [n]

α2

α3 y[n] e5 [n]

Analyzing the above structure, we obtain the noise transfer function for each noise source, 2 − 0.4 z −1 − 0.6 z −2 e i [ n], 1 ≤ i ≤ 7 : G1 ( z) = H ( z) = , 1 + 0.2 z −1 − 0.15z − 2 (α2 d1' + α3 ) + [α1 (1 − d 22 )d1' + α2 ]z −1 + α1 (1 − d 22 )z − 2 G2 ( z ) = 1 + 0.2 z −1 − 0.15z − 2 2.0179 − 0.1950 z −1 − 0.5668z −2 , = 1 + 0.2 z −1 − 0.15z − 2 α2 + [α1 (1 − d 22 ) − α2 d1' d 2 −α3 d 2 ]z −1 + α1 (1 − d 22 )d1' d 2 z − 2 G3 ( z ) = 1 + 0.2 z −1 − 0.15z − 2 − 0.28 − 1.0416 z −1 + 0.02 z −2 , = 1 + 0.2 z −1 − 0.15z − 2 G 4 ( z) = α1 = −0.6, G5 ( z) = G6 ( z) = G7 ( z ) = 1. Using Program 12_4.m, we get σ12, n = 4.6683, σ 22, n = 4.4549, σ 32, n = 1.1199, σ 42, n = 0.36. Hence, the total normalized output noise variances = 4.6683 + 4.459 + 1.1199 + 0.36 + 3(1) = 13.6072. (b) The realization along with the noise model for the quantization of products after addition is the same as in Part (a).

The noise transfer functions from the noise sources e2 [ n], e3 [ n], and e4 [ n], are G2 ( z), G3 ( z ), and G 4 ( z), respectively and are given in Part (a). Their corresponding noise variances were also derived in Part (a) and are given by σ 22, n = 4.4549, σ 32, n = 1.1199, σ 42, n = 0.36. Hence, the total normalized output noise variances = 4.459 + 1.1199 + 0.365 + 1 = 6.9389 12.18 The noise model is shown on the next page. The noise transfer function from e1 [ n] to the

1 − z −2

d z + (1 + d1d 2 ) . Its = 1− 1 z 2 + d1 z + d1d 2 1 + d1 + d1d 2 + d1 z + d1d 2 corresponding output noise variance is given by

output is G1 ( z ) =

z −1

z−2

=

z2 − 1

z2

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_ z 1

_ z 1

+

x[n]

+

y[n]

_1

+

+

+

d1

e 2[n]

e1[n]

σ12, n = 1 +

+

d2

[ d12 + (1 + d1d 2 ) 2 ](1 − d12 d 22 ) − 2[ d1 (1 + d1d 2 )(1 − d1d 2 )d 1] (1 − d12 d 22 ) 2

+ 2d13 d 2

− (1 + d12 d 22 )d12

=

2 . 1 − d1d 2

The noise transfer function from e 2 [ n] to the output is d12 z + (1 + d1d 2 )d1 d1 (1 − z − 2 ) G1 ( z ) = = d1 − . Its corresponding z 2 + d1 z + d1d 2 1 + d1 z −1 + d1d 2 z − 2 normalized output noise variance is given by 2d12 σ 22, n = d12 σ12, n = . 1 − d 1d 2 Total output noise variance is therefore

σ o2, n

=

σ12, n

+ σ 22, n

=

2(1 + d12 ) 1 − d1d 2

.

12.19 (a) Quantization of products before addition The noise model of the second-order coupled form structure is shown below; E1(z) V(z)

α

_ z 1

Y(z)

γ E2 (z)

δ

E 3(z) U(z) E4 (z)

_ z 1

β

We observe from discussion on Page 675 that α = δ = r cos θ , β = − γ = sin θ and hence, α + δ = 2r cos θ , α δ − β γ = r 2 .

Analysis yields U ( z) = βz −1Y ( z ) + αV ( z) + E 3 ( z), V ( z) = z −1U ( z ), and Y ( z ) = γV ( z ) + δz −1Y ( z ) + E1 ( z ). Eliminating the variables U(z) and V(z) from these equations

1 − (α + δ ) z −1 + (α δ − β γ ) z −2

Y ( z) =

γz −1

E 3 ( z) + E1 ( z ). Hence, the noise 1 − αz − 1 1 − αz − 1 transfer function from the noise sources e1[ n] and e2 [ n] to the output is given by we arrive at

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G1 ( z) =

Y ( z) 1 − αz − 1 z 2 − αz = = E1 ( z ) E ( z ) = E ( z ) = 0 1 − (α + δ )z −1 + (α δ − β γ)z − 2 z 2 − (α + δ )z + (α δ − β γ) 3 4

δz − ( α δ − β γ )

= 1+

r cos(θ )z − r 2

= 1+

, and the noise transfer function from z 2 − (α + δ ) z + (α δ − β γ ) z 2 − 2r cos(θ )z + r 2 the noise sources e3 [ n] and e4 [ n] to the output is given by Y (z) γ z −1 = = E 3 ( z ) E ( z ) = E ( z ) = 0 1 − ( α + δ ) z − 1 + (α δ − β γ ) z − 2 1 2 − r sin(θ )z = . z 2 − 2r cos(θ )z + r 2 The output roundoff noise variance due each of the noise sources G3 ( z ) =

σ12, n = 1 +

= 1+

γz z2

− (α + δ ) z + (α δ − β γ )

e1[ n] and e2 [ n] is (r 2 cos θ + r 4 )(1 − r 4 ) − 2(−r 3 cos θ + r 3 cos θ ⋅ r 2 )(−2r cos θ ) (1 − r 4 ) 2 + 2r 2 (4r 2 cos 2 θ ) − (1 + r 4 )(4r 2 cos 2 θ )

r 2 (1 + r 2 )(r 2 + cos 2 θ ) − 4r 4 cos 2 θ (1 − r 2 )[(1 + r 2 ) 2 − 4r 2 cos 2 θ ]

, and the output roundoff noise variance due each of

the noise sources e3 [ n] and e4 [ n] is σ 32, n =

r 2 (1 + r 2 ) sin 2 θ (1 − r 2 )[(1 + r 2 ) 2 − 4r 2 cos 2 θ ]

. Hence, the total

roundoff noise variance at the output is σ n2 = 2(σ12, n + σ 22, n ) ⎛ ⎞ r 2 (1 + r 2 )(r 2 + cos 2 θ ) − 4r 4 cos 2 θ r 2 (1 + r 2 ) sin 2 θ ⎟. = 2⎜⎜ 1 + + 2 )[(1 + r 2 ) 2 − 4r 2 cos 2 θ ] 2 )[(1 + r 2 ) 2 − 4r 2 cos 2 θ ] ⎟ − r − r ( 1 ( 1 ⎝ ⎠ Quantization of products after addition Here, the total roundoff noise variance at the output is σ n2 = σ12, n + σ 22, n r 2 (1 + r 2 )(r 2 + cos 2 θ ) − 4r 4 cos 2 θ r 2 (1 + r 2 ) sin 2 θ = 1+ + . (1 − r 2 )[(1 + r 2 ) 2 − 4r 2 cos 2 θ ] (1 − r 2 )[(1 + r 2 ) 2 − 4r 2 cos 2 θ ] (b) From the solution of Problem 12.4 we observe that the transfer function of the modified c2d , where c = r sin θ and d = cot θ. coupled-form structure is H ( z) = z 2 − 2c z + c 2 (1 + d 2 )

The noise model is shown on top of the next page. Analysis yields

Y ( z) =

E 4 ( z ) + cz −1 E 2 ( z) + cz −1 E 3 ( z) − cdz −1 E 4 ( z) + c 2 z −2 E1 ( z ) − c 2 dz −2 E 3 ( z) 1 − 2cdz −1 + c 2 (1 + d 2 )z − 2

.

The noise transfer function from e1[ n] to the output is

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E 2(z) c

_ z 1

E 4(z) c

_ z 1

d

Y(z)

d

E 1(z)

E 3(z) _1

G1 ( z) =

Y ( z) c 2 z −2 r 2 sin 2 θ . = = E1 ( z ) E ( z ) = E ( z ) = E ( z ) = 0 1 − 2cdz −1 + c 2 (1 + d 2 )z − 2 z 2 − 2r cos θ ⋅ z + r 2 2 3 4

The normalized output roundoff noise variance due to e1[ n] is thus

σ12, n =

(r 4 sin 4 θ )(1 − r 4 ) (1 − r 4 ) 2 + 8r 4 cos 2 θ − (1 + 4r 2 cos 2 θ )r 4

.

The noise transfer function from e2 [ n] to the output is (r sin θ )z Y ( z) cz −1 G2 ( z ) = . = = − 1 2 2 − 2 2 E 2 ( z ) E ( z ) = E ( z ) = E ( z ) = 0 1 − 2cdz + c (1 + d )z z − 2r cos θ ⋅ z + r 2 1

3

4

The normalized output roundoff noise variance due to e2 [ n] is thus

(r 2 sin 2 θ )(1 − r 4 )

σ 22, n =

. (1 − r 4 ) 2 + 8r 4 cos 2 θ − (1 + 4r 2 cos 2 θ )r 4 The noise transfer function from e3 [ n] to the output is (r sin θ )z − r 2 sin 2 θ ⋅ cot θ Y ( z) cz −1 − c 2 dz −2 . = = E 3 ( z) E ( z ) = E ( z ) = E ( z ) = 0 1 − 2cdz −1 + c 2 (1 + d 2 )z − 2 z 2 − 2r cos θ ⋅ z + r 2 1 2 4 The corresponding normalized output roundoff noise variance is thus (r 2 sin 2 θ − r 4 sin 4 θ cot 2 θ )(1 − r 4 ) − 4(r 3 sin 3 θ cot θ )(1 − r 2 )r cos θ . σ 32, n = (1 − r 4 ) 2 + 8r 4 cos 2 θ − (1 + 4r 2 cos 2 θ )r 4 The noise transfer function from e4 [ n] to the output is G3 ( z ) =

G4 (z) = = 1+

Y ( z) 1 − cdz −1 z 2 − (r cos θ )z = = E 4 ( z) E ( z ) = E ( z ) = E ( z ) = 0 1 − 2cdz −1 + c 2 (1 + d 2 )z − 2 z 2 − 2r cos θ ⋅ z + r 2 1 2 3

(r cos θ )z − r 2

. The corresponding normalized output roundoff noise variance is z 2 − 2r cos θ ⋅ z + r 2 (r 2 cos 2 θ − r 4 )(1 − r 4 ) + 4(r 3 cos θ )(1 − r 2 )r cos θ . thus σ 42, n = 1 + (1 − r 4 ) 2 + 8r 4 cos 2 θ − (1 + 4r 2 cos 2 θ )r 4

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12.20 The noise model of the Kingsbury structure is shown below: E 1(z) U(z)

k1

E 3(z) E 2(z) _k V(z) 1 1 z–

1 z–

Y(z)

k2

Analysis yields U ( z) = k1Y ( z) + E1 ( z), V ( z ) =

1

U ( z) + k 2Y ( z ) + E 2 ( z), and 1 − z −1 k z −1 z −1 z −1 Y ( z) = [ − k1V ( z) + E 3 ( z )] V ( z) + E 3 ( z ). Eliminating U (z ) and V (z ) =− 1 1 − z −1 1 − z −1 1 − z −1 from these equations we arrive at 1 − [2 − k1 (k1 + k 2 )]z −1 + (1 − k1k 2 )z −2 k z −1 k z −1 z −1 Y ( z) = − 1 E1 ( z) − 1 E 2 (z) + E 3 ( z). (1 − z −1 ) 2 (1 − z −1 ) 2 1 − z −1 1 − z −1 Hence, the noise transfer function from e1[ n] to the output is

− k 1 z −1 − k1 z Y ( z) . = = E1 ( z ) E ( z ) = E ( z ) = 0 1 − [2 − k1 (k1 + k 2 )]z −1 + (1 − k1k 2 )z − 2 z 2 − (2r cos θ )z + r 2 2 3 The corresponding normalized output roundoff noise variance is thus given by k12 (1 − r 4 ) k12 (1 − r 2 )(1 + r 2 ) = σ12, n = (1 − r 4 ) 2 + 2r 2 (−2r cos θ ) 2 − (1 + r 4 )(−2r cos θ ) 2 (1 − r 2 ) 2 (1 + r 2 ) 2 − 4r 2 cos 2 θ (1 − r 2 ) 2

G1 ( z) =

=

k12 (1 + r 2 )

. (1 − r 2 )[(1 + r 2 ) 2 − 4r 2 cos 2 θ ] The noise transfer function from e2 [ n] to the output is

− k 1z −1 (1 − z −1 ) − k1 z + k1 Y ( z) . = = E 2 ( z ) E ( z ) = E ( z ) = 0 1 − [2 − k1 (k1 + k 2 )]z −1 + (1 − k1k 2 )z − 2 z 2 − (2r cos θ )z + r 2 1 3 Its corresponding normalized output roundoff noise variance is thus given by 2 k12 (1 + r 2 ) σ 22, n = . (1 − r 2 )[(1 + r 2 ) 2 − 4r 2 cos 2 θ ] Finally, the noise transfer function from e3 [ n] to the output is

G2 ( z ) =

Y (z) z −1 (1 − z −1 ) z −1 . = = − 1 − 2 2 E 3 ( z) E ( z ) = E ( z ) = 0 1 − [2 − k1 (k1 + k 2 )]z + (1 − k1k 2 )z z − (2r cos θ )z + r 2 1 2 Its corresponding normalized output roundoff noise variance is thus given by 2(1 + r 2 ) . Hence, the total normalized noise variance due to σ 22, n = (1 − r 2 )[(1 + r 2 ) 2 − 4r 2 cos 2 θ ]

G3 ( z ) =

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product roundoff before (after) summation is given by (3k12 + 2)(1 + r 2 ) σ o2, n = σ12, n + σ 22, n + σ 32, n = . (1 − r 2 )[(1 + r 2 ) 2 − 4r 2 cos 2 θ ] 12.21 Let e[n] be a noise source due to product roundoff inside G (z ) with an associated noise transfer function Ge (z ) to the output of G (z ) . Then the normalized noise power at the output of the cascade structure due to e[n] is given by 1 1 σ e2, n = Ge ( z)A2 ( z)Ge ( z −1 ) A2 ( z −1 )z −1dz = Ge ( z)Ge ( z −1 )z −1dz since 2πj 2πj

∫

∫

C C − 1 A 2 ( z ) A 2 ( z ) = 1. Assuming G (z ) is realized with the lowest product roundoff noise power σ g2,n , the normalized noise power at the output of the cascade due to product roundoff in G (z ) is still σ g2,n . From the solution of Problem 12.14, we note that the lowest roundoff noise power (assuming quantization before addition) is σ g2, n = 4.4031 for cascade structure #3. Next, from

the solution of Problem 12.18, we note that the normalized output noise power due to noise 2d12 sources in A2 ( z) is σ 22, n = . Substituting d1 = 0.2 and d1d 2 = −0.5, we get 1 − d1d 2 σ 22, n = 0.0533 . Hence, the total output noise variance of the cascade is given by σ g2, n + σ 22, n = 4.4031 + 0.0533 = 4.4564.

12.21 The noise model for the structure is as shown below: x[n]

y[n] _1 z _ 0.3 e 1[n]

_ z 1 0.4 e 2[n]

The noise transfer function from e1[ n] to the output is G1 ( z) =

1 (1 + 0.3z −1 )(1 − 0.4 z −1 )

0.2286 − 0.1286 . Thus, the normalized noise power at the output due to e1[ n] is + z − 0.4 z + 0.3 (0.2286) 2 (−0.1286) 2 2 × 0.2286 × (−0.1286) σ12 = 1 + + + =1.0279. 1 − (−0.3)(0.4) 1 − (0.4) 2 1 − (−0.3) 2 0.4 1 . = 1+ The noise transfer function from e2 [ n] to the output is G2 ( z) = − 1 z − 0.4 1 − 0.4 z (0.4) 2 = 1.1905. Thus, the normalized noise power at the output due to e2 [ n] is σ 22 = 1 + 1 − (0.4) 2 Therefore, the total normalized noise power at the output = 1.0279 + 1.1905 = 2.2184.

= 1+

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For a 9-bit signed two’s-complement number representation, the quantization level is δ = 2 −8 = 0.0039062. Hence, the total output noise power due to product roundoff is δ2 σ o2 = (2.2184 ) = 2.8208 × 10 −6. 12 12.22 The unscaled structure is shown below: F1 (z) y[n]

x[n] _ z 1

_ 0.4

Now, F1 ( z) =

1 1 + 0.4 z −1

1.1905. Next, H ( z) =

= 1+

0.5

(−0.4) 2 − 0.4 2 = . Using Table 12.4 we obtain F1 2 = 1 + z + 0.4 1 − (−0.4) 2

1 + 0.5z −1 1 − 0.4 z −1

= 1+

0.9 . Using Table 12.4 we obtain z − 0.4

(0.9) 2 2 H 2 = 1+ = 1.9643. From Eq. (12.128a), F1 2 = α1 = 1.0911, and from Eq. 1 − (−0.4) 2 (12.128b), H 2 = α 2 = 1.0415. The scaled structure is shown below, where b01 = β1 and b11 = 0.5 β1 . From Eq. (12.129) 1 and (12.132a), K = β 0 K = β 0 = = 0.9165 and from Eq. (12.132b), α1 α 1.0911 β1 = 1 = = 0.7785. Therefore, b01 = β1 = 0.7785 and b11 = 0.5 β1 = 0.3893. α 2 1.0415 _ b01

_ K

x[n] _ 0.4

y[n]

_ z 1

_ b11

12.24 The unscaled structure is shown below: F1 (z)

y[n]

x[n] _ 0.4

z–1

z–1 0.7

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0.8

_ 0.6

455

1

Now, F1 ( z) =

z2

=

= 1+

− 0.45 0.05 . Using Table + z + 0.6 z − 0.2

1 + 0.4 z −1 − 0.12 z − 2 z 2 + 0.4 z − 0.12 (−0.45) 2 (0.05) 2 2 × (−0.45) × 0.05 2 = 1.2788. Next, + + 12.4 we get F1 2 = 1 + 1 − (−0.6) × 0.2 1 − (−0.6) 2 1 − (0.2) 2 H (z) =

1 + 0.8z −1 − 0.6 z −2 1 + 0.4 z −1

− 0.12 z − 2

− 0.5 0.9 . Using Table 12.4 we get + z + 0.6 z − 0.2

= 1+

(0.9) 2 (−0.5) 2 2 × 0.9 × (−0.5) 2 H 2 = 1+ + + = 1.7225. 2 2 1 − (−0.6) × 0.2 1 − (−0.6) 1 − (0.2) From Eq. (12.128a) we get F1 2 = α1 = 1.1308 and from Eq. (12.28b) we get 1 H 2 = α 2 = 1.3124. Hence, from Eq. (12.132a) we have β 0 = = 0.8843 and from Eq. α1 α (12.132b) we have β1 = 1 = 0.8616. The scaled structure is as shown below, where α2 K = β 0 K = β0 = 0.8843, b 01 = 1 × β1 = 0.8616, b 11 = 0.8 × β1 = 0.6893, and b 21 = (−0.6) × β1 = −0.5170. _ b01

_ K

x[n] _ 0.4

z–1

_ b11

–1

_ b21

z 0.7

12.25 H ( z) = 3 +

− 4 z −1

1 + 0.6 z −1

+

2 z −1 1 − 0.8 z −1

y[n]

. The unscaled structure is shown below: A=3

F 1(z)

_

z1

x[n]

y[n]

B =_4 _ 0.6

F 2(z) _1

z

C=2 0.8

Here, F1 ( z) =

− 4 z −1

1 + 0.6 z −1

= 1+

− 4.6 ( − 4 .6 ) 2 2 . Using Table 12.4 we get F1 2 = 1 + z + 0.6 1 − ( − 0 .6 ) 2

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= 34.0625. Likewise, F2 ( z) =

2 z −1 1 − 0.8z −1

= 1+

2.8 . Using Table 12.4 we get z − 0.8

(2.8) 2 −4 2 2 F2 2 = 1 + = 22.7778. Also, H ( z ) = 3 + . Using Table 12.4 we get + z + 0.6 z − 0.8 1 − (0.8) 2 ( −4) 2 22 2 × ( −4) × 2 2 H 2 = 32 + + + = 34.3003. 2 2 1 − (−0.6) × 0.8 1 − (−0.6) 1 − (0.8) Denote γ0 = H 2 = 5.8566, γ1 = F 1 2 = 5.8363, γ2 = F 2 2 = 4.7726. −4 A 3 B C 2 Hence, A = = = 0.5140, B = = = −0.8381, C = = = 0.3415, γ1 5.8363 γ2 4.7726 γ0 5.8566 γ γ K1 = 1 = 0.9965, K 2 = 2 = 0.8149. γ0 γ0 The scaled structure is shown below: _ A _ B

_1

x[n]

K1

y[n]

z

_ C

_ 0.6 _1

z

K2

0.8

The noise model is _ A _ eB[n] B

_

z1

x[n] e [n] β _ eC[n] C

_1

y[n]

eK [n] 1

_ 0.6

z

e γ [n]

e [n] K1 A

K2 eK [n] 2

0.8

The noise transfer function from the noise sources eB [n] and eβ [n] is given by: − 0.5979 0.9965z = 0.9965 + G1 ( z) = . Using Table 12.4, we obtain z + 0.6 z + 0.6

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σ12 = (0.9965) 2 +

(−0.5979) 2

= 1.5516. 1 − (−0.6) 2 Likewise, the noise transfer function from the noise sources eC [n] and e γ [n] is given by: 0.8149z 1.6519 = 0.8149 + . Using Table 12.4, we obtain G2 ( z ) = z − 0.8 z − 0.8 σ 22 = (0.8149) 2 +

(1.6519) 2

= 1.8446. 1 − (0.8) 2 The noise transfer function from the remaining three noise sources is G3 ( z ) = 1. Hence, the normalized output noise variance due to these three noise sources is given by σ 32 = 3. Therefore, the total normalized output noise variance is σ 20 = 2σ12 + 2σ 22 + σ 32 = 9.7924. 12.26 The scaled structure is shown below. k1

k2

+

+

k3

+ _ z 1

_ z 1

a1

+

b 1k 2

a2

b 2k 3

⎛ 1 + 0 .4 z − 1 ⎞ ⎛ 1 − 0 .3 z − 1 ⎞ ⎟⎜ ⎟ Here, a1 = −0.2, b1 = 0.4, (a) Cascade Structure # 1 – H ( z ) = ⎜ ⎜ 1 + 0.2 z −1 ⎟⎜ 1 − 0.6 z −1 ⎟. ⎝ ⎠⎝ ⎠ a 2 = 0.6, b2 = −0.3. The values of the scaling constants are determined below. 1 1 2 . Using Table 12.4 we get F1 2 = = 1.0417. Hence, F1 ( z ) = z + 0.2 1 − (−0.2) 2

γ1 = F1 2 = 1.0206. Likewise, ⎛ 1 + 0 .4 z − 1 ⎞ ⎛ ⎞ − 0.25 1.25 z −1 z + 0 .4 ⎟⎜ ⎟= = + . Using Table 12.4 F2 ( z ) = ⎜⎜ − 1 ⎟⎜ 1 − 0.6 z −1 ⎟ ( z + 0.2)( z − 0.6) − z z + 0 . 6 0 . 2 + 1 0 . 2 z ⎝ ⎠⎝ ⎠ 2 (1.25) (−0.25) 2 2 × 1.25 × (−0.2) 2 + + = 1.9485. Hence, we get F2 2 = 1 − (0.6) 2 1 − (−0.2) 2 1 − 0.6 × (−0.2) z 2 + 0.1z − 0.12 0.375 0.125 γ2 = F2 2 = 1.3959. Next, H ( z ) = . Using Table 12.4 = 1+ + 2 z − 0.6 z + 0.2 z − 0.4 z − 0.12 (0.375) 2 (0.125) 2 2 × 0.375 × 0.125 2 + + = 1.3197. Hence, we get H 2 = 1 + 2 2 1 − 0.6 × (−0.2) 1 − (0.6) 1 − (−0.2) γ0 = H 2 = 1.1488.

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The scaling multipliers are therefore given by k1 =

γ 1 = 0.9798, k 2 = 1 = 0.7311, and γ1 γ2

γ k3 = 2 = 1.2151. Therefore, b1 k2 = 0.2925 and b 2 k3 = −0.3645. γ0 The noise at the output due to the input multiplier k1 and multiplier a1 have a variance σ12 = γ12 = 1.0417. The noise at the output due to a1 , k2 , and b1k2 have variance σ 22 which is calculated below. The noise transfer function for these noise sources is 1.2151 − 0.3645z −1 1.2151z − 0.3645 0.3646 G2 ( z ) = . Using Table 12.4 we get = = 1.2151 + − 1 z − 0.6 z − 0.6 1 − 0.6 z

σ 22 = (1.2151) 2 +

(0.3646) 2

= 1.6841. 1 − (0.6) 2 Hence the total noise power (variance) at the output = 2 × 1.0417 + 3 × 1.6841 + 2 = 9.1357. In case the quantization is carried out after addition, then the total noise power at the output = 1.0417 + 1.6841 + 1 = 3.7258. ⎛ 1 − 0 .3 z − 1 ⎞ ⎛ 1 + 0 .4 z − 1 ⎞ ⎟⎜ ⎟. Here, a1 = −0.2, b1 = −0.3, (b) Cascade Structure # 2 - H ( z ) = ⎜⎜ − 1 ⎟ ⎜ 1 − 0 .6 z − 1 ⎟ + z 1 0 . 2 ⎝ ⎠⎝ ⎠ a2 = 0.6, b2 = 0.4. The values of the scaling constants are determined below. 1 2 . From Part (a) we have F1 2 = 1.0417 and γ1 = F1 2 = 1.0206. Likewise, F1 ( z ) = z + 0.2 ⎛ 1 − 0.3z −1 ⎞⎛ ⎞ 0.375 0.625 z −1 z − 0.3 ⎟⎜ ⎟= = + . Using Table 12.4 F2 ( z ) = ⎜⎜ ⎟ ⎜ ⎟ − 1 − 1 ⎝ 1 + 0.2 z ⎠⎝ 1 − 0.6 z ⎠ ( z + 0.2)( z − 0.6) z − 0.6 z + 0.2 (0.375) 2 (0.625) 2 2 × 0.375 × 0.625 2 + + = 1.0452. Hence, we get F2 2 = 1 − 0.6 × (−0.2) 1 − (0.6) 2 1 − (−0.2) 2

z 2 + 0.1z − 0.12 γ2 = F2 2 = 1.0224. Next, H ( z ) = and from Part (a) we have z 2 − 0.4 z − 0.12 2

H 2 = 1.3197. Hence, γ0 = H 2 = 1.1488.


γ 1 = 0.9798, k 2 = 1 = 0.9982, and γ1 γ2

γ k3 = 2 = 0.89. Therefore, b1 k2 = −0.2995 and b 2 k3 = 0.356. γ0 The noise at the output due to the input multiplier k1 and multiplier a1 have a variance σ12 = γ12 = 1.0417. The noise at the output due to a1 , k2 , and b1k2 have variance σ 22 which is calculated below. The noise transfer function for these noise sources is

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G2 ( z ) =

0.89 + 0.356 z −1

=

0.89z + 0.356 0.89 . Using Table 12.4 we get = 0.89 + z − 0.6 z − 0.6

1 − 0.6 z −1 (0.89) 2 σ 22 = (0.89) 2 + = 2.0298. 1 − (0.6) 2 Hence the total noise power (variance) at the output = 2 × 1.0417 + 3 × 2.0298 + 2 = 10.1728. In case the quantization is carried out after addition, then the total noise power at the output = 1.0417 + 2.0298 + 1 = 4.0715. ⎛ 1 − 0 .3 z − 1 ⎞ ⎛ 1 + 0 .4 z − 1 ⎞ ⎟ Here, a1 = 0.6, b1 = −0.3, ⎟⎜ (c) Cascade Structure # 3 – H ( z ) = ⎜ ⎜ 1 − 0.6 z −1 ⎟⎜ 1 + 0.2 z −1 ⎟. ⎠ ⎠⎝ ⎝ a 2 = −0.2, b2 = 0.4. The values of the scaling constants are determined below. 1 1 2 F1 ( z ) = . Using Table 12.4 we get F1 2 = = 1.5625. Hence, z − 0.6 1 − (0.6) 2 ⎛ 1 − 0 .3 z − 1 ⎞ ⎛ ⎞ z −1 z − 0 .3 ⎟⎜ ⎟= γ1 = F1 2 = 1.25. Likewise, F2 ( z ) = ⎜⎜ . From ⎟ ⎜ ⎟ − 1 − 1 ⎝ 1 − 0.6 z ⎠⎝ 1 + 0.2 z ⎠ ( z − 0.6)( z + 0.2) z 2 + 0.1z − 0.12 2 Part (b) we have F2 2 = 1.0452 and γ2 = F2 2 = 1.0224 . Next, H ( z ) = z 2 − 0.4 z − 0.12 2

and from Part (a) we have H 2 = 1.3197. Hence, γ0 = H 2 = 1.1488. γ 1 = 0.8, k 2 = 1 = 1.2226, and The scaling multipliers are therefore given by k1 = γ1 γ2 γ k3 = 2 = 0.89. Therefore, b1 k2 = −0.3668 and b 2 k3 = 0.356. γ0 The noise at the output due to the input multiplier k1 and multiplier a1 have a variance

σ12 = γ12 = 1.5625. The noise at the output due to a1 , k2 , and b1k2 have variance σ 22 which is calculated below. The noise transfer function for these noise sources is 0.89 + 0.356 z −1 0.89z + 0.356 0.178 G2 ( z ) = . Using Table 12.4 we get = = 0.89 + z + 0.2 z + 0.2 1 + 0.2 z −1

σ 22 = (0.89) 2 +

(0.178) 2

= 0.8251. 1 + (−0.2) 2 Hence the total noise power (variance) at the output = 2 × 1.5625 + 3 × 0.8251 + 2 = 7.6003. In case the quantization is carried out after addition, then the total noise power at the output = 1.5625 + 0.8251 + 1 = 3.3876. ⎛ 1 + 0 .4 z − 1 ⎞ ⎛ 1 − 0 .3 z − 1 ⎞ ⎟ Here, a1 = 0.6, b1 = 0.4, ⎟⎜ (d) Cascade Structure # 4 – H ( z ) = ⎜ ⎜ 1 − 0.6 z −1 ⎟⎜ 1 + 0.2 z −1 ⎟. ⎠ ⎠⎝ ⎝ a 2 = −0.2, b2 = −0.3. The values of the scaling constants are determined below.

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1 2 . From Part (c) we have F1 2 = 1.5625 and γ1 = F1 2 = 1.25. Likewise, z − 0.6 ⎞ ⎛ 1 + 0 .4 z − 1 ⎞ ⎛ z −1 z + 0 .4 2 ⎟= ⎟⎜ . From Part (a) we have F2 2 = 1.9485 F2 ( z ) = ⎜⎜ ⎟ ⎜ ⎟ − 1 − 1 ⎝ 1 − 0.6 z ⎠⎝ 1 + 0.2 z ⎠ ( z − 0.6)( z + 0.2) z 2 + 0.1z − 0.12 and from Part (a) we have and γ2 = F2 2 = 1.3959. Next, H ( z ) = z 2 − 0.4 z − 0.12 F1 ( z ) =

2

H 2 = 1.3197 and γ0 = H 2 = 1.1488.


γ 1 = 0.8, k 2 = 1 = 0.8955, and γ1 γ2

γ k3 = 2 = 1.2151. Therefore, b1 k2 = 0.3582 and b 2 k3 = −0.3645. γ0 The noise at the output due to the input multiplier k1 and multiplier a1 have a variance σ12 = γ12 = 1.5625. The noise at the output due to a1 , k2 , and b1k2 have variance σ 22 which is calculated below. The noise transfer function for these noise sources is − 0.6075 1.2151 − 0.3645z −1 1.2151z − 0.3645 G2 ( z ) = . Using Table 12.4 we = = 1.2151 + − 1 z + 0.2 z + 0.2 1 + 0.2 z get σ 22 = (0.89) 2 +

(−0.6075) 2

= 1.8609. 1 + (−0.2) 2 Hence the total noise power (variance) at the output = 2 × 1.5625 + 3 × 1.8609 + 2 = 10.7077. In case the quantization is carried out after addition, then the total noise power at the output = 1.5625 + 1.8609 + 1 = 5.4234. 12.27 (a) Parallel Form #1 - H ( z) =

( z + 0.4)( z − 0.3) 0.625 0.6 . The unscaled = 1+ + − 1 ( z + 0.2)( z − 0.6) 1 − 0.6 z 1 + 0.2 z −1

structure is shown below: F1 (z)

+

+

x[n]

y[n]

_ z 1

0.625 0.6 F2 (z)

+

0.6

_ z 1 _ 0.2

Here the scaling transfer function F1 ( z ) for the middle branch is given by F1 ( z ) =

Not for sale

0.625z z − 0.6

461

(0.375) 2 0.375 2 = 0.6104. Hence, . From Table 12.4 we get F1 2 = (0.625) 2 + z − 0.6 1 − (0.6) 2 γ1 = F 1 2 = 0.7813. Similarly the scaling transfer function F2 ( z) for the bottom branch is − 0.12 0.6 z = 0.6 + . From Table 12.4 we get given by F2 ( z) = z + 0.2 z + 0.2

= 0.625 +

(−0.12) 2 2 F2 2 = (0.6) 2 + = 0.375. Hence, γ2 = F 2 2 = 0.6124. 1 − (−0.2) 2 0.375 0.125 + . From Table 12.4 we get Finally, H ( z) = 1 + z − 0.6 z + 0.2 (0.375) 2 (0.125) 2 2 × 0.375 × 0.125 2 H 2 = 1+ + + = 1.3197. Hence, γ 0 = H 2 = 1.1488. 2 2 1 − 0.6 × (−0.2) 1 − (0.6) 1 − (−0.2) The scaled structure with noise sources is shown below where e [n] - C C

+

G 1(z)

A x[n]

+

+ _ z 1

eA[n] ea [n] 1 B

+

ea [n] 2

G2 (z)

+ _ z 1 _

+

+

+

y[n]

e k [n] 1

0.6

+

eB[n]

k1

k2

+ e k [n] 2

0.2

γ γ 0.625 0.6 1 = 0.7999, B = = 0.9798, C = , k1 = 1 = 0.6801, k 2 = 2 = 0.5331. γ1 γ2 γ0 γ0 γ0 The noise transfer function from each of the noise sources eC [ n], ek1 [ n], and ek 2 [ n] is

A=

G 3 ( z ) = 1. The corresponding output noise variance is σ 32 = 1. The noise transfer function from each of the noise sources e A [n] and ea1 [ n] is k1 0.4081 0.6801z G1 ( z ) = = = 0.6801 + . From Table 12.4 we get z − 0 .6 z − 0 .6 1 − 0 .6 z − 1

σ12 = (0.6081) 2 +

(0.4081) 2

= 0.7227. 1 − (0.6) 2 Likewise, the noise transfer function from each of the noise sources e B [n] and ea 2 [ n] is

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G2 ( z ) =

k2

1 + 0 .2 z − 1

σ 22 = (0.5331) 2 +

=

− 0.1066 0.5331z . From Table 12.4 we get = 0.53301 + z + 0 .2 z + 0 .2

(−0.1066) 2 1 − (−0.2) 2

= 0.296.

Hence the total noise power (variance) at the output, σ o2 = 2σ12 + 2σ 22 + 3σ 32 = 5.0374 . In case the quantization is carried out after addition, then the total noise power at the output, σ o2 = σ12 + σ 22 + σ 32 = 2.0187. (b) Parallel Form #2 - H ( z ) = 1 +

0.375 0.125 0.375z −1 0.125z −1 . + = 1+ + z − 0.6 z + 0.2 1 − 0.6 z −1 1 + 0.2 z −1

The unscaled structure is shown below: F 1(z)

_ z 1

+

x[n]

+

y[n]

0.375

0.6 F 2(z)

0.1

_ z 1

+ _

0.2

Here the scaling transfer function F1 ( z ) for the middle branch is given by F1 ( z ) =

0.375 . z − 0.6

(0.375) 2 2 = 0.2197. Hence, γ1 = F 1 2 = 0.4687. Similarly From Table 12.4 we get F1 2 = 1 − (0.6) 2 0.125 . From the scaling transfer function F2 ( z) for the bottom branch is given by F2 ( z) = z + 0.2 (0.125) 2 2 = 0.0163. Hence, γ2 = F 2 2 = 0.1277. Table 12.4 we get F2 2 = 1 − (−0.2) 2 0.375 0.125 2 + . From Part (a) we get H 2 = 1.3197 and Finally, H ( z) = 1 + z − 0.6 z + 0.2 γ 0 = H 2 = 1.1488. The scaled structure with noise sources is shown on the next page where

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γ γ 0.375 0.125 1 = 0.7999, B = = 0.9796, C = = 0.8705, k1 = 1 = 0.4081, k 2 = 2 = 0.1111. γ1 γ2 γ0 γ0 γ0 The noise transfer function from each of the noise sources eC [ n], ek1 [ n], and ek 2 [ n] is

A=

G 3 ( z ) = 1. The corresponding output noise variance is σ 32 = 1. e [n] - C C

+

G 1(z)

A x[n]

+

_1

+

z

+

2

+

y[n]

1

0.6

+

_1

z

+

G2 (z)

k2

+ e k [n]

eB[n] ea [n]

+ e k [n]

eA[n] ea [n] 1 B

k1

_

+

2

0.2

The noise transfer function from each of the noise sources e A [n] and ea1 [ n] is

k1 z −1

=

k 2 z −1

=

(0.4081) 2 0.4081 = 0.2602. . From Table 12.4 we get σ12 = 1 − (0.6) 2 1 − 0.6 z −1 z − 0.6 Likewise, the noise transfer function from each of the noise sources e B [n] and ea 2 [ n] is G1 ( z) =

G2 ( z ) =

1 + 0.2 z −1

(0.1111) 2 0.1111 0.1111 = 0.0129. . From Table 12.4 we get σ 22 = = z + 0.2 z + 0.2 1 − (−0.2) 2

Hence the total noise power (variance) at the output, σ o2 = 2σ12 + 2σ 22 + 3σ 32 = 3.5462. In case the quantization is carried out after addition, then the total noise power at the output, σ o2 = σ12 + σ 22 + σ 32 = 1.2731. In either case, Parallel form II structure (after scaling) has the lowest roundoff noise variance. 2 − 0 .4 z − 1 − 0 .6 z − 2

⎛ z 2 − 0 .2 z − 0 .3 ⎞ ⎟. = 2⎜⎜ ⎟ 2 1 + 0.2 z −1 − 0.15z − 2 ⎝ z + 0.2 z − 0.15 ⎠ (a) Direct Form: The unscaled structure is shown on top of next page. 2 2z 2 − 0.625 0.225 . From Table 12.4 we = =2+ + F1 ( z) = z + 0.5 z − 0.3 1 + 0.2 z −1 − 0.15z − 2 z 2 + 0.2 z − 0.15

12.28 H ( z ) =

(−0.625) 2 (0.225) 2 2 × (−0.625) × 0.225 2 + + = 4.3319. Hence, get F1 2 = 2 2 + 1 − (−0.5) × 0.3 1 − (−0.5) 2 1 − (0.3) 2 β0 =

2 z 2 − 0.4 z − 0.6 − 0.125 − 0.675 1 . From Table = 2+ + = 0.4805. Next, H ( z) = 2 z + 0.5 z − 0.3 F1 2 z + 0.2 z − 0.15 Not for sale

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F 1(z) 2

x[n]

y[n] z

_ 0.2

z

–1 _ 0.2

–1 _ 0.3

0.15

(−0.125) 2 (−0.675) 2 2 × (−0.125) × (−0.675) 2 + + = 4.6683. Hence, 12.4 we get H 2 = 2 2 + 1 − (−0.5) × 0.3 1 − (−0.5) 2 1 − (0.3) 2 F1 2 β1 = = 0.9632. H2 The scaling multipliers are therefore K = β 0 K = 0.4805 × 2 = 0.961, b01 = β1 = 0.9632, b11 = −0.2 × β1 = −0.1926, b21 = −0.3 × β1 = −0.289. The scaled structure with the noise sources is shown below: G 1(z)

G4 (z) 0.9632

2

y[n]

x[n] e 1[n]

_ 0.2 e 2[n] 0.15

–1

e 4[n] _ 0.1926

–1

e 5[n] _ 0.289

z

z

e 6[n]

e 3[n]

Quantization before addition: The noise transfer function associated with the noise sources b z 2 + b11 z + b21 0.9632 z 2 − 0.1926 z − 0.289 e1[ n], e2 [ n], and e3 [ n] is given by G1 ( z ) = 01 = z 2 + 0.2 z − 0.15 z 2 + 0.2 z − 0.15 − 0.0601 − 0.3251 = 0.9632 + + . From Table 12.4 we get z + 0.5 z − 0.3

σ12 = (0.9632) 2 +

(−0.0601) 2

+

(−0.3251) 2

+

2 × (−0.0601) × (−0.3251) = 1.0827. The noise 1 − (−0.5) × 0.3

1 − (−0.5) 2 1 − (0.3) 2 transfer function associated with the noise sources e4 [ n], e5 [ n], and e6 [ n] is given by G 4 ( z) = 1. Hence, σ 42 = 1. Hence, total normalized output noise power is σ o2 = 3σ12 + 3σ 42 = 3 × 1.0827 + 3 = 6.2481. Not for sale

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Quantization after addition: Here, total normalized output noise power is σ o2 = σ12 + σ 42 = 1.0827 + 1 = 2.0827. ⎛ 1 − 0.6568 z −1 ⎞⎛ 1 + 0.4568 z −1 ⎞ ⎟ ⎟⎜ (b) Cascade Structure #1 - H ( z ) = 2⎜ ⎜ 1 − 0.3z −1 ⎟⎜ 1 + 0.5z −1 ⎟. ⎠ ⎠⎝ ⎝ The unscaled structure is shown below. F 2(z)

2

x[n]

F 1(z)

y[n] _ z 1

_

0.3

F1 ( z) =

2 1 − 0.3z −1

=

0.6568

_ z 1

_ 0.5

0.4568

2z 0.6 . From Table 12.4 we get = 2+ z − 0.3 z − 0.3

(0.6) 2 1 2 F1 2 = 2 2 + = 4.3956. Hence, β0 = = 0.477. F1 2 1 − (0.3) 2 2(1 − 0.6568z −1 ) − 1.446 − 0.2676 F2 ( z) = . From Table 12.4 we get =2+ + z + 0.5 z − 0.3 (1 − 0.3z −1 )(1 + 0.5z −1 ) σ 22 = 2 2 + β1 =

(−1.446) 2 1 − (−0.5) 2

+

(−0.2676) 2 1 − (0.3) 2

+

2 × (−1.446) × (−0.2676) = 7.5395. Hence, 1 − (−0.5) × 0.3

F1 2 2 − 0.4 z −1 − 0.6 z −2 2 = 0.1737. Next, H ( z ) = . From Part (a), H 2 = 4.6683. F2 2 1 + 0.2 z −1 − 0.15z − 2

Hence, β1 =

F2 2 = 1.2708. H2

The scaling multipliers are therefore K = β 0 K = 0.477 × 2 = 0.954, b01 = β1 = 0.1737, b11 = −0.6568 × β1 = −0.1141, b02 = β 2 = 1.2708, b12 = 0.4568 × β2 = 0.5805. The scaled structure with the noise sources is shown below: 0.954

1.2708

0.1737 _ z 1

e 1[n]

0.3 e 2[n]

e 3[n] _

0.1141 e 4[n] e5 [n]

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_ 0.5

_ z 1

e 6[n]

0.5805 e 7[n]

466

Quantization before addition - The noise transfer function associated with the noise sources e1[ n] and e2 [ n] is ⎛ 0.1737 − 0.1141z −1 ⎞⎛ 1.2708 + 0.5805 z −1 ⎞ − 0.0138 − 0.0745 ⎟ = 0.2207 + ⎟⎜ . From + G1 ( z ) = ⎜⎜ ⎟ ⎟⎜ −1 −1 z + z − 0 . 5 0 . 3 1 0 . 3 z 1 0 . 5 z + − ⎠ ⎠⎝ ⎝ Table 12.4 we get (−0.0138) 2 (−0.0745) 2 2 × (−0.0138) × (−0.0745) σ12 = (0.2207) 2 + + + = 0.0569. 1 − (−0.5) × 0.3 1 − (−0.5) 2 1 − (0.3) 2

The noise transfer function associated with the noise sources e3 [ n], e4 [ n], and e5 [ n] is

G3 ( z ) =

1.2708 + 0.5805z −1

= 1.2708 +

1 + 0.5z −1

σ 32 = (1.2708) 2 +

(−0.0549) 2 1 − (−0.5) 2

− 0.0549 . From Table 12.4 we get z + 0.5

= 1.619. The noise transfer function associated with the noise

sources e6 [ n] and e7 [ n] is given by G6 ( z ) = 1. Hence, σ 62 = 1. Hence, total normalized output noise power is σ o2 = 2σ12 + 3σ 32 + 2σ 62 = 2 × 0.0569 + 3 × 1.619 + 2 = 6.9708. Quantization after addition Here, total normalized output noise power is σ o2 = σ12 + σ 32 + σ 62 = 0.0569 + 1.619 + 1 = 2.6759. ⎛ 1 + 0.4568 z −1 ⎞⎛ 1 − 0.6568 z −1 ⎞ ⎟. ⎟⎜ (c) Cascade Structure #2 - H ( z ) = 2⎜ ⎜ 1 − 0.3 z −1 ⎟⎜ 1 + 0.5 z −1 ⎟ ⎠ ⎠⎝ ⎝ The unscaled structure is shown below:

2

F2 (z)

F1 (z) _ z 1

0.3

0.4568

_ 0.5

_ z 1 _ 0.6568

0.6 (0.6) 2 2 2 = 4.3956. F1 ( z ) = =2+ . Using Table 12.4 we get F1 2 = 2 + z − 0.3 1 − (0.3) 2 1 − 0.3 z −1 1 = 0.477. Likewise, Hence, β 0 = F1 2 2

F2 ( z ) =

2(1 + 0.4568 z −1 )

(1 − 0.3 z

−1

)(1 + 0.5 z

−1

)

= 2+

− 0.054 0.5676 + . Using Table 12.4 we get z + 0.5 z − 0.3

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F2 = 2 2 +

(−0.054) 2

1 − ( −0.5) 2

+

(0.5676 ) 2

1 − (0.3) 2

+

2 × (−0.054) × 0.5676 = 4.3046. Hence, 1 − ( −0.5) × 0.3

β2 =

⎛ 1 + 0.4568 z −1 ⎞⎛ 1 − 0.6568 z −1 ⎞ F1 2 ⎟ ⎟⎜ = 0.482. Finally, H ( z ) = 2⎜ ⎜ 1 − 0.3 z −1 ⎟⎜ 1 + 0.5 z −1 ⎟ F2 2 ⎠ ⎠⎝ ⎝

=2+

− 0.1249 − 0.6751 . Using Table 12.4 we get + z + 0 .5 z − 0 .3

H 2 = 22 + β2 =

(−0.1249) 2 1 − (−0.5) 2

+

(−0.6751) 2 1 − (0.3) 2

+

2 × (−0.1249) × (−0.6751) = 4.6683. Hence, 1 − (−0.5) × 0.3

F2 2 = 0.9603. H 2

The scaling multipliers are therefore K = β 0 K = 0.477 × 2 = 0.954, b01 = β1 = 0.482,

b11 = −0.6568 × β1 = −0.3166, b02 = β 2 = 0.9603, b12 = 0.4568 × β 2 = 0.4387. The scaled structure with the noise sources is shown below: 0.954

0.9603

0.482 _ z 1

e 1[n] 0.3

e 3[n] _

_ 0.5

0.3166

e 2[n]

_ z 1

e 6[n] 0.4387

e 4[n] e5 [n]

e 7[n]

Quantization before addition – The noise transfer function from the noise sources e1 [n] and e 2 [n] is given by ⎛ 0.482 − 0.3166 z −1 ⎞⎛ 0.9603 + 0.4387 z −1 ⎞ ⎟ = 0.4629 + − 0.0289 + − 0.1563 . Using ⎟⎜ G1 ( z ) = ⎜ −1 −1 ⎟ ⎜ ⎟ ⎜ z + 0 .5 z − 0 .3 1 − 0.3 z 1 + 0.5 z ⎠ ⎠⎝ ⎝ Table 12.4 we get σ12 = 0.25. Likewise, the noise transfer function from the noise sources

e 3 [n], e 4 [n], and e 5 [n] is given by G 3 ( z ) =

0.9603 + 0.4387 z −1 1 + 0.5 z −1

= 0.9603 +

− 0.0415 . z + 0.5

Using Table 12.4 we get σ 22 = 0.9245. Finally, the noise transfer function from the noise sources e 6 [n] and e 7 [n] is given by G6 ( z ) = 1. Hence, σ 62 = 1. Therefore, the total normalized output noise power is σ 2o = 2σ12 + 3σ 32 + 2σ 62 = 2 × 0.25 + 3 × 0.9245 + 2 = 5.2735.

Quantization after addition Here, total normalized output noise power is σ 2o = σ12 + σ 32 + σ 62 = 0.25 + 0.9245 + 1 = 2.1745.

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0.25

(d) Parallel Form I Realization - H ( z ) = 2 +

1 + 0 .5 z − 1

+

− 2.25 1 − 0 .3 z − 1

.

The unscaled structure is shown below: 2 F 1(z)

+

x[n]

+

0.25 _ 0.5

y[n]

_ z 1

F 2(z)

+ _ 2.25

0.3

F1 ( z ) =

0.25 1 + 0 .5 z

= 0.25 +

−1

_ z 1

− 0.125 . Using Table 12.4 we get z + 0 .5

( −0.125) 2

1 = 3.4641. = 0.0833. Hence, β1 = F1 2 1 − ( −0.5) 2 − 0.675 − 2.25 F2 ( z ) = = −2.25 + . Using Table 12.4 we get z − 0.3 1 − 0 .3 z − 1 2

F1 2 = (0.25) 2 +

( −0.675) 2

1 = 0.424. = 5.5632. Hence, β 2 = F2 2 1 − (0.3) 2 − 0.125 − 0.675 H ( z) = 2 + . Using Table 12.4 we get + z + 0 .5 z − 0 .3 2

F1 2 = ( −2.25) 2 +

2

H 2 = 22 +

β0 =

( −0.125) 2

1 − ( −0.5) 2

+

( −0.675) 2

1 − (0.3) 2

+

2 × ( −0.125) × ( −0.675) = 4.6683. Hence, 1 − ( −0.5) × 0.3

1 = 0.4628. H 2

The scaled structure with the noise sources is shown on top of the next page. The noise transfer function associated with the noise sources e1[n] and e2 [n] is given by 1 /β 0 β1 − 0.3119 = 0.6238 + G1 ( z ) = . Using Table 12.4 we obtain, z + 0.5 1 + 0.5 z −1 σ12 = (0.6238) 2 +

( −0.3119) 2

= 0.5188. Likewise, the noise transfer function associated with 1 − ( −0.5) 2 the noise sources e3 [n] and e4 [n] is given by

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e 5[n] 2β 0 _ 0.125β

x[n]

1

+

G 1(z)

e 2[n]

+

+

+

0.675β 2

1 /β 0 β 2 1 − 0.3 z

−1

σ 32 = (0.9162 ) 2 +

= 0.9162 +

(0.2749) 2

1 − (0.3) 2

1/β 1β 0

_1

z

_ 0.5

y[n]

e 6[n]

+ 1/β 2β 0

_1

+

+

G2 (z)

e 3[n]

e 4[n]

G3 ( z ) =

+

+

e 1[n]

_

+

z

0.3

e 7[n]

0.2749 . Using Table 12.4 we obtain, z − 0.3

= 0.224.

The noise transfer function associated with the remaining three noise sources is G5 ( z ) = 1. Hence, σ 25 = 1. Quantization before addition Hence, total normalized output noise power is σ o2 = 2σ12 + 2σ 32 + 3σ 25 = 5.8824. Quantization after addition Here, total normalized output noise power is σ 2o = σ12 + σ 32 + σ 25 = 2.4412. 12.28 (a) The noise model for the allpass structure of Figure P12.9(a) is shown below: R(z) _ z1

X(z) _1

V(z) di

W(z) E(z)

Y(z)

Analysis yields W ( z ) = E ( z ) + di V ( z), R( z) = X ( z) − W ( z), V ( z) = X ( z) + z −1 R( z), and Y ( z ) = W ( z ) + z −1 R( z ). To determine the noise transfer function G (z ) we set X ( z ) = 0 in the

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470

above equations. This leads to R( z ) = −W ( z ), V ( z ) = z −1 R( z ) = − z −1W ( z ), and hence,

W ( z) = E ( z ) − di z −1W ( z ) or, E ( z) = (1 + di z −1 )W ( z). As a result, Y ( z ) = W ( z ) − z −1W ( z ) = (1 − z −1 )W ( z ). Consequently, the noise transfer function is given by

G( z) =

− (1 + di ) Y ( z) z −1 1 − z −1 . Using Table 12.4 we get = = = 1+ X ( z ) 1 + d i z −1 z + d i z + di

σ e2 = 1 +

(1 + di ) 2 1 − di2

=

2 . 1 − di

(b) Let Gi (z ) be the noise transfer function for di , 1 ≤ i ≤ 3. Then, z −1 z −1 z −1 ⋅ A2 ( z ), G 2 ( z ) = , and G3 ( z ) = . From the results of Part (a), it G1 ( z ) = z + d2 z + d3 z + d1 2 for k = 1, 2, 3. follows that the noise variances due to d1 ,d2 , and d3 are given by σ k2 = 1 − dk Hence the total noise power at the output of the digital filter structure of Figure P12.9 is given 2 2 2 by σ o2 = + + . 1 − d1 1 − d 2 1 − d3 2. 12.30 Let the total noise power at the output of G (z ) due to product round-off be given by σ G Assuming a total of L multipliers in the realization of G (z ) we get

⎛ 1 2π ⎞ 2 ⎜ j ω = kl ⎜ Gl (e ) dω ⎟⎟, where Gl (z ) denotes the noise transfer function due the 2π l =1 ⎝ 0 ⎠ l –th noise source in G (z ). Now if each delay is replaced by two delays, then each of the noise L

2 σG

∑

∫

transfer function becomes Gl ( z 2 ). Thus, the total noise power at the output due to noise

⎞ ⎛ 1 2π 2 sources in = is given by k l ⎜⎜ Gl (e j 2ω ) dω ⎟⎟. Replacing ω by ωˆ / 2 in 2π l =1 ⎝ 0 ⎠ L ⎞ ⎞ 1 L ⎛ 1 4π ⎛ 1 4π 2 2⎛ 1⎞ ⎜ ˆ ω 2. j kl ⎜ Gl (e ) ⎜ ⎟ dωˆ ⎟⎟ = k l ⎜⎜ Gl (e jωˆ ) dωˆ ⎟⎟ = σ G the integral we get π 2 2 π 2 2 ⎝ ⎠ l =1 ⎝ l =1 ⎝ 0 0 ⎠ ⎠ L

∑

1 Since, 2π

2π

∫ 0

∑

2 σˆ G

G(z 2 )

G (e j 2 ω )

2

∫

∑

∫

A(e jω )

2

1 dω = 2π

2π

∫

∫

2

G (e j 2ω ) dω, the total noise power at the output

0

of the cascade is still equal to σ o2 ; 12.31 For the first factor in the numerator, there are R possible choices of factors. Once this factor has been chosen, there are R − 1 choices for the next factor and continuing further we get that the total number of possible ways in which the factors in the numerator can be generated equal to R( R − 1)( R − 2) L 2 × 1 = R! . Similarly, the total number of ways in which the factors in the

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denominator can be generated is equal to R! . Since the numerator and the denominator are factored independent of each other, hence the total number of possible realizations are N = ( R! )( R! ) = ( R! ) 2 . 12.32 (i) A pole-zero plot of H 1( z) is shown below: 1

1 2

0.5

Imaginary Part

3

0 3

0. 5 2

1

1

1

0. 5

0 Real Part

0.5

1

First we pair the poles closest to the unit circle with their nearest zeros resulting in the secondorder section z 2 + 1.0094 z + 1 H a (z) = . z 2 + 0.6081z + 0.9208 Next, the poles that are closest to the poles of H a (z) are matched with their nearest zeros resulting in the second-order section z 2 + 1.2914 z + 1 H b ( z) = . z 2 + 0.2749z + 0.6673 Finally, the remaining poles and zeros are matched yielding the second-order section z 2 + 1.8606 z + 1 H c (z) = . z 2 − 0.41591z + 0.2066 For ordering the sections to yield the smallest peak output noise due to product round-off under an L2 -scaling rule, the sections should be placed from most peaked to least peaked as shown below. Ha (z)

Hb (z)

Hc (z)

For ordering the sections to yield the smallest peak output noise power due to product round-off under an L∞ -scaling rule, the sections should be placed from least peaked to most peaked as shown below. Hc (z)

Hb (z)

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Ha (z)

472

(ii) (iii)

A pole-zero plot of H 2 ( z) is shown below: 1

1

2

Imaginary Part

0.5

3

0

3

0. 5 2 1

1 1. 5

1

0. 5

0 Real Part

0.5

1

1.5

Following the same procedure as in Part (i), we get z 2 − 0.8508z + 1 z 2 − 1.1141z + 1 z 2 − 1.7988z + 1 H a (z) = , H b (z) = , H a ( z) = . z 2 − 0.6120 z + 0.9208 z 2 − 0.3474 z + 0.7127 z 2 − 0.2993z + 0.20358 For ordering the sections to yield the smallest peak output noise due to product round-off under an L2 -scaling rule, the sections should be placed from most peaked to least peaked as shown below. Ha (z)

Hb (z)

Hc (z)

For ordering the sections to yield the smallest peak output noise power due to product round-off under an L∞ -scaling rule, the sections should be placed from least peaked to most peaked as shown below. Hc (z)

Hb (z)

Ha (z)

σ2 12.33 SNR = x . After scaling, σ x changes to Kσ x where K is as given by Eq. (12.151). σ o2

Therefore, SNR =

σ 2x (1 − α ) 2 σ 2o

. 1

(i) For uniform density function

σ x2

=

∫

1 2 x dx 2 −1

1 3

= . Thus, SNR =

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(1 − α ) 2 3σ o2

. With

473

b = 12, σ o2 =

2 −24 12

= 4.967 × 10 − 9 and α = 0.95. Hence,

⎛ (1 − α ) 2 ⎞ ⎟ = 52.24 dB. SNR dB = 10 log10 ⎜ ⎜ 3σ 2 ⎟ o ⎠ ⎝ 1

(ii) For Gaussian input with σ x2 = , SNR = 9

(1 − α ) 2

. Again with b = 12,

9σ o2

⎛ (1 − α ) 2 ⎞ ⎟ = 47.97 dB. SNR dB = 10 log10 ⎜ ⎜ 9σ 2 ⎟ o ⎠ ⎝ (iii) For a sinusoidal input of known frequency, i.e., x[ n] = sin(ωo n). In this case, the average 1

power = σ x2 = . Hence, SNR = 2

(1 − α ) 2 σ 2x σ o2

=

(1 − α ) 2 2σ o2

. Therefore,

⎛ (1 − α ) 2 ⎞ ⎟ = 69.91 dB. SNR dB = 10 log10 ⎜ ⎜ 2σ 2 ⎟ o ⎠ ⎝ 1

12.34 (a) H LP ( z) = [1 + A1 ( z )], where A1 ( z) = 2

H LP

( e jω )

proving

2

− α + z −1 1 − α z −1

. Hence,

2 ∂ H LP (e jω ) 2(1 + cos ω) ⎛1− α ⎞ . Proving =⎜ ⎟ ∂α ⎝ 2 ⎠ 1 − 2α cos ω + α 2

∂ H LP (e jω )

= 0 is equivalent to ω=0

2

= 0. Now,

∂α ω=0

∂ H LP (e jω )

2

∂α Thus,

2

1− α 2(1 + cos ω) ⎛ 1 − α ⎞ 2(1 + cos ω)(α − 2 cos ω) . =− ⋅ −⎜ ⎟ ⋅ 2 2 1 − 2α cos ω + α 1 − 2α cos ω + α 2 ⎝ 2 ⎠

∂ H LP (e jω )

2

=

∂α

−2 2 + = 0. 1− α 1− α

ω=0 1 2

(b) H HP ( z ) = [1 − A1 ( z )], where A1 ( z) =

H HP

( e jω )

2

∂ H HP (e jω ) ∂α

− α + z −1 1 − α z −1

. Hence,

2

2(1 − cos ω) ⎛1+ α ⎞ . Now, =⎜ ⎟ ⎝ 2 ⎠ 1 − 2α cos ω + α 2 2

2

=−

1+ α 2(1 − cos ω) ⎛ 1 + α ⎞ 2(1 − cos ω)(α − 2 cos ω) . ⋅ −⎜ ⎟ ⋅ 2 1 − 2α cos ω + α 2 ⎝ 2 ⎠ 1 − 2α cos ω + α 2

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Thus,

2

∂ H HP (e jω )

=

∂α

−2 2 + = 0. 1+ α 1+ α

ω=0 1

12.35 (a) H BP ( z ) = [1 − A2 ( z)], where A2 ( z ) = 2 2 ⎛1− α ⎞ real. Hence, H BP (e jω ) = ⎜ ⎟ ⎝ 2 ⎠

Now,

∂ H BP (e jω )

2

∂α

α + − β (1 + α )z −1 + z −2

1 − β (1 + α ) z −1 + αz − 2

, with α and β being

2⎛

⎞ 2(1 − cos ω) ⎜ ⎟ ⎜ 1 + β 2 (1 + α ) 2 + α 2 + 2α cos(2ω) − 2 β (1 + α ) 2 cos ω ⎟. ⎝ ⎠

⎛ − 2 ⎞⎛ 1 − α ⎞ =⎜ ⎟⎜ ⎟ ⎝ 1 − α ⎠⎝ 2 ⎠

⎞ 2(1 − cos ω) ⎟ ⎜ ⎜ 1 + β 2 (1 + α ) 2 + α 2 + 2α cos(2ω) − 2 β (1 + α ) 2 cos ω ⎟ ⎠ ⎝

2⎛

2⎛

⎞ 2(1 − cos ω) ⎜ ⎟ ⎜ 1 + β 2 (1 + α ) 2 + α 2 + 2α cos(2ω) − 2 β (1 + α ) 2 cos ω ⎟ × ⎝ ⎠ ⎛ 2 β 2 (1 + α ) + 2α + 2 cos(2ω) − 4 β (1 + α ) cos ω ⎞⎟ ⎜ ⎜ 1 + β 2 (1 + α ) 2 + α 2 + 2α cos(2ω) − 2 β (1 + α ) 2 cos ω ⎟. ⎝ ⎠ Using the fact β = cos(ωo ), we get ⎛1− α ⎞ +⎜ ⎟ ⎝ 2 ⎠

∂ H BP (e jω )

2

=

∂α

−2 −2 (2 β 2 + 2α − 2 − 2αβ 2 ) 2 − = + = 0. 2 2 2 2 2 1 − α 1 + α − 2α + 2αβ − β − α β 1− α 1− α

ω = ωo

Similarly,

∂ H BP (e jω ) ∂β

2

2⎛

⎞ 2(1 − cos ω) ⎜ ⎟ ⎜ 1 + β 2 (1 + α ) 2 + α 2 + 2α cos(2ω) − 2 β (1 + α ) 2 cos ω ⎟ × ⎝ ⎠ ⎛ ⎞ 2 β (1 + α ) − 2(1 + α ) 2 cos ω ⎜− ⎟ ⎜ 1 + β 2 (1 + α ) 2 + α 2 + 2α cos(2ω) − 2 β (1 + α ) 2 cos ω ⎟. ⎝ ⎠ ⎛1− α ⎞ =⎜ ⎟ ⎝ 2 ⎠

Again, using the fact β = cos(ωo ), it can be seen that

∂ H BP (e jω )

2

= 0.

∂β ω = ωo

1

(b) Here, H BS ( z ) = [1 + A2 ( z )], where A2 ( z) is as given in Part (a). Hence, 2

H BS

( e jω )

2

∂ H BS (e jω ) ∂α

⎛1+ α ⎞ =⎜ ⎟ ⎝ 2 ⎠ 2

2⎛

⎞ 4 β 2 + 2 + 2 cos(2ω) − 8 β cos ω ⎜ ⎟ ⎜ 1 + β 2 (1 + α ) 2 + α 2 + 2α cos(2ω) − 2 β (1 + α ) 2 cos ω ⎟. Thus, ⎝ ⎠

⎞ 4 β 2 + 2 + 2 cos(2ω) − 8 β cos ω ⎛ 1 + α ⎞⎛⎜ ⎟ =⎜ ⎟⎜ ⎝ 2 ⎠⎝ 1 + β 2 (1 + α ) 2 + α 2 + 2α cos(2ω) − 2 β (1 + α ) 2 cos ω ⎟⎠

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⎞ 4 β 2 + 2 + 2 cos(2ω) − 8 β cos ω ⎜ ⎟ ⎜ 1 + β 2 (1 + α ) 2 + α 2 + 2α cos(2ω) − 2 β (1 + α ) 2 cos ω ⎟ × ⎝ ⎠ ⎛ 2(1 + α ) β 2 + 2α + 2 cos(2ω) − 4 β (1 + α ) cos ω ⎞⎟ ⎜ ⎜ 1 + β 2 (1 + α ) 2 + α 2 + 2α cos(2ω) − 2 β (1 + α ) 2 cos ω ⎟. ⎝ ⎠ Substituting β = cos(ωo ), it can be seen that ⎛1+ α ⎞ −⎜ ⎟ ⎝ 2 ⎠

∂ H BS (e jω )

2

=

∂α

2⎛

2 ⎛1+ α ⎞ −⎜ ⎟ 1− α ⎝ 2 ⎠

ω=0

∂ H BS (e jω )

Similarly,

2

=

∂α

2⎛

2⎞ ⎜ 2(1 + α )( β + 1) ⎟ = 0. ⎜ (1 + α ) 2 ( β + 1) 2 ⎟ ⎝ ⎠

2 ⎛1+ α ⎞ −⎜ ⎟ 1− α ⎝ 2 ⎠

2⎛

ω=π

Now,

∂ H BS (e jω )

2

⎛1+ α ⎞ =⎜ ⎟ ⎝ 2 ⎠

∂β

2⎞ ⎜ 2(1 + α )( β + 1) ⎟ = 0. ⎜ (1 + α ) 2 ( β + 1) 2 ⎟ ⎝ ⎠

⎞ 8 β − 8 β cos ω ⎟ ⎜ ⎜ 1 + β 2 (1 + α ) 2 + α 2 + 2α cos(2ω) − 2 β (1 + α ) 2 cos ω ⎟ ⎠ ⎝

2⎛

⎛1+ α ⎞ −⎜ ⎟ ⎝ 2 ⎠

2⎛

⎞ 4 β 2 + 2 + 2 cos(2ω) − 8 β cos ω ⎜ ⎟ ⎜ 1 + β 2 (1 + α ) 2 + α 2 + 2α cos(2ω) − 2 β (1 + α ) 2 cos ω ⎟ × ⎝ ⎠

⎛ ⎞ (1 + α ) 2 2( β − cos ω) ⎜ ⎟ ⎜ 1 + β 2 (1 + α ) 2 + α 2 + 2α cos(2ω) − 2 β (1 + α ) 2 cos ω ⎟. ⎝ ⎠

Again, substituting β = cos(ωo ), it can be seen that

∂ H BS (e jω )

2

= 0 and

∂β ω=0

∂ H BS (e jω )

2

= 0.

∂β ω =π

12.36 For a BR transfer function G (z ) realized in a parallel allpass form, its power-complementary 2

2

transfer function H (z ) is also BR satisfying the condition G(e jω ) = 1 − H (e jω ) . Let

ω = ωo be a frequency where G(e jω ) is a maximum, i.e., G(e jωo ) = 1. Then, it follows that

H (e jωo ) = 0. From the power- complementary condition it follows that 2 G ( e jω )

⋅

∂ G ( e jω ) ∂ω

=

− 2 H ( e jω )

⋅

∂ H ( e jω ) ∂ω

. Therefore, at ω = ωo ,

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2 G(e jω )

⋅

∂ G ( e jω ) ∂ω

whether or not

=

− 2 H (e jω )

⋅

∂ H ( e jω ) ∂ω

ω = ωo

∂ H (e jω ) ∂ω

, and thus,

∂ G ( e jω ) ∂ω

ω = ωo

=0 ω =ω o

= 0. Hence, low passband sensitivity of G (z ) does not ω =ω o

necessarily imply low passband sensitivity of H (z ) . 12.37 Without error feedback – The transfer function of the structure without error feedback is 1 1 = , where r = 1 − ε. The given by H ( z ) = 1 + α1 z −1 + α2 z − 2 1 − 2r cos θ z −1 + r 2 z − 2

corresponding impulse response h[n] is given by h[ n] =

r n sin( n + 1)θ µ[ n]. To keep the sin θ

output y[n] from overflowing we must insert a multiplier of value

L=

∞

∑ n = 0 h[n] .

1 L

at the input where

From Eq. (12.164) we get

1

16

≤ L2 ≤

. π 2 (1 − r ) 2 sin 2 θ (1 − r ) 2 (1 − 2r cos θ + r 2 ) The quantization noise model for H (z ) is as shown below: (1) :

e[n] y[n]

x[n] _ z 1 _α

_α

The output noise power is given by σ n2 = σ e2

1

_ z 1

2

∞

∑

2

h[ n] =

n=0

1+ r2 1− r2

⋅

1 r4

− 2r 2

cos θ + 1

⋅ σ e2 .

The output signal power, assuming an input signal of variance σ 2x is given by σ 2y =

σ y2 ∞

∑

L2 n = 0

2

h[ n] . Hence, the SNR is given by SNR =

signed representation, σ e2 =

2 σ2 S σy = = x . For a (b + 1) -bit N σ n2 L2 σ e2

σ x2 S 2 −2 b . Therefore, from the . Hence, here SNR = = 12 N L2 2 − 2b

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inequality of Eq. (1) given above, we get 2 −2 b 16 × 2 −2b N . ≤ ≤ 12σ x2 (1 − r ) 2 (1 − 2r cos θ + r 2 ) S 12σ x2 π 2 (1 − r ) 2 sin 2 θ 1 3

For an WSS uniformly distributed input between [ −1, 1], σ x2 = . Hence,

(a)

2 −2 b 4(1 − r ) 2 (1 − 2r cos θ + r 2 )

≤

N 4 2 −2 b ≤ ⋅ . S π 2 (1 − r ) 2 sin 2 θ

If ε → 0, and θ → 0, then (1 − r ) 2 → ε 2 , cos(2θ ) = 1 − 2 sin 2 θ ≅ 1 − 2θ 2 , and sin 2 θ ≅ θ 2 . In

2 −2 b

this case we have

4ε 2 (ε 2 + 4θ 2 )

N 2 −2 b ≤ . S π 2 ε 2θ 2

≤

1 3

(b) For an input with a Gaussian distribution between [ −1, 1], σ x = . In this case, we have

3⋅

2 −2 b 4ε 2 (ε 2 + 4θ 2 )

≤

N 3 2 −2 b ≤ ⋅ . S π 2 ε 2θ 2

(c) For a sinusoidal input between [−1, 1] of known frequency ωo , σ 2x =

variance here is therefore σ n2 = σ e2

∞

∑

2

h[ n] =

n=0

1 . The output noise 2

2 − 2b 1 + r 2 1 . Thus, ⋅ ⋅ 12 1 − r 2 r 4 − 2r 2 cos θ + 1

N 1+ r2 1 2 −2 b 2 −2 b = ⋅ = ≅ . S 1 − r 2 6 r 4 − 2r 2 cos(2θ ) + 1 24ε(ε 2 + θ 2 ) 24εθ 2 With error feedback z 2 − 2z + 1 1 − 2 z − 1 + z −2 2(r cos θ − 1) z + 1 − r 2 G( z) = = = 1+ . z 2 − 2r cos θ z + 1 1 − 2r cos θ z −1 + r 2 z − 2 z 2 − 2r cos θ z + 1

(

Thus, G

2

)

= 1+

(4(1 − r cos θ )2 + (1 − r 2 )2 )(1 − r 4 ) + 8r cos θ ⋅ (1 − r 2 )2 (r cos θ − 1) .

(1 − r 2 ) 2 + 2r 2 (4r 2 cos2 θ ) − 4(1 + r 4 )r 2 cos 2 θ For r = 1 − ε with ε → 0, and θ → 0, we get after some manipulation G

2

= 1+

θ4

4ε(ε 2

+ θ2)

2

, and σ n2 = σ e2 G .

Now, remains the same as before since it depends only upon the denominator. Also, the overall transfer function of the structure remains the same as before. The output noise power with

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2 error feedback is thus Nˆ = σ e2 G , whereas, the output noise power without error feedback is

N=

σ e2

2 G Nˆ H . Hence, = . 2 N H

Now, G

2

2

= 1+

θ4

4 ε (ε 2 + θ 2 )

. Since θ >> ε, G

2

≅

θ4 4εθ 2

θ2 . 4ε

=

1+ r2 1 1 + (1 − ε ) 2 1 ⋅ = ⋅ Also, H 2 = 1 − r 2 r 4 − 2r 2 cos 2 θ + 1 1 − (1 − ε ) 2 (1 − ε ) 4 − 2(1 − ε ) 2 (1 − 2θ 2 ) + 1 1 1 ≈ ≈ . 4ε(θ 2 + ε 2 ) 4εθ 2 Nˆ N ≅ θ 4 . As a result, with error feedback, the ratio gets multiplied by θ 4 . Thus, N S 2 −2b θ 2 Nˆ θ 2 2 −2b (a) input with uniform density: ≤ ≤ ⋅ . S π2 ε2 16 ε 2 2 −2b 3θ 2 Nˆ 3θ 2 2 −2b (b) wide-sense stationary, Gaussian density, white: ≤ ≤ ⋅ . S 16ε 2 π2 ε2 Nˆ 2 −2 b θ 2 (c) sinusoid with known frequency: . = S 24 ε 12.38 The coupled form with error feedback is shown below: _ z 1

λ1 γ _ z 1

α

u[n] + e 2[n]

Q _

_ v1 [n]

e1 [n] + y[n]

Q

δ

_ z 1

v 2[n] β λ2

_ z 1

Analysis yields V1 ( z ) = λ1 z −1 E1 ( z ) + δz −1Y ( z ) + γz −1U ( z ), E1 ( z) = Y ( z) − V1 ( z),

V2 ( z ) = λ2 z −1 E 2 ( z ) + βz −1Y ( z ) + αz −1U ( z ), and E 2 ( z) = Y ( z) − V2 ( z). Eliminating U1 ( z), V1 ( z), U 2 ( z), and V2 ( z) from these equations we arrive at the noise transfer functions

G1 ( z ) =

(1 − αz −1 )(1 + λ1 z −1 ) Y ( z) = and E1 ( z) E ( z ) = 0 1 − (α + δ )z −1 + (α δ − β γ)z − 2 2

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G 2 (z) =

γ z −1 (1 + λ2 z −1 ) Y ( z) = . The total output noise power is thus E 2 ( z ) E ( z ) = 0 1 − (α + δ ) z − 1 + ( α δ − β γ ) z − 2 2 2

2

given by σ o2 = G1 2 σ e21 + G2 2 σ e22 . Hence, the total output noise power, for a lowpass filter design, can be reduced by placing the zeros of the noise transfer functions in the passband. For each of the noise transfer functions given above, we can place only a zero at z = 1 by choosing λi = −1, i = 1, 2. Using the notations b = −(α + δ ) and d = α δ − β γ, we rewrite the noise transfer functions as

z 2 − (1 + α )z + α

γz − γ and G2 ( z) = . Using Table z 2 + bz + d z 2 + bz + d z 2 + bz + d [(1 + α + b) 2 + (α − d ) 2 ](1 − d 2 ) + 2(1 + α + b)(α − d )(1 − d ) 2 12.4 we obtain G1 2 = 1 + and (1 − d 2 ) 2 + 2db 2 − (1 + d 2 )b 2 G1 ( z) =

= 1+

− (1 + α + b)z + (α − d )

2γ 2 (1 − d 2 ) 2 G2 2 = . (1 − d 2 ) 2 + 2db 2 − (1 + d 2 )b 2 12.39 The Kingsbury structure with error feedback is shown below. 1 z–

1 z–

_

λ1

k1

Q

1 z–

_

λ2 +

v 1[n]

x[n]

1 z–

v 2[n]

u 1[n]

+ Q

_

λ3

u 2[n]

_k

1

v 3[n]

+ Q

u 3[n]

1 z–

y[n]

k2

Analysis yields: V1 ( z ) = k1Y ( z ) + z −1U1 ( z ) + λ1 z −1 E1 ( z ), E1 ( z) = U1 ( z) − V1 ( z),

V2 ( z ) = k 2Y ( z ) + U1 ( z ) + λ2 z −1 E 2 ( z ), E 2 ( z) = U 2 ( z) − V2 ( z), V3 ( z) = Y ( z) − k1U1 ( z ) + λ3 z −1E 3 ( z), E 3 ( z ) = U 3 ( z ) − V3 ( z ), and Y ( z ) = z −1U1 ( z ).

Eliminating U1 ( z), V1 ( z),U 2 ( z ), V2 ( z ),U 3 ( z), and V3 ( z) from these equations we get − k1 z −1 (1 + λ1 z −1 ) Y ( z) G1 ( z) = = , E1 ( z ) E ( z ) = E ( z ) = 0 1 − [2 − k1 (k1 + k 2 )]z −1 + (1 − k1k2 )z − 2 2

3

G2 ( z ) =

− k1 z −1 (1 − z −1 )(1 + λ2 z −1 ) Y ( z) = , E 2 ( z ) E ( z ) = E ( z ) = 0 1 − [2 − k1 (k1 + k 2 )]z −1 + (1 − k1k 2 )z − 2 1 3

G3 ( z ) =

z −1 (1 − z −1 )(1 + λ3 z −1 ) Y ( z) = . E 3 ( z ) E ( z ) = E ( z ) = 0 1 − [2 − k1 (k1 + k 2 )]z −1 + (1 − k1k 2 )z − 2 1 2

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2

2

2

The total output noise power is thus given by σ 2o = G1 2 σ 2e1+ G2 2 σ e22 + G3 2 σ 2e3 . Hence, the output noise power for a lowpass filter design can be reduced by placing zeros of the noise transfer functions in the passband. For each of the noise transfer functions given above, we can only place one zero at z = 1 by choosing λi = −1, i = 1, 2, 3, in which case, using the notations

b = −[2 − k1 (k1 + k2 )] and d = 1 − k1k2 , the noise transfer functions reduce to G1 ( z ) = G2 ( z ) = G3 ( z ) =

− k1 z −1 (1 − z −1 )

1 + b z −1 + d z − 2

− k1 z −1 (1 − z −1 ) 2 1 + b z −1

+ d z−2

z −1 (1 − z −1 ) 2

− k1 z + k1

=

z2 + b z + d =

=

,

− k1 ( z 2 + z + 1) z( z 2

+ b z + d)

z2 + z + 1

=

=

− k1 A Cz + D , + 2 z z + bz + d

A Cz + D 1 1 + , where A = , C = 1 − , z z2 + b z + d d d

1 + b z − 1 + d z − 2 z( z 2 + b z + d ) b⎞ ⎛ and D = −⎜ 2 + ⎟. Using Table 12.4 we then obtain d⎠ ⎝ 2(1 − d 2 ) − 2(1 − d )b 2 G1 2 = k12 ⋅ , (1 − d 2 ) 2 + 2db 2 − (1 + d 2 )b 2

(C 2 + D 2 )(1 − d 2 ) − 2CD(1 − d )b 2 2 2 G3 2 = A 2 + 2 BC + , and G 2 2 = k12 G3 2 . 2 2 2 2 2 (1 − d ) + 2db − (1 + d )b 12.40 The coupled-form structure with state-variables labeled is shown below: s 2[n+1]

x[n]

γ _ z 1

s 1[n+1]

δ

α

y[n]

_ z 1

β

Its transfer function is given by Eq. (12.19) where α = δ = r cos θ and β = − γ = r sin θ . Analysis of the above structure yields s1[ n + 1] = α(x[ n] + s1[ n]) + βs2 [ n] and s 2 [ n + 1] = γ(x[ n] + s1[ n]) + δs2 [ n]. Rewriting these two equations in matrix form we get ⎡ s1[ n + 1] ⎤ ⎡α β ⎤ ⎡ s1[ n] ⎤ ⎡α ⎤ ⎡α β ⎤ ⎢⎣ s 2 [ n + 1]⎥⎦ = ⎢⎣ γ δ ⎥⎦ ⎢⎣ s 2 [ n]⎥⎦ + ⎢⎣ γ ⎥⎦ x[ n]. Thus, A = ⎢⎣ γ δ ⎥⎦. Therefore, α γ ⎤ ⎡α β ⎤ ⎡α 2 + γ 2 αβ + γδ ⎤ ⎡r 2 0 ⎤ =⎢ A t A = ⎡⎢ . Likewise, ⎥= ⎣ β δ ⎥⎦ ⎢⎣ γ δ ⎥⎦ ⎣ αβ + γδ β 2 + δ 2 ⎦ ⎢⎣ 0 r 2 ⎥⎦

α β ⎤ ⎡ α γ ⎤ ⎡α 2 + β 2 αγ + βδ ⎤ ⎡r 2 0 ⎤ =⎢ AA t = ⎡⎢ . Since for stability r < 1, A is a ⎥= ⎣ γ δ ⎥⎦ ⎢⎣ β δ ⎥⎦ ⎣ αγ + βδ γ 2 + δ 2 ⎦ ⎢⎣ 0 r 2 ⎥⎦ normal form matrix, and hence the coupled-form structure does not support limit cycles. Not for sale

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12.41 The modified coupled-form structure with state-variables labeled is shown on top of the next page. Its transfer function determined in the solution of Problem 12.4 is given by c2 H ( z) = . Now, a second-order transfer function with poles at z 2 − 2cdz + c 2 (1 + d 2 ) x[n]

s 1[n+1]

_ z 1

s 2[n+1] _ z 1

c

d

c

y[n]

d _1

z = r e ± jθ has a denominator given by z 2 − 2r cos θ ⋅ z + r 2 . Comparing this denominator with that of H(z), we get c = r sin θ and d = cot θ. Analyzing the modified coupled-form structure we arrive at the following equations: s1[ n + 1] = c d s1[ n] − c s2 [ n] + x[ n] and s2 [ n + 1] = c s1[ n] + c d s2 [ n] . Rewriting these two

⎡ s [ n + 1] ⎤ ⎡cd − c ⎤ ⎡ s1[ n] ⎤ ⎡1 ⎤ equations in matrix form we get ⎢ 1 = + x[ n]. Thus, ⎣ s 2 [ n + 1]⎥⎦ ⎢⎣ c cd ⎥⎦ ⎢⎣ s 2 [ n]⎥⎦ ⎢⎣0 ⎥⎦ A = ⎡cd − c ⎤ = ⎡r cos θ − r sin θ ⎤. Therefore, ⎢⎣ c cd ⎥⎦ ⎢⎣ r sin θ r cos θ ⎥⎦ ⎡ 2 0⎤ A t A = ⎡ r cos θ r sin θ ⎤ ⎡r cos θ − r sin θ ⎤ = ⎢r . Likewise, ⎢⎣ − r sin θ r cos θ ⎥⎦ ⎢⎣ r sin θ r cos θ ⎥⎦ ⎣ 0 r 2 ⎥⎦ ⎡ 2 0⎤ AA t = ⎡r cos θ − r sin θ ⎤ ⎡ r cos θ r sin θ ⎤ = ⎢r . Since for stability r < 1, A is a ⎢⎣ r sin θ r cos θ ⎥⎦ ⎢⎣ − r sin θ r cos θ ⎥⎦ ⎣ 0 r 2 ⎥⎦ normal form matrix, and hence the modified coupled-form structure does not support limit cycles. z −1 ⇒ z = δ Γ + 1. Γ β + β1 z −1 + β2 z −2 β0 z 2 + β1 z + β2 . Therefore, H ( z) = 0 = 1 + α1 z −1 + α2 z − 2 z 2 + α1 z + α2

12.42 δ =

β z 2 + β1 z + β2 β (δΓ + 1) 2 + β1 (δΓ + 1) + β2 H ( δ ) = H ( z ) z = δΓ + 1 = 0 = 0 z 2 + α1 z + α 2 z = δΓ + 1 (δΓ + 1) 2 + α1 (δΓ + 1) + α 2 β (δ 2 Γ 2 + 2δΓ + 1) + β1 (δΓ + 1) + β2 ( β0 Γ 2 )δ 2 + (2 β0 Γ + β1Γ)δ + ( β0 + β1 + β2 ) = 0 = (δ 2 Γ 2 + 2δΓ + 1) + α1 (δΓ + 1) + α2 Γ 2 δ 2 + (2Γ + α1Γ)δ + (1 + α1 + α2 ) =

( β0 Γ 2 ) + (2 β0 Γ + β1Γ)δ −1 + ( β0 + β1 + β2 )δ −2 Γ 2 + (2Γ + α1Γ)δ −1 + (1 + α1 + α 2 )δ − 2

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12.43 From Section 11.3.1 we know that computation of each DFT sample requires 2 N + 4 real multiplications. Assuming that the quantization noise generated from each multiplier is independent of the noise generated from other multipliers, we get 2 −2 b ( N + 2) σ r2 = (2 N + 4)σ o2 = . 6 2 2b 2 2b 12.44 SNR = . Hence, an SNR of 30 dB implies = 10 3 , or 2 2 N N

(

1 2

)

b = log 2 10 3 × 512 2 = 12.9829. We choose b + 1 = 14 bits per sample to get an SNR of 30 dB. 12.45 Let N = 2 ν. Consider the m -th stage. The output sees 4(2 ν − m ) noise sources from the m -th 1 stage. Each noise source has a variance reduction by a factor of ⎛⎜ ⎞⎟ ⎝ 4⎠

by

1 2

ν−m

due to multiplication

at each stage till the output. Hence the total noise variance at the output due to the noises

(

)( ) ν ν output = σ 2 = ∑ 4(2 ν − m )(2 − 2(ν − m ) )σ o2 = 4σ o2 2 − ν ∑ 2 m

injected in the m -th stage is 4 2 ν − m 2 − 2(ν − m) σ o2 . Therefore, the total noise variance at the

m =1

⎛ 2 − 2b

⎞ −ν ⎟2 ⎜ = 4⎜⎜ ⎟ ⎜ 12 ⎝ ⎠ ⎝

m =1

⎛ 2( 2 ν

− 1) ⎞⎟ 2 − 2b − ν ) ≈ 2 2 − 2 b for large N . = 2 ( 1 − 2 2 − 1 ⎟⎠ 3 3

(

)

2b 1 12.46 SNR = 2 . Hence, b = log 2 10 3 × 2 × 256 = 9.4829. We choose b + 1 = 11 bits per 2 2N sample to get an SNR of 30 dB. 22b 1 2.5 SNR = . Hence b = log 2 (10 ) × 5 × 256 = 10.1439 . We choose b+1 = 12 bits per 2N 2 sample to get an SNR of 30dB. CHECK

(

)

M12.1 %POLE_PLOT % POLE_PLOT(B) obtains the pole distribution % plot of a second order transfer function with a denominator % of the form Z*Z - KZ + L. For stability 0 < L < 1 and % abs(K) < 1+L so the range of K is (-2,2). % % for a B-bit wordlength, 1 bit is reserved for the sign % as the coefficients are sorted in the sign magnitude form. % Both L and the K are quantizd to B-1 bits. % function[]=pole_plot(bits); point = 'o'; % Not for sale

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% This part prints the unit circle for reference % One bit is kept for the sign so effectively for the % quantization. The remaining number of bits is bits-1 % bits = bits-1; % % sets up the axis zplane(2,2); % plotting queue of roots using zplane.m with a % dummy zero and pole axis([-1 1 -1 1]); % adjusting axis to "hide" dummy zero/pole, % calling zplane without a zero is not legal in MATLAB release % 12 and 13 title(['Second order Pole distribution for ',num2str(bits+1),' bits']); hold on; % % The quantization step. % step = power(2,-bits); for index_1 = 0:1:(power(2,bits)-1) % 0 < index1 < 2^bits-1 L = index_1*step; % 0 < L < 1 for index_2 = -(power(2,bits)-1):1:(power(2,bits)-1) % -2^bits-1 < index2 < 2^bits-1 K = 2*index_2*step; % -2 < K < 2 p = roots([1 -K L]); % finding roots if abs(p(1)) < 1 % testing root 1 if (imag(p(1)) == 0) % real root plot(p(1),0,point); % plotting else % complex root plot(p(1),point); % plotting end end if abs(p(2)) < 1 % testing root 2 if (imag(p(1)) == 0) % real root plot(p(2),0,point); % plotting else % complex root plot(p(2),point); % plotting end end % title(['L = ',num2str(L),' K = ',num2str(K)]); % above 3 lines for debugging/demo purposes % place break here and step thru to see poles for each L and K end end title(['Second order Pole distribution for ',num2str(bits+1),' bits']); hold off;

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M12.2 The pole-distribution plot of the second-order direct form structure is obtained by running the MATLAB program given in Exercise M12.1. Running this program using the statement pole_plot(4) yields the following plot:

Imaginary Part

Second order Pole distribution for 4 bits 1 0.5 0 -0.5 -1 -1

-0.5

0 0.5 Real Part

1

For plotting the pole distribution of the coupled form structure, we use the following MATLAB program: % POLE_PLOT_COUPLED(B) obtains the pole distribution % plot of a second order transfer function in coupled form % with the transfor function rZ^2 / Z^2 - (a + d)Z + (ad - br) % For stability -1 < a,b,d,r < 1 % % for a B-bit wordlength, 1 bit is reserved for the sign % as the coefficients are sorted in the sign magnitude form. % Both L and the K are quantizd to B-1 bits. % function[]=pole_plot_coupled(bits); bits = 4; point = '.'; % % This part prints the unit circle for referenc % % One bit is kept for the sign so effectively for the % quantization, the remaining number of bits is bits-1 % bits = bits-1; % % sets up the axis zplane(2,2); % plotting queue of roots using zplane.m with a % dummy zero and pole axis([-1 1 -1 1]); % adjusting axis to "hide" dummy zero/pole, % calling zplane without a zero is not legal in MATLAB release % 12 and 13 title(['Second-order coupled form pole distribution for ', num2str(bits+1),' bits']); hold on; %

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% The quantization step. % step = power(2,-bits); for index_1 = -(power(2,bits)-1):1:(power(2,bits)-1) a = index_1*step; for index_2 = -(power(2,bits)-1):1:(power(2,bits)-1) % -2^bits-1 < index2 < 2^bits-1 b = index_2*step; for index_3 = -(power(2,bits)-1):1:(power(2,bits)-1) d = index_3*step; for index_4 = -(power(2,bits)-1):1:(power(2,bits)-1) r = index_4*step; p = roots([1 -(a+d) (a*d-b*r)]); % finding roots if abs(p(1)) < 1 % testing root 1 if (imag(p(1)) == 0) % real root plot(p(1),0,point); % plotting else % complex root plot(p(1),point); % plotting end end % testing root 2 if (imag(p(1)) == 0) % real root plot(p(2),0,point); % plotting else % complex root plot(p(2),point); % plotting end end end end % title(['a = ',num2str(a),' b = ',num2str(b),'d = ',num2str(d),' r = ',num2str(r)]); end end title(['Second-order coupled form pole distribution for ',num2str(bits+1),' bits']); hold off; M12.3 % Modified Program 12_1 % Coefficient Quantization Effects on the frequency % response of a direct form IIR filter clf; [b,a] = ellip(7,0.02,55,0.7,'high'); [h,w] = freqz(b,a,512); g = 20*log10(abs(h)); % Truncate the filter coefficients to 5 bits bq = a2dT(b,6); aq = a2dT(a,6); [hq,w] = freqz(bq,aq,512); gq = 20*log10(abs(hq));

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plot(w/pi,g,'b',w/pi,gq,'r:'); grid; axis([0 1 -80 5]); xlabel('\omega/\pi'); ylabel('Gain, dB'); title('original - solid line, quantized - dashed line'); pause % The call to the ZPLANE function below sets the axes % properties to those corresponding to a pole-zero plot zplane(b,a); % The HOLD ON command retains the axes properties as % they are and does not create new axes for the next plot hold on; plotzp(bq,aq)

M12.4 MATLAB code fragments are as follows: num = [0.06891875 0.13808186 0.18636107 0.13808186 0.06891875]; den = [1 -1.30613249 1.48301305 -0.77709026 0.2361457]; nvar = filternorm(num,den)^2; disp('Normalized Output Noise Variance = ');disp(nvar)

The computed noise variance is Normalized Output Noise Variance = 0.4026 which is seen to be the same as that determined in Example 12.5.. M 12.5 MATLAB code is shown below: N = 7; wn = 0.45; Rp = 0.2; Rs = 50; [B,A] = ellip(N,Rp,Rs,wn); zplane(B,A); z = cplxpair(roots(B)); p = cplxpair(roots(A)); const = B(1)/A(1); disp('Numerator Factors'); factor = factorize(B);disp(factor) disp('Denominator Factors'); factor = factorize(A);disp(factor) disp('Scale factor = ');disp(const) sos = zp2sos(z,p,const)

The above program yields Not for sale

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Numerator Factors 1.0000 1.0000 1.0000 1.0500 1.0000 0.2577 1.0000 0.0072

0 1.0000 1.0000 1.0000

Denominator Factors 1.0000 -0.2756 1.0000 -0.3951 1.0000 -0.6694 1.0000 -0.4353

0.9379 0.7545 0.4245 0

Scale factor = 0.03485569511749 ⎛ ⎞⎛ 1 + 0.2577 z −1 + z − 2 ⎞ 1 + 1.05z −1 + z − 2 ⎟⎜ ⎟× H ( z ) = 0.0349⎜⎜ −1 + 0.9379 z − 2 ⎟⎜ 1 − 0.3951 + 0.7545 z − 2 ⎟ − 1 0 . 2756 z ⎝ ⎠⎝ ⎠ ⎛ 1 + 0.0072 z −1 + z − 2 ⎞⎛ 1 + z −1 ⎞ ⎜ ⎟⎜ ⎟. ⎜ 1 − 0.6694 + 0.4245z − 2 ⎟⎜ 1 − 0.4353z −1 ⎟ ⎝ ⎠⎝ ⎠

Note that the ordering has no effect if L∞ − scaling is used.

Imaginary Part

1 0.5 0 -0.5 -1 -1

-0.5

0 0.5 Real Part

1

sos = 0.0349 1.0000 1.0000 1.0000

0.0349 1.0500 0.2577 0.0072

0 1.0000 1.0000 1.0000

1.0000 1.0000 1.0000 1.0000

-0.4353 -0.6694 -0.3951 -0.2756

0 0.4245 0.7545 0.9379

M 12.6 %Program_M12_6.m [B,A] = ellip(7,0.2,50,0.45); [d0,d1]=tf2ca(B,A); num1 = 0.5*(conv(fliplr(d1),d0)+conv(d1,fliplr(d0))); den = conv(d0,d1); num2 = 0.5*(-conv(fliplr(d1),d0)+conv(d1,fliplr(d0))); [h1,w]=freqz(num1,den,512); [h2,w]=freqz(num2,den,512); plot(w/pi,20*log10(abs(h1)),'-r',w/pi,20*log10(abs(h2)),'--b');

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xlabel('\omega/\pi');ylabel('Gain, dB');grid Passband Details

HLP(z)

0

-20

Gain, dB

Gain, dB

0

0.05

HHP(z)

-40

-0.05 -0.1 -0.15 -0.2

-60 0

0.2

0.4

0.6

0.8

1

-0.25

0

ω/ π

0.1

0.2

0.3

0.4

0.5

ω/ π

M12.7 num1 num2 den1 den2

= = = =

input('Numerator first factor = '); input('Numerator second factor = '); input('Denominator first factor = '); input('Denominator second factor = ');

% The scaling functions are f1num = 1; f1den = [den1]; f2num = num1; f2den = conv(den1,den2); f3num = conv(num1,num2); f3den = conv(den1,den2); x = [1 zeros([1,511])]; % Impulse Responses % Sufficient length for impulse response to have % decayed to nearly zero f1 = filter(f1num,f1den,x); f2 = filter(f2num,f2den,x); f3 = filter(f3num,f3den,x); figure(1),stem(f1(1:50)); ylabel('Amplitude'); xlabel('Samples'); title('Impulse Response of F_1(z)') figure(2),stem(f2(1:50)); Not for sale

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ylabel('Amplitude'); xlabel('Samples'); title('Impulse Response of F_2(z)') figure(3),stem(f3(1:50)); ylabel('Amplitude'); xlabel('Samples'); title('Impulse Response of H(z)') k1 = sum(f1.*f1); k2 = sum(f2.*f2); k3 = sum(f3.*f3); disp(['L_2 norm of F_1(z) = ' num2str(k1)]); disp(['L_2 norm of F_2(z) = ' num2str(k2)]); disp(['L_2 norm of H(z) = ' num2str(k3)]); b_0 = 1/sqrt(k1); b_1 = sqrt(k1)/sqrt(k2); b_2 = sqrt(k2)/sqrt(k3); disp(['First scaling factor, b_0 = ' num2str(b_0)]); disp(['Second scaling factor, b_1 = ' num2str(b_1)]); disp(['Third scaling factor, b_2 = ' num2str(b_2)]); % The scaled transfer functions disp(['Scaled gain = ' num2str(k1)]); disp('The scaled second order sections are ') disp([k2*num1 k3*num2 ; den1 den2]) % The noise transfer functions g1num = conv(num1,num2)*(k2*k3); g1den = conv(den1,den2)*k3; g2num = num2; g2den = den2; g1 = filter(g1num,g1den,x); g2 = filter(g2num,g2den,x); var = sum(f1.*f1)*3+sum(g2.*g2)*5+3; disp('The normalized noise variance'); disp(var);

For the first cascade realization we have L_2 norm of F_1(z) = 4.9808 L_2 norm of F_2(z) = 88.2071 L_2 norm of H(z) = 64.2444 First scaling factor, b_0 = 0.44807 Second scaling factor, b_1 = 0.23763 Third scaling factor, b_2 = 1.1717 Scaled gain = 4.9808 The scaled second order sections are 88.2071 -66.9403 88.2071 64.2444 1.0000 1.0462 0.8385 1.0000

42.1700 1.0657

64.2444 0.4046

The normalized noise variance 35.5077

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For the second cascade realization we have L_2 norm of F_1(z) = 2.8179 L_2 norm of F_2(z) = 11.375 L_2 norm of H(z) = 64.2444 First scaling factor, b_0 = 0.59572 Second scaling factor, b_1 = 0.49772 Third scaling factor, b_2 = 0.42078 Scaled gain = 2.8179 The scaled second order sections are 11.3750 7.4666 11.3750 64.2444 1.0000 1.0657 0.4046 1.0000

-48.7551 1.0462

64.2444 0.8385

The normalized noise variance 106.5131

The L_2 norm of F_1(z) The L_2 norm of F_2(z) The L_2 norm of H(z)

= 4.9808 = 11.375 = 64.2444

The first scaling factor, b_0 The second scaling factor, b_1 The third scaling factor, b_2

= 0.44807 = 0.66172 = 0.42078

The scaled gain = 4.9808 The scaled second order sections are 11.3750 7.4666 11.3750 64.2444 -48.7551 64.2444 1.0000 1.0462 0.8385 1.0000 1.0657 0.4046 The normalized noise variance = 98.0695 M12.8 The parallel form I and II structures used for the simulation are shown on top of the next page:

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k4 k0

+

k4

_ 1.4379k 1

k5

+

+

k0

+ _ z 1

_ z 1

_ z 1

0.9462k1

k5

+

+

+

+

_ z 1

0.8385

0.8385

k2

2.4525k 1

1.0462

1.0462

_ 0.5079k

k6

3

+

+

k2

1.0657

1.0657 _ 5.2081k _ z 1

+ _ z 1

_ z 1

+

1.2072k 1

+ 3

_ 4.6669k 3

k6

+

_ z 1

0.4046

0.4046

_ 0.2055k

3

The MATLAB program below can be used to simulate a 4-th order IIR transfer function in both parallel forms. num1 num2 den1 den2

= = = =

input('Numerator first factor ='); input('Numerator Second factor ='); input('Denominator first factor ='); input('Denominator second factor =');

% Parallel Form I num = conv(num1,num2); den = conv(den1,den2); [r1,p1,k11] = residuez(num,den); R1 = [r1(1) r1(2)]; P1 = [p1(1) p1(2)]; R2 =[r1(3) r1(4)]; P2 = [p1(3) p1(4)]; [num11,den11] = residuez(R1,P1,0); [num12,den12] = residuez(R2,P2,0); disp('Parallel Form I'); disp('The numerators are'); disp(k11); disp(real(num11)); disp(real(num12)); disp('The denominators are'); disp(real(den11)); disp(real(den12)); imp = [1 zeros([1,2000])];

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f0 = filter([1 0 0],den11,imp); f1 = filter(num11,den11,imp); f2 = filter([1 0 0],den12,imp); f3 = filter(num12,den12,imp); figure(1),stem(real(f0(1:50))); ylabel('Amplitude'); xlabel('Samples'); title('Impulse Response of F_0(z)') saveas(gcf,'M12_8_P1_0.tif') figure(2),stem(real(f1(1:50))); ylabel('Amplitude'); xlabel('Samples'); title('Impulse Response of F_1(z)') saveas(gcf,'M12_8_P1_1.tif') figure(3),stem(real(f2(1:50))); ylabel('Amplitude'); xlabel('Samples'); title('Impulse Response of F_2(z)') saveas(gcf,'M12_8_P1_2.tif') figure(4),stem(real(f3(1:50))); ylabel('Amplitude'); xlabel('Samples'); title('Impulse Response of F_3(z)') saveas(gcf,'M12_8_P1_3.tif') g0 = sum(f0.*conj(f0)); g1 = sum(f1.*conj(f1)); g2 = sum(f2.*conj(f2)); g3 = sum(f3.*conj(f3)); disp(['L_2 norm of F_0(z) disp(['L_2 norm of F_1(z) disp(['L_2 norm of F_2(z) disp(['L_2 norm of F_3(z)

= = = =

' ' ' '

num2str(g0)]); num2str(g1)]); num2str(g2)]); num2str(g3)]);

k0 = sqrt(1/g0); k1 = sqrt(g0/g1); k2 = sqrt(1/g2); k3 = sqrt(g2/g3); f = filter(num,den,imp); g = sum(f.*conj(f)); k4 = sqrt(1/g); k5 = k4/(k0*k1); k6 = k4/(k2*k3); disp('The disp(['k0 disp(['k2 disp(['k4 disp(['k6

scaling constants are'); = ' num2str(k0)]); disp(['k1 = ' num2str(k1)]); = ' num2str(k2)]); disp(['k3 = ' num2str(k3)]); = ' num2str(k4)]); disp(['k5 = ' num2str(k5)]); = ' num2str(k6)]);

noise = 3*(k5/k0)^2+3*(k6/k2)^2+2*k5^2+2*k6^2+3; disp(['Product roundoff noise variance = ' num2str(noise)]); % Parallel From II Not for sale

493

num = conv(num1,num2); den = conv(den1,den2); [r1,p1,k11] = residue(num,den); R1 = [r1(1) r1(2)]; P1 = [p1(1) p1(2)]; R2 =[r1(3) r1(4)]; P2 = [p1(3) p1(4)]; [num11,den11] = residue(R1,P1,0); [num12,den12] = residue(R2,P2,0); disp('Parallel Form II'); disp('The numerators are'); disp(k11); disp(real(num11)); disp(real(num12)); disp('The denominators are'); disp(real(den11)); disp(real(den12)); imp = [1 zeros([1,2000])]; f0 = filter([1 0 0],den11,imp); f1 = filter(num11,den11,imp); f2 = filter([1 0 0],den12,imp); f3 = filter(num12,den12,imp); figure(1),stem(real(f0(1:50))); ylabel('Amplitude'); xlabel('Samples'); title('Impulse Response of F_0(z)') saveas(gcf,'M12_8_P2_0.tif') figure(2),stem(real(f1(1:50))); ylabel('Amplitude'); xlabel('Samples'); title('Impulse Response of F_1(z)') saveas(gcf,'M12_8_P2_1.tif') figure(3),stem(real(f2(1:50))); ylabel('Amplitude'); xlabel('Samples'); title('Impulse Response of F_2(z)') saveas(gcf,'M12_8_P2_2.tif') figure(4),stem(real(f3(1:50))); ylabel('Amplitude'); xlabel('Samples'); title('Impulse Response of F_3(z)') saveas(gcf,'M12_8_P2_3.tif') g0 = sum(f0.*conj(f0)); g1 = sum(f1.*conj(f1)); g2 = sum(f2.*conj(f2)); g3 = sum(f3.*conj(f3)); disp(['L_2 norm of F_0(z) disp(['L_2 norm of F_1(z) disp(['L_2 norm of F_2(z) disp(['L_2 norm of F_3(z)

= = = =

' ' ' '

num2str(g0)]); num2str(g1)]); num2str(g2)]); num2str(g3)]);

k0 = sqrt(1/g0); k1 = sqrt(g0/g1); k2 = sqrt(1/g2); Not for sale

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k3 = sqrt(g2/g3); f = filter(num,den,imp); g = sum(f.*conj(f)); k4 = sqrt(1/g); k5 = k4/(k0*k1); k6 = k4/(k2*k3); disp('The scaling constants are'); disp(['k0 = ' num2str(k0)]); disp(['k1 = ' num2str(k1)]); disp(['k2 = ' num2str(k2)]); disp(['k3 = ' num2str(k3)]); disp(['k4 = ' num2str(k4)]); disp(['k5 = ' num2str(k5)]); disp(['k6 = ' num2str(k6)]); noise = 3*(k5/k0)^2+3*(k6/k2)^2+2*k5^2+2*k6^2+3; disp(['Product roundoff noise variance = ' num2str(noise)]); Parallel Form I The numerators are 2.9476 -1.4397 0.9462 -0.5079 -5.2081

0 0

The denominators are 1.0000 1.0462 1.0000 1.0657

0.8385 0.4046

L_2 norm of F_0(z) = 4.9808 L_2 norm of F_1(z) = 22.5061 L_2 norm of F_2(z) = 2.8179 L_2 norm of F_3(z) = 65.8497 The scaling constants are k0 = 0.44807 k1 = 0.47044 k2 = 0.59572 k3 = 0.20686 k4 = 0.12476 k5 = 0.59188 k6 = 1.0124 Product roundoff noise variance = 19.6501 Parallel Form II The numerators are 1 2.4525 1.2072 -4.6669 0.2055 The denominators are 1.0000 1.0462 1.0000 1.0657

0.8385 0.4046

L_2 norm of F_0(z) = 4.9808 L_2 norm of F_1(z) = 20.4333

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L_2 norm of F_2(z) = 2.8179 L_2 norm of F_3(z) = 65.5918 The scaling constants are k0 = 0.44807 k1 = 0.49372 k2 = 0.59572 k3 = 0.20727 k4 = 0.12476 k5 = 0.56396 k6 = 1.0104 Product roundoff noise variance = 19.0615

Parallel Form II has the lowest output round off noise. M 12.9 The scaled Gray-Markel cascaded lattice structure used for simulation is shown below: W4 _d

_ '' d2

_d ' 3

4

d4 z

α 1C ____ k4

d1''' _1

_1

α 2C ____ k3 k 4

_ ''' d1

d 2''

d 3' _1

W1

W2

W3

z

α 3C ____ k 2 k3 k 4

z

_

α 4C _______ k 1k 2k 3k 4

z1

α 5C _______ k 1k 2k 3k 4

The MATLAB program that can be used to simulate the Gray-Markel realization of a 4-th order IIR transfer function is as follows: num1 num2 den1 den2

= = = =

input('Numerator first factor = '); input('Numerator second factor = '); input('Denominator first factor = '); input('Denominator second factor = ');

num = conv(num1,num2); den = conv(den1,den2); num = num/den(1); den = den/den(1); % Gray-Markel realization [d,alpha] = tf2latc(num,den); disp('Lattice parameters are'); disp(['d1 = ' num2str(d(1))]); disp(['d2 = ' num2str(d(2))]); disp(['d3 = ' num2str(d(3))]); disp(['d4 = ' num2str(d(4))]); disp('Feed-forward multipliers are'); disp(['alpha1 = ' num2str(alpha(1))]); disp(['alpha2 = ' num2str(alpha(2))]); disp(['alpha3 = ' num2str(alpha(3))]); disp(['alpha4 = ' num2str(alpha(4))]); imp = [1 zeros([1,499])]; qold1 = 0; for k = 1:500 w1 = imp(k)-d(1)*qold1;

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y1(k) = w1; qnew1 = w1; qold1 = qnew1; end k1 = sqrt(1/(sum(y1.*conj(y1)))); figure(1),stem(real(y1(1:50))); ylabel('Amplitude'); xlabel('Samples'); title('Impulse Response of W1') saveas(gcf,'M12_9_1.tif') imp = [1 zeros([1,499])]; qold1 = 0;qold2 = 0; for k = 1:500 w2 = imp(k)-d(2)*qold2*1/k1; w1 = k1*w2-d(1)*qold1; y2(k)=w2; qnew1 = w1; qnew2 = w1*d(1)+qold1; qold1 = qnew1;qold2 = qnew2; end k2 = sqrt(1/(sum(y2.*conj(y2)))); figure(2),stem(real(y2(1:50))); ylabel('Amplitude'); xlabel('Samples'); title('Impulse Response of W2') saveas(gcf,'M12_9_2.tif') qold1 = 0;qold2 = 0;qold3 = 0; for k = 1:500 w3 = imp(k)-d(3)*qold3*1/k2; w2 = k2*w3-d(2)*qold2*1/k1; w1 = k1*w2-d(1)*qold1; y3(k) = w3; qnew1 = w1; qnew2 = w1*d(1)+qold1; qnew3 = w2*d(2)+qold2*1/k1; qold1 = qnew1;qold2 = qnew2;qold3 = qnew3; end k3 = sqrt(1/sum(y3.*conj(y3))); figure(3),stem(real(y3(1:50))); ylabel('Amplitude'); xlabel('Samples'); title('Impulse Response of W3') saveas(gcf,'M12_9_3.tif') qold1 = 0;qold2 = 0;qold3 = 0;qold4 = 0; for k = 1:500 w4 = imp(k)-d(4)*qold4/k3; w3 = k3*w4-d(3)*qold3/k2; w2 = k2*w3-d(2)*qold2/k1; w1 = k1*w2-d(1)*qold1; y4(k) = w4; qnew1 = w1; qnew2 = w1*d(1)+qold1; qnew3 = w2*d(2)+qold2*1/k1; qnew4 = w3*d(3)+qold3*1/k2; qold1 = qnew1;qold2 = qnew2;qold3 = qnew3;qold4 = qnew4; end

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k4 = sqrt(1/sum(y4.*conj(y4))); figure(4),stem(real(y3(1:50))); ylabel('Amplitude'); xlabel('Samples'); title('Impulse Response of W4') saveas(gcf,'M12_9_4.tif') disp('The scaling parameters are'); disp(['k1 = ' num2str(k1)]); disp(['k2 = ' num2str(k2)]); disp(['k3 = ' num2str(k3)]); disp(['k4 = ' num2str(k4)]); alpha(5) = alpha(5)/(k1*k2*k3*k4); alpha(4) = alpha(4)/(k1*k2*k3*k4); alpha(3) = alpha(3)/(k2*k3*k4); alpha(2) = alpha(2)/(k3*k4); alpha(1) = alpha(1)/k4; %%%% Computation of noise variance %%%%%% imp = [1 zeros([1,499])]; for k = 1:500 w4 = imp(k)-d(4)*qold4/k3; w3 = k3*w4-d(3)*qold3/k2; w2 = k2*w3-d(2)*qold2/k1; w1 = k1*w2-d(1)*qold1; qnew1 = w1; qnew2 = w1*d(1)+qold1; qnew3 = w2*d(2)+qold2*1/k1; qnew4 = w3*d(3)+qold3*1/k2; y11 = w4*d(4)+qold4/k3; y0(k) = alpha(1)*y11+alpha(2)*qnew4+alpha(3)*qnew3 + alpha(4)*qnew2+alpha(5)*qnew1; qold1 = qnew1;qold2 = qnew2;qold3 = qnew3;qold4 = qnew4; end nv = sum(y0.*conj(y0)); disp(['Product roundoff noise variance = ' num2str(nv)]);

The output data generated by this program are as follows: Lattice parameters are d1 = 0.19149 d2 = 0.75953 d3 = 0.44349 d4 = 0.27506 Feed-forward multipliers are alpha1 = -0.9001 alpha2 = -1.7069 alpha3 = 4.6151 alpha4 = -3.5028 alpha5 = 1.8177 The scaling parameters are k1 = 0.98149 k2 = 0.65047 k3 = 0.89628 k4 = 0.96143

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The product roundoff noise variance = 81.1431 M12 .10 (a)

y[ −1] = 0.3, x[ 0] = 0.08

(b) y[ −1] = 0.9, x[ 0] = 0.03 α = 0.6

0.25

0.5

0.2

0.4

Amplitude

Amplitude

α = 0.6

0.15 0.1 0.05 0

0.3 0.2 0.1

0

5

10 Time index n

15

0

20

0

5

10 Time index n

15

20

(c) y[ −1] = 8, x[ 0] = 5 α = 0.6

α = 0.6

0.04

10

0.03 Amplitude

Amplitude

8 6 4

0.01

2 0

0.02

0

5

10 Time index n

15

0

20

0

5

10 Time index n

15

20

In all three cases, the condition of Eq. (12.187) is satisfied and hence, the structure exhibits zero input granular limit cycles. M 12.11 Plot of the output samples generated by the modified program is shown below: α 1 = -0.875

α 2 = 0.875

0.4

Amplitude

0.2 0 -0.2 -0.4

0

10

20 Time index n

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30

40

499

As can be seen from the above plot, output goes to zero values and hence the structure of Figure 12.49 does not support overflow limit cycles if sign-magnitude truncation is used to truncate the sum of products of Eq. (12.195).

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Chapter 13 13.1 Up-sampler – Let x1[ n] and x 2 [ n] be the inputs to a factor-of-L up-sampler with corresponding outputs given by y1[ n] and y2 [ n] , respectively: ⎧ x [ n / L ], n = 0, ± L, ± 2 L,K, and ⎧ x [ n / L ], n = 0, ± L, ± 2 L,K, y1[ n] = ⎨ 1 y 2 [ n] = ⎨ 2 otherwise, 0, otherwise. ⎩ 0, ⎩ Let x 3 [ n] = αx1[ n] + βx 2 [ n] be the input to the up-sampler. Then, the corresponding ⎧ x [ n / L ], n = 0, ± L, ± 2 L,K, output y3 [ n] is given by y3 [ n] = ⎨ 3 0, otherwise, ⎩ ⎧αx [ n / L ] + βx 2 [ n / L ], n = 0, ± L, ± 2 L,K, =⎨ 1 0, otherwise, ⎩ ⎧αx [ n / L ], n = 0, ± L, ± 2 L,K, ⎧ βx 2 [ n / L ], n = 0, ± L, ± 2 L,K, =⎨ 1 +⎨ 0, otherwise, 0, otherwise, ⎩ ⎩ = αy1[ n] + βy2 [ n]. Hence, the up-sampler is a linear system.

Down-sampler - Let x1[ n] and x 2 [ n] be the inputs to a factor-of-M down-sampler with corresponding outputs given by y1[ n] and y2 [ n] , respectively: y1[ n] = x1[ Mn] and y2 [ n] = x 2 [ Mn]. Let x 3 [ n] = αx1[ n] + βx 2 [ n] be the input to the up-sampler. Then, the corresponding output y3 [ n] is given by y3 [ n] = x 3 [ Mn] = αx1[ Mn] + βx 2 [ Mn] = αy1[ n] + βy2 [ n]. Hence, the down-sampler is a linear system. 13.2 Up-Sampler – For inputs x1[ n] and x 2 [ n], the outputs of the factor-of-L up-sampler are, respectively given by ⎧ x [ n / L ], n = 0, ± L, ± 2 L,K, ⎧ x [ n / L ], n = 0, ± L, ± 2 L,K and x u 2 [ n] = ⎨ 2 . x u1 [ n] = ⎨ 1 0 , otherwise, 0, otherwise, ⎩ ⎩ Let x 2 [ n] = x1[ n − no ], where no is an integer. Then, x 2 [ n / L ] = x1[(n / L ) − no ]. Hence, ⎧ x [(n − n o ) / L ], n = 0, ± L, ± 2 L,K, But, x u 2 [ n] = ⎨ 1 0, otherwise. ⎩ ⎧ x [(n − n o ) / L ], n − n o = 0, ± L, ± 2 L,K, x u1 [ n] = ⎨ 1 0, otherwise. ⎩

Since xu2 [ n] ≠ xu1 [ n − no ], the up-sampler is a time-varying system. Down-Sampler – For inputs x1[ n] and x 2 [ n], the outputs of the factor-of-M downsampler are, respectively given by y1[ n] = x1[ Mn] and y2 [ n] = x 2 [ Mn]. Let x 2 [ n] = x1[ n − n o ], where no is an integer. Then, y2 [ n] = x 2 [ Mn] = x1[ Mn − no ]. However, y1[ n − no ] = x1[ M (n − no )] = x1[ Mn − Mno ]. Since y 2 [ n] ≠ y1[ n − n o ], the down-sampler is a time-varying system.

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13.3 X(z)

2

V(z)

Vu (z)

2

_ z1

_ z1

2

Analysis yields V ( z ) = W ( z) =

z −1 / 2 2

Wu ( z ) =

z −1 2

1 2

X ( z1 / 2 ) − X ( z) −

z −1 2

W(z)

2

Y(z)

Wu (z)

1 2

X ( z1 / 2 ) + X (− z1 / 2 ), z −1 / 2 2

X (− z1 / 2 ), Vu ( z ) =

1 2

1 2

X ( z ) + X (− z ),

X (− z ). Hence, Y ( z ) = z −1V u( z) +W u( z ) = z −1 X ( z), or in other

words y[ n] = x[ n − 1]. 13.4 c[ n] =

M −1 ⎛ 1 − W nM 1 M kn = 1 ⎜ WM n ⎜ M k =0 M ⎝ 1 − WM

∑

⎛ ⎞ ⎟. Hence, if n ≠ rM , c[ n] = 1 ⎜ 1 − 1 n ⎟ M ⎜⎝ 1 − WM ⎠

the other hand, if n = rM , then c[ n] =

M −1

M −1

∑

⎞ ⎟ = 0. On ⎟ ⎠

M −1

∑

∑

1 M krM = 1 kn = 1 1= = 1. WM WM M k =0 M k =0 M k =0 M

13.5 (a) For M = 6 and L = 5, − k } = {W − k } = {W 0 , W −1 , W −2 , W −3 , W −4 , W −5} , and {WM 6 6 6 6 6 6 6 − Lk } = {W −5k } = {W 0 , W −5 , W −10 , W −15 , W −20 , W −25} {WM 6 6 6 6 6 6 6

= {W60 , W6−5 , W6−4 , W6−3 , W6−2 , W6−1} = {W6− k }. − k } to have same set of values for 0 ≤ k ≤ M − 1, as {W − kL } , each should (b) For {WM M − kL − nL have unique values for each k. Therefore WM ≠ WM for all k , n ∈ [0, M − 1], or

( k − n) L ≠ rM for any positive integer r, which implies that L and M should be relatively prime.

13.6 X(z)

M

V 1(z)

H(z)

Y 1(z)

X(z)

H(zM )

For the figure on the left-hand side we have V1 ( z ) =

V 2(z)

M

Y 2(z)

M −1

∑

1 k ), and X ( z 1 / M WM M k =0

M −1

∑

1 k ). For the figure on the right-hand side we have Y1 ( z ) = H ( z ) X ( z1 / M WM M k =0

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V2 ( z ) =

M −1

M −1

∑

∑

1 kM ) X ( z1 / M W k ) = 1 k ). H ( zWM H ( z ) X ( z1 / M WM M M k =0 M k =0

H ( z M ) X ( z ), Y2 ( z ) =

Hence, Y1 ( z) = Y2 ( z). X(z)

V 1(z)

L

H(z L )

Y 1(z)

V 2(z)

H(z)

X(z)

Y 2(z)

L

For the figure on the left-hand side we have V1 ( z ) = X ( z L ), Y1 ( z ) = H ( z L ) X ( z L ). For the figure on the right-hand side we have V2 ( z ) = H ( z ) X ( z ), Y2 ( z ) = H ( z L ) X ( z L ). Hence, Y1 ( z) = Y2 ( z). 13.7 (a) The system of Figure P13.1 with internal variables labeled is shown below: X(z)

L

V(z)

G(z)

U(z)

L

Y(z)

Analysis yields V ( z ) = X ( z L ), U ( z ) = G ( z )V ( z ), and Y ( z ) =

L −1

∑

1 U ( z1 / L WL− k ). L k =0

Substituting the first equation in the second equation we get U ( z ) = G ( z ) X ( z L ). Substituting this equation in the expression for Y (z ) in the above we get Y ( z) =

L −1

L −1

∑

∑

1 1 G( z1 / L ⋅ WL− k ) X ( z ⋅ WL− kL ) = G( z1 / L ⋅ WL− k ) X ( z ), since WL−kL = 1. L k =0 L k =0 L −1

∑

Y (z) 1 Therefore, H ( z ) = = G( z1 / L WL− k ). Hence, Figure P13.1 is a LTI system. X ( z) L k = 0 L −1

(b) It follows from the last equation given above, if

∑

1 G( z1 / L WL− k ) = 1, then L k =0

H ( z ) = 1, i.e., Y ( z ) = X ( z ), or y[ n] = x[ n]. Or, in other words, the system of Figure L −1

∑

1 P13.1 is an identity system for G( z1 / L WL− k ) = 1. L k =0 13.8 Consider the multirate structure shown below. Analysis yields x[n]

L

H(z)

1 U ( z ) = H ( z ) X ( z ), and Y ( z ) = L L

L −1

u[n]

∑ G( z

G(z)

L

y[n]

1/ L

W Lk )U ( z 1 / LW Lk ). Substituting the first

k =0

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⎡ 1 L −1 ⎤ equation into the second we get Y ( z ) = ⎢ G ( z 1 / LWLk ) H ( z 1 / LW Lk )⎥ X ( z ). Hence, ⎢L ⎥ ⎣ k =0 ⎦

∑

1 if L

L −1

∑ G( z

1/ L

W Lk ) H ( z 1 / LW Lk ) = 1, we have Y ( z ) = X ( z ), or in other words, the

k =0

above multirate structure is an identity system. We break the system as shown below into two parts with the transfer functions and satisfying the relation L

x[n]

1 L

L −1

∑ G( z

G(z)

u[n] u[n]

H(z)

L

x[n]

1/ L

W Lk ) H ( z 1 / LW Lk ) = 1. If we now place the second system in front of the

k =0

first we arrive at the system shown in Figure P13.2 which is an identity system 1 provided L

L −1

∑ G( z

1/ L

W Lk ) H ( z 1 / LW Lk ) = 1.

k =0

13.9 Making use of the multirate identities we simplify the structure of Figure P13.3 as indicated below: x[n]

x[n]

5

15

3

3

3

x[n]

5

y[n]

5

5 v[n]

5

y[n]

y[n]

Analysis of the last structure yields, v[ n] = x[5n], and v[ n / 5], n = 0, ± 5, ± 10,K, x[ n], n = 0, ± 5, ± 10,K, y[ n] = ⎧⎨ or, y[ n] = ⎧⎨ otherwise, otherwise. ⎩ 0, ⎩ 0, 13.10 x[n]

L

v[n]

L

y[n]

⎧v[ n / L ], n = 0, ± L, ± 2 L,K, or, Analysis yields v[ n] = x[ Ln], and y[ n] = ⎨ otherwise, ⎩ 0, x[ n], n = 0, ± L, ± 2 L,K, y[ n] = ⎧⎨ otherwise. ⎩ 0,

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13.11 Since 3 and 4 are relatively prime, we can interchange the positions of the factor-of-3 down-sampler and the factor-of-4 up-sampler as indicated below: _ z 6

x[n]

2

3

y[n]

4

which simplifies to the structure shown below: x[n]

_ z 6

6

y[n]

4

Using the Noble identity of Figure 13.14(a) we redraw the above structure as indicated below: x[n]

6

u[n]

_ z 1

v[n]

y[n]

4

Analysis yields u[ n] = x[6 n], v[ n] = u[ n − 1] = x[6 n − 1], and v[ n / 4], n = 0, ± 4, ± 8,K, ⎧ x[(3n / 2) − 1], n = 0, ± 4, ± 8,K, =⎨ y[ n] = ⎧⎨ otherwise, otherwise. 0, ⎩ 0, ⎩ 13.12 As outlined in Section 8.2, the transpose of a digital filter structure is obtained by reversing all paths, replacing the pick-off node with an adder and vice-versa, and interchanging the input and the output nodes. Moreover, in a multirate structure, the transpose of a factor-of-M down-sampler is a factor-of-M up-sampler and vice-versa. Applying these operations to the factor-of-M decimator shown below on the left-hand side, we arrive at a factor-of-M up-sampler shown below on the right-hand side. x[n]

H(z)

M

y[n]

x[n]

M

H(z)

y[n]

13.13 (a) To prove Eq. (13.20), consider the fractional-rate sampling rate converter of Figure 13.16(b) with internal variables labeled as shown below: x[n]

L

x u[n]

H(z)

v[n]

M

y[n]

Analysis yields x u [ Ln] = x[ n], n = 0, ± 1, ± 2,K and v[ n] = Substituting the first equation in the second we get v[ n] =

∞

∑ h[n − l]xu [l].

l = −∞ ∞

∑ h[n − Lm]x[m]. Finally,

m = −∞

y[ n] = v[ Mn] =

∞

∑ h[ Mn − Lm]x[m].

m = −∞

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503

(b) Next, to prove Eq. (13.21), we make use of the z-domain relations of the downsampler and the up-sampler. From Eq. (13.17) we have Y ( z) =

M −1

∑

1 − k ) X ( z1 / M W − k ). But X ( z ) = X ( z L ). Hence, H ( z1 / M WM u u M M k =0 M −1

∑

1 − k ) X ( z L / M W − Lk ). Y (z) = H ( z1 / M WM M M k =0 13.14 x[n]

E 00 (z)

2

+

3

_ z1

E 01 (z)

_ z1

+

3

_1

z

2

E 02 (z)

3

E 10 (z)

3

+

_ z1

+

y[n]

_1

z

E 11 (z)

3

+ _ z1

E 12 (z)

3

13.15 x[n]

8

400 Hz

x u [n]

v[n ]

H(z)

3200 Hz

15

y[n] 213.3333 Hz

3200 Hz

(a) FT = 400 Hz, L = 8, M = 15. Now, the sampling rate of x u [n] and v[n] is 400L = 400 × 8 = 3.2 kHz. Hence, the sampling rate of y[n] is 3200 / M = 3200 / 15 = 213.33 Hz. (b) The normalized stopband edge angular frequency of H (z ) (for no aliasing) 1 π ⎛π π ⎞ π Hz. ωs = min⎜ , ⎟ = = . Hence, the stopband edge frequency is Fs = 30 ⎝ L M ⎠ M 15 13.16 x[n] 650 Hz

8

x u [n] 3.25 kHz

H(z)

v[n ] 3.25 kHz

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15

y[n] 361.1 Hz

504

(a) FT = 650 Hz, L = 5, M = 9. Now, the sampling rate of x u [n] and v[n] is 650L = 650 × 5 = 3.25 kHz. Hence, the sampling rate of y[n] is 3250 / M = 3250 / 9 = 361.1 Hz. (b) The normalized stopband edge angular frequency of H (z ) (for no aliasing) π ⎛π π ⎞ π = . Hence, the stopband edge frequency is Hz. ωs = min ⎜ , ⎟ = ⎝L M⎠ M 9 13.17 Applying the transpose operation to the M-channel analysis filter bank shown below on the left-hand side, we arrive at the M-channel synthesis filter bank shown below on the right-hand side. x[n]

H0 (z)

M

v 0[n]

H1 (z)

M

v 1[n]

H M _1(z)

M

vM _ 1[n]

y[n]

+

H0 (z)

M

v 0[n]

+

H1 (z)

M

v 1[n]

H M _1(z)

M

vM _ 1[n]

v 0[n]

M

H0 (z)

+

v 1[n]

M

H1 (z)

+

vM _ 1[n]

M

H M _1(z)

y[n]

13.18 Specifications for H(z) are: Fp = 180 Hz, Fs = 200 Hz, δ p = 0.002, δs = 0.001. H(z)

12 kHz

H (z) =

30 12 kHz 400 Hz

I ( z)F ( z 5 ) F(z5 )

I(z)

12 kHz

I(z)

F(z5 )

I(z)

5

30 5

6

F(z)

12 kHz 2.4 kHz

6

2.4 kHz

400 Hz

Specifications for F(z) are: Fp = 900 Hz, Fs = 1000 Hz, δ p = 0.001, δs = 0.001.

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505

100 . Using Eq. (10.3) we arrive at the order of F(z) given by 12000 − 20 log10 0.001 × 0.001 − 13 NF = = 386.3 ⇒ N F = 387 ⎛ 100 ⎞ 14.6⎜ ⎟ ⎝ 12000 ⎠ Specifications for I(z) are: Fp = 180 Hz, Fs = 2200 Hz, δ p = 0.001, δs = 0.001.

Here ∆f =

2020 . Using Eq. (10.3) we arrive at the order of I(z) given by 12000 − 20 log10 0.001 × 0.001 − 13 NI = = 19.12 ⇒ N F = 207 ⎛ 2020 ⎞ 14.6⎜ ⎟ ⎝ 12000 ⎠ 2400 12000 Hence, R M , F = (387 + 1) = 155,200 mps, R M , I = (20 + 1) = 50,400 mps. 6 5 The above realization requires a total of 205,600 mps. As a result, the computational complexity is slightly higher than in Example 13.10.

Here ∆f =

13.19 Specifications for H(z) are: Fp = 180 Hz, Fs = 200 Hz, δ p = 0.002, δs = 0.001. H ( z ) = I ( z )F ( z 3 ) F(z3 )

I(z)

12 kHz

I(z)

F(z3 )

I(z)

3 12 kHz

30 3

10

F(z) 4 kHz

10

4 kHz

400 Hz

Specifications for F(z) are: Fp = 540 Hz, Fs = 600 Hz, δ p = 0.001, δs = 0.001. 60 Here ∆f = . Using Eq. (10.3) we arrive at the order of F(z) given by 12000 − 20 log10 0.001 × 0.001 − 13 NF = = 643.8 ⇒ N F = 644 ⎛ 60 ⎞ 14.6⎜ ⎟ ⎝ 12000 ⎠ Specifications for I(z) are: Fp = 180 Hz, Fs = 3800 Hz, δ p = 0.001, δs = 0.001. 3260 . Using Eq. (10.3) we arrive at the order of I(z) given by 12000 − 20 log10 0.001 × 0.001 − 13 NI = = 11.85 ⇒ N F = 12 ⎛ 3260 ⎞ 14.6⎜ ⎟ ⎝ 12000 ⎠

Here ∆f =

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506

4000 12000 = 262,000 mps, R M , I = (12 + 1) = 52,000 mps. 10 3 The above realization requires a total of 314,000 mps. As a result, the computational complexity is higher than in Example 13.10 and that in Problem 13.18.

Hence, R M , F = (644 + 1)

40000 = 20. The general structure of 2,000

13.20 (a) The desired down-sampling factor is M = the desired decimator is thus as shown below: H(z) 40 kHz

20 2 kHz

40 kHz

Now, the normalized stopband edge angular frequency of a factor-of-20 decimator is π . Hence, the desired stopband edge frequency in this case is ωs = 20 20000 Fs = = 1000 Hz. The specifications of the decimation filter H(z) is thus as 20 200 follows: Fp = 800 Hz, Fs = 1000 Hz, δ p = 0.002, δs = 0.002. Here ∆f = . 40000 Using Eq. (10.3) we arrive at the order of H(z) given by − 20 log10 0.002 × 0.002 − 13 NH = = 561.4 ⇒ 562. Therefore, the computational ⎛ 200 ⎞ 14.6⎜ ⎟ ⎝ 40000 ⎠ 40000 complexity is given by R M , H = (562 + 1) = 1126,000 mps. 20 (b) For a two-stage realization of the decimator, there are 4 possible realizations of the decimation filter: Realization #1 – H ( z ) = F ( z 2 ) I ( z ) F(z 2)

I(z) 40 kHz

40 kHz I(z) 40 kHz

2 40 kHz F(z)

2 40 kHz

20 kHz

10 10 kHz

2 kHz

10 20 kHz

2 kHz

Specifications for F (z ) and I (z ) are as follows: 400 . 40000 18200 I ( z) : Fp = 800 Hz, Fs = 19,000 Hz, δ p = 0.001, δs = 0.002, and thus, ∆f = . 40000 Orders of F (z ) and I (z ) are given by

F ( z) : Fp = 1600 Hz, Fs = 2000 Hz, δ p = 0.001, δs = 0.002, and thus, ∆f =

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507

− 20 log10 0.001 × 0.002 − 13 = 301.29 ⇒ 302 ⎛ 400 ⎞ 14.6⎜ ⎟ ⎝ 40000 ⎠ − 20 log10 0.001 × 0.002 − 13 NI = = 6.62 ⇒ 7 ⎛ 18200 ⎞ 14.6⎜ ⎟ ⎝ 40000 ⎠ Computational complexities of the two sections are: 20000 40000 R M , F = (302 + 1) = 606,000 mps and R M , I = (7 + 1) = 160,000 mps. 10 2 Hence, the total computational complexity of the two-stage realization is R M = R M , F + R M , I = 766,000 mps. NF =

Realization #2 – H ( z ) = F ( z 4 ) I ( z ) F(z 4)

I(z) 40 kHz

40 kHz I(z) 40 kHz

4 40 kHz F(z)

4 40 kHz

10 kHz

5 10 kHz

2 kHz

5 10 kHz

2 kHz


F ( z ) : Fp = 3200 Hz, Fs = 4000 Hz, δ p = 0.001, δs = 0.002, and thus, ∆f =


Realization #3 – H ( z ) = F ( z 5 ) I ( z )

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508

F(z 5)

I(z) 40 kHz

40 kHz

40 kHz

I(z)

4 kHz

F(z)

5 8 kHz

40 kHz

40 kHz

4

5

2 kHz 4

8 kHz

2 kHz

Specifications for F (z ) and I (z ) are as follows: F ( z ) : Fp = 4000 Hz, Fs = 5000 Hz, δ p = 0.001, δs = 0.002, and thus, ∆f =

1000 . 40000

I ( z) : Fp = 800 Hz, Fs = 7,000 Hz, δ p = 0.001, δs = 0.002, and thus, ∆f =

6200 . 40000

Orders of F (z ) and I (z ) are given by


Realization #4 – H ( z ) = F ( z 10 ) I ( z ) I(z)

40 kHz

40 kHz

I(z)

40 kHz

F(z 10 )

40 kHz

4 kHz

F(z)

10

40 kHz

2

10

4 kHz

2 kHz

2

4 kHz

2 kHz


F ( z) : Fp = 8000 Hz, Fs = 10,000 Hz, δ p = 0.001, δs = 0.002, and thus, ∆f =

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509


Hence, the optimum two-stage design with the lowest computational complexity is the Realization #4. 13.21 (a) The desired up-sampling factor is L =

24000 = 50. The general structure of the 480

desired interpolator is thus as shown below: H(z)

50 480 Hz

24 kHz

24 kHz

Now, the normalized stopband edge angular frequency of a factor-of-50 interpolator is π . Hence, the desired stopband edge frequency in this case is ωs = 50 12000 Fs = = 240 Hz. The specifications of the decimation filter H(z) is thus as 50 50 follows: Fp = 190 Hz, Fs = 240 Hz, δ p = 0.002, δs = 0.002. Here ∆f = . 24000 Using Eq. (10.3) we arrive at the order of H(z) given by − 20 log10 0.002 × 0.002 − 13 NH = = 1347.27 ⇒ 1348. ⎛ 50 ⎞ 14.6⎜ ⎟ ⎝ 24000 ⎠ 24000 Computational complexity is thus R M , H = (1348 + 1) = 647,520 mps. 50 (b) H ( z ) = F ( z10 ) I ( z ). Specifications for F (z ) and I (z ) are as follows: 500 F ( z) : Fp = 1900 Hz, Fs = 2400 Hz, δ p = 0.001, δs = 0.002, and thus, ∆f = . 24000

Not for sale

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50 480 Hz

F(z 10 ) 24 kHz

I(z)

24 kHz

F(z)

5

480 Hz 2400 Hz

24 kHz

I(z)

10 2400 Hz 24 kHz

24 kHz

I ( z) : Fp = 190 Hz, Fs = 2160 Hz, δ p = 0.001, δs = 0.002, and thus, ∆f =

1970 . 24000

Orders of F (z ) and I (z ) are given by


Therefore, the complexity in a two-stage design with H ( z ) = F ( z10 ) I ( z ). is approximately 25% of that of the single-stage design. 13.22 A computationally efficient realization of a factor-of-3 interpolator as shown below 3

H(z)

is obtained from the 3-band polyphase decomposition of H(z) given by H ( z) = E 0 ( z 3 ) + z −1E1 ( z 3 ) + z −2 E 2 ( z 3 ). The general form of the polyphase representation of the interpolator is as shown below: E 0(z)

3

+ _ z 1

E 1(z)

3

+ _ z 1

E 2(z)

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3

511

Since H(z) is a length-15 linear-phase transfer function, H ( z ) = h[ 0] + h[1]z −1 + h[2]z −2 + h[3]z −3 + h[ 4]z −4 + h[5]z −5 + h[6]z −6 + h[7]z −7 + h[6]z −8 + h[5]z −9 + h[ 4]z −10 + h[3]z −11 + h[2]z −12 + h[1]z −13 + h[0]z −14 , the transfer functions of the sub-filters are as follows: E 0 ( z) = h[0] + h[3]z −1 + h[6]z −2 + h[5]z −3 + h[2]z −4 , E1 ( z ) = h[1] + h[ 4]z −1 + h[ 7]z −2 + h[ 4]z −3 + h[1]z −4 ,

E 2 ( z ) = h[ 2] + h[ 5]z −1 + h[6]z −2 + h[3]z −3 + h[ 0]z −4 . A computationally efficient realization of the factor-of-3 interpolator is obtained by sharing common multipliers as shown below: h[2]

_ z 1

+

_ z 1

+

_ z 1

_ z 1

+

+

3

h[5]

+

y[n]

_ z 1

h[6]

+ h[3] h[0] x[n]

_ z 1 _ z 1

+

_ z 1

+

_ z 1

+

_ z 1

+

3

_ z 1

+

_ z 1

+

_ z 1

+

_ z 1

+

3

h[4] h[1] h[7]

13.23 A computationally efficient realization of a factor-of-4 decimator as shown below H(z)

4

is obtained from the 4-band polyphase decomposition of H(z) given by H ( z ) = E 0 ( z 3 ) + z −1E1 ( z 3 ) + z −2 E 2 ( z 3 ) + z −3 E 3 ( z 3 ). The general form of the polyphase representation of the interpolator is as shown below:

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512

4

E0 (z)

+

4

E 1(z)

+

4

E2 (z)

+

4

E3 (z)

_1

z

_1

z

_1

z

Since H(z) is a length-16 linear-phase transfer function, H ( z ) = h[ 0] + h[1]z −1 + h[2]z −2 + h[3]z −3 + h[ 4]z −4 + h[5]z −5 + h[6]z −6 + h[7]z −7 + h[7]z −8 + h[6]z −9 + h[5]z −10 + h[ 4]z −11 + h[3]z −12 + h[2]z −13 + h[1]z −14 + h[0]z −15 , the transfer functions of the sub-filters are as follows: E 0 ( z ) = h[0] + h[ 4]z −1 + h[7]z −2 + h[3]z −3 , E1 ( z ) = h[1] + h[ 5]z −1 + h[6]z −2 + h[ 2]z −3 ,

E 2 ( z ) = h[ 2] + h[6]z −1 + h[ 5]z −2 + h[1]z −3 .

E 3 ( z) = h[3] + h[7]z −1 + h[ 4]z −2 + h[0]z −3 . A computationally efficient realization of the factor-of-4 decimator is obtained by sharing common multipliers as shown below h[7] +

4

_ z1

_ z1

_ z1

_ z1

h[4]

+ _ z1

_1

z

_ z1

+

_ z1

h[0]

+

+

_ z1

h[3] +

4 +

4

_1

z

_1

h[5]

_1

z

+

z

h[6] +

+ _ z1

_ z1

+

_ z1

h[1] +

+

h[2] +

4

13.24 H ( z ) =

N −1

∑ z − i = (1 + z −1 ) + (z − 2 + z − 3 ) + L + (z − ( N − 2) + z − ( N −1) )

i=0

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= (1 +

z −1 )(1 + z −2

+

z −2

+L+

z −( N − 2) )

= (1 +

z −1 )G ( z 2 ),

F ( z) =

( N / 4) −1 z −i .

∑

∑

i=0

Using a similar technique we can show that G ( z ) = (1 +

can write H ( z ) = (1 +

where G ( z ) =

( N / 2) −1 z −i .

⎛ ( N / 4 ) −1 ⎞ z − 2i ⎟. Therefore we ⎜ ⎟ ⎝ i=0 ⎠

z −1 )⎜

∑

⎛ ( N / 4) −1 ⎞ z − 4i ⎟ = (1 + z −1 )(1 + z − 2 )F ( z 4 ), where ⎜ ⎟ ⎝ i=0 ⎠

z −1 )(1 + z − 2 )⎜

∑

Continuing this decomposition process further we arrive at

i=0 K −1

H ( z ) = (1 + z −1 )(1 + z − 2 ) L (1 + z − 2 ), where N = 2 K . The transfer function of a box-car decimation filter of length-16 can be expressed as:

H (z) =

15

∑

z −i =

i=0

1 − z −16 1 − z −1

.

As a result, a computationally efficient realization of a factor-of-16 decimator using a length-16 boxcar decimation filter is as shown below: +

+

16 _ z1

_ z1

13.25 Let u[n] denote the output of the factor-of-L interpolator. Then ∞

∑ (u[n] − u[n − 1])2

E = n = −∞

,

∞

(13-1)

∑ u 2 [ n]

n = −∞

and ∞

∑ u[n]u[n − 1]

C = n = −∞

∞

.

(13-2)

∑ u 2 [ n]

n = −∞

Substituting Eq. (13-2) in Eq. (13-1) we get E = 2(1 − C ). Hence, as C → 1, i.e., as the signal u[n] becomes highly correlated, E → 0. ∞

∑

1 Now, by Parsevals’relation, u[ n]v[ n] = 2π n = −∞

Not for sale

π

∫ U (e jω ) V * (e jω )dω, where U (e jω )

−π

514

and V (e jω ) are the DTFTs of u[n] and v[n] , respectively. If we let v[ n] = u[ n − 1] in the numerator of Eq. (13-1), and v[ n] = u[ n] in the denominator of Eq. (13-1), then we can write

E=

1 2π

π

∫

U (e jω )

2

π

−π

1 2π

∫ U ( e jω )

e jω dω

= 0

π

∫ U ( e jω )

2

2

cos(ω) dω ,

π

∫ U (e jω )

dω

−π

2

dω

0

assuming u[n] to be a real sequence. If x[n] is assumed to be a broadband signal with a flat magnitude spectrum, i.e., X (e jω ) = 1 for 0 ≤ ω ≤ π, then the magnitude spectrum of is bandlimited to the range 0 ≤ ω ≤ π / L, i.e., π/L

∫ cos(ω)dω

⎧1, 0 ≤ ω < π / L, Therefore, C = 0 U ( e jω ) = ⎨ π/L 0 , otherwise. ⎩ dω

∫

=

sin(π / L ) . Hence, as (π / L )

0

L → ∞ , C → 1.

13.26 X(e jω )

M

W(e jω )

H(z)

R(e jω)

– z L

S(e jω )

M

Y(e jω)

Analysis of the above structure yields W (e jω ) = X (e jωM ), R (e jω ) = H (e jω ) X (e jωM ), S (e jω ) = e − jωL R(e jω ) = e − jωL H (e jω ) X (e jωM ), Y (e

jω

1 )= M

M −1

∑ S (e

− jωk / M

e jω / M ). If the filter H (e jω ) is assumed to be close to

k =0

an ideal lowpass filter with a cutoff at ω / M , we can assume that all images of X (e jω ) are suppressed leaving only the term in the expression for Y (e jω ). Hence, we 1 1 can write Y (e jω ) = S ( e jω / M ) = H (e jω / M )e − jωL / M X (e jω ). Since H (z ) is a M M Type 1 FIR filter with exact linear phase and a delay of ( N − 1) / 2 = KM samples and a magnitude response equal to M in the passband, we have Y (e jω ) = e − jωK e − jωL / M X (e jω ). Thus, the structure of Figure P13.7 Is approximately an allpass filter with a fixed delay of K samples and a variable noninteger delay of L/M samples. 13.27 An ideal M-th band lowpass filter H (z ) is characterized by a frequency response

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515

1, − π / M ≤ ω ≤ π / M , H (e jω ) = ⎧⎨ The transfer function H (z ) can be expressed in otherwise. ⎩0, M −1

∑z

an M-branch polyphase form as H ( z ) =

−k

H k ( z M ). From the above we observe

k =0

M −1

∑

H ( zWMr ) = M H 0 ( z M ). Therefore, H 0 (e jωM ) =

r =0

Or in other words, H 0 ( z

M

M −1

∑ H (e

1 M

j ( ω − 2 πr / M

r =0

)=

1 . M

) is an allpass function.

13.28 An equivalent realization of the structure of Figure P13.38 obtained by realizing the filter in a Type 1 polyphase form is shown below on the left. By moving the downsampler through the system and invoking the cascade equivalence #1 of Figure 13.14 we arrive at the structure shown below on the right. E 0(z L )

L

L

+

L

L

E 0(z)

+

L

E 1(z)

+

L

E 2 (z)

+

L

E L _ 1 (z)

_ z1

_ z1

E 1(z L )

+ _ z1

_ z1

E 2 (z L )

+

_ z1

_ z1

E L _ 1 (z L )

The structure on the right hand side reduces to the one shown below on the left from which we arrive at the simplified equivalent structure shown below on the right. L

13.29 (a) H ( z ) =

N −1

∑

h[ n]z − n .

L

( N / 2 ) −1

Let H 0 ( z ) =

n=0 ( N / 2) −1

H1 ( z ) =

E0 (z)

E0 (z)

∑ (h[2i] + h[2i + 1])z − 2i

and

i=0

∑ (h[2i] − h[2i + 1])z − 2i . Then

i=0

(1 + z −1 )H 0 ( z 2 ) + (1 − z −1 )H 1( z 2 ) =

( N / 2 ) −1

∑

h[2i ]z − 2i +

i=0 − 1 2 − 1 (b) H ( z) = (1 + z )H 0 ( z ) + (1 − z )H 1( z 2 )

Not for sale

( N / 2) −1

∑ h[2i + 1]z − 2i + 1 = H (z).

i=0

516

(

)

(

)

= H 0 ( z 2 ) + H1 ( z 2 ) + z −1 H 0 ( z 2 ) − H1 ( z 2 ) = E 0 ( z 2 ) + z −1E1 ( z 2 ). Therefore, E 0 ( z ) = H 0 ( z ) + H 1( z ) and E1 ( z ) = H 0 ( z ) − H 1( z ).

[

]

[

]

⎡ H ( z 2 ) + H1 ( z 2 )⎤ ⎡ H ( z 2 )⎤ (c) Now, H ( z ) = 1 z −1 ⎡1 1 ⎤ ⎢ 0 2 ⎥ = 1 z −1 ⎢ 0 2 2 ⎥ ⎢⎣1 − 1⎥⎦ H ( z ) ⎦ ⎣ 1 ⎣ H 0 ( z ) − H1 ( z ) ⎦

(

)

(

)

= H 0 ( z 2 ) + H1 ( z 2 ) + z −1 H 0 ( z 2 ) − H1 ( z 2 ) = (1 + z −1 )H 0 ( z 2 ) + (1 − z −1 )H1 ( z 2 ). (d) If L = 2, i.e., N = 2 2 , then we can express

H 0 ( z 2 ) = (1 + z −2 )H 00 ( z 4 ) + (1 − z −2 )H 01 ( z 4 ) and H1 ( z 2 ) = (1 + z −2 )H10 ( z 4 ) + (1 − z −2 )H11 ( z 4 ). Substituting these expressions in H ( z) = (1 + z −1 )H 0 ( z 2 ) + (1 − z −1 )H1 ( z 2 ) we get

[

H ( z ) = (1 + z −1 ) (1 + z − 2 )H 00 ( z 4 ) + (1 − z − 2 )H 01 ( z 4 )

[

]

+ (1 − z −1 ) (1 + z − 2 )H10 ( z 4 ) + (1 − z − 2 )H11 ( z 4 )

]

= (1 + z −1 )(1 + z −2 )H 00 ( z 4 ) + (1 + z −1 )(1 − z −2 )H 01 ( z 4 ) + (1 − z −1 )(1 + z −2 )H10 ( z 4 ) + (1 − z −1 )(1 − z −2 )H11 ( z 4 ) ⎡ H ( z 4 )⎤ 1 1 ⎤ ⎢ 00 ⎡1 1 ⎥ H 01( z 4 ) ⎥ − − 1 1 1 1 ⎢ ⎥ − − − 1 2 3 ⎢ =1 z = 1 z −1 z − 2 z z ⎢1 1 − 1 − 1⎥ ⎢ H ( z 4 ) ⎥ ⎢⎣1 − 1 − 1 1 ⎥⎦ ⎢ 10 4 ⎥ ⎣ H 11( z ) ⎦ Continuing this process it is easy to establish that for we have ⎡ Hˆ 0 ( z L ) ⎤ ⎢ ˆ L ⎥ H ( z) = 1 z −1 L z − ( L −1) R L ⎢ H 1( z ) ⎥ . M ⎢ ⎥ ⎢⎣ Hˆ L −1( z L )⎥⎦

[

]

[

[

[

13.30 Now H ( z) = 1 z −1

z−2

⎡ Hˆ 0 ( z 4 )⎤ ⎢ˆ 4 ⎥ H (z ) − 3 z R 4 ⎢ 1 4 ⎥. ⎢ Hˆ 2 ( z )⎥ ⎢ˆ 4 ⎥ ⎣ H 3( z ) ⎦

]

]

⎡ Hˆ 0 ( z 4 )⎤ ⎢ˆ 4 ⎥ H (z ) z − 3 R 4 ⎢ 1 4 ⎥ = 1 z −1 ⎢ Hˆ 2 ( z )⎥ ⎢ˆ 4 ⎥ ⎣ H 3( z )⎦

]

[

z−2

⎡ E 0 ( z 4 )⎤ ⎥ ⎢ E (z 4 ) z − 3 ⎢ 1 4 ⎥. ⎢ E 2 ( z )⎥ ⎢ 4 ⎥ ⎣ E 3 ( z )⎦

]

⎡ Hˆ 0 ( z )⎤ ⎡ E 0 ( z )⎤ ⎡ E 0 ( z )⎤ 1 1 ⎤ ⎡ E 0 ( z )⎤ ⎡1 1 ⎥ ⎢ ˆ ⎢ ⎥ ⎢ ⎥ 1 1 E z E z H z ( ) ( ) ( ) − − 1 1 1 1⎥ ⎢ E 1( z ) ⎥. Therefore, ⎢ 1 ⎥ = R −4 1 ⎢ 1 ⎥ = R 4 ⎢ 1 ⎥ = ⎢ ˆ ⎢ − − 1 1 1 1⎥ ⎢ E ( z )⎥ E ( z) E ( z) ⎢ H 2 ( z )⎥ ⎢ 2 ⎥ 4 ⎢ 2 ⎥ 4 ⎢1 − 1 − 1 1 ⎥ ⎢ 2 ⎥ ⎣ ⎦ ⎣ E 3 ( z)⎦ ⎢ Hˆ ( z )⎥ ⎣ E 3 ( z)⎦ ⎣ E 3 ( z)⎦ ⎣ 3 ⎦ A length-16 Type 1 linear-phase FIR transfer function is of the form H ( z) =h 0 + h1 z −1 + h 2 z −2 + h 3 z −3 + h 4 z −4 + h 5 z −5 + h 6 z −6 + h 7 z −7 + h 7 z −8 + h 6 z −9

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+ h 5 z −10 + h 4 z −11 + h 3 z −12 + h 2 z −13 + h1 z −14 + h 0 z −15 . Hence, E 0 ( z) = h 0 + h 4 z −1 + h 7 z −2 + h 3 z −3 , E1 ( z) = h1 + h 5 z −1 + h 6 z −2 + h 2 z −3 , E 2 ( z) = h 2 + h 6 z −1 + h 5 z −2 + h1 z −3 , E 3 ( z) = h 3 + h 7 z −1 + h 4 z −2 + h 0 z −3 . Thus, Hˆ ( z) = g + g z −1 + g z −2 + g z −3 , Hˆ ( z) = g + g z −1 − g z −2 − g z −3 , 0

0

1

1

0

1

2

3

3

2

Hˆ 2 ( z) = g 4 + g 5 z −1 − g 5 z −2 − g 4 z −3 , Hˆ 3 ( z) = g 6 + g 7 z −1 + g 7 z −2 + g 6 z −3 , where 1 1 1 g 0 = (h0 + h1 + h2 + h3 ), g1 = (h4 + h5 + h6 + h7 ), g 2 = (h0 − h1 + h2 − h3 ), 4 4 4 1 1 1 g3 = (h4 − h5 + h6 − h7 ), g 4 = (h0 + h1 − h2 − h3 ), g 5 = (h4 + h5 − h6 − h7 ), 4 4 4 1 1 g6 = (h0 − h1 − h2 + h3 ), g 7 = (h4 − h5 − h6 + h7 ). Note that Hˆ 0 ( z) and Hˆ 3 ( z ) 4 4 are Type 1 linear-phase FIR transfer functions, whereas, Hˆ 1 ( z ) and Hˆ 2 ( z ) are Type 2 linear-phase FIR transfer functions. A computationally efficient realization of a factorof-4 decimator using a four-band structural subband decomposition of the decimation filter H (z ) is shown below: M

^ H0 (z)

+

M

^ H1 (z)

+

M

^ H 2 (z)

+

M

^ H 3 (z)

_ z1

_1

z

R4

_ z1

Because of the symmetry or anti-symmetry in the impulse responses of the subband filters, each subband filter can be realized using only 2 multipliers. Hence, the final realization uses only 8 multipliers. Note also that by delay-sharing, the total number of delays in implementing the four subband filters can be reduced to 3.

13.31 y[ n] = P− 2 (α ) x[ n − 2] + P−1 (α ) x[ n − 1] + P0 (α ) x[ n] + P1 (α ) x[ n + 1] + P2 (α ) x[ n + 2] (α + 1)α(α − 1)(α − 2) 1 where P− 2 (α ) = = (α 4 − 2α 3 − α 2 + 2α ), (−2 + 1)(−2)(−2 − 1)(−2 − 2) 24 (α + 2)α(α − 1)(α − 2) 1 P−1 (α ) = = − (α 4 − α 3 − 4α 2 + 4α ), 6 (−1 + 2)(−1)(−1 − 1)(−1 − 2) (α + 2)(α + 1)(α − 1)(α − 2) 1 4 P0 (α ) = = (α − 5α 2 + 4), (0 + 2)(0 + 1)(0 − 1)(0 − 2) 4

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(α + 2)(α + 1)α(α − 2) 1 = − (α 4 + α 3 − 4α 2 − 4α ), 6 (1 + 2)(1 + 1)(1 − 0)(1 − 2) (α + 2)(α + 1)α(α − 1) 1 = (α 4 + 2α 3 − α 2 − 2α ). P2 (α ) = 24 (2 + 2)(2 + 1)(2 − 0)(2 − 1) P1 (α ) =

We consider the computation of y[ n], y[ n + 1], y[ n + 2], y[ n + 3] using 5 input samples: x[ n − 2] through x[ n + 2]. For α 0 = 0, P− 2 (α 0 ) = 0, P−1 (α 0 ) = 0, P0 (α 0 ) = 1, P1 (α 0 ) = 0, and P2 (α 0 ) = 0. For α1 = 5 / 4, P− 2 (α1 ) = −0.022, P−1 (α1 ) = 0.127, P0 (α1 ) = −0.3428, P1(α1) = 1.1426, and P2 (α1 ) = 0.0952. For α 2 = 10 / 4, P− 2 (α 2 ) = 0.2734, P−1 (α 2 ) = −1.4062, P0 (α 2 ) = 2.9531,

P1 (α 2 ) = −3.2812, and P2 (α 2 ) = 2.4609. For α 3 = 15 / 4, P− 2 (α 3 ) = 3.5718, P−1 (α 3 ) = −17.2949, P0 (α 3 ) = 32.86, P1 (α 3 ) = −29.873, and P2 (α 3 ) = 11.7358. The block filter implementation is thus given by

⎡ x[ n − 2]⎤ ⎡ y[ n] ⎤ ⎡ 0 0 1 0 0 ⎤ ⎢ x[ n − 1] ⎥ ⎢ y[ n + 1] ⎥ ⎢ − 0.022 − 0.3428 1.1426 0.127 0.0952 ⎥ ⎢ ⎥. Another ⎢ y[ n + 2]⎥ = ⎢ 0.2734 − 1.4062 2.9531 − 3.2812 2.4609 ⎥ ⎢ x[ n] ⎥ ⎥ ⎢ ⎢ 32.86 − 29.873 11.7358⎦⎥ ⎢ x[ n + 1] ⎥ ⎣ y[ n + 3]⎦ ⎣ 3.5718 − 17.2949 ⎣ x[ n + 2]⎦ implementation is given by 1 1 1 1 1 y[ n] = α 4 ⎛⎜ x[ n − 2] − x[ n − 1] + x[ n] − x[ n + 1] + x[ n + 2]⎞⎟ 24 6 4 6 24 ⎝ ⎠ 1 1 1 1 + α 3 ⎛⎜ − x[ n − 2] + x[ n − 1] − x[ n + 1] + x[ n + 2]⎞⎟ 6 6 12 ⎝ 12 ⎠ 1 4 5 4 1 + α 2 ⎛⎜ − x[ n − 2] + x[ n − 1] − x[ n] + x[ n + 1] − x[ n + 2]⎞⎟ 24 6 4 6 24 ⎝ ⎠ 1 4 4 1 + α⎛⎜ x[ n − 2] − x[ n − 1] + x[ n + 1] − x[ n + 2]⎞⎟ + x[ n]. 6 6 12 ⎝ 12 ⎠ The Farrow structure implementation of the interpolator is shown below: x[n]

α

α

H3(z)

H2(z)

H1(z)

H0(z)

α

y[n]

α

where H 0 ( z) = z − 2 − z −1 + − z + z 2 , H1 ( z) = − z − 2 + z −1 + z − z 2 , 6 24 6 4 6 24 12 6 24 1

1

1

1

1

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1

1

1

519

H 2 ( z) = −

1 −2 z 24

+ z −1 − 4 6

5 4 1 2 + z− z , 4 6 24

13.32 From Eq. (13.75), we have a 0 = − a 4

and H 3 ( z ) =

1 t4 t 42

1 t1 t12

1 t2 t 22

1 t3 t 32

t 43

t13

t 23

t 33

1 t0 t 02

1 t1 t12

1 t2 t 22

1 t3 t 32

t 03

t13

t 23

t 33

1 −2 z 12

− z −1 + z − 4 6

4 6

1 2 z . 12

. Both the numerator and the

denominator are determinants of Vandermonde matrices and have a nonzero value if ti ≠ t j , i ≠ j. From the solution of Problem 6.28, we get

(t −t )(t −t )(t −t )(t −t )(t −t )(t −t ) (t −t )(t −t )(t −t ) a0 = a 4 1 4 2 4 3 4 2 1 3 1 3 2 = a 4 1 4 2 4 3 4 , or (t 1−t 0 )(t 2 −t 0 )(t 3 −t 0 )(t 2 −t1 )(t 3 −t1 )(t 3 −t 2 ) (t 1−t 0 )(t 2 −t 0 )(t 3 −t 0 ) (t −t )(t −t )(t −t ) a0 = −a4 4 1 4 2 4 3 . (t 0 −t1 )(t 0 −t 2 )(t 0 −t 3 ) 13.33 From Eq. (13.74) we get a1 = −a 4

(t 4 −t 0 )(t 2 −t 0 )(t 3 −t 0 )(t 2 −t 4 )(t 3 −t 4 )(t 3 −t 2 ) (t 1−t 0 )(t 2 −t 0 )(t 3 −t 0 )(t 2 −t1 )(t 3 −t1 )(t 3 −t 2 )

(t 4 −t 0 )(t 2 −t 4 )(t 3 −t 4 ) . Substituting the value of a 4 given by Eq. (13.77) in (t 1−t 0 )(t 2 −t1 )(t 3 −t1 ) 1 the above we arrive at a1 = . Likewise, from Eq. (13.74) (t 1−t 0 )(t 1−t 2 )(t 1−t 3 )(t 1−t 4 ) (t −t )(t −t )(t −t )(t −t )(t −t )(t −t ) we get a2 = a 4 1 0 4 0 3 0 4 1 3 1 3 4 (t 1−t 0 )(t 2 −t 0 )(t 3 −t 0 )(t 2 −t1 )(t 3 −t1 )(t 3 −t 2 ) (t −t )(t −t )(t −t ) = a 4 4 0 4 1 3 4 . Substituting the value of a 4 given by Eq. (13.77) in the (t 2 −t 0 )(t 2 −t1 )(t 3 −t 2 ) 1 above we arrive at a 0 = . Finally, from Eq. (13.74) we (t 2 −t 0 )(t 2 −t1 )(t 2 −t 3 )(t 2 −t 4 ) (t −t )(t −t )(t −t )(t −t )(t −t )(t −t ) get a3 = −a 4 1 0 2 0 4 0 2 1 4 1 4 2 (t 1−t 0 )(t 2 −t 0 )(t 3 −t 0 )(t 2 −t1 )(t 3 −t1 )(t 3 −t 2 ) (t −t )(t −t )(t −t ) = −a 4 4 0 4 1 4 2 . Substituting the value of a 4 given by Eq. (13.77) in the (t 3 −t 0 )(t 3 −t1 )(t 3 −t 2 ) 1 above we arrive at a3 = . (t 3 −t 0 )(t 3 −t1 )(t 3 −t 2 )(t 3 −t 4 ) = −a4

13.34 t i = i, m ≤ i ≤ m + 4.

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1 1 = , (m − m − 1)(m − m − 2)(m − m − 3)(m − m − 4) 24 1 1 a m +1 = =− , (m + 1 − m)(m + 1 − m − 2)(m + 1 − m − 3)(m + 1 − m − 4) 6 1 1 am + 2 = = , (m + 2 − m)(m + 2 − m − 1)(m + 2 − m − 3)(m + 2 − m − 4) 4 1 1 am + 3 = =− , (m + 3 − m)(m + 3 − m − 1)(m + 3 − m − 2)(m + 3 − m − 4) 6 1 1 am + 4 = = . (m + 4 − m)(m + 4 − m − 1)(m + 4 − m − 2)(m + 4 − m − 3) 24 am =

( 3) From Eq. (13.69), we have the expressions for Bm (t ) given by

0, t < m, ⎧ 1 ⎪ m ≤ t < m + 1, (t − m) 3 , ⎪ 24 ⎪ 1 1 m + 1 ≤ t < m + 2, (t − m) 3 − (t − m − 1) 3 , ⎪ (3) 6 24 Bm (t ) = ⎨ 1 1 1 m + 2 ≤ t < m + 3, (t − m) 3 − (t − m − 1) 3 + (t − m − 2) 3 , ⎪ 4 6 24 ⎪1 3 3 1 3 1 3 1 ⎪ (t − m) − (t − m − 1) + (t − m − 2) − (t − m − 3) , m + 3 ≤ t < m + 4, 6 4 6 24 ⎪ 0, t ≥ m + 4. ⎩ ( 3) The normalized 3rd order B-spline is then given by β (m3) (t ) = ( m + 4 − m) Bm (t )

0, t < m, ⎧ 1 ⎪ 3 m ≤ t < m + 1, (t − m) , ⎪ 6 ⎪ 2 1 m + 1 ≤ t < m + 2, (t − m) 3 − (t − m − 1) 3 , ⎪ (3) 3 6 = 4 Bm (t ) = ⎨ 2 1 m + 2 ≤ t < m + 3, (t − m) 3 − (t − m − 1) 3 + (t − m − 2) 3 , ⎪ 3 6 ⎪1 2 2 ⎪ (t − m) 3 − (t − m − 1) 3 + (t − m − 2) 3 − (t − m − 3) 3 , m + 3 ≤ t < m + 4, 3 3 6 ⎪ 0 t ≥ m + 4. , ⎩ Substituting t = 1 + α and evaluating β (m3) (t ) for m = −1, 0, 1 2, we have

β (−31) (α) = −

α3 α2 α 1 α3 2 + − + , β 0(3) (α) = + α2 + , 6 2 2 6 2 3

β1(3) (α) = −

α3 α2 α 1 α3 + + + , β 2(3) (α) = . 2 2 2 6 6 2

Substituting back in Eq. (13.91), we have y[ n] =

β 3 (α) x[n + k ] ∑ = −1 ( ) k

k

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⎛ α3 α2 α 1 ⎞ ⎛ α3 ⎞ ⎛ α3 α2 α 1 ⎞ 2 2⎟ ⎜ ⎟ ⎜ = − + − + x[n − 1] + + α + x[n] + ⎜ − + + + ⎟ x[n + 1] ⎜ 6 ⎟ ⎜ 2 ⎟ ⎜ 2 2 2 6 3 2 2 6 ⎟⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎛ α3 ⎞ ⎟ x[n + 2] = ⎛⎜ 1 x[ n − 1] + 2 x[ n] + 1 x[ n] ⎞⎟ + ⎛⎜ − 1 x[ n − 1] + 1 x[ n + 1] ⎞⎟α +⎜ ⎜ 6 ⎟ 3 6 2 ⎠ ⎝6 ⎠ ⎝ 2 ⎝ ⎠ 1 1 1 1 ⎞ ⎛ 1 ⎞ ⎛1 + ⎜ x[n − 1] − x[ n] + x[ n + 1] ⎟α 2 + ⎜ − x[n − 1] + x[ n] − x[ n + 1] + x[ n + 2] ⎟α 3 . 2 2 2 6 ⎠ ⎝ 6 ⎠ ⎝2 In the z-domain, the input-output relation is thus given by Y ( z ) = H 3 ( z ) + H 2 ( z )α + H1 ( z )α 2 + H 0 ( z )α 3 , where 1 1 1 1 1 1 1 1 H 0 ( z ) = − z −1 + − z + z 2 , H 1 ( z ) = z −1 − 1 + z , H 2 ( z ) = − z −1 + z , 6 2 2 6 2 2 2 2 1 −1 2 1 H 3 ( z ) = z + + z. The corresponding Farrow structure is shown on top of the 6 3 6 next page: x[n]

α

α

H3(z)

H2(z)

H1(z)

H0(z)

y[n]

α

13.35 For the factor-of-4/3 interpolator design, if we use cubic B-spline with uniformly spaced knots at the problem reduces exactly to the design given in the solution of Problem 13.34. 13.36 From Eq. (13.94) with k = 0 we have H ( z) = α +

L −1

∑ z − i Ei (z L ).

Substituting the

i =1

expression for on the left-hand side Eq. (13.97) we get INCOMPLETE

13.37 For a half-band zero-phase lowpass filter, the transfer function is of the form

H ( z ) = h[0] + z −1

∞

∑ h[2n]z − 2n , where h[0] = 2 . 1

If the half-band filter has a zero at

n = −∞ n≠0

z = −1, then H (−1) = h[0] −

∞

∑

h[2 n] = 0, or h[0] =

n = −∞ n≠0

∞

∑ h[2n].

n = −∞ n≠0

13.38 From Eq. (13.99), a zero-phase half-band filter H (z ) satisfies the condition H ( z ) + H ( − z ) = a constant.

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(a) The zero-phase equivalent here is given by H1 ( z ) = z + 2 + z −1 . Hence, H1 ( z ) + H1 ( − z ) = z + 2 + z −1 − z + 2 − z −1 = 4. A plot of the scaled magnitude response of H1 ( z) is given below: H1(z) 1

Magnitude

0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

(b) The zero-phase equivalent here is given by H 2 ( z ) = − z 3 + 9 z + 16 + 9 z −1 − z −3 . Hence, H 2 ( z ) + H 2 ( − z ) = − z 3 + 9 z + 16 + 9 z −1 − z −3 + z 3 − 9 z + 16 − 9 z −1 + z −3 = 32. A plot of the scaled magnitude response of H 2 ( z) is given below: H2(z) 1

Magnitude

0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/ π

(c) The zero-phase equivalent here is given by H 3 ( z ) = −3z 3 + 19z + 326 + 19z −1 − 3z −3 . Hence,

H 3 ( z) + H 3 (− z) = −3z 3 + 19z + 32 + 19z −1 − 3z −3 + 3z 3 − 19z + 32 − 19z −1 + 3z −3 = 64. A plot of the scaled magnitude response of H 3 ( z) is given below: H3(z) 1

Magnitude

0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

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(d) The zero-phase equivalent here is given by H 4 ( z ) = 3z 5 − 25z 3 + 150 z + 256 + 150 z −1 − 25z −3 + 3z −5 . Hence, H 4 ( z) + H 4 (− z) = 3z 5 − 25z 3 + 150z + 256 + 150z −1 − 25z −3 + 3z −5 − 3z 5 + 25z 3 − 150z + 256 − 150z −1 + 25z −3 − 3z −5 = 512. A plot of the scaled magnitude response of H 4 ( z) is given below: H4(z) 1

Magnitude

0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

13.39 (a) A function F (ω) of has p-th order zero at a frequency ω = ωi if d p F (ω) dω

p

Hence,

= 0. The function H (e jω ) has p zeros at cos ω = 1 , i.e., at ω = π . ω = ωi

d p H ( e jω ) dω p

= 0. Moreover, the order of the highest power of ω= π

(1 − cos ω) / 2 is p − 1. As a result,

d p −1 H ( e j ω ) dω p −1

= 0. ω= 0

2

2

− jω ⎞ ⎛ 1 + e − jω ⎞ ⎛ 1 + cos ω ⎟ and 1 − cos ω = −e jω ⎜ 1 − e ⎟ . (b) Now, = e jω ⎜ ⎜ ⎟ ⎜ ⎟ 2 2 2 2 ⎝ ⎠ ⎝ ⎠

Substituting these expressions in Eq. (13.120) we arrive at H (e

jω

)=e

⎛ ⎜ ⎝

jpω ⎜ 1 + e

− jω

2

p p −1

l

− jω ⎞ ⎞ ⎛ p − 1 + l ⎞(−e − jω ) l ⎛⎜ 1 − e ⎟ . Replacing with ⎟ ⎜ ⎟ ⎜ ⎟ ⎟ l 2 ⎝ ⎠ ⎝ ⎠ ⎠ l=0

∑

e jω in z the above we get the expression for the zero-phase transfer function as

⎛ H ( z) = z ⎜ ⎝

p ⎜1+

z −1 ⎞⎟ 2 ⎟⎠

p l −1

∑ l

l

−1 ⎛ p − 1 + l ⎞(− z ) l ⎛⎜ 1 − z ⎞⎟ . ⎜ ⎟ ⎜ 2 ⎟ l ⎝ ⎠ ⎝ ⎠

13.40 (a)

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X(z)

_ z1

V0 (z)

_ 1+z 1

V1 (z) V2 (z)

2

2

2

W0 (z) W1 (z)

H0 (z) H0 (z) + H 1(z)

W2 (z)

H 1(z)

U0 (z) U 1(z) U 2(z)

2

R 0(z)

_ 1_ z 1

2

R 1(z) _ z1

2

R 2(z) _ _ z 2_ z 1

+

Y(z)

⎡V0 ( z )⎤ ⎡ z −1 ⎤ Analysis of the above structure yields V( z) = ⎢ V1 ( z ) ⎥ = ⎢1 + z −1 ⎥ X ( z), ⎢V ( z )⎥ ⎢ 1 ⎥ ⎥⎦ ⎣ 2 ⎦ ⎢⎣ ⎡ − z −1 / 2 ⎤ ⎡W0 ( z )⎤ ⎡ z −1 / 2 ⎤ ⎢ ⎥ − 1 / 2 1 / 2 W( z) = ⎢ W1 ( z ) ⎥ = 1 + z X (z ) + ⎢1 − z −1 / 2 ⎥ X ( z −1 / 2 ), ⎢ ⎥ ⎢ ⎥ ⎢W ( z )⎥ 1 1 ⎥⎦ ⎢⎣ ⎥⎦ ⎣ 2 ⎦ ⎢⎣ ⎤ ⎡ ⎤ − z −1 / 2 H 0 ( z ) z −1 / 2 H 0 ( z ) ⎡U 0 ( z )⎤ ⎡ ⎢ ⎥ ⎢ ⎥ − 1 / 2 1 / 2 − 1 / 2 U( z) = ⎢U1 ( z) ⎥ = ⎢(1 + z )(H 0 ( z) + H1 ( z ))⎥ X ( z ) + ⎢(1 − z )(H 0 ( z ) + H1 ( z))⎥ X (− z1 / 2 ), ⎢U ( z )⎥ H1 ( z ) H1 ( z ) ⎥ ⎢ ⎥ ⎣ 2 ⎦ ⎢⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎤ − z −1 H 0 ( z ) z −1 H 0 ( z 2 ) ⎡ R0 ( z ) ⎤ ⎡ ⎢ ⎥ ⎢ ⎥ R( z) = ⎢ R1 ( z) ⎥ = ⎢(1 + z −1 ) H 0 ( z 2 ) + H1 ( z 2 ) ⎥ X ( z) + ⎢(1 − z −1 ) H 0 ( z 2 ) + H1 ( z 2 ) ⎥ X (− z). ⎢ R ( z )⎥ ⎢ ⎥ ⎥ H1 ( z 2 ) H1 ( z 2 ) ⎣ 2 ⎦ ⎢⎣ ⎣ ⎦ ⎦

(

)

(

Y ( z ) = (1 − z −1 ) R0 ( z ) + z −1 R1 ( z ) + ( z −2 − z −1 ) R2 ( z )

[ ( ) [ ( ) = [2 z −1H 0 ( z 2 ) + 2 z − 2 H1 ( z 2 )]X ( z) = 2 z −1[ H 0 ( z 2 ) + z −1H1 ( z 2 )]X ( z).

)

]

= (1 − z −1 )z −1H 0 ( z 2 ) + (1 + z −1 )z −1 H 0 ( z 2 ) + H1 ( z 2 ) + ( z − 2 − z −1 )H1 ( z 2 ) X ( z) + − (1 − z −1 )z −1H 0 ( z 2 ) + (1 − z −1 )z −1 H 0 ( z 2 ) + H1 ( z 2 ) + ( z − 2 − z −1 )H1 ( z 2 ) X (− z)

]

Hence,

T ( z) = 2 z −1[ H 0 ( z 2 ) + z −1 H1 ( z 2 )]. 1 1 1 ⎡1 ⎤ (b) T ( z ) = 2 z −1 ⎢ H ( z ) + H (− z ) + H ( z ) − H (− z )⎥ = 2 z −1 H ( z ). 2 2 2 ⎣2 ⎦ H ( z ) = K . (c) Length of H 0 ( z ) = K and length of 1

F (d) The total computational complexity of the above structure is 3K T 2 multiplications per second, where is the sampling frequency in Hz. On the other hand, a direct implementation of H (z ) requires 2 KFT multiplications per second. 13.41

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R(z)

X 1(z)

2

X 2(z)

Y 1(z)

2

Y 2(z)

z

+

2

2

_1

_ z1

Analysis yields R( z ) = z −1 X1 ( z 2 ) + X 2 ( z 2 ), Y 1( z ) = R( z1 / 2 ) + R( − z1 / 2 ) = [ z −1 / 2 X1 ( z ) + X 2 ( z )] + [ − z −1 / 2 X1 ( z ) + X 2 ( z )] = 2 X 2 ( z ), Y 2( z ) = z −1 / 2 R( z1 / 2 ) − z −1 / 2 R( − z1 / 2 )

= [ z −1 X1 ( z ) + z −1 / 2 X 2 ( z )] + [ z −1 X1 ( z ) − z −1 / 2 X 2 ( z )] = 2 z −1 X1 ( z ). Thus, the output y1[ n] is a scaled replica of the input x 2 [ n] while the output y2 [ n] is a scaled replica of the delayed input x1[ n − 1].

13.42 X 1(z)

W(z)

2 – z 1

X2 (z)

V(z)

H(z2)

2

Y 1(z)

2

Y 2(z)

– z 1

2

Analysis yields W ( z ) = X1 ( z 2 ) + z −1 X 2 ( z 2 ), V ( z ) = H ( z 2 ) X1 ( z 2 ) + z −1 H ( z 2 ) X 2 ( z 2 ), 1 Y1 ( z ) = V ( z1 / 2 ) + V (− z1 / 2 ) = H ( z ) X1 ( z ), 2 1 Y2 ( z ) = z −1 / 2V ( z1 / 2 ) − z −1 / 2V (− z1 / 2 ) = z −1 H ( z ) X 2 ( z ). 2 Y (z) Y (z) = z −1 H ( z). Hence, the system is time-invariant. Therefore, 1 = H ( z), and 2 X 1( z) X 2 ( z)

( (

)

)

13.43 X 1(z)

H(z2)

2 –1 z

X 2(z)

2 – z 1

2

2

Y1 (z)

Y2 (z)

Y 1( z ) Y (z) = H ( z), and 2 = z −1 H ( z). Here now, X 1( z) X 2 ( z) X 2 ( z) = Y1 (z), and hence, Y2 ( z ) = z −1 H ( z ) X 2 ( z ) = z −1 H ( z )Y1 ( z ) = z −1 H 2 ( z ) X1 ( z ). Y (z) Thus, 2 = z −1 H 2 ( z). Hence, the system is time-invariant. X 1( z) 13.44 From the solution of Problem 13.42,

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X 1(z)

W(z)

3 – z 1

X2 (z)

V(z)

3

Y1 (z)

– z 1 3

3 – z 1

X 3(z)

H(z3)

– z 1 3

3

Y2 (z)

Y3 (z)

–2

Y(z) – (C+1) z

3

Analysis yields W ( z) = X1 ( z 3 ) + z −1Y1 ( z 3 ) + z −2 X 3 ( z 3 ),

V ( z) = H ( z 3 )W ( z ) = H ( z 3 ) X1 ( z 3 ) + z −1H ( z 3 ) X 2 ( z 3 ) + z −2 H ( z 3 ) X 3 ( z 3 ), 1 Y1 ( z ) = V ( z1 / 3 ) + V ( z1 / 3 e − j 2 π / 3 ) + V ( z1 / 3 e − j 4π / 3 ) 3 1 = H ( z ) X1 ( z ) + H ( z e − j 2 π ) X1 ( z e − j 2 π ) + H ( z e − j 4 π ) X1 ( z e − j 4 π ) 3

[

]

[

1 3 1 + 3

+

]

[z −1 / 3 H (z) X 2 (z) + z −1 / 3 e − j 2π / 3 H (z e − j 2π ) X 2 (z e − j 2π ) + z −1 / 3 e − j 4π / 3 H (z e − j 4π ) X 2 (z e − j 4π )] [z −2 / 3 H (z) X 3 (z) + z −2 / 3 e − j 4π / 3 H (z e − j 4π ) X 3 (z e − j 4π ) + z −2 / 3 e − j8π / 3 H (z e − j8π ) X 3 (z e − j8π )]

1 [H ( z) X1 ( z) + H ( z) X1 ( z) + H ( z) X1 ( z)] 3 1 + z −1 / 3 H ( z ) X 2 ( z ) + z −1 / 3 e − j 2π / 3 H ( z ) X 2 ( z ) + z −1 / 3 e − j 4π / 3 H ( z ) X 2 ( z ) 3 1 −2 / 3 + z H ( z ) X 3 ( z ) + z − 2 / 3 e − j 4 π / 3 H ( z ) X 3 ( z ) + z − 2 / 3 e − j 8π / 3 H ( z ) X 3 ( z ) 3 = H ( z) X1 (z), 1 Y2 ( z ) = z −1 / 3V ( z1 / 3 ) + z −1 / 3 e j 2 π / 3V ( z1 / 3 e − j 2π / 3 ) + z −1 / 3 e j 4 π / 3V ( z1 / 3 e − j 4π / 3 ) 3 1 − = z H ( z) X 3 ( z), 1 Y3 ( z ) = z −1 / 3V ( z1 / 3 ) + z −1 / 3 e j 4 π / 3V ( z1 / 3 e − j 2π / 3 ) + z −1 / 3 e j8π / 3V ( z1 / 3 e − j 4 π / 3 ) 3 − 2 = z H ( z ) X 2 ( z ). Now, X 3 ( z) = Y3 ( z) and X 2 ( z) = Y1 ( z). Hence, =

[ [

] ]

[

]

[

]

Y2 ( z ) = z −1H ( z) X 3 ( z ) = z −1H ( z)Y3 ( z ) = z −1 H ( z )[ z −1 H ( z ) X 2 ( z )] = z −2 H 2 ( z )Y1 ( z ) = z −2 H 3 ( z ) X1 ( z ). Therefore, Y ( z) = −2Y2 ( z) + 3z −(C +1)Y3 ( z) = −2 z −2 H 3 ( z ) X1 ( z ) + 3z − (C + 1) z −1 H 2 ( z ) X1 ( z ) = z −2 [3H 2 ( z ) − 2 H 3 ( z )]X1 ( z ) for C = 0. Thus, the transfer function of the system of Figure P13.12 is G ( z ) = z −2 [3H 2 ( z ) − 2 H 3 ( z )].

M13.1 (a) (i) Not for sale

527

Input Sequence

Output sequence up-sampled by5

2

2 1.5 1

Amplitude

Amplitude

1 0

0.5 0

-1 -0.5 -2

0

20

40 60 Time index n

80

-1

100

0

20

40 60 Time index n

80

100

(ii) Input Sequence

Output sequence up-sampled by5

1

0.2 0.15 Amplitude

Amplitude

0.8 0.6 0.4

0.05

0.2 0

0.1

0

10

20 30 Time index n

40

0

50

0

10

20 30 Time index n

40

50

M13.2 (a) (i) Output sequence down-sampled by 5 2

1

1 Amplitude

Amplitude

Input Sequence 2

0 -1 -2

0 -1

0

10

20 30 Time index n

40

50

-2

0

10

20 30 Time index n

40

50

(ii)

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Output sequence down-sampled by 5 1

0.8

0.8 Amplitude

Amplitude

Input Sequence 1

0.6 0.4 0.2 0

0.6 0.4 0.2

0

10

20 30 Time index n

40

0

50

0

2

4 6 Time index n

8

10

M13.3 (a) Output spectrum 1

0.8

0.8 Magnitude

Magnitude

Input spectrum 1

0.6 0.4

0.6 0.4

0.2 0

0.2

0

0.2

0.4

0.6

0.8

0

1

0

0.2

0.4

ω/ π

0.6

0.8

1

ω/ π

(b) Output spectrum 1

0.8

0.8 Magnitude

Magnitude

Input spectrum 1

0.6 0.4 0.2 0

0.6 0.4 0.2

0

0.2

0.4

0.6

0.8

0

1

0

0.2

ω/ π

0.4

0.6

0.8

1

ω/ π

M13.4 (a) Output spectrum 1

0.8

0.8 Magnitude

Magnitude

Input spectrum 1

0.6 0.4 0.2 0

0.6 0.4 0.2

0

0.2

0.4

0.6

0.8

1

ω/ π

0

0

0.2

0.4

0.6

0.8

1

ω/ π

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(b) Output spectrum 1

0.8

0.8 Magnitude

Magnitude

Input spectrum 1

0.6 0.4

0.6 0.4

0.2 0

0.2

0

0.2

0.4

0.6

0.8

0

1

0

0.2

0.4

ω/ π

0.6

0.8

1

ω/ π

M13.5 (a) Output spectrum 0.25

0.8

0.2 Magnitude

Magnitude

Input spectrum 1

0.6 0.4 0.2 0

0.15 0.1 0.05

0

0.2

0.4

0.6

0.8

0

1

0

0.2

0.4

ω/ π

0.6

0.8

1

ω/ π

(b) Output spectrum 0.14

0.8

0.12 Magnitude

Magnitude

Input spectrum 1

0.6 0.4 0.2 0

0.1 0.08 0.06

0

0.2

0.4

0.6

0.8

1

0.04

0

ω/ π

0.2

0.4

0.6

0.8

1

ω/ π

M13.6 (a)

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Output sequence 2

1

1 Amplitude

Amplitude

Input sequence 2

0

0

-1 -2

-1

0

20


100

-2

120

0

10

20 Time index n

30

40

M13.7 (a) Output sequence 2

1

1 Amplitude

Amplitude

Input sequence 2

0 -1 -2

0 -1

0

10

20 30 Time index n

40

-2

50

0

50 100 Time index n

150

M13.8 (a) Output sequence 2

1

1 Amplitude

Amplitude

Input sequence 2

0 -1 -2

0 -1

0

10

20 Time index n

30

40

-2

0

5

10 15 Time index n

20

25

M13.9 Using Program 13_9.m we arrive at the transfer function of the desired elliptic half1 band lowpass filter: H 0 ( z ) = A0 ( z 2 ) + z −1A1 ( z 2 ) , where 2 − 0.0192 + 0.3903z 2 + 1.2456 z −4 + z −6 A0 ( z 2 ) = and 1 + 1.2456 z − 2 + 0.3903z − 4 + 0.0192 z − 6 0.1206 + 0.8884 z −2 + 1.7442 z −4 + z −6 A1 ( z 2 ) = . The power-complentary half1 + 1.7442 z − 2 + 0.8884 z − 4 + 0.1206 z − 6 1 band highpass transfer function is given by H 0 ( z ) = A0 ( z 2 ) − z −1A1 ( z 2 ) . A plot 2

[

]

[

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]

531

of the magnitude responses of the above half-band lowpass and highpass filters is shown below: H0(z)

Magnitude

1

H1(z)

0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

M13.10 (a) A digital lowpass half-band filter can be designed by applying a bilinear transformation to an analog lowpass Butterworth transfer function with a 3-dB cutoff frequency at 1 rad/sec. The 3-dB cutoff frequency of the digital lowpass Butterworth half-band filter is therefore at ωc = 2 tan −1 (1) / π = 0.5. To design a 3rd-order digital lowpass Butterworth half-band filter we use the MATLAB statement [num,den] = butter(3,0.5); which yields 0.1667(1 + z −1 ) 3 H ( z) = . As can be seen from the pole-zero plot of given below all 1 + 0.3333z − 2 poles are on the imaginary axis:

Imaginary Part

1 0.5 3

0 -0.5 -1 -1.5

-1

-0.5

0 0.5 Real Part

1

1.5

Using the MATLAB statement [d1,d2] = tf2ca([1 3 3 1]/6, [1 0 1/3 0]); we arrive at the parallel allpass decomposition of H (z ) as

H ( z) =

1 [ A ( z 2 ) + z −1 A1 ( z 2 )], 2 0

where A0

(z 2 ) =

1 3

+ z−2

1+

1 −2 z 3

and A1 ( z 2 ) = 1. Hence,

the power-complementary highpass transfer function is given by 0.1667(1 − 3z −1 + 3z −2 − z −3 ) 1 G( z) = [ A0 ( z 2 ) − z −1 A1 ( z 2 )] = . 2 1 + 0.3333z − 2

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1

Magnitude

0.8 0.6 0.4 G(z)

H(z)

0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

(b) To design a 5th-order digital lowpass Butterworth half-band filter we use the MATLAB statement [num,den] = butter(5,0.5); which yields 0.0528(1 + z −1 ) 5 H (z) = . As can be seen from the pole-zero plot of given 1 + 0.6334 z − 2 + 0.0557z − 4 below all poles are on the imaginary axis:

Imaginary Part

1 0.5 0 -0.5 -1 -1

-0.5

0 0.5 Real Part

1

Using the MATLAB statement [d1,d2] = tf2ca(0.0528*[1 5 10 10 5 1], [1 0 0.6334 0 0.0557]); we arrive at the parallel allpass decomposition of H (z ) as

H ( z) = [ A0 ( z 2 ) + z −1 A1 ( z 2 )], where A0 ( z 2 ) = 2 1

A1 ( z 2 ) =

0.5279 + z −2 1 + 0.5279z − 2

0.1056 + z −2 1 + 0.1056 z − 2

and

. Hence, the power-complementary highpass transfer function

is given by G( z) = [ A0 ( z 2 ) − z −1 A1 ( z 2 )] = 2

0.0528(1 − z −1 ) 5

1

1 + 0.6334 z − 2 + 0.0557 z − 4

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533

1

Magnitude

0.8 0.6 G(z)

0.4

H(z)

0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

M13.11 M13.12 L-th band Nyquist Filter, L = 5 0

Gain, dB

-20 -40 -60 -80 0

0.2

0.4

0.6

0.8

1

ω/π

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Chapter 14 14.1 (a) 2-band polyphase decomposition – Using P ( z ) = 1 − c z −1 , we express H (z) =

H ( z) =

a + bz −1

as 1 + cz −1 (a + bz −1 )(1 − cz −1 )

(1 + cz −1 )(1 − cz −1 )

=

a + (b − ac)z −1 − bcz −2 1 − c2 z−2

−2 ⎛ a − bcz − 2 ⎞ ⎟ + z −1 ⎛⎜ b − ac ⎞⎟. Hence, E 0 ( z 2 ) = a − bcz and = ⎜⎜ ⎜ ⎟ 2 −2 ⎟ 1 − c2 z−2 ⎝ 1 − c2z−2 ⎠ ⎝ 1− c z ⎠ b − ac E1 ( z 2 ) = . 1 − c2z −2 (b) 3-band polyphase decomposition – Using P ( z ) = 1 − cz −1 + c 2 z −2 , we rewrite H (z ) as (a + bz −1 )(1 − cz −1 + c 2 z −2 ) a + (b − ac)z −1 + c(ac − b)z −2 + bc 2 z −3 H (z) = = (1 + cz −1 )(1 − cz −1 + c 2 z − 2 ) 1 − c3z −3 2 −3 ⎛ a + bc 2 z − 3 ⎞ ⎟ + z −1 ⎛⎜ b − ac ⎞⎟ + z − 2 ⎛⎜ c(ac − b) ⎞⎟. Hence, E 0 ( z 3 ) = a + bc z , = ⎜⎜ ⎜ ⎟ ⎜ ⎟ 3 −3 ⎟ 1 − c3z −3 ⎝ 1 − c3z −3 ⎠ ⎝ 1 − c3z −3 ⎠ ⎝ 1− c z ⎠ b − ac c(ac − b) E1 ( z 3 ) = , E 2 (z 3 ) = . 3 − 3 1− c z 1 − c3z −3

a − bz −1 . H a ( zW21 ) = H a (− z ) = . Thus, 1 + cz −1 1 − cz −1 1 1 ⎛ a + bz −1 a − bz −1 ⎞⎟ a − bcz − 2 and E 0 ( z 2 ) = [ H a ( z ) + H a (− z )] = ⎜⎜ + = 2 2 ⎝ 1 + cz −1 1 − cz −1 ⎟⎠ 1 − c 2 z − 2 1 1 ⎛ a + bz −1 a − bz −1 ⎞⎟ (b − ac)z −1 z −1 E 1( z 2 ) = [ H a ( z ) − H a (− z )] = ⎜⎜ − . = 2 2 ⎝ 1 + cz −1 1 − cz −1 ⎟⎠ 1 − c 2 z − 2 Hence, a 2-band polyphase decomposition of H a (z ) is given by

14.2 (a) H a ( z ) =

a + bz −1

⎛ a − bcz − 2 ⎞ ⎟ + z −1 ⎛⎜⎜ b − ac ⎞⎟⎟. H a ( z ) = ⎜⎜ ⎟ 2 − 2 ⎝ 1 − c2z −2 ⎠ ⎝ 1− c z ⎠ 3 − 4 z −1 + 2.1z −2 3 + 4 z −1 + 2.1z −2 (b) H b ( z ) = . H b ( zW21 ) = H b (− z ) = . 1 − 0.8 z −1 + 0.7 z − 2 1 + 0.8 z −1 + 0.7 z − 2 Thus, 1 1 ⎛ 3 − 4 z −1 + 2.1z − 2 3 + 4 z −1 + 2.1z − 2 ⎞⎟ + E 0 ( z 2 ) = [ H b ( z ) + H b (− z )] = ⎜⎜ 2 2 ⎝ 1 − 0.8 z −1 + 0.7 z − 2 1 + 0.8 z −1 + 0.7 z − 2 ⎟⎠

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=

3 + z −2 + 1.47 z −4 1 + 0.76 z − 2 + 0.49 z − 4

.

1 1 ⎛ 3 − 4 z −1 + 2.1z − 2 3 + 4 z −1 + 2.1z − 2 ⎞⎟ − [ H b ( z ) − H b (− z )] = ⎜⎜ 2 2 ⎝ 1 − 0.8 z −1 + 0.7 z − 2 1 + 0.8 z −1 + 0.7 z − 2 ⎟⎠ − 1.6 z −1 − 1.12 z −3 − 1.6 − 1.12 z −2 . Hence, E1 ( z 2 ) = . = 1 + 0.76 z − 2 + 0.49 z − 4 1 + 0.76 z − 2 + 0.49 z − 4 z −1 E1 ( z 2 ) =

Hence, a 2-band polyphase decomposition of H b (z ) is given by ⎛ 3 + z − 2 + 1.47 z − 4 ⎞ ⎛ ⎞ − 1.6 − 1.12 z − 2 ⎟ + z −1 ⎜ ⎟ H b ( z ) = ⎜⎜ ⎜ 1 + 0.76 z − 2 + 0.49 z − 4 ⎟. −2 −4 ⎟ ⎝ 1 + 0.76 z + 0.49 z ⎠ ⎝ ⎠ (c) H c ( z ) = E 0 (z 2 ) = =

=

4 + 2.5z −1 − 3.5z −2 + 2 z −3 1 − 0.4 z −1 + 0.78z − 2 + 0.18z

. H c (− z) = −3

4 − 2.5z −1 − 3.5z −2 − 2 z −3 1 + 0.4 z −1 + 0.78z − 2 − 0.18z − 3

.

1 [ H c ( z ) + H c (− z )] 2

1 ⎛⎜ 4 + 2.5z −1 − 3.5z − 2 + 2 z − 3 4 − 2.5z −1 − 3.5z − 2 − 2 z − 3 ⎞⎟ + 2 ⎜⎝ 1 − 0.4 z −1 + 0.78 z − 2 + 0.18 z − 3 1 + 0.4 z −1 + 0.78 z − 2 − 0.18 z − 3 ⎟⎠

4 + 0.62 z −2 − 2.38 z −4 − 0.36 z −6

. 1 + 1.4 z − 2 + 0.7524 z − 4 − 0.0324 z − 6 1 z −1 E1 ( z 2 ) = [ H c ( z ) − H c (− z )] = 2 1 ⎛ 4 + 2 .5 z − 1 − 3 .5 z − 2 + 2 z − 3 4 − 2.5z −1 − 3.5z − 2 − 2 z − 3 ⎞⎟ = ⎜⎜ − 2 ⎝ 1 − 0.4 z −1 + 0.78 z − 2 + 0.18 z − 3 1 + 0.4 z −1 + 0.78 z − 2 − 0.18 z − 3 ⎟⎠ =

4.1z −1 + 1.83z −3 + 2.19z −5

. Thus, 1 + 1.4 z − 2 + 0.7524 z − 4 − 0.0324 z − 6 4.1 + 1.83z −2 + 2.19z −4 . E1 ( z 2 ) = 1 + 1.4 z − 2 + 0.7524 z − 4 − 0.0324 z − 6

Hence, a 2-band polyphase decomposition of H c (z ) is given by ⎞ ⎛ ⎛ 4 + 0.62 z − 2 − 2.38 z − 4 − 0.36 z − 6 ⎞ 4.1 + 1.83z − 2 + 2.19 z − 4 ⎟ ⎟ + z −1 ⎜ H c ( z ) = ⎜⎜ ⎜ 1 + 1.4 z − 2 + 0.7524 z − 4 − 0.0324 z − 6 ⎟. − 2 + 0.7524 z − 4 − 0.0324 z − 6 ⎟ z 1 1 . 4 + ⎠ ⎝ ⎠ ⎝ 14.3 (a) H1 ( z ) =

3 − 4 z −1 1 − 0.5z −1

=

2

∑ z − k E k (z 3 ).

We can write,

k =0

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536

⎡ H1 ( z ) ⎤ ⎡1 1 ⎢ ⎥ ⎢ 1 −1 ⎢ H1 (W3 z ) ⎥ = ⎢1 W3 ⎢ H (W 2 z )⎥ ⎢1 W − 2 3 ⎣ 1 3 ⎦ ⎣

3 1 ⎤ ⎡ E 0 (z ) ⎤ ⎥⎢ ⎥ W3− 2 ⎥ ⎢ z −1 E1 ( z 3 ) ⎥. Thus, W3−1 ⎥⎦ ⎢⎣ z − 2 E 2 ( z 3 )⎥⎦

⎡ E 0 ( z 3 ) ⎤ ⎡1 1 ⎢ −1 ⎥ ⎢ −1 3 ⎢ z E1 ( z ) ⎥ = ⎢1 W3 ⎢ z − 2 E 2 ( z 3 )⎥ ⎢1 W − 2 3 ⎣ ⎦ ⎣ Therefore,

1 ⎤ ⎥ W3− 2 ⎥ W3−1 ⎥⎦

−1

⎡ H1 ( z ) ⎤ ⎡1 1 ⎥ ⎢ 1 ⎢ 1 1 ⎢ H1 (W3 z ) ⎥ = 3 ⎢1 W3 ⎢ H (W 2 z )⎥ ⎢1 W 2 3 ⎣ 1 3 ⎦ ⎣

1 ⎤ ⎡ H1 ( z ) ⎤ ⎥ ⎥⎢ W32 ⎥ ⎢ H1 (W31 z ) ⎥. W31 ⎥⎦ ⎢⎣ H1 (W32 z )⎥⎦

1

E 0 ( z 3 ) = [ H1 ( z ) + H1 ( zW31 ) + H1 ( zW32 )] 3 =

1 ⎛⎜ 3 − 4 z −1 3 − 4e j 2 π / 3 z −1 3 − 4e j 4 π / 3 z −1 ⎞⎟ 3 − z −3 = , + + 3 ⎜⎝ 1 − 0.5z −1 1 − 0.5e j 2 π / 3 z −1 1 − 0.5e j 4 π / 3 z −1 ⎟⎠ 1 − (0.5) 3 z − 3

z −1 E1 ( z 3 ) = [ H1 ( z ) + W31 H1 ( zW31 ) + W32 H1 ( zW32 )] 3 1

=

j 4π / 3 z −1 ⎞ j 2π / 3 z −1 1 ⎛⎜ 3 − 4 z −1 j 4 π / 3 3 − 4e j 2 π / 3 3 − 4e ⎟ e e + + 3 ⎜⎝ 1 − 0.5z −1 1 − 0.5e j 2π / 3 z −1 1 − 0.5e j 4 π / 3 z −1 ⎟⎠

⎞ ⎛ − 2.5 ⎟, = z −1⎜⎜ ⎟ 3 − 3 ⎝ 1 − (0.5) z ⎠ z − 2 E 2 ( z 3 ) = [ H1 ( z ) + W32 H1 ( zW31 ) + W31 H1 ( zW32 )] 3 1

=

j 4π / 3 z −1 ⎞ j 2 π / 3 z −1 1 ⎛⎜ 3 − 4 z −1 j 2 π / 3 3 − 4e j 4 π / 3 3 − 4e ⎟ e e + + 3 ⎜⎝ 1 − 0.5z −1 1 − 0.5e j 2π / 3 z −1 1 − 0.5e j 4 π / 3 z −1 ⎟⎠

⎛ ⎞ − 1.25 ⎟. = z − 2 ⎜⎜ 3 z −3 ⎟ − 1 ( 0 . 5 ) ⎝ ⎠

Hence, E 0 ( z ) = (b) H 2 ( z ) =

3 − z −1 1 − 0.125z −1

, E1 ( z ) =

4 + 2.1z −1 − 3.4 z − 2 1 − 0.8z −1

+ 0.6 z − 2

⎡ H 2 ( z ) ⎤ ⎡1 1 ⎢ 1 z ) ⎥ = ⎢1 W −1 H W ( ⎢ 2 3 ⎥ ⎢ 3 ⎢ H (W 2 z )⎥ ⎢1 W − 2 3 ⎣ 2 3 ⎦ ⎣ ⎡ E 0 ( z 3 ) ⎤ ⎡1 1 ⎢ −1 3 ) ⎥ = ⎢1 W −1 ( z E z 1 ⎢ ⎥ ⎢ 3 ⎢ z − 2 E 2 ( z 3 )⎥ ⎢1 W − 2 3 ⎣ ⎦ ⎣

=

− 2.5 1 − 0.125z

, E 2 ( z) = −1

− 1.25 1 − 0.125z −1

.

2

∑ z − k E k (z 3 ).

We can write,

k =0

3 1 ⎤ ⎡ E0 (z ) ⎤ ⎥ ⎢ ⎥ W3− 2 ⎥ ⎢ z −1 E1 ( z 3 ) ⎥. Thus, W3−1 ⎥⎦ ⎢⎣ z − 2 E 2 ( z 3 )⎥⎦

1 ⎤ ⎥ W3− 2 ⎥ W3−1 ⎥⎦

−1

⎡ H 2 ( z) ⎤ ⎡1 1 ⎥ ⎢ 1 ⎢ 1 1 ⎢ H 2 (W3 z ) ⎥ = 3 ⎢1 W3 ⎢ H (W 2 z )⎥ ⎢1 W 2 3 ⎣ 2 3 ⎦ ⎣

1 ⎤ ⎡ H 2 (z) ⎤ ⎥ ⎥⎢ W32 ⎥ ⎢ H 2 (W31 z ) ⎥. W31 ⎥⎦ ⎢⎣ H 2 (W32 z )⎥⎦

1

Therefore, E 0 ( z 3 ) = [ H1 ( z ) + H1 ( zW31 ) + H1 ( zW32 )] 3 Not for sale

537

=

=

1 ⎛⎜ 4 + 2.1z −1 − 3.4 z − 2 4 + 2.1e j 2 π / 3 z −1 − 3.4e j 4π / 3 z − 2 4 + 2.1e j 4 π / 3 z −1 − 3.4e j 2 π / 3 z − 2 + + 3 ⎜⎝ 1 − 0.8 z −1 + 0.6 z − 2 1 − 0.8e j 2π / 3 z −1 + 0.6e j 4 π / 3 z − 2 1 − 0.8e j 4 π / 3 z −1 + 0.6e j 2π / 3 z − 2

4 − 0.716 z −3 − 1.224 z −6 1 + 0.928z − 3 + 0.216 z − 6

,

1 1 4 + 2.1z −1 − 3.4 z −2 z −1 E1 ( z 3 ) = [ H 2 ( z ) + W31 H 2 ( zW31 ) + W32 H 2 ( zW32 )] = ( 3 3 1 − 0.8 z −1 + 0.6 z − 2 4 + 2.1e j 2π / 3 z −1 − 3.4e j 4π / 3 z −2 4 + 2.1e j 4π / 3 z −1 − 3.4e j 2π / 3 z −2 + e j 2π / 3 + e j 4π / 3 ) 1 − 0.8e j 2π / 3 z −1 + 0.6e j 4π / 3 z − 2 1 − 0.8e j 4π / 3 z −1 + 0.6e j 2π / 3 z − 2 − 1.56 z −3 − 0.876 z −6 = , 1 + 0.928z − 3 + 0.216 z − 6 1 1 4 + 2.1z −1 − 3.4 z −2 z −1 E 2 ( z 3 ) = [ H 2 ( z ) + W32 H 2 ( zW31 ) + W31 H 2 ( zW32 )] = ( 3 3 1 − 0.8 z −1 + 0.6 z − 2 4 + 2.1e j 2π / 3 z −1 − 3.4e j 4π / 3 z −2 4 + 2.1e j 4π / 3 z −1 − 3.4e j 2π / 3 z −2 + e j 4π / 3 + e j 2π / 3 ) 1 − 0.8e j 2π / 3 z −1 + 0.6e j 4π / 3 z − 2 1 − 0.8e j 4π / 3 z −1 + 0.6e j 2π / 3 z − 2 5.3z −3 + 2.312 z −6 = . 1 + 0.928z − 3 + 0.216 z − 6 4 − 0.716 z −1 − 1.224 z −2 − 1.56 z −1 − 0.876 z −2 Hence, E 0 ( z ) = , E1 ( z ) = , 1 + 0.928 z −1 + 0.216 z − 2 1 + 0.928 z −1 + 0.216 z − 2 5.3z −1 + 2.312 z −2 E 2 (z) = . 1 + 0.928z −1 + 0.216 z − 2

14.4 H 0 ( z) = h[0] + h[1]z −1 + h[2]z −2 + h[3]z −3 + h[ 4]z −4 + h[5]z −5 , and

H 1( z ) = H 0 (− z ) = h[0] − h[1]z −1 + h[2]z −2 − h[3]z −3 + h[ 4]z −4 − h[5]z −5 . A realization of H 0 ( z ) and H1 ( z) in the form of Figure P14.1 using 5 delays and 6 multipliers is shown below: z–1 h[5]

h[4]

z–1 h[3]

z–1 h[2]

z–1 h[1]

z

–1

y[n]

h[0]

x 0[n] _1

_1

_1

x 1[n]

14.5 H 0 ( z) = h[0] + h[1]z −1 + h[2]z −2 + h[3]z −3 + h[ 4]z −4 + h[5]z −5 , and H 1( z) = z −5 H 0 ( z −1 ) = h[5] + h[ 4]z −1 + h[3]z −2 + h[2]z −3 + h[1]z −4 + h[0]z −5 .

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⎞ ⎟ ⎟ ⎠

H 1( z ) = H 0 (− z ) = h[0] − h[1]z −1 + h[2]z −2 − h[3]z −3 + h[ 4]z −4 − h[5]z −5 . A realization of H 0 ( z ) and H1 ( z) in the form of Figure P14.1 using 6 multipliers is shown below: y[n] h[0]

x 0[n]

h[1]

h[2]

z–1

z

–1

z–1

z

–1

h[3]

h[5]

–1

z

–1

z–1

–1

z–1

z–1

z z

h[4]

x 1[n]

14.6 (a) The structure of Figure P14.2(a) with internal variables labeled is shown below: V0 (z)

X 0(z)

V 1(z)

X 1(z) D X2 (z) X 3(z)

V 2(z) V3 (z)

R 0(z 4 ) R 1(z 4 ) R 2(z 4 ) R3 (z 4 )

_ z1 _ z1

_ z1

Y(z)

⎡V0 ( z )⎤ ⎡1 1 1 1 ⎤ ⎡ X 0 ( z )⎤ ⎢ V1 ( z ) ⎥ ⎢1 − j − 1 j ⎥ ⎢ X1 ( z ) ⎥ = , Analyzing the above structure we arrive at ⎢ V ( z )⎥ ⎢1 − 1 1 − 1⎥ ⎢ X ( z )⎥ ⎢ 2 ⎥ ⎢1 j − 1 − j ⎥ ⎢ 2 ⎥ ⎦ ⎣ X 3 (z)⎦ ⎣V3 ( z ) ⎦ ⎣ and Y ( z ) = R3 ( z 4 )V3 ( z) + z −1 R2 ( z 4 )V2 ( z) + z −2 R1 ( z 4 )V1 ( z ) + z −3 R0 ( z 4 )V0 ( z). The first equation leads to V0 ( z) = X 0 ( z) + X1 ( z) + X 2 ( z) + X 3 ( z), V1 ( z) = X 0 ( z ) − jX1 ( z ) − X 2 ( z) + jX 3 ( z), V2 ( z) = X 0 ( z) − X1 ( z) + X 2 ( z) − X 3 ( z), V4 ( z) = X 0 ( z) + jX1 ( z ) − X 2 ( z) − jX 3 ( z). Substituting these 4 equations in the equation for Y (z ) and solving for the transfer functions Gi ( z ) = Y ( z) / X i ( z ), 0 ≤ i ≤ 3, we arrive at

G0 ( z) = R3 ( z 4 ) + z −1 R2 ( z 4 ) + z −2 R1 ( z 4 ) + z −3 R0 ( z 4 ), G1 ( z ) = jR3 ( z 4 ) − z −1 R2 ( z 4 ) − jz −2 R1 ( z 4 ) + z −3 R0 ( z 4 ), G2 ( z) = − R3 ( z 4 ) + z −1 R2 ( z 4 ) − z −2 R1 ( z 4 ) + z −3 R0 ( z 4 ), G3 ( z) = − jR3 ( z 4 ) − z −1 R2 ( z 4 ) + jz −2 R1 ( z 4 ) + z −3 R0 ( z 4 ).

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Substituting the expressions for Ri ( z), 0 ≤ i ≤ 3, we finally arrive at

G0 ( z) = 2 + z −1 + 4 z −2 + z −3 − 1.5z −4 + 0.3z −5 − 0.9z −6 + 3.7z −7 + 3.1z −8 − 0.8z −9 + 2.3z −10 + 1.7 z −11 , G1 ( z ) = j 2 − z −1 − j 4 z −2 + z −3 − j1.5z −4 − 0.3z −5 + j 0.9 z −6 + 3.7 z −7 + j 3.1z −8 + 0.8 z −9 − j 2.3z −10 + 1.7 z −11 ,

G 2 ( z ) = −2 + z −1 − 4 z −2 + z −3 + 1.5z −4 + 0.3z −5 + 0.9 z −6 + 3.7 z −7 − 3.1z −8 − 0.8 z −9 − 2.3z −10 + 1.7 z −11 ,

G3 ( z ) = − j 2 − z −1 + j 4 z −2 + z −3 + j1.5z −4 − 0.3z −5 − j 0.9 z −6 + 3.7 z −7 − j 3.1z −8

+ 0.8 z −9 + j 2.3z −10 + 1.7 z −11 .

(b) The magnitude responses of all 4 analysis filters are shown below: G1(e jω )

G0(e jω )

1

1

0

π_ 4

3π __ 4

π

7π _ 4

2π

ω

0

π_ 4

3π __ 4

π

π_ 4

3π __ 4

π

2π

ω

G3(e jω )

G2(e jω )

1

1

0

π_ 4

3π __ 4

π

7π _ 4

2π

ω

0

14.7 Y ( z ) = [H 0 ( z )G0 ( z) + H1 ( z )G1 ( z)]X ( z). Now, H 0 ( z ) =

5π _ 4

7π _ 4

2π

1 + z −1 1 − z −1 and H1 ( z ) = . 2 2

1 − z −1 1 + z −1 and G1 ( z ) = − . Then, 2 2 1 1 1 Y ( z) = ⎛⎜ (1 + z −1 ) 2 − (1 − z −1 ) 2 ⎞⎟ X ( z) = (1 + 2 z −1 + z − 2 − 1 + 2 z −1 − z − 2 ) X ( z ) 4 4 4 ⎝ ⎠ = z −1 X ( z ). Or in other words, y[ n] = x[ n − 1] indicating that the structure of Figure P14.3 is a perfect reconstruction system with the above choices of filters.

Choose G0 ( z ) =

14.8 (a) Since H 0 ( z) and H1 ( z) are power-complementary, H 0 ( z)H 0 ( z −1 ) + H1 ( z )H1 ( z −1 ) = 1. Now,

Y ( z) = (H 0 ( z)G0 ( z) + H1 ( z )G1 ( z))X ( z)

(

)

= z − N H 0 ( z )H 0 ( z −1 ) + H1 ( z)H1 ( z −1 ) X ( z) = z − N X ( z). Or in other words, y[ n] = x[ n − N ] indicating that the structure of Figure P14.3 is a perfect reconstruction system with the above choices of filters.

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ω

(b) If H 0 ( z) and H1 ( z) are causal FIR transfer functions of order N each, H 0 ( z)

and H1 ( z) are polynomials in z −1. As a result, H 0 ( z −1 ) and H1 ( z −1 ) are polynomials in z with the highest power being z N . Hence, z − N H 0 ( z −1 ) and z − N H1 ( z −1 ) are polynomials in z −1 , making the synthesis filters G 0 ( z ) and G1 ( z) causal FIR transfer functions of order N each.

(c) From Figure P14.3, for perfect reconstruction we require H 0 ( z)G 0 ( z) + H 1( z)G1( z ) = z − N . From Part (a) we note that the perfect

reconstruction condition is satisfied with G0 ( z) = z − N H 0 ( z −1 ) and G1 ( z ) = z − N H1 ( z −1 ), if H 0 ( z )H 0 ( z −1 ) + H1 ( z )H1 ( z −1 ) = 1, i.e., if H 0 ( z ) and H1 ( z) are power-complementary. The last condition is satisfied if and only z − N / 2 − no z − N / 2 − no (z + z n o ) and H1 ( z ) = (z − z n o ). As a result, 2 2 z − N / 2 − no G 0 ( z ) and G1 ( z) are also of the form G0 ( z ) = (z + z n o ) and 2 z − N / 2 − no G1 ( z ) = (z − z n o ). 2

if H 0 ( z ) =

14.9 (a) G ( z ) =

=

0.0985(1 + z −1 ) 3

(1 − 0.1584 z −1 )(1 − 0.4189 z −1 + 0.3554 z − 2 )

1 0.3554 − 0.4189 z −1 + z −2 − 0.1584 + z −1 [ ]. Hence, + 2 1 − 0.4189 z −1 + 0.3554 z − 2 1 − 0.1584 z −1

A0 ( z ) =

0.3554 − 0.4189 z −1 + z −2 1 − 0.4189 z −1 + 0.3554 z − 2

and A1 ( z ) =

− 0.1584 + z −1 1 − 0.1584 z −1

.

(b) G (z ) can be realized with only 3 multipliers as a parallel connection of two allpass filters in the form of Figure 8.43, where the allpass filter A0 ( z ) is realized using any one of the Type 2 or Type 3 second-order allpass structures requiring 2 multipliers and the allpass filter A1 (z ) is realized using any one of the Type 1 first-order allpass structures requiring 1 multiplier. (c) H ( z ) =

1 2

[A0 ( z ) − A1 ( z )] =

0.2564 (1 − z −1 ) 3

(1 − 0.1584 z −1 )(1 − 0.4189 z −1 + 0.3554 z − 2 )

.

(d) A plot of the magnitude responses of G (z ) and H (z ) is shown on top of the next page:

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jω

Magnitude

jω

|G(e )|

1

|H(e )|

0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

14.10 (a) G ( z ) =

=

0.1868(1 + 1.0902 z −1 + 1.0902 z −2 + z −3 )

(1 − 0.3628 z −1 )(1 − 0.5111z −1 + 0.7363 z − 2 )

1 0.7363 − 0.5111z −1 + z −2 − 0.3628 + z −1 + [ ]. Hence, 2 1 − 0.5111z −1 + 0.7363 z − 2 1 − 0.3628 z −1

A0 ( z ) =

0.7363 − 0.5111z −1 + z −2 1 − 0.5111z −1 + 0.7363 z − 2

and A1 ( z ) =

− 0.3628 + z −1 1 − 0.3628 z −1

.

(b) G (z ) can be realized with only 3 multipliers as a parallel connection of two allpass filters in the form of Figure 8.43, where the allpass filter A0 ( z ) is realized using any one of the Type 2 or Type 3 second-order allpass structures requiring 2 multipliers and the allpass filter A1 (z ) is realized using any one of the Type 1 first-order allpass structures requiring 1 multiplier. (c) H ( z ) =

1 2

[A0 ( z ) − A1 ( z )] =

0.5495(1 − 1.7866 z −1 − 1.7866 z −2 + z −3 )

(1 − 0.3628 z −1 )(1 − 0.5111z −1 + 0.7363 z − 2 )

.

(d) A plot of the magnitude responses of G (z ) and H (z ) is shown below: |G(e ω )|

|H(e ω )|

j

Magnitude

1

j

0.8 0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

ω/π

14.11 An N-th order, with N odd, elliptic lowpass transfer function H (z ) satisfies the condition

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H ( z ) H ( z −1 ) =

1 1 + ε 2 R N ( z ) R N ( z −1 )

,

(A)

where R N ( z ) is a rational function of the form N −1 −1 ⎞ 2

⎛1− z (1 − z −1e jφ l )(1 − z −1e − jφ l ) ⎟ . RN ( z) = ⎜ − 1 jξ l −1 − jξ l ⎜ 1 − z −1 ⎟ ( )( ) 1 − z e 1 − z e ⎝ ⎠ l =0 In the above equation, the frequencies ξ l are the transmission zeros at which

∏

(B)

H (e jω ) is equal to 0, i.e., H (e jξ l ) = 0, and the frequencies φ l are the reflection

zeros at which H (e jω ) is equal to the maximum value of 1, i.e., H (e jφ l ) = 1. From Eq. (B) it follows that R N ( z ) = − R N ( z −1 ). Now, as H (z ) satisfies the power-symmetric condition, we have

(C)

H ( z ) H ( z −1 ) + H (− z ) H (− z −1 ) = 1, i.e., H ( z ) H ( z −1 ) = 1 − H (− z ) H (− z −1 ). (D) From the above equation it follows that the transmission zeros of H ( − z ) are at frequencies π − φ l and its reflection zeros are at π − ξ l . As a result, φ l + ξ l = π. Hence, it follows that R N (− z ) = 1 / R N ( z ), or equivalently, R N ( z ) R N (−z ) = 1. (E) 1 From Eq. (A) we have H ( − z ) H ( − z −1 ) = , which when 2 1 + ε R N ( − z ) R N ( − z −1 ) substituted in Eq. (D) yields 1 1 . In view of = H ( z ) H ( z −1 ) = 1 − 1 1 + ε 2 R N ( − z ) R N ( − z −1 ) 1 + ε 2 R N ( − z ) R N ( − z −1 ) Eq. (E), the above equation reduces to 1 (F) H ( z ) H ( z −1 ) = . 1 −1 1+ RN ( z)RN ( z ) ε2 Comparing Eqs. (A) and (F) we thus conclude that ε 2 = 1, and hence, 1 . Now at a pole z = λ of H (z ), H ( z ) H ( z −1 ) = 1 + R N ( z ) R N ( z −1 ) R N (λ ) R N (λ−1 ) = −1. From this relation, and Eqs. (C) and (E), it follows then that

R N (λ ) = 1. Also, for a lowpass power-symmetric transfer function, ξ l > π / 2 R N ( −λ )

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and φ l < π / 2. Consequently, the poles of the rational function R N ( z ) / R N (− z ) must lie on the left half of the z-plane. Since, the magnitude of the rational function R N ( z ) / R N (− z ) on the imaginary axis is 1, using maximum-modulus theorem it can be shown that R N ( z ) / R N (− z ) < 1 for Re z > 0. In a similar manner by replacing z with − z , it can be shown that R N (− z ) / R N ( z ) < 1 for Re z > 0, or, equivalently, R N ( z ) / R N (− z ) > 1 for Re z < 0. Thus, ⎧< 1, Re z > 0, ⎪ ⎨= 1, Re z = 0, ⎪⎩> 1, Re z < 0, or, in other words, all poles of H (z ) lie on the imaginary axis of the z-plane. RN ( z) R N (− z )

14.12 Now the magnitude-square function of an N-th order analog lowpass Butterworth 1 2 , where Ω c is the 3-dB function G a (s) is given by G a ( jΩ) = 1 + (Ω / Ω c ) 2 N 1 2 . The cutoff angular frequency. For Ω c = 1, then G a ( jΩ) = 1 + Ω2N corresponding transfer function of of an N-th order analog highpass Butterworth function is simply G a (1 / s ), whose the magnitude-square function is given by

⎛ 1 ⎞ ⎟⎟ G a ⎜⎜ ⎝ jΩ ⎠

2

=

Ω2N 1 + Ω2N

. As a result,

2

1 Ω2N 1 G a ( jΩ) + G a ⎛⎜ ⎞⎟ = + = 1. Now the bilinear transformation ⎝ jΩ ⎠ 1 + Ω2N 1 + Ω2N maps the analog angular frequency Ω to the digital angular frequency ω through the 1 − jΩ 1 − (1 / jΩ) relation e jω = . As − e jω = , the analog angular frequency 1/Ω is 1 + jΩ 1 + (1 / jΩ) mapped to the digital angular frequency π + ω. Hence, the relation 2

2 1 G a ( jΩ) + G a ⎛⎜ ⎞⎟ j ⎝ Ω⎠

2

2

2

2

= 1, becomes H 0 (e jω ) + H 0 (e j (π + ω) ) = 2

2

2

H 0 (e jω ) + H 0 (−e jω ) = H 0 (e jω ) + H1 (e jω ) = 1, where H0 (e jω ) is the frequency response of the digital lowpass filter H 0 ( z) obtained by applying the bilinear transformation to G a (s) and H1 (e jω ) is the frequency response of the digital highpass filter H 0 ( z) obtained by applying the bilinear transformation to Ga (1 / s ). Note that a 2

2

digital transfer function satisfying the condition H 0 (e jω ) + H1(e jω ) = 1, is called power-symmetric.

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Moreover, from the relation it follows that the analog 3-dB cutoff angular frequency Ω c = 1 is mapped into the digital 3-dB cutoff angular frequency ωc = π / 2. Hence, H 0 ( z) is a digital half-band lowpass filter.

P (z) , where P0 ( z ) and D0 ( z ) are polynomials in z −1. Hence, Let H 0 ( z) = 0 D0 ( z ) H 0 ( e jω ) = H 0 ( − e jω )

=

P0 (e jω )

P ( − e jω ) and H 0 (− e jω ) = 0 . Now, D 0 ( e jω ) D 0 ( − e jω ) 2

=

P0 (− e jω )P0* (− e jω )

= 1 − H 0 ( e jω ) j ω j ω * D0 ( − e ) D0 ( − e )

D0 (e jω ) D0* (e jω ) − P0 (e jω )P0* (e jω ) D0 (e jω ) D0* (e jω )

2

= 1−

P0 (e jω )P0* (e jω ) D0 (e jω ) D0* (e jω )

. Note that there are no common factors between

P0 (e jω ) and D0 (e jω ) , and between P0* (e jω ) and D0* (e jω ). As a result, there are no common factors between P0 (−e jω )P0* (−e jω ) and D0 (−e jω ) D0* (−e jω ). This implies then D0 (e jω ) D0* (e jω ) = D0 (−e jω ) D0* (−e jω ). Consequently, D0 (e jω ) = D0 (−e jω ), or P (z) P (z) D0 (e jω ) = d 0 (e j 2ω ). Hence, D0 ( z ) = d 0 ( z 2 ). Since H 0 ( z ) = 0 = 0 , it D0 ( z ) d 0 ( z 2 ) P (− z ) follows then H1 ( z ) = 0 . We have shown earlier that H 0 ( z) and H1( z) are powerd0 (z 2 )

complementary. Also, P0 ( z ) is a symmetric polynomial of odd order and P1 ( z) is an anti-symmetric polynomial of odd order. As a result, we can express H 0 (z) =

1 2

[ A0 (z) + A1 (z)] and H1 (z) = 12 [ A0 (z) − A1 (z)], where

A0 ( z) and A1 ( z) are

stable allpass functions for stable H 0 ( z) and H1 ( z). But, H1 ( z) = H 0 (− z). Hence, H 0 (z) =

1 2

[ A0 (− z) − A1 (− z)]. It therefore follows that

A0 ( z ) = A0 (− z ) = A0 ( z 2 ), and

1 A1 ( z ) = − A1 ( − z ) = z −1A1 ( z 2 ). Thus, H 0 ( z ) = [A0 ( z 2 ) + z −1A1 ( z 2 )]. 2

14.13 From Eq. (14.18), Y ( z ) = T ( z ) X ( z ) + A( z ) X ( − z ). Let Z −1{T ( z )} = t[ n], and Z −1{A( z )} = a[ n]. Then, an inverse z-transform of Eq. (14.18) yields

y[ n] =

∞

∑

t[l]x[ n − l] +

l = −∞

Define f 0 [ n] =

∞

∑

a[l](−1) n − l x[ n − l] =

l = −∞ t[ n] + (−1) − n a[ n]

and

∞

∑ (t[l] + (−1) n − l a[l])x[n − l].

l = −∞ f1[ n] = t[ n] − (−1) − n a[ n].

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Then we can write

545

⎧ ∞ f 0 [l]x[n − l], for n even, ⎪ The corresponding equivalent y[n] = ⎨ l∞= −∞ f1[l]x[n − l], for n odd. ⎪ ⎩ l = −∞ realization of the 2-channel QMF bank is therefore as indicated below:

∑ ∑

n even f 0[n] y[n]

x[n] f 1[n]

n odd

As can be seen from the above representation, the QMF bank is, in general, a timevarying system with a period 2. Note that A( z ) = 0, if then it becomes a linear, time-invariant system.

1

14.14 G a (s) =

. The digital transfer function H 0 ( z ) obtained by a (s + 1)(s 2 + s + 1) bilinear transformation is given by ( z + 1) 3 ( z + 1) 3 = H 0 ( z ) = G a ( s ) s = z −1 = ( z − 1 + z + 1)[( z − 1) 2 + ( z 2 − 1) + ( z + 1) 2 ] 2 z(3z 2 + 1) z +1 =

(1 + z −1 ) 3 6 + 2z − 2

=

1 + 3z −1 + 3z − 2 + z − 3 6 + 2z − 2

=

⎞ 1 1 ⎛⎜ 1 + 3z − 2 + z −1 ⎟⎟ = [A0 ( z 2 ) + z −1A1 ( z 2 )], ⎜ 2 ⎝ 3 + z−2 ⎠ 2

1 + 3 z −1

and A1 ( z) = 1. 3 + z −1 The corresponding power-complementary transfer function is given by ⎞ (1 − z −1 ) 3 1 1 ⎛ 1 + 3z − 2 . H1 ( z ) = [A0 ( z 2 ) − z −1A1 ( z 2 )] = ⎜⎜ − z −1 ⎟⎟ = −2 2 2 ⎝ 3 + z−2 6 + 2 z ⎠ A realization of the analysis part of the QMF bank is as shown in Figure 14.11 where the first-order allpass transfer function A0 (z ) is realized using any one of the single-multiplier structure of Figure 8.24 and the zero-th order allpass transfer function A1 (z ) is replaced with a direct connection between the input and the output. where A0 ( z) =

1 . The digital 14.15 G a (s) = s 5 + 3.2361 s 4 + 5.2361 s 3 + 5.2361 s 2 + 3.2361 s + 1 transfer function H 0 ( z) obtained by a bilinear transformation is given by H 0 (z) =

(1 + z −1 ) 5

18.9443 + 12 z − 2 + 1.0557 z − 4

=

0.0528(1 + z −1 ) 5 1 + 0.6334 z − 2 + 0.0557 z − 4

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= = =

0.0528 + 0.5279 z −2 + 0.2639 z −4 1 + 0.6334 z − 2 + 0.0557 z − 4 0.0528(1 + 9.4704 z −2 ) 1 + 0.1056 z − 2 1 2

+ z −1

+

+ 0.2639 z −1 + +0.5279 z −3 + 0.0528 z −5 1 + 0.6334 z − 2 + 0.0557 z − 4

0.2639(1 + 1.8948 z −2 ) 1 + 0.5278 z − 2

[A0 (z 2 ) + z −1A1 (z 2 )], where A0 (z) =

A1 ( z ) =

0.1056 + z −1 1 + 0.1056 z −1

and

0.5278 + z −1

. 1 + 0.5278 z −1 Its power-complementary transfer function is given by 0.5278 + z −2 1 0.1056 + z −2 1 − z −1 ]. H1 ( z) = [A0 ( z 2 ) − z −1A1 ( z 2 )] = [ 2 2 1 + 0.1056 z − 2 1 + 0.5278 z − 2 In the realization of a magnitude-preserving QMF bank as shown in Figure 14.11, the realization of the allpass filters A0 ( z ) and A1 (z) require 1 multiplier each, and hence, the realization of the analysis (and the synthesis) filter bank requires a total of 2 multipliers. 14.16 (a) Total number of multipliers required is 4(2 N − 1). Hence, the total number of multiplications per second is equal to 4(2 N − 1) FT = 4(2 N − 1) / T , where FT = 1 / T is the sampling frequency in Hz.

[A0 (z 2 ) + z −1A1 (z 2 )] and 1 H1 ( z) = [A0 ( z 2 ) − z −1A1 ( z 2 )]. If the order of A0 ( z ) is K and the order of 2

(b) In Figure 14.11, H 0 ( z ) =

1 2

A1 (z) is L, then the order of H 0 ( z) is 2 K + 2 L + 1 = N . Hence, K + L = ( N − 1) / 2. The total number of multipliers needed to implement is K, while the total number of multipliers needed to implement is L. Hence, the total number of multipliers required to implement the QMF bank of Figure 14.11 is 2( K + L ) = N − 1. However, the multipliers here are operating t half of the sampling rate of the input x[ n]. As a result, the total number of multiplications per second in this case is ( N − 1) FT / 2 = ( N − 1) / 2T . 14.17 The 4 filters of the 2-channel QMF bank are: H 0 ( z) = 4 z −2 , H1 ( z ) = z −1 ,

G0 ( z) = 0.25z −1 , G1 ( z) = z −1. Substituting these transfer functions in Eq. (14.21) we get H 0 (− z)G0 ( z) + H1 (− z)G1 ( z ) = 4 z −2 × 0.25z −1 − z −1 × z −2 = z −3 − z −3 = 0, implying aliasing cancellation condition holds. Next, substituting the transfer functions in Eq. (14.27) we get

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H 0 ( z)G0 ( z ) + H1 ( z)G1 ( z) = 4 z −2 × 0.25z −1 + z −1 × z −2 = 2 z −3 . Hence, it is a perfect reconstruction QMF bank. 14.18 The aliasing cancellation condition is satisfied if the synthesis filters satisfy Eq. (14.37). Hence, we choose H1 ( z) = G0 (− z ) = d − e z −1 + f z −2 − g z −3 + h z −4 ,

and G1 ( z ) = − H 0 (− z) = −(a − bz −1 + cz −2 ) = − a − bz −1 − cz −2 . 14.19 The 4 filters satisfy the aliasing cancellation condition of Eq. (14.21) and the perfect reconstruction condition of Eq. (14.27). Interchanging H 0 ( z) and G 0 ( z ),

and interchanging H1 ( z) and G1 ( z) in Eq. (14.21) we arrive at the new aliasing term A( z ) = G0 (− z)H 0 ( z) + G1 (− z)H1 ( z). Hence, A(− z ) = G0 ( z )H 0 (− z) + G1 ( z)H1 (− z ) = 0, implying that the aliasing condition is still satisfied after interchanging the analysis and synthesis filters. It also follows from Eq. (14.27), an interchange of the analysis and synthesis filters does not change the expression for the distortion transfer function and as a result the perfect reconstruction condition is still satisfied. 14.20 From Eq, (14.73), it can be seen that the normalized product filter P ( z ) is of odd length 2 L + 1. Let the degrees of H 0 ( z ) and G0 ( z) be N and K , respectively. Hence, it follows from Eq. (14.76), N + K = 2 L. Therefore, either N and K are both even or both odd. As a result, H 0 ( z ) and G0 ( z) cannot be of even and odd lengths, respectively.

Moreover, P (z ) is a symmetric polynomial. If H 0 ( z ) is a symmetric polynomial and G0 ( z) is an antisymmetric polynomial, their product cannot be a symmetric polynomial. 14.21 P ( z ) = (1 + z −1 ) 3 (1 + z ) 3 ( az 2 + bz + c + bz −1 + az −2 ) = az 5 + (6 a + b) z 4 + (15a + 6b + c ) z 3 + (20 a + 16b + 6c) z 2 + (16 a + 26 b + 15c) z + (12 a + 30b + 20c ) + (16 a + 26 b + 15c ) z −1 + (20 a + 16 b + 6c ) z −2 + (15a + 6b + c ) z −3 + (6 a + b) z −4 + az −5 .

Since the even powers of P ( z ) must be zeros and the coefficient of z 0 be equal to 1, we must have 6 a + b = 0, 20 a + 16 b + 6c = 0, and 12a + 30b + 20c = 1. Solving these 3 equations we arrive at a = 0.01171875, b = −0.0703125, c = 0.1484375 . Hence, P ( z ) = (1 + z −1 ) 3 (1 + z ) 3 (0.1171875 z 2 − 0.0703125 z + 0.1484375 − 0.0703125 z −1 + 0.1171875 z 2 ) = 0.1171875 z 5 (1 + z −1 ) 6 (1 − 5.4255 z −1 + 9.4438 z −2 ) × (1 − 0.5745 z −1 + 0.1059 z −2 ) .

One possible factorization of P ( z ) is given by

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H 0 ( z) = 0.01171875(1 + z −1 )(1 − 5.4255z −1 + 9.4438z −2 )(1 − 0.5745z −1 + 0.1059z −2 ), and G0 ( z) = (1 + z −1 ) 5 . Thus, the highpass analysis filter is given by H1 ( z) = G0 (− z) = (1 − z −1 ) 5 . 14.22 Without any loss of generality, let N = 4. Let H 0( z ) = h[0] + h[1] z −1 + h[2] z −2 + h[3] z −3 + h[ 4] z −4 . Then,

H 1( z ) = z −4 H 0(− z −1 ) = z −4 (h[0] − h[1]z + h[2]z 2 − h[3]z 3 + h[4]z 4 )

= h[4] − h[3]z −1 + h[2]z −2 − h[1]z −3 + h[0]z −4 . A realization of the analysis filter bank using N + 1 = 5 multipliers and 2 N = 8 two-input adders is shown on top of the next page: z–1

z–1 h[4]

h[3]

z–1

z–1

h[2]

h[1]

y 0[n] h[0]

x[n] _1

_1

z–1

y 1[n]

z–1

z–1

z–1

14.23 For an orthogonal filter bank, the two analysis filters H 0 ( z ) =

and H 1 ( z ) =

∑

N n=0

∑= N

n 0

h0 [n]z − n

h1[ n] z − n are causal FIR filters satisfy the power

complementary property, i.e., H 0 ( z ) H 0 ( z −1 ) + H 1 ( z ) H 1 ( z −1 ) = γ , where γ > 0. We assume h0 [0] ≠ 0 and h1[0] ≠ 0. It follows from Section 7.3 that a linearphase FIR filter has either a symmetric or an anti-symmetric impulse response. As a result, if H 0 (z ) and H 1 (z ) are linear-phase transfer functions, then they are of the form H 0 ( z ) = e − jβ 0 z − N H 0 ( z −1 ) and H 1 ( z ) = e − jβ 0 z − N H 1 ( z −1 ). Substituting the expressions for H 0 ( z −1 ) and H1 ( z −1 ) in the powercomplementary condition we arrive at e jβ 0 z N H 02 ( z ) + e jβ1 z N H 12 ( z ) = γ, which can be rewritten as

[e jβ 0 / 2 H 0 ( z ) + je jβ1 / 2 H 1 ( z )][e jβ 0 / 2 H 0 ( z ) − je jβ1 / 2 H 1 ( z )] = γ z − N . Both quantities inside the square brackets are causal FIR transfer functions. Hence, we have e jβ 0 / 2 H 0 ( z ) + je jβ1 / 2 H 1 ( z ) = α 0 z − N 0 ,

(G)

e jβ 0 / 2 H 0 ( z ) − je jβ1 / 2 H 1 ( z ) = α1 z − N 1 ,

(H)

and

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where α 0 α1 = γ and N 0 + N1 = N . Adding Eq. (H) from Eq. (G) we get H 0 ( z ) = e − jβ 0 / 2 (α 0 z − N 0 + α1 z − N 1 ) and H 1 ( z ) = − je − jβ1 / 2 (α 0 z − N 0 − α1 z − N 1 ). Hence, as the analysis filters have transfer functions that are weighted sum of two delays, they can not be used to design filters with any practical frequency response specifications.

14.24 From Eq. (14.90) we have H 0 ( z) = − z −3 H1 (− z −1 ) = − z −3 (a − bz + cz 2 − dz 3 ) = d − cz −1 + bz −2 − az −3 . Next, from Eq. (14.92) we get G0 ( z ) = z −3 H 0 ( z −1 ) = z −3 (d − cz + bz 2 − az 3 ) = − a + bz −1 − cz −2 + dz −3 and G1 ( z ) = z −3 H1 ( z −1 ) = z −3 ( a + bz + cz 2 + dz 3 ) = d + cz −1 + bz −2 + az −3 .

14.25 Substituting the transfer functions in Eq. (14.21) we get H 0 (− z )G0 ( z) + H1 (− z)G1 ( z) = (3 − 4 z −1 )(−0.5 + z −1 ) + (1 − 2 z −1 )(1.5 − 2 z −1 ) = ( −1.5 + 5z −1 − 4 z −2 ) + (1.5 − 5z −1 + 4 z −2 ) = 0 implying that the aliasing cancellation condition is satisfied. Next, substituting the transfer functions in Eq. (14.27) we get H 0 ( z)G0 ( z ) + H1 ( z)G1 ( z) = (3 + 4 z −1 )(−0.5 + z −1 ) + (1 + 2 z −1 )(1.5 − 2 z −1 ) = (−1.5 + z −1 + 4 z −2 ) + (1.5 + z −1 − 4 z −2 ) = 2 z −1 implying that the perfect reconstruction condition is satisfied.

14.26 H 0 ( z ) = a + bz −1 + cz −2 + dz −3 + ez −4 + fz −5 . From Eq. (14.90) we get H 1 ( z ) = z −5 H 0 ( − z −1 ) = z −5 ( a − bz + cz 2 − dz 3 + ez 4 − fz 5 )

= − f + ez −1 − dz −2 + cz −3 − bz −4 + az −5 . From Eq. (14.92), we arrive at G0 ( z ) = z −5 H 0 ( z −1 ) = f + ez −1 + dz −2 + cz −3 + bz −4 + az −5 and

G1 ( z ) = z −5 H 1 ( z −1 ) = a − bz −1 + cz −2 − dz −3 + ez −4 − fz −5 . 14.27 To develop the realization of the synthesis filter bank we redraw Figure P14.4 as shown below: 2 _

_ z 1

k1 2

_1

_k

k1 _ z 1

k3

_

3

_ z 1

kN

kN

The i-th stage of the above analysis filter bank is of the form shown below:

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(i)

V0 (z)

(i)

X 0 (z)

(i)

_

(i) X 1 (z)

_1

z

Y0 (z) ki

ki

(i)

Y1 (z)

(i) V1 (z)

The input-output relation of the lattice part of the above figure is given by (i ) ⎡Y (i ) ( z )⎤ ⎡ 1 k i ⎤ ⎡V0 ( z )⎤ 0 ⎥ which can be solved for the input variables ⎢ ⎥= ⎢ ⎢⎣Y1(i ) ( z )⎥⎦ ⎢⎣− k i 1 ⎥⎦ ⎢⎣V1(i ) ( z )⎥⎦ (i ) (i ) −1 ⎡U (i ) ( z )⎤ ⎡ 1 1 ⎡ 1 − k i ⎤ ⎡Y0 ( z )⎤ k i ⎤ ⎡Y0 ( z )⎤ ⎥, a ⎢ ⎥ ⎢ leading to ⎢ 0(i ) ⎥ = ⎢ = 1 ⎥⎦ ⎢Y (i ) ( z )⎥ ⎢⎣U 1 ( z )⎥⎦ ⎣− k i 1 ⎥⎦ ⎢⎣Y1(i ) ( z )⎥⎦ 1 + k i2 ⎢⎣k i ⎦ ⎣ 1 realization of which is indicated below: (i)

(i)

Y0 (z) si (i)

_s i

Y1 (z)

U0 (z) s i = k i /(1+k i2) (i)

U1 (z)

where si = k i /(1 + k i2 ). A cascade of the above two lattice structures thus has an input-output relation given by (i ) (i ) ⎡U (i ) ( z )⎤ 1 ⎡ 1 − ki ⎤ ⎡ 1 k i ⎤ ⎡V0 ( z )⎤ ⎡1 0 ⎤ ⎡V0 ( z )⎤ 0 ⎥. ⎢ ⎥= ⎢ ⎥= ⎢ 1 ⎥⎦ ⎢⎣− k i 1 ⎥⎦ ⎢V (i ) ( z )⎥ ⎢⎣0 1 ⎥⎦ ⎢V (i ) ( z )⎥ ⎢⎣U 1(i ) ( z )⎥⎦ 1 + k i2 ⎢⎣k i ⎦ ⎣ 1 ⎦ ⎣ 1 Likewise, the input-output relation of the delay part of the i-th stage of the analysis ⎡V (i ) ( z )⎤ ⎡1 0 ⎤ ⎡ X (i ) ( z )⎤ 0 ⎥ which can be solved for the filter bank is given by ⎢ 0(i ) ⎥ = ⎢ −1 ⎥ ⎢ ( i ) 0 z ⎣ ⎦ ⎢⎣ X 1 ( z )⎥⎦ ⎢⎣V1 ( z )⎥⎦ ⎡ (i ) ⎤ ⎡W (i ) ( z )⎤ ⎡ −1 0 ⎤ ⎢V0 ( z )⎥, a realization of which is input variables leading to ⎢ 0(i ) ⎥ = ⎢ z ⎢⎣W1 ( z )⎥⎦ ⎣ 0 1 ⎥⎦ ⎢⎣V1(i ) ( z )⎥⎦ indicated below: (i)

V0 (z)

_ z 1

(i)

V1 (z)

(i)

W0 (z)

(i)

W1 (z)

Hence, the cascaded-lattice realization of the synthesis filter bank of the twochannel orthogonal filter bank is as shown below where si = k i /(1 + k i2 ). :

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_ z 2

_ z 2

2

2

sN

s3

s1

_s N

_s

_s

3

_ z 1 1 _1

14.28 Figure P14.4 with the down-samplers removed and internal variables labeled is shown below: (N _ 2)

Y0

X(z) _

_ z 1

_k

k1

k1

(N)

Y 0 (z) _

3

k3

_2

z

_1

(z)

(N _ 2) Y1 (z)

kN

kN

_2

(N)

Y 1 (z)

z

(r ) (r ) (r ) (r ) Denoting the transfer functions H 0 ( z ) = Y0 ( z ) / X ( z ) and H 1 ( z ) = Y1 ( z ) / X ( z ), we arrive at an equivalent representation of the above analysis filter bank as indicated below: X(z)

(N _ 2)

H0

(N)

H 0 (z)

(z) _

(N _ 2) H1 (z)

_ z 2

kN

kN

(N)

H 1 (z)

(N ) ( N − 2) ( z ) + k N z − 2 H1( N − 2) ( z ) and Analysis yields H 0 ( z ) = H 0

H1( N ) ( z ) = −k N H 0( N − 2) ( z ) + z − 2 H1( N − 2) ( z ). Solving these two equations, we get 2 (1 + k N ) H 0( N − 2) ( z ) = H 0( N ) ( z ) − k N H1( N ) ( z ) and 2 − 2 ( N − 2) (1 + k N ) z H1 ( z ) = k N H 0( N ) ( z ) + H 1( N ) ( z ). Choose k N such that the highest power of

z −1 in H 0( N ) ( z ) − k N H1( N ) ( z ) is eliminated. By construction, the next highest power of z −1 ( N − 2) ( z ) to N − 2. Moreover, the coefficients of z 0 also is removed reducing the order of H 0

and z −1 in k N H 0( N ) ( z ) + H1( N ) ( z ) are also zero for the above choice of k N , resulting in a ( N − 2)

( z ) of order N − 2. Continuing this process, we arrive at a cascaded lattice causal H1 realization of the analysis filter bank of an orthogonal QMF bank. (3) From Eq. (14.87) we have H 0 ( z ) = 0.3415 + 0.5915z −1 + 0.1585z − 2 − 0.0915z − 3 and (3) from Eq. (14.89a) we have H1 ( z ) = −0.0915 − 0.1585z −1 + 0.5915z − 2 − 0.3415z − 3 (1) (3) (3) We now form (1 + k 32 ) H 0 ( z ) = H 0 ( z ) − k 3 H1 ( z ) = (0.3415 + 0.0915k 3 )

+ (0.5915 + 0.1585 k3 ) z −1 + (0.1585 − 0.5915 k 3 ) z −2 + ( −0.0915 + 0.3415k 3 ) z −3 .

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(1) (3) (3) Choose k 3 = 0.0915 / 0.3415 = 0.2679. Then 1.0718H 0 ( z ) = H 0 ( z ) − k 3 H1 ( z ) (1) = 0.366 + 0.634 z −1 , implying, H 0 ( z ) = 0.3415 + 0.5915z −1 = 0.3415(1 + 1.7321z −1 ). (1) (3) (3) We next form (1 + k 32 ) z − 2 H1 ( z ) = [k 3 H 0 ( z ) + H1 ( z )] or,

1.0718 z − 2 H 1(1) ( z ) = [0.2679H 0(3) ( z ) + H1(3) ( z )] = 0.634 − 0.366 z −1 , implying, H1(1) ( z ) = 0.5915 − 0.3415z −1 = 0.3415(1.7321 − z −1 ). Hence, k1 = 1.7321. A cascaded lattice realization of the analysis filter bank is as shown below: 2 _ z 1

_1.7321

_ 0.2679

1.7321

0.2679

_ z 2

_1

2

From the solution of Problem 14.27 we arrive at the structure of the synthesis filter bank shown on the next page to ensure the realization of the orthogonal perfect reconstruction QMF bank. _ z 2

2

0.25

0.433

_

_

0.25

2

_ z 1

0.433 _1

14.29 (a) H 0 ( z ) = 1 + 3 z −1 + 14 z −2 + 22 z −3 − 12 z −4 + 4 z −5 . Hence, H 1 ( z ) = − z −5 H 0 ( − z −1 ) = 4 + 12 z −1 + 22 z −2 − 14 z −3 + 3 z −4 − z −5 . (b) To develop the cascade lattice realization of the analysis filter bank, we rewrite (5) (5) the analysis transfer functions as H 0 ( z ) = H 0 ( z ) and H1 ( z ) = H1 ( z ). Next, (3) (3) (5) (5) we determine H 0 ( z ) using the relation (1 + k 52 ) H 0 ( z ) = H 0 ( z ) − k 5 H1 ( z )

and choose k5 so that the coefficient of z −5 is eliminated. Now,

(1 + k 52 ) H 0(3) ( z ) = (1 − 4k 5 ) + (3 − 12k 5 ) z −1 + (14 − 22k 5 ) z − 2 + (22 + 14k 5 ) z − 3 + (−12 − 3k 5 ) z −4 + (4 + k 5 ) z −5 . We choose k 5 = −4. Then (1 + k 52 ) H 0 ( z ) = (3)

= 17 H 0(3) ( z ) = 17 + 51z −1 + 102 z − 2 − 34 z − 3 . Note that as expected the above choice of k 5 also cancels the coefficient of z −4 . We thus have,

H 0(3) ( z ) = 1 + 3 z −1 + 6 z − 2 − 2 z − 3 . Next we determine H1(3) ( z ) using the relation (1 + k 52 ) z − 2 H1(3) ( z ) = k 5 H 0(5) ( z ) + H1(5) ( z ) which leads to 17 z − 2 H 1(3) ( z ) = −34 z − 2 − 102 z − 3 + 51z − 4 − 17 z − 5 . It should be noted that here,

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as expected, the choice of k 5 has resulted in zero-valued coefficients of z 0 and z −1 . Hence, H1 ( z ) = −2 − 6 z −1 + 3 z − 2 − z − 3 . It can be verified that the same (3)

(3) (3) expression for H1(3) ( z ) is obtained from H 1 ( z ) = − z − 3 H 0 (− z −1 ). (1) Next, we determine H 0 ( z ) using the relation

(1 + k 32 ) H 0(1) ( z ) = H 0(3) ( z ) − k 3 H1(3) ( z ) and choose k 3 = 2 so that the coefficient (1) (1) of z −3 is eliminated. This results in H 0 ( z ) = 1 + 3 z −1 and H1 ( z ) = 3 − z −1. (1) Analysis of the structure of Figure P14.4 yields H 0 ( z ) = 1 + k1 z −1 and

H1(1) ( z ) = k1 − z −1. Hence, k1 = 3. The final realization of the analysis filter bank is shown below: 2 _2

_3

_ z 1

3

4 _

2 _ z 2

_1

_ z 2

4

2

From the solution of Problem 14.27 we arrive at the structure of the synthesis filter bank shown below to ensure the realization of a paraunitary perfect reconstruction QMF bank. _ z 2

_ z 2

2

4/17

_ 2/5

_ 3/10

_ 4/17

2/5

3/10

2

_ z 1

_1

14.30 (a) H 0 ( z ) = 0.5 − z −1 + 10.5 z −2 − 13.5 z −3 − 5 z −4 − 2.5 z −5 . Hence, H 1 ( z ) = − z −5 H 0 ( − z −1 ) = −2.5 + 5 z −1 − 13.5 z −2 − 10.5 z −3 − z −4 − 0.5 z −5 .

(b) To develop the cascade lattice realization of the analysis filter bank, we rewrite (5) (5) the analysis transfer functions as H 0 ( z ) = H 0 ( z ) and H1 ( z ) = H1 ( z ). Next, (3) (3) (5) (5) we determine H 0 ( z ) using the relation (1 + k 52 ) H 0 ( z ) = H 0 ( z ) − k 5 H1 ( z )

and choose k5 so that the coefficient of z −5 is eliminated. Now,

(1 + k 52 ) H 0(3) ( z ) = (0.5 + 2.5k 5 ) + (−1 − 5k 5 ) z −1 + (10.5 + 13.5k 5 ) z − 2 + (−13.5 + 10.5k 5 ) z − 3 + ( −5 + k 5 ) z −4 + ( −2.5 + 0.5k 5 ) z −5 . We choose k 5 = 5. Then (1 + k 52 ) H 0 ( z ) = (3)

= 26 H 0(3) ( z ) = 13 − 26 z −1 + 78 z − 2 + 39 z − 3 . Note that as expected the above choice of k 5 also cancels the coefficient of z −4 . We thus have,

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H 0(3) ( z ) = 0.5 − z −1 + 3 z − 2 + 1.5 z − 3 . Next we determine H1(3) ( z ) using the (3) (5) (5) relation (1 + k 52 ) z − 2 H1 ( z ) = k 5 H 0 ( z ) + H1 ( z ) which leads to

26 z − 2 H1(3) ( z ) = 39 z − 2 − 78 z − 3 − 26 z − 4 − 13 z − 5 . It should be noted that here, as expected, the choice of k 5 has resulted in zero-valued coefficients of z 0 and z −1 . (3) Hence, H1 ( z ) = 1.5 − 3 z −1 − z − 2 − 0.5 z − 3 . It can be verified that the same (3) (3) expression for H1(3) ( z ) is obtained from H 1 ( z ) = − z − 3 H 0 (− z −1 ). Next, we (1) (1) (3) (3) determine H 0 ( z ) using the relation (1 + k 32 ) H 0 ( z ) = H 0 ( z ) − k 3 H1 ( z ) and

choose k 3 = −3 so that the coefficient of z −3 is eliminated. This results in

H 0(1) ( z ) = 0.5 − z −1 = 0.5(1 − 2 z −1 ) and H1(1) ( z ) = −1 − 0.5 z −1 = 0.5(−2 − z −1 ). (1) Analysis of the structure of Figure P14.4 yields H 0 ( z ) = 1 + k1 z −1 and

H1(1) ( z ) = k1 − z −1. Hence, k1 = −2. The final realization of the analysis filter bank is shown on top of next page: 2 3

2

_ z 1

_5

_2 _1

_3

_ z 2

5 _ z 2

2

From the solution of Problem 14.27 we arrive at the structure of the synthesis filter bank shown below to ensure the realization of a paraunitary perfect reconstruction QMF bank. _ z 2

_ z 2

2

_ 2/5

5/26 _

_ 3/10

2/5

5/26

_ z 1

3/10

2

_1

14.31 (a) The input-output relation of the l –th two-pair is given by ⎡ Rl ( z ) ⎤ ⎡ 1 k l ⎤ ⎡ Rl − 1 ( z ) ⎤ ⎢⎣ S l ( z ) ⎥⎦ = ⎢⎣ kl 1 ⎥⎦ ⎢ z − 2 S ( z)⎥ , from which we obtain l −1 ⎦ ⎣

Rl ( z) = Rl −1 ( z ) + kl [ z −2 S l −1 ( z)] and S l ( z) = kl Rl −1 ( z) + [ z −2 S l −1 ( z)]. The realization of the two-pair is thus as shown below: Rl _ 1(z)

R l (z) kl

_ z 2Sl _ 1(z)

kl S l (z)

(b) The portion of Figure P14.5 upto the m-th stage is then as shown below:

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Pm _ 1 (z)

Pm(z) km

Q m _ 1(z)

km

_2

Qm(z)

z

Analysis of the above structure yields Pm ( z ) = Pm −1 ( z ) + k m z −2 Qm −1 ( z ) and Qm ( z ) = k m Pm −1 ( z ) + z −2 Qm −1 ( z ).

(c) Analysis of the structure of Figure P14.5 yields H 0 ( z) = PM ( z ) + Q M ( z ) = PM ( z ) + z − (2 M + 1) PM ( z −1 ). Hence, H 0 ( z −1 ) = PM ( z −1 ) + z (2 M +1) PM ( z ).

Therefore, z −(2 M +1) H 0 ( z −1 ) = z − (2 M +1) PM ( z −1 ) + PM ( z) = H 0 ( z). Since the order of H 0 ( z) is 2 M + 1, H 0 ( z) is a Type 2 linear-phase FIR transfer function. Similarly, analysis of the structure of Figure P14.5 yields H1 ( z) = PM ( z) − Q M ( z)

= PM ( z ) − z − (2 M + 1) PM ( z −1 ). Hence, H1 ( z −1 ) = PM ( z −1 ) − z (2 M + 1) PM ( z ). Therefore, z − (2 M + 1) H1 ( z −1 ) = z − (2 M + 1) PM ( z −1 ) − PM ( z ) = − H1 ( z ). Since the order of H1 ( z) is 2 M + 1, H1 ( z) is a Type 4 linear-phase FIR transfer function.

It follows from Figure P14.5 that ⎡ 1 k M ⎤ ⎡1 0 ⎤ ⎡ 1 k1 ⎤ ⎡1 0 ⎤ ⎡ 1 k 0 ⎤ E( z ) = ⎡1 1 ⎤ ⎢ L ⎢⎣1 − 1⎥⎦ ⎣ k M 1 ⎥⎦ ⎢⎣0 z −1 ⎥⎦ ⎢⎣ k1 1 ⎥⎦ ⎢⎣0 z −1 ⎥⎦ ⎢⎣ k 0 1 ⎥⎦ 1 0 ⎤ = D 2 Γ (z)TM KT1 Γ (z)T0 , where D 2 = ⎡1 1 ⎤ and Γ ( z) = ⎡⎢ −1 . We ⎢⎣1 − 1⎥⎦ ⎣0 z ⎥⎦ determine the synthesis bank for perfect reconstruction by choosing −1 K T −1 Γ −1 ( z )T −1 . The final R ( z ) = αz − M E −1 ( z ) = αz − M D 2−1 Γ −1 ( z)TM 1 0 realization of the synthesis filter bank is shown below which satisfies the perfect reconstruction condition within a scale factor. (c)

_ z 2

_ z 2

2 _

_

2

_

kM

k1

_

kM

k1

_

k0

_

_ z 1

k0

_1

14.32 From Figure P14.5 we observe that H 0 ( z ) = PM ( z ) + QM ( z ) and

H 1 ( z ) = PM ( z ) − QM ( z ). Solving these equations we get 1 2

1 2

PM ( z ) = [ H 0 ( z ) + H 1 ( z )] and QM ( z ) = [ H 0 ( z ) − H 1 ( z )]. Next, from the

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solution of Problem 14.31, Part (b), we have Pm ( z ) = Pm −1 ( z ) + k m z −2 Qm −1 ( z ) and Qm ( z ) = k m Pm −1 ( z ) + z −2 Qm −1 ( z ). Solving these two equations we get 1 1 Pm −1 ( z ) = [ Pm ( z ) − k m Qm ( z )] and Qm −1 ( z ) = [Qm ( z ) − k m Pm ( z )]. 2 2 −2 1 − km (1 − k m )z To synthesize the analysis filter bank, choose k m so that the coefficient of the highest power of Pm −1 (z ) in z −1 is equal to 0. It can be shown that this choice of also guarantees that the coefficient of the second highest power of is also 0. Hence, the order of Pm −1 (z ) is 2 less that of Pm (z ). (a) Here M = 2. Hence, P2 ( z ) = [ H 0 ( z ) + H 1 ( z )] = 1 + 2 z −1 + 12 z −2 + 15 z −3 2 1

+ 4 z −4 + z −5 and Q2 ( z ) = [ H 0 ( z ) − H1 ( z )] = 2 + 4 z −1 + 15z −2 + 12z −3 + 2 z −4 + z −5 . 1 2

We form P2 ( z ) − k 2 Q2 ( z ) = (1 − 2 k 2 ) + (2 − 4 k 2 ) z −1 + (12 − 15k 2 ) z −2 + (15 − 12 k 2 ) z −3 + ( 4 − 2 k 2 ) z −4 + (2 − k 2 ) z −5 . Choose k2 = 2. Then

P2 ( z ) − k 2 Q2 ( z ) = −3 − 6 z −1 − 18 z −2 − 9 z −3 . Thus, 1 P1 ( z ) = [ P2 ( z ) − k 2 Q2 ( z )] = 1 + 2 z −1 + 6 z −2 + 3z −3 . Likewise, 2 1 −k 2

Q1 ( z ) =

1 (1 − k 22 ) z − 2

[Q2 ( z ) − k 2 P2 ( z )] = 3 + 6 z −1 + 2 z − 2 + z − 3 . Next, we form

P1 ( z ) − k1Q1 ( z ) = (1 − 3k1 ) + (2 − 6 k1 ) z −1 + (6 − 2 k1 ) z −2 + (3 − k1 ) z −3 . Choose k1 = 3. 1 Then, P0 ( z ) = [ P1 ( z ) − k1Q1 ( z )] = 1 + 2 z −1 . Likewise, 2 1 −k 1

Q0 ( z ) =

1 (1 − k 21) z − 2

[Q1 ( z ) − k1 P1 ( z )] = 2 + z −1 . From Figure P14.5 we have

P0 ( z ) = 1 + k0 z −1 and Q0 ( z) = k0 + z −1 . Hence, it follows that k0 = 2. As a result, the analysis filter bank is as shown below: P 0(z)

X(z) _ z 1

3

2 _ z 2

Q 0 (z)

H0 (z)

2

2

3

2

P 2(z)

P 1(z)

2 _ z 2

Q 1 (z)

Q 2 (z)

_1

H1 (z)

2

The corresponding synthesis filter bank for a perfect reconstruction filter bank is as shown below:

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_ _

2

_ z 2

_ z 2

2 2

_3

_

2

2

_3

_

2

_ z 1

_1

(b) Here M = 2. Hence, P2 ( z ) = [ H 0 ( z ) + H 1 ( z )] = 1 − 2.5z −1 − 13.5z −2 + 18z −3 2 1

+ 5z −4 − 2 z −5 and Q2 ( z ) = [ H 0 ( z ) − H1 ( z )] = −2 + 5z −1 + 18z −2 − 13.5z −3 − 2.5z −4 + z −5 . We form 2 1

P2 ( z ) − k 2 Q2 ( z ) = (1 + 2 k 2 ) − (2.5 + 5k 2 ) z −1 − (13.5 + 18k 2 ) z −2 + (18 + 13.5k 2 ) z −3 1 + (5 + 2.5k 2 ) z −4 − (2 + k 2 ) z −5 . Choose k 2 = −2. Then P1 ( z ) = [ P2 ( z ) − k 2 Q2 ( z )] 1 − k 22

= 1 − 2.5z −1 − 7.5z −2 + 3z −3 . Likewise, 1 Q1 ( z ) = [Q2 ( z ) − k 2 P2 ( z )] = 3 − 7.5z −1 − 2.5z − 2 + z − 3 . Next, we form (1 − k 22 ) z − 2 P1 ( z ) − k1Q1 ( z ) = (1 − 3k1 ) − (2.5 − 7.5k1 ) z −1 − (7.5 − 2.5k1 ) z −2 + (3 − k1 ) z −3 . Choose 1 k1 = 3. Then, P0 ( z ) = [ P1 ( z ) − k1Q1 ( z )] = 1 − 2.5z −1 . Likewise, 1 − k 21

Q0 ( z ) =

1 (1 − k 21) z − 2

[Q1 ( z ) − k1 P1 ( z )] = −2.5 + z −1 . Hence, k 0 = −2.5. As a result, the

analysis filter bank is as shown below: P 0(z)

X(z)

P 1(z)

_ 2.5

_ z 1

P 2(z)

_2

3

_ z 2

_ z 2

Q 0 (z)

2

_2

3

_ 2.5

H0 (z)

Q 2 (z)

Q 1 (z)

_1

H1 (z)

2


2

_ z 2

_ z 2

2 2

_3

2.5

2

_3

2.5

_ z 1

_1

(c) Here M = 2. Hence, P2 ( z ) = [ H 0 ( z ) + H 1 ( z )] = 1 − 2.5z −1 + 0.75z −2 + 6 z −3 2 1

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− 5z −4 + 2 z −5 and Q2 ( z ) = [ H 0 ( z ) − H1 ( z )] 1 2

= 2 − 5z −1 + 6 z −2 + 0.75z −3 − 2.5z −4 + z −5 . We form P2 ( z ) − k 2 Q2 ( z ) = (1 − 2 k 2 ) − (2.5 − 5k 2 ) z −1 − (0.75 − 6 k 2 ) z −2 + (6 − 0.75k 2 ) z −3 1 − (5 − 2.5k 2 ) z −4 + (2 − k 2 ) z −5 . Choose k2 = 2. Then P1 ( z ) = [ P2 ( z ) − k 2 Q2 ( z )] 1 − k 22 = 1 − 2.5z −1 + 3.75z −2 − 1.5z −3 . Likewise, 1 Q1 ( z ) = [Q2 ( z ) − k 2 P2 ( z )] = −1.5 + 3.75z −1 − 2.5z − 2 + z − 3 . Next, we form 2 2 − (1 − k 2 ) z P1 ( z ) − k1Q1 ( z ) = (1 + 1.5k1 ) − (2.5 − 3.75k1 ) z −1 + (3.75 + 2.5k1 ) z −2 − (1.5 + k1 ) z −3 . 1 Choose k1 = −1.5. Then, P0 ( z ) = [ P1 ( z ) − k1Q1 ( z )] = 1 − 2.5z −1 . Likewise, 2 1 −k 1

Q0 ( z ) =

1 (1 − k 21) z − 2

[Q1 ( z ) − k1 P1 ( z )] = −2.5 + z −1 . Hence, k 0 = −2.5. As a result, the

analysis filter bank is as shown below: P 0(z)

X(z) _ z 1

P 2(z)

P 1(z)

_ 2.5

_ 1.5

_ 2.5

_ 1.5 _ z 2

2

2 2 _ z 2

Q 0 (z)

H0 (z)

Q 2 (z)

Q 1 (z)

_1

H1 (z)

2


2

_ z 2

_ z 2

2 _2

1.5

2.5

_2

1.5

2.5

_ z 1

_1

14.33 To show the system of Figure 14.18 is, in general, periodic with a period L, we need to show that if Xˆ 1 ( z ) is the output for an input X1 ( z), and Xˆ 2 ( z ) is the output for an input

X 2 ( z), then if X 2 ( z ) = z − L X1 ( z ), i.e., x 2 [n] = x1[n − L ], then the corresponding output satisfies Xˆ 2 ( z ) = z − L Xˆ 1 ( z ), i.e., xˆ 2 [n] = xˆ1[n − L ]. Now,

⎡ X (z) ⎤ ⎢ X ( zW ) ⎥ Xˆ ( z ) = g T ( z)H ( m ) ( z) ⎢ ⎥ , where g (z ) is the vector composed of the synthesis M ⎢ X ( zW L −1 )⎥ ⎣ ⎦ bank filters given by Eq. (14.98) and H ( m ) ( z ) is the analysis filter bank modulation

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⎡ X 2 ( z) ⎤ ⎢ X 2 ( zW ) ⎥ matrix given Eq. (14.102). Now, Xˆ 2 ( z) = g T ( z)H ( m ) ( z ) ⎢ ⎥ M ⎢ ⎥ L − 1 )⎦ ⎣ X 2 ( zW ⎡ z − L X1 ( z ) ⎤ ⎡ X1 ( z ) ⎤ ⎢ −L ⎥ ⎢ X1 ( zW ) ⎥ −L = g T ( z )H ( m ) ( z ) ⎢ z X1 ( zW ) ⎥ = z − L g T ( z)H ( m ) ( z ) ⎢ ⎥ = z Xˆ 1 ( z ). As a M M ⎢ ⎥ ⎢ L −1 ) ⎥ ⎢⎣ z − L X1 ( zW L −1 )⎥⎦ ⎣ X1 ( zW ⎦ result, the structure of Figure 14.18 is, in general, a time-varying system with a period of L. 14.34 H 0 ( z) = E 00 ( z 2 ) + z −1E 01 ( z 2 ), H1 ( z) = E10 ( z 2 ) + z −1E11 ( z 2 ). Thus, ⎡ E ( z ) E 01 ( z )⎤ E( z ) = ⎢ 00 . ⎣ E10 ( z ) E11 ( z ) ⎥⎦ (a) H 0 ( z) = E 0 ( z 2 ) + z −1E1 ( z 2 ). Hence, H1 ( z ) = H 0 (− z ) = E 0 ( z 2 ) − z −1E1 ( z 2 ).

Thus, E 00 ( z ) = E 0 ( z ), E 01 ( z ) = E1 ( z), E10 ( z) = E 0 ( z ), and E11 ( z) = − E1 ( z). ⎡ E ( z ) E1 ( z ) ⎤ Therefore, E( z ) = ⎢ 0 . ⎣ E 0 ( z ) − E1 ( z )⎥⎦ (b) Two cases need to be considered – N odd and N even. Consider first the N odd case. For simplicity let H 0 ( z) = a 0 + a1z −1 + a2 z −2 + a3 z −3 . Then,

H1 ( z) = z −3 H 0 ( z −1 ) = z −3 (a 0 + a1z + a2 z 2 + a3 z 3 ) = a3 + a 2 z −1 + a1z −2 + a0 z −3 . Thus, E 00 ( z) = E 0 ( z) = a0 + a 2 z −1 , E 01 ( z ) = E1 ( z) = a1 + a3 z −1 , and E10 ( z ) = a3 + a1z −1 , E11 ( z) = a2 + a0 z −1 ⇒ E10 ( z) = z −1E1 ( z −1 ) and E11 ( z) = z −1E 0 ( z −1 ). Next, consider the N even case. Assume H 0 ( z ) = a0 + a1z −1 + a2 z −2 + a3 z −3 + a 4 z −4 . Then, H1 ( z) = z −4 H 0 ( z −1 ) = z −4 (a0 + a1z + a 2 z 2 + a3 z 3 + a 4 z 4 )

= a 4 + a3 z −1 + a2 z −2 + a1z −3 + a 0 z −4 . Thus, E 00 ( z) = E 0 ( z) = a0 + a2 z −1 + a 4 z −2 , E 01 ( z) = E1 ( z) = a1 + a3 z −1 , and

E10 ( z ) = a 4 + a2 z −1 + a 0 z −2 , E11 ( z) = a3 + a1z −1 ⇒ E10 ( z ) = z −2 E 0 ( z −1 ) and E11 ( z ) = z −1 E1 ( z −1 ). In the general case, for N odd with N = 2 K + 1, E11 ( z) = z − K E 0 ( z −1 ) and E (z) E1 ( z ) ⎡ ⎤ E10 ( z ) = z − K E1 ( z −1 ) ⇒ E( z) = ⎢ − ( N −1) 0/ 2 . − 1 − ( N − 1 ) / 2 − 1 E1 ( z ) z E 0 ( z )⎥⎦ ⎣z

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Likewise, for N even with N = 2 K , E10 ( z) = z − K E 0 ( z −1 ) and E ( z) E1 ( z) ⎡ ⎤ . E11 ( z ) = z − ( K −1) E1 ( z −1 ) ⇒ E( z ) = ⎢ − N / 20 − 1 − ( N − 2 ) / 2 − 1 E 0 (z ) z E1 ( z )⎥⎦ ⎣z 14.35 (a) H 0 ( z ) = [A0 ( z 2 ) + z −1A1 ( z 2 )] and H1 ( z) = [A0 ( z 2 ) − z −1A1 ( z 2 )]. 4 4 ⎡A ( z ) A1 ( z ) ⎤ ⎡ A ( z ) A1 ( z ) ⎤ Hence, E(z ) = ⎢ 0 , and R (z ) = ⎢ 1 . ⎥ − A z A z ( ) ( ) 1 ⎣A0 ( z ) − A0 ( z )⎥⎦ ⎣ 0 ⎦ (b) To prove E(z ) is lossless, we need to show E† (e jω )E(e jω ) = c 2 I for some 1

1

constant c: E† (e jω )E(e jω ) = ⎡ A0* (e jω ) A0* (e jω )⎤ ⎡A0 (e jω ) A1 (e jω ) ⎤ ⎢ jω jω ⎥ ⎢ jω jω ⎥ ⎢⎣ A 1 (e ) − A1 (e ) ⎥⎦ ⎣A0 (e ) − A1 (e )⎦ ⎤ ⎡ jω ) 2 + A (e jω ) 2 A ( e 0 0 0 ⎥ ⎢ =⎢ ⎥ = 2I. Hence, E(z) is 2 2 j ω j ω 0 A1 (e ) + A1 (e ) ⎥ ⎢⎣ ⎦ lossless. (c) R ( z )E( z ) =

R ( z) =

1 ⎡ 2A 1( z )A 0 ( z ) 0 4 ⎢⎣

0 ⎤ A 1( z )A 0 ( z ) = I. Therefore, 2A 1( z )A 0 ( z )⎥⎦ 2

A1 ( z)A0 ( z) −1 E ( z). 2

(d) As in part (c), R ( z)E( z ) =

A1 ( z )A0 ( z) I. 2

⎡ E ( z) E 01 ( z)⎤ ⎡ − 2 z − 2 + 8z −1 − 6 2 − z −1 ⎤ . Hence, =⎢ 14.36 E( z ) = ⎢ 00 − 3 + 2 z −1 1 ⎥⎦ ⎣ E10 ( z) E11 ( z) ⎥⎦ ⎣

H 0 ( z) = E 00 ( z 2 ) + z −1E 01 ( z 2 ) = −6 + 2 z −1 + 8z −2 − z −3 − 2 z −4 and H1 ( z) = E10 ( z 2 ) + z −1E11 ( z 2 ) = −3 + z −1 + 2 z −2 . Now, det(E(z)) = ( −2 z −2 + 8 z −1 − 6) − (2 − z −1 )( −3 + 2 z −1 ) = z −1 . From Eq. (14.133) we have −K ⎡ ⎡ R ( z ) R01 ( z )⎤ − 2 z − 2 + 8 z −1 − 6 − 2 + z −1 ⎤ − K E −1 ( z ) = cz cz R ( z ) = ⎢ 00 = ⎥ ⎢ 3 − 2 z −1 1 ⎣ R10 ( z ) R11 ( z ) ⎥⎦ z −1 ⎣ ⎦ −2 −1 −1 ⎤ ⎡ = ⎢ − 2 z + 8z−1 − 6 − 2 + z ⎥ for K = 1 and c = 1. Therefore, 3 − 2z 1 ⎦ ⎣

G0 ( z) = R00 ( z 2 ) + z −1 R01 ( z 2 ) = −6 − 2 z −1 + 8z −2 + z −3 − 2 z −4 and G1 ( z) = R10 ( z 2 ) + z −1 R11 ( z 2 ) = 3 + z −1 − 2 z −2 .

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⎡2 − 2 1 ⎤ 14.37 P = ⎢3 1 − 2⎥. From Eqs. (14.121) and (14.122a) to (14.122c), we thus get ⎢⎣2 4 2 ⎥⎦ ⎡ H 0 ( z )⎤ ⎡2 − 2 1 ⎤ ⎡ 1 ⎤ ⎢ H1 ( z ) ⎥ = ⎢ 3 1 − 2 ⎥ ⎢ z −1 ⎥ , resulting in the analysis filters ⎢ H ( z )⎥ ⎢⎣2 4 2 ⎥⎦ ⎢⎣ z − 2 ⎥⎦ ⎣ 2 ⎦ H 0 ( z ) = 2 − 2 z −1 + z −2 , H1 ( z) = 3 + z −1 − 2 z −2 , H 2 ( z) = 2 + 4 z −1 + 2 z −2 . 0.64 0.24⎤ ⎡ 0.8 Now, 4P −1 = ⎢ − 0.8 0.16 0.56⎥. Using Eqs. (14.124) and (14.125a) to (14.125c), ⎢⎣ 0.8 − 0.96 0.64⎥⎦ 0.64 0.24⎤ ⎡ 0.8 we arrive at [G0 ( z ) G1 ( z ) G2 ( z)] = z − 2 z −1 1 ⎢ − 0.8 0.16 0.56⎥ , leading to ⎢⎣ 0.8 − 0.96 0.64⎥⎦ the synthesis filters G0 ( z ) = 0.8 − 0.8z −1 + 0.8z −2 , G1 ( z) = −0.96 + 0.16 z −1 + 0.64 z −2 ,

[

]

G 2 ( z ) = 0.64 + 0.56 z −1 + 0.24 z −2 .

⎡1 − 2 3 − 1⎤ 1 0 ⎥. For y[ n] = 4 x[ n − 3] to hold, we require 14.38 R ( z) = ⎢2 1 ⎢0 3 − 1 1 ⎥ ⎢⎣1 1 − 1 2 ⎥⎦ R ( z )E( z ) = 4I. Hence,

⎡ − 1.4118 2.5882 0.2353 = ⎢ 0.2353 ⎢ 2.5882 − 1.4118 ⎢⎣ 1.8824 − 2.1176

14.39

⎡1 − 2 3 − 1⎤ 1 0⎥ E( z ) = 4[R ( z )] −1 = 4 ⎢2 1 ⎢0 3 − 1 1 ⎥ ⎢⎣1 1 − 1 2 ⎥⎦ − 1.8824 0.2353 ⎤ 1.6471 − 0.7059⎥. 2.1176 0.2353 ⎥ 1.1765 2.3529 ⎥⎦

−1

⎡1 2 2 ⎤ z −1 1 ⎢3 − 3 1⎥. Therefore, ⎢⎣2 1 2⎥⎦ H 0 ( z ) = 2 + 3z −1 + z −2 , H1 ( z) = 1 − 3z −1 + 2 z −2 , H 2 ( z) = 2 + z −1 + 2 z −2 . The

[H 0 (z)

[

H1 ( z) H 2 ( z)] = z − 2

]

⎡2 3 1 ⎤ corresponding Type 1 polyphase component matrix is thus given by E( z) = ⎢1 − 3 2⎥. ⎢⎣2 1 2⎥⎦ For perfect reconstruction we require the Type 2 polyphase component matrix to be given 5 / 3 − 3⎤ ⎡ 8/3 by R ( z ) = [E( z)]−1 = ⎢ − 2 / 3 − 2 / 3 1 ⎥. Hence, the synthesis filters are obtained from ⎢⎣ − 8 / 3 4 / 3 3 ⎥⎦

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[G 0 (z)

5 / 3 − 3⎤ ⎡ 8/3 z −1 1 ⎢ − 2 / 3 − 2 / 3 1 ⎥ resulting in ⎢⎣ − 8 / 3 4 / 3 3 ⎥⎦

[

]

G1 ( z ) G 2 ( z )] = z − 2

G 0 ( z) = − − z −1 + z − 2 , G1 ( z ) = 8 3

2 3

8 3

4 3

− z −1 + z − 2 , and G 2 ( z ) = 3 + z −1 + 3z −2 . 3 3 2

5

1 1 2 ⎤ 1 − 1 − 2⎥. For perfect reconstruction we require 1 2 − 1⎥ 3 1 2 ⎥⎦ 0.5 0 − 0.5 ⎤ ⎡ 1 1 1 0 25 0 1.25 ⎥. Thus, . − ⎢ R ( z ) = E −1 ( z ) = ⎢ 0 4 0 95 1 2 0.05 ⎥ . . . − − 3 ⎢⎣ 1.2 − 0.15 − 0.6 − 0.15⎥⎦

⎡2 14.40 E( z) = ⎢ 3 ⎢3 ⎢⎣1

[G 0 (z)

G 1 ( z ) G 2 ( z ) G 3 ( z )] =

1 3

[

z−3

z−2

z −1

0.5 0 − 0.5 ⎤ ⎡ 1 1 0 25 0 1.25 ⎥. . − ⎢ 1 ⎢ − 0.4 − 0.95 1.2 0.05 ⎥ ⎢⎣ 1.2 − 0.15 − 0.6 − 0.15⎥⎦

]

Hence, G0 ( z) = (1.2 − 0.4 z −1 − z − 2 + z − 3 ), G1 ( z ) = (−0.15 − 0.95z −1 + 0.25z − 2 + 0.5z − 3 ), G2 ( z )

1 3 1 = (−0.6 + 1.2 z −1 ), G3 ( z ) 3

1 3

= (−0.15 + 0.05z −1 + 1.25z − 2 − 0.5z − 3 ). 1 3

14.41 G0 ( z) = −2 + 3z −1 + 2 z −2 , G0 ( z ) = 3 + z −1 − 2 z −2 , G0 ( z ) = 2 − z −1 + z −2 . Rewriting these equations in an matrix form we get ⎡ 2 −2 2 ⎤ [G 0 (z) G1(z) G 2 (z)] = z − 2 z −1 1 ⎢ 3 1 − 1⎥. This implies, ⎢⎣ − 2 3 1 ⎥⎦ ⎡ 2 −2 2 ⎤ R ( z) = ⎢ 3 1 − 1⎥. For perfect reconstruction we require 1 ⎦⎥ ⎣⎢ − 2 3

[

]

0.25 0 ⎤ ⎡ 0.125 E( z ) = R −1 ( z) = ⎢ − 0.0313 0.1875 0.25⎥. Hence, ⎣⎢ 0.3438 − 0.0625 0.25⎥⎦ ⎡ H 0 ( z)⎤ ⎡ 0.125 0.25 0 ⎤⎡ 1 ⎤ ⎢ H1 ( z) ⎥ = ⎢ − 0.0313 0.1875 0.25⎥ ⎢ z −1 ⎥ implying ⎢ H ( z)⎥ ⎢⎣ 0.3438 − 0.0625 0.25⎥⎦ ⎢ z − 2 ⎥ ⎣ 2 ⎦ ⎦ ⎣

H 0 ( z) = 0.125 + 0.25z −1 , H1 ( z ) = −0.0313 + 0.1875z −1 + 0.25z −2 , H1 ( z ) = 0.3438 − 0.0625 z −1 + 0.25z −2 .

14.42 If the filter bank is alias-free, then

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H1 ( z ) H L −1 ( z) ⎤ ⎡ G0 ( z) ⎤ ⎡T ( z )⎤ L ⎡ H 0 ( z) ⎢ H 0 ( zW ) H1 ( zW ) H L −1 ( zW ) ⎥ ⎢ G1 ( z) ⎥ ⎢ 0 ⎥ L ⎥⎢ ⎢ ⎥ = ⎢ M ⎥ The above M M O M M ⎢ ⎥ ⎢ ⎥ L −1 ) H ( zW L −1 ) L H L − 1 ) ⎥ ⎢G 1 L −1 ( zW ⎦ ⎣ L −1 ( z )⎦ ⎣ 0 ⎦ ⎣ H 0 ( zW L equations are (1): H 0 ( z)G0 ( z) + H1 ( z)G1 ( z) + K + H L −1 ( z )G L −1 ( z) = T ( z), (2): H 0 ( zW r )G0 ( z) + H1 ( zW r )G1 ( z) + K + H L −1 ( zW r )G L −1 ( z) = 0, 1 ≤ r ≤ L − 1.

Replacing z by zW L − r in the above equation we get H 0 ( zW L )G0 ( zW L − r ) + H1 ( zW L )G1 ( zW L − r ) + K + H L −1 ( zW L )G L −1 ( zW L − r ) = 0, or equivalently, (3): H 0 ( z )G0 ( zW L − r ) + H1 ( z)G1 ( zW L − r ) + K + H L −1 ( z )G L −1 ( zW L − r ) = 0, 0 ≤ r ≤ L − 1. Rewriting Eqs. (1) and (3) in matrix form we get G1 ( z ) G L −1 ( z) ⎤ ⎡ H 0 ( z ) ⎤ ⎡T ( z)⎤ L ⎡ G0 ( z ) ⎢ G0 ( zW ) G1 ( zW ) G L −1 ( zW ) ⎥ ⎢ H1 ( z ) ⎥ ⎢ 0 ⎥ L ⎢ ⎥⎢ ⎥ = ⎢ M ⎥. M M O M M ⎢ ⎥ ⎢ ⎥ L −1 ) G ( zW L −1 ) L G L −1 )⎥ ⎢ H L −1 ( zW 1 ⎣G0 ( zW ⎦ ⎣ L −1 ( z ) ⎦ ⎣ 0 ⎦ Thus, if an L-channel filter bank is alias-free with a given set of analysis and synthesis filters, then the filter bank is still alias-free if the analysis and synthesis filters are interchanged. 14.43 An equivalent representation of the structure of Figure P14.6 is shown below ⎡1 0 L 0 ⎤ ⎢0 1 L 0 ⎥ where Γ m = ⎢ M M O M ⎥ . ⎢⎣0 0 L z −1 ⎥⎦ X(z)

L _ z1

L _1

z

_1

Γ1

S0

S1

Γ2

ΓK

SK

L

z

L

The corresponding synthesis bank for designing an L-channel perfect −1 and reconstruction bank is thus as shown below where Tm = S m

⎡ z −1 ⎢ Λ m = z −2 Γ −m1 = ⎢ 0 ⎢ M ⎣ 0

0 z

−1

M 0

L L O L

0⎤ ⎥ 0⎥. M⎥ 1⎦

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L

_1

z

+

L ΛK

TK

_1

Λ2

Λ1

T1

z

T0

+

L

_1

z

L

+

⎡ H 0 ( z ) ⎤ ⎡1 − 2 3 ⎤ ⎡1 0 0 ⎤ ⎡ 3 − 5 3 ⎤ ⎡ 1 ⎤ 14.44 ⎢ H 1 ( z ) ⎥ = ⎢0 1 − 2⎥ ⎢0 1 0 ⎥ ⎢− 1 3 − 2⎥ ⎢ z −1 ⎥ ⎢ H ( z )⎥ ⎣⎢0 0 1 ⎦⎥ ⎢0 0 z −1 ⎥ ⎣⎢ 0 − 1 1 ⎦⎥ ⎢ z − 2 ⎥ ⎣ 2 ⎦ ⎦ ⎣ ⎣ ⎦ ⎡ 5 − 11 − 3 z −1 7 + 3 z −1 ⎤ ⎡ 1 ⎤ ⎢ ⎥ = ⎢ − 1 3 + 2 z −1 − 2 − 2 z −1 ⎥ ⎢ z −1 ⎥. Hence, H 0 ( z ) = 5 − 11z −1 + 4 z −2 + 3 z −3 , ⎢ ⎥ − z −1 z −1 ⎥ ⎣ z − 2 ⎦ ⎢⎣ 0 ⎦ H 1 ( z ) = −1 + 3 z −1 − 2 z −3 , H 2 ( z ) = − z −2 + z −3 .

[G0 ( z )

[

= z −2

[

G1 ( z ) G2 ( z )] = z

]

⎡1 − 2 3 ⎤ z −1 1 ⎢0 1 − 2 ⎥ 1 ⎥⎦ ⎢⎣0 0

−2

− 1 ⎡ z −1

⎢ ⎢ 0 ⎢⎣ 0

0 z

−1

0

0 ⎤ ⎡ 3 − 5 3 ⎤ −1 ⎥ 0⎥ ⎢− 1 3 − 2⎥ 1 ⎥ ⎢⎣ 0 − 1 1 ⎥⎦ ⎦

⎡3 z −1 + 1 8 z −1 + 3 7 z −1 + 4 ⎤ ⎥ ⎢ z −1 1 ⎢ z −1 + 2 3 z −1 + 6 3 z −1 + 8 ⎥. Hence, 3 4 ⎥⎦ ⎢⎣ 1

]

G0 ( z ) = 1 + 2 z −1 + 2 z −2 + 3 z −3 , G1 ( z ) = 3 + 6 z −1 + 6 z −2 + 8 z −3 ,

G 2 ( z ) = 4 + 8 z −1 + 7 z − 2 + 7 z −3 . 14.45 Figure 14.20 with internal variables labeled is shown below: x[n]

L _1

z

L

c0 [n]

b 0[n]

c1 [n]

b 1[n]

L L

_1

z

L

cL _ 1[n]

+

_ z1

P(z)

_ z1

_ z1

bL _ 1[n]

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_ z1

L

+

y[n]

565

From the above figure it follows that we can express the z-transforms of {cl [ n]} as

C l ( z) =

L −1

∑

1 ( z1 / L W k ) − l X (( z1 / L W k ), 0 ≤ l ≤ L − 1, where W = e − j 2π / L . L l=0

Likewise, the z-transforms of {b s [n]} can be expressed as

Bs ( z ) =

L −1

∑ Ps, l (z)Cl (z), 0 ≤ s ≤ L − 1, where Ps,l (z) denotes the z-transforms the

s=0

( s, l ) –th element of P (z ). Finally, the z-transform of the output y[n] is given by

Y (z) =

L −1

∑

z − ( L −1− s ) B

s

(z L )

=

L −1

∑

z − ( L −1− s )

L −1

∑ Ps, l (z L )Cl (z L )

s=0 s=0 l=0 L −1 L −1 L −1 1 = z − ( L −1− s ) Ps , l ( z L ) ( zW k ) − l X ( zW k ) L s=0 k =0 l=0 L −1 L −1 L −1 1 = X ( zW k ) W − kl z − ( L −1− s ) z − l Ps, l ( z L ). L k =0 l=0 s=0 In the above expression, terms of the form X ( zW k ), k ≠ 0,

∑

∑

∑

∑

∑

∑

represent the contribution coming from aliasing. Hence, the expression for Y (z ) is free from these aliasing terms for any arbitrary input x[n ] if and only if L −1

L −1

l=0

s=0

∑

W − kl

∑ z − ( L −1− s) z − l Ps, l (z L ) = 0, k ≠ 0.

⎡ V0 ( z ) ⎤ ⎡T ( z)⎤ ⎥ ⎢ ⎥ The above expression can be written in a matrix form as D ⎢ V1 ( z ) ⎥ = ⎢ 0 ⎥ , M M ⎢V ⎥ ⎢ ⎥ ⎣ L −1 ( z )⎦ ⎣ 0 ⎦ where D is the DFT matrix, T (z ) is transfer function of the alias-free system, and †⎢

Vl ( z) =

L −1

∑ z − l z − (L −1− s) Ps, l (z L ). Since DD

†

= LI, the above matrix equation can

s=0

⎡ V0 ( z ) ⎤ ⎡T ( z)⎤ ⎢ V1 ( z) ⎥ ⎥ ⎢ be alternately written as ⎢ = D⎢ 0 ⎥. This implies that M M ⎥ ⎢V ⎥ ⎢⎣ 0 ⎥⎦ ⎣ L −1 ( z )⎦

Vl ( z) = V ( z ), 0 ≤ l ≤ L − 1, as the first column of D has all elements equal to 1. As a result, the L-channel QMF bank is alias-free if and only if Vl ( z) is the same for all l. The two figures below show the polyphase realizations of V0 ( z ) and V1 ( z) :

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P 0,0 (z L )

P 0,1 (z L )

_1

z P1,0 (z L )

_ z1

P1,1 (z L )

_1

z

_ z1

_ z1

_ z1

PL _ 1,0(z L )

y[n]

PL _ 1,1(z L )

y[n]

(b) V1 ( z)

(a) V0 ( z)

The realization of V1 ( z) can be redrawn as indicated below: P1,1 (z L )

_ z1

P 2,1 (z L ) _ z1

_ z1 _ z LP 0,1 (z L )

y[n]

(c) V1 ( z) Because of the constraint V0 ( z) = V1 ( z ), the polyphase components in Figures (a) and (c) should be the same. From these two figures it follows that the first column of P (z ) is an upwards-shifted version of the second column, with the topmost

element appearing with a z −1 attached. This type of relation holds for the k-th column and the (k + 1) –th column of P (z ) . As a result, P (z ) is a pseudo-circulant matrix of the form of Eq. (14.155). 14.46 H 0( z ) = E 0 ( z 2 ) + z −1 E 1( z 2 ), H 1( z ) = E 0 ( z 2 ) − z −1 E 1( z 2 ). Hence, ⎡ E ( z ) E1 ( z ) ⎤ E( z ) = ⎢ 0 . Likewise, ⎣ E 0 ( z ) − E1 ( z )⎥⎦ G 0 ( z ) = E 0 ( z 2 ) + z −1 E 1( z 2 ) = z −1 R 00 ( z 2 ) + R10 ( z 2 ), G 1 ( z ) = − E 0 ( z 2 ) + z −1 E 1( z 2 ) = z −1 R 01 ( z 2 ) + R11 ( z 2 ). Hence, ⎡ R ( z ) R01 ( z )⎤ ⎡ E1 ( z ) E1 ( z ) ⎤ R ( z ) = ⎢ 00 . Therefore, = ⎣ R10 ( z ) R11 ( z ) ⎥⎦ ⎢⎣ E0 ( z ) − E 0 ( z )⎥⎦

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0 ⎤ ⎡ E ( z ) E1 ( z ) ⎤ ⎡ E0 ( z ) E1 ( z ) ⎤ ⎡2 E0 ( z ) E1 ( z ) =⎢ P ( z ) = R ( z ) E( z ) = ⎢ 1 ⎢ ⎥ ⎥⎦. ⎥ E z E z ( ) ( ) − − E z E z 0 2 E z E z ( ) ( ) ( ) ( ) 0 1 0 1 ⎣ 0 ⎦⎣ 0 ⎦ ⎣ Here, P0 ( z ) = 2 E0 ( z ) E1 ( z ) and P1 ( z ) = 0. As a result, P(z) is pseudo-circuluant.

14.47 (a) H 0 ( z ) = 1 − 2 z −1 + 4.5 z −2 + 6 z −3 + z −4 + 0.5 z −5 , H 0 ( z −1 ) = 0.5 z 5 + z 4 + 6 z 3 + 4.5 z 2 − 2 z + 1. Therefore,

H 0 ( z ) H 0 ( z −1 ) = 0.5 z 5 + 6.25 z 3 + 22.5 z + 62.5 + 22.5 z −1 + 6.25 z −3 + 0.5 z −5 . Thus, H 0 ( z ) H 0 ( z −1 ) + H 0 ( − z ) H 0 ( − z −1 ) = 125. Hence, H 0 (z ) is a power-symmetric function. The highpass analysis filter is given by

H 1 ( z ) = z −5 H 0 (− z −1 ) = −0.5 + z −1 − 6 z −2 + 4.5 z −3 + 2 z −4 + z −5 . The two synthesis filters are time-reversed versions of the analysis filters as per Eq. (14.92): G0 ( z ) = 2 z −5 H 0 ( z −1 ) = 1 + 2 z −1 + 12 z −2 + 9 z −3 − 4 z −4 + 2 z −5 and

G1 ( z ) = 2 z −5 H 1 ( z −1 ) = 2 + 4 z −1 + 9 z −2 − 12 z −3 + 2 z −4 − z −5 . 1 15 15 1 (b) H 0 ( z ) = 1 + z −1 + z − 2 − z − 4 + 2 z − 5 , H 0 ( z −1 ) = 2 z 5 − z 4 + z 2 + z + 1. 2 4 4 2 Therefore,

H 0 ( z ) H 0 ( z −1 ) = 2 z 5 + 10.75 z 3 − 3.375 z + 20.3125 − 3.375 z −1 + 10.75 z −3 + 2 z −5 . Thus, H 0 ( z ) H 0 ( z −1 ) + H 0 (− z ) H 0 (− z −1 ) = 40.625. Hence, H 0 (z ) is a power-symmetric function. The highpass analysis filter is given by 1 15 H 1 ( z ) = z − 5 H 0 (− z −1 ) = −2 − z −1 + z − 3 − z − 4 + z − 5 . The two synthesis filters are 2 4 time-reversed versions of the analysis filters as per Eq. (14.92):

G0 ( z ) = 2 z − 5 H 0 ( z − 1 ) = 4 − 2 z − 1 + G1 ( z ) = 2 z − 5 H 1 ( z −1 ) = 2 − z −1 +

15 − 3 z 2

+ z − 4 + 2 z − 5 and

15 − 2 z − 2z −4 − 4z −5. 2

14.48 H 0 ( z) = 1 + az −1 + z −2 and H1 ( z ) = 1 + az −1 + bz −2 + az −3 + z −4 . The corresponding synthesis filters are given by G0 ( z) = H1 (− z) = 1 − az −1 + bz −2 − az −3 + z −4 , and

G1 ( z) = − H 0 (− z) = −1 + az −1 − z −2 . To show that the filter bank is alias-free and satisfies the perfect reconstruction property we ⎡ H ( z ) H 0 (− z )⎤ ⎡G0 ( z )⎤ ⎡cz − K ⎤ need to show that ⎢ 0 = , where c ≠ 0. Now, ⎣ H 0 (− z ) H1 (− z ) ⎥⎦ ⎢⎣ G1 ( z ) ⎥⎦ ⎢⎣ 0 ⎥⎦ H 0 ( z )G0 ( z) + H1 ( z)G1 ( z) = (1 + az −1 + z −2 )(1 − az −1 + z −2 − az −3 + z −4 ) + (−1 + az −1 − z −2 )(1 + az −1 + z −2 + az −3 + z −4 ) = 2 a (b − 2) z −3 , and H 0 (− z)G0 ( z ) + H1 ( z)G1 (− z) = H 0 (− z ) H1 (− z) − H 0 (− z ) H1 (− z) = 0. Thus, if a ≠ 0 and

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b ≠ 2, the above filter bank is alias-free and also satisfies the perfect reconstruction property.

14.49 If H 0 ( z) is required to have 2 zeros at z = −1, then it is of the form H 0 ( z) = (1 + z −1 ) 2 C ( z), where C (z ) is a first-order polynomial. Now,

P ( z) = H 0 ( z) H 0 ( z −1 ) = (1 + z −1 ) 2 (1 + z) 2 R( z), where R (z ) is a zero-phase polynomial of the form R ( z ) = az + b + az −1 . For perfect reconstruction we require, P ( z ) + P (− z ) = 2, i.e., (1 + z −1 ) 2 (1 + z ) 2 ( az + b + az −1 ) + (1 − z −1 ) 2 (1 − z ) 2 ( − az + b − az −1 ) = 2. Since the above equation must hold for all z, we observe that at z = 1, we get 1 8

1 4

2a + b = . Likewise, at z = j , we get b = . Hence, a = −

P ( z) = (1 + z −1 ) 2 (1 + z) 2 (−

1 16

z+

1 4

−

1 −1 z ). 16

1 . 16

Therefore,

The analysis filter H 0 ( z) is

obtained by a spectral factorization of P(z). Three choices of spectral factorization resulting in linear-phase analysis filters are given in Section 14.3.3. 14.50 14.51 (a) An orthogonal perfect reconstruction filter bank is obtained by choosing the minimum-phase spectral factor of P(z) as the lowpass analysis filter H 0 ( z ) and then determining the lowpass synthesis filter G0 ( z ) using the relation

G0 ( z) = z −5 H 0 ( z −1 ). The minimum-phase spectral factor of P(z) is given by H 0 ( z) = 0.33267(1 + z −1 ) 3 (1 − 0.5745z −1 + 0.1059z −2 ). Hence, G0 ( z ) = z −5 H 0 ( z −1 ) = 0.33267(1 + z −1 ) 3 (0.1059 − 0.5745z −1 + z −2 ). (b) A perfect reconstruction filter bank with symmetric lowpass analysis and synthesis filters of even length is given by H 0 ( z ) = (1 + z −1 )(a + bz −1 + cz −2 + bz −3 + az −4 ) and G0 ( z) = (1 + z −1 ) 5 . Another possible design of a perfect reconstruction filter bank with symmetric lowpass analysis and synthesis filters of even length is given by H 0 ( z ) = (1 + z −1 ) 3 (a + bz −1 + cz −2 + bz −3 + az −4 ) and G0 ( z) = (1 + z −1 ) 3 . (c) A perfect reconstruction filter bank with symmetric lowpass analysis and synthesis filters of odd length is given by H 0 ( z ) = (a + bz −1 + cz −2 + bz −3 + az −4 ) and G0 ( z) = (1 + z −1 ) 6 . Another possible design of a perfect reconstruction filter bank with symmetric lowpass analysis and synthesis filters of odd length is given by H 0 ( z) = (1 + z −1 ) 2 (a + bz −1 + cz −2 + bz −3 + az −4 ) and G0 ( z) = (1 + z −1 ) 4 .

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14.52 14.53 14.54 The polyphase matrices for the structure of Figure P14.7(a) are given by 0⎤. Therefore, 0⎤ and R ( z) = ⎡ 1 E( z ) = ⎡ 1 ⎢⎣ − P ( z) 1⎥⎦ ⎢⎣ P ( z) 1⎥⎦ 0⎤ ⎡ 1 0⎤ = ⎡1 0⎤. Hence, the structure of Figure R ( z ) E( z ) = ⎡ 1 ⎢⎣ − P ( z) 1⎥⎦ ⎢⎣ P ( z ) 1⎥⎦ ⎢⎣0 1⎥⎦ P14.7(a) is a perfect reconstruction QMF bank. 14.55 The polyphase matrices for the structure of Figure P14.7(b) are given by 1 Q( z )⎤ ⎡ 1 0⎤ and R ( z) = ⎡ 1 0⎤ ⎡1 − Q( z)⎤. Therefore, E( z ) = ⎡⎢ ⎥ 1 ( ) − P ( z ) P z 1⎥⎦ ⎢⎣0 ⎢ ⎥ ⎢ 0 1 1 ⎥⎦ ⎦ ⎣ ⎣ ⎦⎣ 0 ⎤ ⎡1 − Q( z )⎤ ⎡1 Q( z )⎤ ⎡ 1 0⎤ = ⎡ 1 0⎤ ⎡1 0⎤ ⎡ 1 0⎤ R ( z ) E( z ) = ⎡ 1 ⎢⎣ − P ( z) 1⎥⎦ ⎢⎣0 1 ⎥⎦ ⎢⎣0 1 ⎥⎦ ⎢⎣ P ( z ) 1⎥⎦ ⎢⎣ − P ( z) 1⎥⎦ ⎢⎣0 1⎥⎦ ⎢⎣ P ( z) 1 ⎥⎦ = ⎡1 0⎤ Hence, the structure of Figure P14.7(b) is a perfect reconstruction QMF ⎢⎣0 1⎥⎦ bank. 14.56 If the 2-channel QMF banks in the middle of the structure of Figure 14.24 are of perfect reconstruction type, then each of these two filter banks have a distortion transfer function of the form αz − M , where M is a positive integer. H10 (z)

2

2

G10(z) _

H11(z)

2

2

_ αz M

G11(z)

Likewise, the 2-channel analysis filter bank on the left with the 2-channel synthesis filter bank on the right form a perfect reconstruction QMF bank with a distortion transfer function βz − L , where L is a positive integer: HL (z)

2

2

GL (z) __

HH(z)

2

2

_ βz L

GH(z)

Hence, an equivalent representation of Figure 14.24 is as indicated below:

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HL (z)

2

_ αz M

2

GL (z)

HH(z)

2

_ αz M

2

GH(z)

which reduces to HL (z)

2

GL (z)

2

_ αz 2M

_ αz 2M

2

HH(z)

_ βz L

GH(z)

2

Thus, the overall structure is also of perfect reconstruction type with have a distortion transfer function given by αβz − (2 M + L ) . 14.57 We analyze the 3-channel filter bank of Figure 14.27(b). If the 2-channel QMF bank of Figure 14.27(a) is of perfect reconstruction type with a distortion transfer function βz − L , the structure of Figure 14.27(b) should be implemented as indicated below to ensure perfect reconstruction: H 0 (z) H 0 (z)

H1 (z)

*

2

G 0 (z)

2

+

H1 (z) x[n]

2

2

2

G 0 (z)

2

G1 (z)

G1 (z)

_ βz L

2

2

+

y[n]

An equivalent representation of the above structure is as shown below:

x[n]

H 0 (z)

2

_ βz L

2

G 0 (z)

H1 (z)

2

_ βz L

2

G1 (z)

+

y[n]

+

y[n]

which reduces to

x[n]

_ βz 2L

H 0 (z)

2

2

G 0 (z)

H1 (z)

2

2

G1 (z)

verifying the perfect reconstruction property. In a similar manner, the perfect reconstruction property of Figure 14.27(c) can be proved. Not for sale

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M14.1 The specifications of the corresponding zero-phase half-band filter are as follows: stopband edge ωs = 0.65π and a minimum stopband attenuation of α s = 2 × 30 + 6.02 = 66.02 dB. The desired stopband ripple is therefore

δs = 10 − α s / 20 = 10 −3.301 = 0.0005. The passband edge of the half-band filter is at

ω p = π − 0.6π = 0.4π. Using the function remezord we then estimate the order of F(z) and using the function remez we next design Q(z). To this end the code fragments used are [N,fpts,mag,wt]=remezord([0.4 0.6],[1 0],[0.0005 0.005]); The order of F(z) is found to be 30 which is of the form 4 K + 2 = N for K = 4. The order of H 0 ( z) is therefore 15. The filter Q(z) is designed using the statement [q,err]=remez(N,fpts,mag,wt); To determine the coefficients { f [ n]} of the filter F(z) we add err to the central coefficient q[16]. Next using the statement h0 = firminphase(f); we determine the minimum-phase spectral factor of F(z) which are the coefficients of the lowpass analysis filter H 0 (z ) : Columns 1 through 7 0.2818 0.5076 0.3582 0.0083 0.1047 Columns 8 through 14 -0.0089 -0.0663 0.0122 0.0319 0.0335 Columns 15 through 16 -0.0121 0.0008

-0.0386

-0.1749

0.0418

-0.0134

-

The highpass analysis filter H 1 (z ) is obtained using the code fragments k = 0:15; h1 = ((-1).^k).*h0;

The synthesis filters G0 (z ) and G1 (z ) can be found from the analysis filters using the code fragments G0 = fliplr(h0); G1 = fliplr(h1);

Plots of the roots of and the gain responses of the two analysis filters are shown below:

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H0(z)

0

0.5 Gain, dB

Imaginary Part

1

15

0 -0.5

H1(z)

-20

-40

-1 -1

-0.5

0 0.5 Real Part

1

-60

0

0.2

0.4

0.6

0.8

1

ω/π

M14.2 The following MATLAB programs can be used to design the analysis filters corresponding to a two-channel QMF paraunitary lattice filter bank. (The program uses the function fminu from the MATLAB Optimization Toolbox, which in turn uses the functions cubici2, cubici3, optint, searchq, and updhess.) % Program for the design of a two-channel QMF % lattice filter bank. Len = input('The length of the filter = '); if (mod(Len,2) ~= 0) sprintf('Length has to be an even number') Len = Len+1; end ord = Len/2-1; ws = 0.55*pi; kinit = [1;zeros([ord,1])]; % set the parameters for the optimization routine options = optimset('MaxIter', 2500, 'Display', 'off'); kfin = fminunc('filtopt',kinit,options,Len,ws); e00old = 1; e01old = kfin(1); e10old = -kfin(1); e11old = 1; for k = 2:length(kfin) e00new = [e00old 0]-kfin(k)*[0 e01old]; e01new = kfin(k)*[e00old 0]+[0 e01old]; e10new = [e10old 0]-kfin(k)*[0 e11old]; e11new = kfin(k)*[e10old 0]+[0 e11old]; e00old = e00new; e01old = e01new; e10old = e10new; e11old = e11new; end E1 = [e00old;e01old]; h0 = E1(:); scale_factor = abs(sum(h0)); h0 = h0/scale_factor; E2 = [e10old;e11old]; h1 = E2(:);

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h1 = h1/scale_factor; [H0,W] = freqz(h0,1,1024); [H1,W] = freqz(h1,1,1024); plot(W/pi,20*log10(abs(H0)), W/pi, 20*log10(abs(H1))); grid on title('Gain response of the analysis filters'); xlabel('\omega/\pi'); ylabel('Gain, dB'); function val = filtopt(kval,Len,ws) e00old = 1; e01old = kval(1); e10old = -kval(1); e11old = 1; for k = 2:length(kval) e00new = [e00old 0]-kval(k)*[0 e01old]; e01new = kval(k)*[e00old 0]+[0 e01old]; e10new = [e10old 0]-kval(k)*[0 e11old]; e11new = kval(k)*[e10old 0]+[0 e11old]; e00old = e00new; e01old = e01new; e10old = e10new; e11old = e11new; end E1 = [e00old;e01old]; h0 = E1(:); [H0,W] = freqz(h0,1,1024); val = 0; for k = 1:length(W) if (W(k) > ws) val = val+abs(H0(k))^2; end end

Due to the non-linear nature of the function to be optimized, different values of kinit should be used to optimize the analysis filter's gain response. The gain responses of the two analysis filters is as shown below. From the gain responses, the minimum stopband attenuation of the analysis filters is observed to be about 23 dB. Gain response of the analysis filters 10 H0(z)

Gain, dB

0

H1(z)

-10 -20 -30 -40

0

0.2

0.4

0.6

0.8

1

ω/ π

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M14.3 The MATLAB program used to generate the prototype lowpass filter and the analysis filters of the 4-channel uniform DFT filter bank is given below: L = 19; f = [0 0.15 0.35 1]; m = [1 1 0 0]; w = [10 1]; N = 4; WN = exp(-2*pi*j/N); plottag = ['- ';'--';'-.';': ']; h = zeros(N,L); n = 0:L-1; h(1,:) = remez(L-1, f, m, w); for i = 1:N-1 h(i+1,:) = h(1,:).*(WN.^(-i*n)); end; clf; for i = 1:N [H,w] = freqz(h(i,:), 1, 256, 'whole'); plot(w/pi, abs(H), plottag(i,:)); hold on; end; grid on; hold off; xlabel('\omega/\pi');ylabel('Magnitude'); title('Magnitude responses of the analysis filter bank');

The plots generated by the above program is given below: Magnitude responses of the analysis filter bank

Magnitude

1

H0(z) H1(z)

H2(z)

H3(z)

H0(z)

0.8 0.6 0.4 0.2 0

0

0.5

1

1.5

2

ω/ π

M14.4 The first 8 impulse response coefficients of Johnston's 16A lowpass filter H L (z ) are given by

0.001050167, –0.005054526, –0.002589756, 0.0276414, –0.009666376, –0.09039223, 0.09779817, 0.4810284 The remaining 8 coefficients are given by flipping the coefficients left to right, From Eq. (14.98), the highpass filter in the tree-structured 3-channel filter bank is given by H 2 ( z ) = H H ( z ) = z −15 H L ( z −1 ). The two remaining filters are given by H 0 ( z ) = H L ( z ) H L ( z 2 ) and H 1 ( z ) = H L ( z ) H H ( z 2 ). The MATLAB program used to generate the gain plots of the 3 analysis filters is given by:

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G1 = [0.10501670e-2 -0.50545260e-2 -0.25897560e-2 0.27641400e-1 -0.96663760e-2 -0.90392230e-1 0.97798170e-1 0.48102840]; G = [G1 fliplr(G1)]; n = 0:15; H0 = (-1).^n.*G; Hsqar = zeros(1,31); Gsqar = zeros(1,31); Hsqar(1:2:31) = H0; Gsqar(1:2:31) = G; H1 = conv(Hsqar,G); H2 = conv(Gsqar,G); [h0,w0] = freqz(H0,[1]); [h1,w1] = freqz(H1,[1]); [h2,w2] = freqz(H2,[1]); g0 = 20*log10(abs(h0));g1 = 20*log10(abs(h1)); g2 = 20*log10(abs(h2)); plot(w0/pi,g0,'b-',w1/pi,g1,'r-',w2/pi,g2,'g-.'); axis([0 1 -120 20]); grid on; xlabel('\omega/\pi');ylabel('Gain, dB');

The plots generated are given below: 20 H0(z)

Gain, dB

0

H1(z)

H2(z)

-20 -40 -60 -80 0

0.2

0.4

0.6

0.8

1

ω/π

M14.5

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576

Digital Signal Processing (Solution Manual) - 3rd Edition by Mitra

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