Instant download and all chapters Solutions Manual Digital Control System Analysis Design 4th Edition Charles L. Phillips Troy Nagle Aranya Chakrabortty https://testbankdata.com/download/solutions-manual-digital-control-system-analysisdesign-4th-edition-charles-l-phillips-troy-nagle-aranya-chakrabortty/
CHAPTER 4
4.2-1. (a) Show that a pole of circle. (b) Show that a pole of circle. (c) Show that a pole of
E ( s)
in the left half-plane transforms into a pole of
E ( s)
on the imaginary axis transforms into a pole of
in the right half-plane transforms into a pole of
E ( s)
E ( z)
inside the unit
E ( z)
E ( z)
on the unit
outside the unit
circle. Solution: A pole at s transforms into a pole at z = εsT. (a)
s = −a + jb, a > 0 z = ε( − a + jb)T = ε− aT ε jbT = ε − aT < 1
(b)
s = jω ⇒ z = ε jωT = 1/ ωT
(c)
s = a + jb, a > 0 z = ε( a + jb )T = εaT ε jbT = ε aT > 1
4.2-2. Let T = 0.05 s and E ( s) =
s+2
(s − 1)(s + 1)
(a) Without calculating
E ( z)
, find its poles.
(b) Give the rule that you used in part (a). (c) Verify the results of part (a) by calculating (d) Compare the zero of (e) The poles of
E ( z)
E ( z)
with that of
E ( z)
E ( s)
.
.
are determined by those of
E ( s)
. Does an equivalent rule exist for zeros?
Solution: (a)
s = 1, ∴ z = εT = ε0.05 = 1.051
78
s = −1, ∴ z = ε−T = ε−0.05 = 0.9512
(b) a pole at s transforms into a pole at z = εsT = ε0.05s (c)
E ( s) =
3 −1 s+2 = 2 + 2 (s − 1)(s + 1) s − 1 s + 1
∴ E( z) =
(d)
−0.5 z z 2 − 0.9013 z 1.5 z + = z − 1.051 z − 0.9512 ( z − 1.051)( z − 0.9512)
s = −2; z = 0, 0.9013
(e) No – both poles and zeros of E(s) determine the zeros of E(z).
4.2-3. Find the z-transform of the following functions, using z-transform tables. Compare the pole-zero in the z-plane with those of and in the s-plane (see Problems 3.4-1). locations of E ( z)
E ( s)
Let T = 0.1 s . (a)
(b)
(c)
(d)
(e)
E ( s) =
20 (s + 2)(s + 5)
E ( s) =
s +2 s(s + 1)
E ( s) =
s2 + 5s + 6 s(s + 4)(s + 5)
(f)
E ( s) =
5 s(s + 1)
E ( s) =
s+2 s (s + 1)
E ( s) =
2 s + 2s + 5
E * ( s)
2
2
(g) Verify any partial-fraction expansions required, using MATLAB. Solution: (a)
E( z) =
20(ε−0.2 − ε−0.5 ) z 1.415 z = 3( z − ε−0.2 )( z − ε −0.5 ) ( z − 0.8187)( z − 0.6065)
ωs = 2π T = 20π E(s): s = −2, − 5 E *(s): s = −2 ± jk20π, − 5 ± jk20π E(z): z = 0.8187, 0.6065
(b)
E( z) =
5 z (1 − ε −0.1 ) 0.4758 z = −0.1 ( z − 1)( z − ε ) ( z − 1)( z − 0.9048)
poles: E(s): s = 0, − 1 E *(s): s = ± jk20π, − 1± jk20π
79
E( z): z = 1, 0.9048
(c)
E ( s) =
2 −1 s+2 = + s(s + 1) s s + 1
∴ E( z) =
2z z ( z − 0.8096) −z + = z − 1 z − 0.9048 ( z − 1)( z − 0.9048)
poles: same as (b). (d)
E ( s) =
2 −1 1 s+2 = 2+ + s s +1 s (s + 1) s 2
∴ E( z) =
−z z 0.2 z 0.1048 z 2 − 0.0858 z + + = 2 z − 1 z − 0.9048 ( z − 1)2 ( z − 0.9048) ( z − 1)
poles: same as (b) (e)
E ( s) =
s 2 + 5s + 6 0.3 −0.5 1.2 = + + s( s + 4)( s + 5) s s+4 s+5
∴ E( z) =
=
0.3z 0.5 z 1.2 z − + z − 1 z − 0.6703 z − 0.6065
z 3 − 1.5841z 2 + 0.6231z ( z − 1)( z − 0.6703)( z − 0.6065)
poles: E(s): s = 0, − 4, − 5 E *(s): s = ± jk20π, − 4 ± jk20π, − 5 ± jk20π
E( z): z = 1, 0.6703, 0.6065
(f)
E ( s) =
2 ,∴ ( s + 1)2 + 22
∴ E( z) =
⎤ 2⎡ 0.1798 z z ε−0.1 sin 0.2 = 2 ⎢ 2 −0.1 −0.2 ⎥ 2 ⎣ z − 2 z ε cos 0.2 + ε ⎦ z − 1.7736 z + 0.8187
poles: E(s): s = −1 ± j 2 = 2.236 / ± 116.6° E*(s): s = −1 ± j 2 ± jk 20π = −1 ± j(2 + k 20π)
E( z): z = 0.8868 ± j 0.1798 = 0.9048/ ± 11.46°
80
(g)
num=[0 1 5 6]; den=[1 9 20 0]; [r, p, k]=residue(num,den)
4.3-1. Find the z-transforms of the following functions: (a) E ( s) =
(b)
(ε
s
)
−1
2
ε2s s(s + 1)
, T = 0.5 s
(0.5s + 1)(1 − ε−0.25s ) E ( s) = , 0.5s(s + 0.25)
T = 0.25 s
Solution: (a)
E(z) =
=
(b)
0.3935( z + 1) z 3 ( z − 0.6065)
" 0.5s + 1 % z − 1 " 8 −7 % E(z) = (1− z −1 ) $ '= $ + ' z # s s + 0.25 & # 0.5s(s + 0.25) &
=
4.3-2.
z(1− ε−0.5 ) (z 2 − 1) " 1 % z 2 − 1 = ' $ # s(s + 1) & z4 z 4 (z − 1)(z − ε−0.5 )
z − 1 ⎡ 8z 7z ⎤ z − 0.5152 − = ⎢ ⎥ z ⎣ z − 1 z − 0.9344 ⎦ z − 0.9394
Zero-order hold
1 E(s) = s T=1s
1 - ε-Ts s
Plant M(s)
5s s + 0.1
C(s)
Fig. P4.3-2 (a) Find the system response at the sampling instants to a unit step input for the system of Fig. P4.3-2. Plot versus time. c( nT )
(b) Verify your results of (a) by determining the input to the plant,
m(t )
, and then calculating
c (t )
by continuous-time techniques. (c) Find the steady-state gain for a constant input (dc gain), from both the pulse transfer function and from the plant transfer function. (d) Is the gain in part (c) obvious from the results of parts (a) and (b)? Why? Solution: 81
G(z) =
(a)
z − 1 " 5s % z − 1 5z 5(z − 1) × = '= $ −0.2 z z # s(s + 0.1) & z − 0.8187 z −ε
5z ⎛ z ⎞ 5( z − 1) Y ( z) = ⎜ = ⎟ ⎝ z − 1 ⎠ z − 0.8187 z − 0.8187
∴ c(nT ) = 5(0.8187)n
(b)
M ( s) =
1 s
∴ Y (s) =
5s 5 = , ∴ y(t ) = 5ε−0.1t s(s + 0.1) s + 0.1
∴ y(nT ) = y(2n) = 5ε−0.1(2n) = 5(ε−0.2 )n = 5(0.8187)n dc gain = G p (s)
(c)
s =0
=
dc gain = G( z ) z =1 =
(d)
5(0) =0 0 + 0.1
5(1 − 1) =0 1 − 0.8187
Yes - Css (kT ) = (1)(dcgain) = 0 from (a), ∴ dc gain = 0 Css (t ) = (1)(dcgain) = 0 from (b), ∴ dc gain = 0
4.3-3. Repeat Problem 4.3-2 for the case that T = 0.1 s and the plant transfer function is given by: (a) (b)
G p ( s) =
5 s + 3s + 2
G p ( s) =
5 s + 2s + 2
2
2
Solution:
82
(a)
% z − 1 " 2.5 −5 z −1 " 5 2.5 % + + '= $ $ ' z # s s + 1 s + 2& z # s(s + 1)(s + 2) &
(a)G(z) =
∴
=
z − 1 ⎡ 2.5 z 5z 2.5 z ⎤ − + ⎢ z ⎣ z − 1 z − 0.9048 z − 0.8187 ⎥⎦
=
⎤ z −1 ⎡ 0.0227 z 2 + 0.0205z 0.0227 z + 0.0205 ⎢ ⎥= z ⎣ ( z − 1)( z − 0.9048)( z − 0.8187) ⎦ ( z − 0.9048)( z − 0.8187)
Y ( z) 0.0227 z + 0.0205 2.5 −5 2.5 = = + + z ( z − 1)( z − 0.9048)( z − 0.8187) z − 1 z − 0.9048 z − 0.8187
∴ c(kT ) = 2.5 − 5(0.9048)n + 2.5(0.8187)n 1 5 2.5 5 2.5 (b) M ( s) = , ∴ Y (s) = = − + s s(s + 1)(s + 2) s s +1 s + 2
∴ y(t ) = 2.5 − 5ε−t + 2.5ε−2t ∴ y(nT ) = 2.5 − 5(0.9048)n + 2.5(0.8187)n (c) dcgain = Gp (s)
s =0
= 2.5
dcgain = G( z ) z =1 =
0.0227 + 0.0225 = 2.5 (1 − 0.9048)(1 − 0.8187)
(d) Yes − Css (kT ) = (1)(dc gain) = 2.5, from (a); ∴ dc gain = 2.5 Css (t ) = (1)(dc gain) = 2.5, from (b); ∴ dc gain = 2.5
(b)
% z −1 " 2.5 −2.5(s +1) z −1 " 5 −2.5 % + = $ $ + ' 2 2 ' 2 z # s[(s +1) +1 ] & z # s (s +1) +1 (s +1)2 +1&
(a) G(z) =
=
=
∴
⎤ z − 1 ⎡ 2.5 z −2.5 z 2 + 2.251z −0.2258 + 2 + ⎢ ⎥ z ⎣ z − 1 z − 1.801z + 0.8187 z 2 − 1.801z + 0.8187 ⎦ 0.02337 z + 0.02187 ⇒ poles: z = 0.9048 / ± 5.73° z 2 − 1.801z + 0.8187 = 0.900 ± j 0.09033
Y ( z) 0.02337 z + 0.02187 = z ( z − 1)( z − 0.900 − j 0.0903)( z − 0.900 + j 0.0903) =
k1 k1* 2.50 + + z − 1 z − 0.900 − j 0.0903 z − 0.900 + j 0.0903
83
k1 =
=
0.02337(0.900 + j 0.0903) + 0.02187 (0.900 + j 0.0903 − 1)( j (2)(0.0903)) 0.04295/ 2.82° = 1.766/ − 225.9° (0.1347/ 137.9°)(0.1806/ 90°)
From (2-29), (2-30):
aT = ln(0.9048) = −1 ⇒ a = −1 bT = 5.73° = 0.1 rad ⇒ b = 1
A = 2 k1 = 3.531, θ = 134.1° ∴ y(nT ) = 2.5 + 3.531(0.9048)n cos(0.1n + 134.1°) (b) M ( s) =
∴Y ( s) =
1 s
5 2.5 −2.5( s + 1) −2.5 = + + s ( s + 1)2 + 1 ( s + 1)2 + 1 s(s 2 + 2s + 2)
∴ y (t ) = 2.5 − 2.5(ε −t cos t + ε −t sin t ) = 2.5 + 3.535ε−t cos(t + 135°) ∴ y (nT ) = 2.5 + 3.535ε −0.1n cos(0.1n + 135°) = 2.5 + 3.535(0.9048)n cos(0.1n + 135°)
(c)
G p ( s) G( z)
(d)
s =0
s =1 =
= 5 2 = 2.5 0.02337 + 0.02187 ≈ 2.5 1 − 1.800 + 0.8187
Same as (d) of part (a).
4.3-4. (a) Find the conditions on a transfer function
G ( z)
such that its dc gain is zero. Prove your result.
(b) For ⎡ 1 − ε−Ts ⎤ G ( z ) = ⇥ ⎢⎢ G p (s)⎥⎥ s ⎣ ⎦
find the conditions of
G p ( s)
such that the de gain of
G ( z)
is zero. Prove your result.
(c) Normally, a pole at the origin in the s-plane transforms into a pole at z = 1 in the z-plane. The function in the brackets in part (b) has a pole at the origin in the s-plane. Why does this pole not transform into a pole at z = 1? 84
(d) Find the conditions on (e) For
G ( z)
such that its dc gain is unbounded. Prove your result.
G ( z)
as given in part (b), find the conditions of
G p ( s)
such that the dc gain of
G ( z)
is
unbounded. Prove your result. Solution: (a)
dc gain = G( z)
z =1 =
0, ∴ G( z) has a zero at z = 1.
(b)
dc gain = G( z)
z =1 = Gp (s) s =0 ,∴G(s)
has a zero at s = 0.
∴ let G p (s) = sG p1 (s) ⎧⎪⎛ 1 − t −Ts ⎞ G(z) = ⎨⎜ ⎟ sG p1 (s) ⎩⎪⎝ s ⎠
(
∴ G( z)
⎫
)⎪⎬⎪ = z −z 1 ⎭
⎡ G p1 (s) ⎤ ⎣ ⎦
has a zero at z = 1.
(c)
z −1 Note: ⎡⎣1 − ε −Ts ⎤⎦ = .
(d)
dc gain = lim G(z)
(e)
G p (s) has a pole at s = 0.
∴ this term cancels the pole at z = 1.
z
z→1
∴G(z) =
∴ G(z) has a pole at z = 1.
1 ∴ let G p (s) = G p1 (s) s
& z −1 N (z) z −1 # 1 N (z) = % 2 G p1 (s)( = z $s z (z − 1)2 D1 (z) z(z − 1)D1 (z) '
4.3-5. Find the system response at the sampling instants to a unit-step input for the system of Fig. P4.3-5. 1 E(s) = s
1 s + 0.5
T=1s
1 - ε-Ts s
3s (s + 1)(s + 2)
C(s)
Fig. P4.3-5 Solution: $ z −1 ! $ ! 1 3 C(z) = # & & # " s(s + 0.5) % z " (s + 1)(s+ 2) %
$ 1 ! z(1− ε−0.5 ) 1 0.7869z = &= # −0.5 " s(s + 0.5) % 0.5 (z − 1)(z − ε ) (z − 1)(z − 0.6065) ⎤ z − 1 ⎡ 1.5 −3 z −1 ⎡ 3 1.5 ⎤ + + = z ⎢⎣ s(s+ 1)(s+ 2) ⎥⎦ s + 1 s + 2 ⎥⎦ z ⎢⎣ s
85
=
z − 1 ⎡ 1.5z −3z 1.5z ⎤ 0.5990 z + 0.2184 + + = ⎢ ⎥ z ⎣ z − 1 z − 0.3679 z − 0.1353 ⎦ ( z − 0.3679)( z − 0.1353)
∴
C( z) 0.4714 z + 0.1719 = z ( z − 0.6065)( z − 0.3679)( z − 0.1353)( z − 1)
=
−10.35 9.845 −2.487 2.991 + + + z − 0.6065 z − 0.3679 z − 0.1353 z − 1
∴ C(nT ) = 2.991 − 10.35(0.6065)n + 9.845(0.3679)n − 2.487(0.1353)n
4.3-6. Zero-order hold
1 E(s) = s
1-ε s
T=1s
Plant M(s)
-Ts
5s s + 0.1
C(s)
Fig. P4.3-6 (a) Find the output
c( kT )
(b) What is the effect on
for the system of Fig. P4.3-6, for
c( kT )
e (t )
equal to a unit-step function.
of the sampler and data hold in the upper path? Why?
(c) Sketch the unit-step response
of the system of Fig. P4.3-6. This sketch can be made
c (t )
without mathematically solving for
C ( s)
.
(d) Repeat part (c) for the case that the sampler and data hold in the upper path is removed. Solution: (a)
#1+ ε−Ts & # 2 & # 1 & 2Tz ( % 2(= % 2(= s %$ (' $ s ' $ s ' (z − 1)2
Upper path: Cu (z) = %
Lower path: Cl (z) =
z − 1 " 2 %" z % " 1 % 2Tz $ 2 '$ '= $ 2'= z # s &# z − 1& # s & (z − 1)2
∴ C( z) = Ce ( z) − Cu ( z) = 0
(b) No effect. The Output of the samples and data hold at the sampling instants is equal to the input. (c) cu(t) = Output of data hold. cl(t) = integral of step function = 2t
86
(d)
Cu (t ) = Ce (t ),
∴ C(t ) = 0
4.3-7. G3(s)
D(z)
T
+
E(s)
C(s)
+ G1(s)
G2(s)
T
(a)
G3(s) E(s)
+ G1(s)
G4(s)
C(s)
+ G2(s)
T (b)
G2(s) E(s)
+ G1(s)
G3(s)
Digital filter A/D
D(z)
C(s)
+ D/A
(c)
Fig. P4.3-7 (a) Express each
C ( s)
and
C ( z)
as functions of the input for the systems of Fig. P4.3-7.
(b) List those transfer functions in Fig. P4.3-7 that contain the transfer function of a data hold. Solution: (a)
(i) C( z) = G3 ( z) D( z) E( z) + G1G2 E( z)
87
