Manual for K-Notes Why K-Notes? Towards the end of preparation, a student has lost the time to revise all the chapters from his / her class notes / standard text books. This is the reason why K-Notes is specifically intended for Quick Revision and should not be considered as comprehensive study material. What are K-Notes? A 40 page or less notebook for each subject which contains all concepts covered in GATE Curriculum in a concise manner to aid a student in final stages of his/her preparation. It is highly useful for both the students as well as working professionals who are preparing for GATE as it comes handy while traveling long distances. When do I start using K-Notes? It is highly recommended to use K-Notes in the last 2 months before GATE Exam (November end onwards). How do I use K-Notes? Once you finish the entire K-Notes for a particular subject, you should practice the respective Subject Test / Mixed Question Bag containing questions from all the Chapters to make best use of it.
BASIC CONCEPTS
In continuous time signals independent variable is continuous and thus these signals are defined for a continuum of values of independent variable.
Discrete time signals are only defined at discrete times and consequently for these signals the independent variable takes discrete set of values.
Representation of continuous time signals
We use symbol ‘t’ to denote independent variable for continuous time signal.
These signals can be represented by a wave form as shown below
If possible, these can also be represented by a mathematical function like x(t) = sin t
Representation of discrete time signal
We use symbol ‘n’ to denote independent variable for discrete time signal.
These signals can be represented as a series of numbers like x[n] = [5, 4, 5, 7, 9, 2……] Arrow indicates reference point or x [0]
If possible, we can represent the same by a function like x[n] = sin n 4
Also these signals can be represented by a wave form as shown below
Energy & Power Signals Interval , Energy of continuous time signal E
lim
T
x t T T
2
dt
x t
2
dt
Energy of discrete time signal E
lim N 2 2 x n x n T nN n
Power of continuous time signals P
lim
1 T 2T
T
x t T
2
dt
Power of discrete time signals P
N lim 2 1 x n N 2N 1 nN
Signals having non-zero (finite) power and infinite energy are called as Power Signals. ex.
Signals having finite (non-zero) energy and zero power are called as Energy Signals. ex.
x(t) = sint x[n] = [1, 2, 3, 4]
The bounded signal radiate finite energy and periodic signal radiate finite average power.
Even & Odd signals
A signal is said to be “even” if it satisfies the condition x(t) = x (–t) or x [n] = x[–n]
A signal is said to be “odd” if it satisfies the condition x(t) = –x(–t) or x [n] = –x[–n]
Any signal (even those which are neither odd nor even) can be broken into odd & even parts
Odd Part x0 t
Even Part xe t
x t x t 2 x t x t 2
; x0 n
x n x n
; xe n
2 x n x n 2
Periodic and Aperiodic Signals
A signal is said to be periodic with period “T” or “N” if x(t + T) = x(t) x[n + N] = x[n]
Otherwise, the signals are said to be aperiodic.
Classification of systems (i) Linear & Non-Linear Systems For Linearity if x1 t y1 t x2 t y 2 t
then, this condition must be true
1x1 t 2 x2 t 1y1 t 2 y 2 t
Example : y(t) = t x (t) is linear y[n] = 2x [n] + 3 is non-linear (ii) Time Invariant & Time-variant Systems For system to be time-invariant the following condition must hold true x(t - ) y(t – )
It means that following two realizations must be equivalent
The simplest way to verify this is to check the coefficient of ‘t’ inside x(t) eg.
y(t) = tx(t) is time invariant but y(t) = tx(2t) is time variant as coefficient of ‘t’ in side x(t) is not ‘1’
Otherwise, you need to verify the system equivalence shown above.
(iii) Causal & Non-causal Systems
The output should depend only on present & past values of input.
For discrete time system
ht 0 V t 0
h[n] = 0 V n < 0
(iv) Stable & Unstable Systems
Every Bounded input should produce a bounded output.
DT :
K
h k
; CT :
h d
(v) LTI systems with or without memory
The output at any time should depend only on value of input at the same time.
For discrete time system h[n] = 0 V n 0 h[n] = k [n]
For continuous time system h(t) = 0 V t 0 h [t] = k [t]
(vi) Invertible Systems The system is invertible if there exists h1(t) such that Thus h(t) * h1(t) = t
For discrete time, h[n] * h1[n] = n
Shifting and Scaling operations
Shifting Delay if
shift the waveform right by the amount of delay
Advance
if
shift the waveform left by the amount of advance
Scaling Compression if
Replace upper & lower limit by original limit divided by compression factor
Expansion if
Replace upper & lower limit by original limit multiplied by expansion factor.
Note : If both scaling and shifting are given in the question . Ex. x(3t-2)
1. shift the waveform right by the amount of delay
2. Replace upper & lower limit by original limit divided by compression factor
This method is applicable for both continuous and discrete time signal.
LTI system (Linear Time Invariant Systems)
Any continuous time or discrete time system can be represented in terms of impulses.
x t
x t d
x k n k
x[n]
k
LTI systems are characterized on the basis of Impulse Response h(t) or h[n]
The response of a system with impulse as an input is called as impulse response.
Due to time invariance property of LTI system if n h n n k h n k
since x n
y n
x k
K
n k
x k hn k x n * hn = convolution sum
k
for continuous time domain
y t
x h t x t * h t
k
= convolution integral
The condition for causality of system then becomes h[n] =0 V n < 0 ; h(t) = 0 V t < 0
Calculating convolution sum
Suppose x [n] = u[n] h[n] = [1, 2, 5, 7, 9]
Draw plots of both x[n] & h[n]
Flip either x[n] or h[n] about y-axis Here, we flip x[n]
For calculating y[n], shift x[–k] to right by amount ‘n’ For y[0]
The only overlapping between the two is at k = 0, –1, –2 y [0] = x[0] h [0] + x [1] h [–1] + x [2] h [–2] =1x5+1x2+1x1 =8 For y [1]
y [1] = x [0] h [1] + x [1] h [0] + x [2] h [–1] + x [3] h [–2] = 1 x 7 + 1 x 5 + 1 x 1 x 1 x 2 = 15 Similarly, we can calculate all values of y[n] y[n] = [2, 3, 8, 15, 24, 24……..] Calculating Convolution Integral Assume x (t) = u (t) h (t) =
Step 1 Flip either x(t) or h(t) Here, we flip h(t)
Step 2 Shift h( ) by amount “t” to the right to calculate y(t) by calculating overlapping between h t & x
Overlapping area 1 t
=
0
if t < – 1 so, overlapping area = 0
1.1d 1 t
if t > 1 overlapping area = 2 y (t) is shown in adjoining figure:
Properties of Convolution Sum 1) Commutative Property x[n] * h[n] = h[n] * x[n] 2) Distributive Property y1[n] = x[n] * h1[n] y2[n] = x[n] * h2[n] y [n] = y1[n] + y2[n] = x[n] * h1[n] + x2[n]*h2[n] = x[n] * { h1[n] + h2[n] } 3) Associative Property {x[n] * h1[n] }* h2[n] = { x[n]* h2[n] } * h1 [n] Same properties will apply for continuous time domain for convolution integral. Parallel & Cascade structure of LTI systems Parallel:
y[n] = x[n] * [h1 [n] + h2 [n]]
Cascade:
y[n] = x[n] * ([h1 [n] + h2 [n]]) Frequency Response The frequency response of any LTI system is given by its Fourier Transform.
jw DT: H e
CT: H jw
h n e jwn n
ht e
jwt
dt
Group delay & Phase delay Assuming transfer function of system is H(s) jwt input is x(t)= e j w jwt H jw e e jwt Output: H jw e
= H jw e
j wt w
w ArgH jw
Group Delay, g w Phase Delay, w
d w dw w w
Continuous – Time Fourier series Fourier states that any periodic signal can be represented by a set of complex exponential signals provided that it satisfies Drichlet Conditions. Drichlet conditions (i) Over any period x(t) is absolutely integrable i.e.,
T
0 x t dt
(ii) In a finite time interval, x(t) has a finite number of maxima & minima
(iii) It should have finite number of discontinuities in the given interval Note : for distortion less transmission of the of a signal with some finite frequency content through a continuous time LTI system , the frequency response of the system must satisfy these two conditions. 1. The magnitude response H( j ) must be constant for all frequencies of interest ; that is, we must have
H( j ) C For some constant C
2. For the same frequencies of interest, the phase response argH( j) must be linear in frequency, with slope –to and intercept zero ; that is, we must have
argH( j) to
Fourier series as generally expressed in 2 forms.
Trigonometric
Exponential
Trigonometric Fourier Series Analysis equations T
a0
1 x t dt T 0
ak
2 x t cos k0 t dt T 0
bk
2 x t sin k0 t dt T 0
T
where 0 2 T
T
Synthesis equations x t a0
k - k 0
ak cosk 0t
Exponential Fourier Series Analysis equations T
1 jk0t Ck x t e dt T0
k k 0
bk sink 0t
Synthesis equations
x t
k
CK e
jk0t
where 0 2 T
Relation between T.F.S. and E.F.S.
co a0
Cn
Cn
an jbn 2 an jbn 2
Important facts about Trigonometric Fourier series (i) Any odd signal contains only sine terms in Fourier series. (ii) Any even signal contains only cosine terms in Fourier series. (iii) For half–wave symmetric signal x t T 2 x t
Only odd harmonics are present i.e., k = 1, 3, 5……. Properties of complex exponential Fourier Series (i) Linearity F.S. If x t ak F.S. y t bk F.S. A ak + B bk then Ax (t) + By (t)
(ii)
Time-shifting F.S. if x t ak
-jk0t0
F.S. x t t0 e
(iii) Time-Reversal F.S. if x t ak F.S. x t ak
ak where 0 2 T
For odd signal
For even signal
x(t) = – x(–t) ak ak
x(t) = x (–t) ak ak
(iv) Time – Scaling F.S. if x t ak F.S. x t ak
but 0 is replaced by 0 , though Fourier series coefficients remain same. (v) Multiplication F.S. if x t ak F.S. y t bk F.S. z t x t y t ck
Ck
P
bp ak p = convolution sum
(vi) Parseval’s Relation Energy in time domain = Energy frequency Domain 2 1 2 x t dt ak T T k F.S. where x t ak
Discrete –Time Fourier series For a discrete-time signal, with period ‘N’ the following equations are used for Fourier series. Analysis equations Ck
j2Kn
x n e N
N
0 2 N Ck
x n e N
j0Kn
Synthesis equations x n
CK e N
j0Kn
The properties of Fourier series coefficients are same as continuous time Fourier series with one additional property. CK N CK That is, Fourier series coefficients are periodic
IMPORTANT DUALITY A signal discrete in one domain is periodic in other domain & vice versa. Example: For continuous Time Fourier Series, x (t) is periodic in time domain & hence Fourier Series exists where coefficients exist for frequency integral multiple of " 0 " & hence is discrete. Fourier Transform Fourier series exists only for periodic signals, Fourier series converges to Fourier Transform which is continuous as compared to Fourier series which is discrete. Continuous Time Fourier Transform Analysis equation X jw
x t e
jwt
dt
Synthesis equation
1 x t x jw e jwt dw 2
Properties of Continuous Time Fourier Transform
Signal x(t) y(t) Ax(t)+By(t) x(t-t0)
Fourier Transform X(jw) Y(jw) AX(jw)+BY(jw)
x*(t) x(-t) x(at)
X*(-w) X(-w)
x(t)*y(t) d x(t) dt
X(jw)Y(jw) jwX(jw)
x(t)y(t)
1 X(w) * Y(w) 2
x d tx(t)
x t
Parseval’s Relation
x t
2
X w
1 X jw X 0 w jw
j
Ev{x(t)} Od{x(t)} X(t) j0 t
jt0
1 jw X a a
t
e
e
2 1 x w dw 2
d X jw dw
Re{X(jw)} jIm{X(jw)} 2πx(-w) X(w-w0)
Some common Fourier Transform Pairs
Signal
ak e K e
Fourier Transform
jkw0t
2
k
ak k0
2 0
jkw0t
cos w 0 t
0 0
sin w 0 t
0 0 j
1
2
2 2k T K T 2sin T1
t nT n 1, t T1 x t 0, t T1 (sin wt)/πt
1, x 0,
t
1 j
t t0
jt0
e
eatu t ,Re a 0
1 a j
Discrete Time Fourier Transform Analysis equation
Synthesis Equation x n
w
1
u(t)
X e j
w
x n e jn n
1 j j n X e e d 2 2
Properties of Discrete Time Fourier Transform
Signal x[n] y[n]
Fourier Transform X periodic with Y period 2
ax[n] + by [n] x n n0 j0n
e
jn0
X
X *
x*[n]
e
aX bY
x n
X 0
x [–n]
X
x n | k , if n is multiple of k xk n 0, is n is not multiple of k x [n] * y [n]
X k
n x [n]
j
Od {x [n]}
j Im {X( )}
Parseval’s Relation
2
dx
d Re {X( )}
Ev x n
x n n
X Y
2 1 X d 2 2
Some common Fourier Transform Pairs Signal
K N
ak e
e
j0n
sin 0n x [n] = 1 n N1 1, x n N 0, n N1 , n 2 and x [n + N] = x [n]
k
n kN
1, x n 0,
0 2
2
cos 0n
Fourier Transform 2k 2 ak N k
2 jk n N
n N1
n N1
sin Wn W Wn sinc n
n n0
0 2 0 2
j
0 2 0 2
2
2
2k ak N k
2
2 2k N k N sin N1 1 2 sin 2
1, 0 W x 0, W < jn0
e
Laplace Transform
Laplace Transform is more general than Fourier Transform but can only be computed in Region of Convergence (ROC), so it cannot be computed V s S jw; such that ROC = t dt x t e
Laplace transform becomes Fourier transform for 0 , if it lies in ROC.
Analysis Equations for bilateral Laplace Transform H(s) =
ht e
st
dt
for unilateral Laplace Transform H(s) =
ht e
st
dt
0
Synthesis Equation j
x(t) =
1 x s est ds 2j j
Properties of ROC (i) ROC consists of a collection of lines parallel to jw–axis in s–plane. such that
x t e
t
dt
(ii) If X (s) is rational, then ROC does not contain any poles. (iii) If x(t) is of finite duration & absolutely integrable, then ROC is entire s-plane. (iv) If x(t) is right sided signal (i.e., it is zero before some time) and if Re(s) = 0 is in the ROC, then all values of s for which Re(s) > 0 are also in ROC. (v) If x(t) is left sided, (i.e., if it is zero after some time), and if Re (s) = 0 is in ROC, then all values of s for which Re(s) < 0 are also in ROC. (vi) If x(t) is two–sided signal and if the line Re (S) = 0 is in ROC, then the ROC consists of a strip in s–plane include the line Re (S) = 0 (vii) If X(s) is rational, and x(t) is right sided signal, then ROC is right of right most pole. x(t) is left sided signal, then ROC is left of left most pole.
Properties of Laplace Transform
Signal x(t) x1(t)
Transform X(s) X1(s)
ROC R R1
x2(t)
X2(s)
R2
ax1(t) + bx2(t)
aX1(s) + bX2(s)
At least R1 Ռ R2
x t t0
e
e 0 x t st
X s s0
x (at)
1 X s a a
x1 t * x2 t
X1 s X2 s
At least R1 Ռ R2
d x t dt tx(t)
sX s
At least R
d x s ds 1 X s s
R
t
x d
st0
X s
R Shifted version R [i.e., s is in ROC if s s0 is in R] Scaled ROC i.e., s is ROC if s is in R a
At least R
Some common Laplace Transform Pairs Signal t u(t) –u(–t) tn1 ut n 1 !
Transform 1 1 s 1 s 1 n s 1
tn1 u t n 1 ! eatu t
- eatu t tn1 eat u t n 1 ! tn1 at e ut n 1 !
t T
cos 0 t u t sin 0 t u t eat cos 0 t u t eat sin 0 t u t
Re {s} > 0 Re {s} < 0 Re {s} > 0 Re {s} < 0
sn
1 sa 1 sa 1
ROC All s
Re {s} > –a Re {s} < –a
n
Re {s} < –a
s a 1
n
Re {s} > –a
s a
esT s 2 s 20 0 2
All s Re {s} > 0 Re {s} > 0
2
s 0 sa
Re {s} > –a
2
s a 20 0
s a2 20
Re {s} > –a
Initial and Final Value Theorem
x 0 x
lim sX s initial value s
lim s X s Final value, first stability should be ensured, else final value does s0 not exist.
Analysis of LTI system using Laplace Transform Stability
h t dt
; ROC of H(s) should include 0 .
Causality h(t) = 0, t < 0 i.e., right sided signal ROC should be right sided ROC should include Right half plane. but converse is not true.
Z – Transform It is generalization of Discrete Time Fourier Transform Analysis Equation
H z
h k z k
k
Synthesis Equation h[n]
1 H z zn1dz 2j
Indicates integration around counter clockwise circular contour centered at origin & with radius r.
ROC for Z-Transform Z – Transform also exists only inside ROC
x n r n n
is the condition for ROC.
Mapping from s-plane from z–plane
The jw-axis is mapped to unit circle in z–plane.
Right Half plane is mapped to exterior of unit circle.
Left Half plane is mapped to interior of unit circle.
Properties of ROC (i) The ROC x(z) consists of a ring in the z – plane centered about the origin. (ii) The ROC does not contain any poles. (iii) If x[n] is of finite duration, then ROC is the entire z – plane except possibility at z = 0 and/or z = (iv) If x[n] is a right sided sequence and if the circle, | z | = r0 is in the ROC, then all finite values of z, for which | z | > r0 will also be in ROC. (v) If x[n] is a left sided sequence, and the circle | z | = r0 is in ROC, then all finite value of z, for which 0 < | z | < r0 will be in ROC. (vi) If x[n] is two sided sequence and if circle | z | = r0 is in the ROC. Then ROC will consist of a ring in z-plane which consist of ring | z | = r0. (vii) If X (z) is rational and x[n] is right sided than ROC is outside of outer most pole. x[n] is left sided then ROC is inside of inner most pole. (viii) If x[n] is causal, ROC includes z = provided x[n] = 0, n < 0. If x [n] is anti – causal, ROC includes z = 0 provided x [n] = 0, n > 0. (ix) A causal LTI system with rational system function is stable if all poles inside the unit circle that is have magnitude, | z | < 1.
Properties of z–Transform Signal x[n] x1 n
Transform X(z)
x2 n
X2 z
ax1 n bx2 n
x n n0 e
j0n
x n
X1 z
R1 R2
aX1 z bX2 z z
n0
X e
j 0
z
x[–n]
X z 1
n
x k
k
Rx with addition or deletion of origin Rx
X z z0
nx[n]
At least R1 Ռ R2
X z
zn0 x n
x r , n=rk w n 0, n rk for some r x1 n * x2 n
ROC Rx
X zk
z0R x z 1 s.t z R x
Rx
1
k
1 k i.e., z s.t z R x
X1 z X2 z
At least R1 Ռ R2
dz
Rx except addition or deletion of zero
zdX z
1 1 z
1
X z
Rx
z 1
Some common Z –Transform pairs Signal n
Transform 1
ROC All z
u n
1
|z|>1
1z u n 1 n m anun n
a u n 1 n
na u n
1
1
|z|<1 1
1z z m
All z except 0 (if m > 0) or (if m < 0) |z|>|a|
1 1 az 1 1 1 az
|z|<|a|
1
az 1
|z|>|a| 2
1 az 1 nanu n 1
az 1
|z|<|a| 2
1 az 1 Initial & Final value Theorem
x 0 x
lim z
X z Initial value
lim 1 1 X z Final value z 1 z
In z – transform also, stability must be verified before using final value theorem. Sampling Continuous Time signal
Discrete Time signal
Nyquist Sampling Theorem It states that if sampling frequency is greater than twice the maximum frequency in the
signal for the signal to be recovered from its samples. wS 2wM Note: For this condition signal spectrum should be centered around y-axis. Band-pass Sampling Theorem If the signal spectrum is band-pass which means it has minimum & maximum frequency fL = lower frequency
; fu = upper frequency
f K u , where indicates Greatest Integer function fu fL 2f wS u K
xp (t) = x(t) p(t)
p t
t nT n T = sampling interval ; xp t Sampled signal
x(t) = continuous time signal
xp t
x t t nT
n
1 X w * P w 2 2 P w w kws T k XP w
XP w
1 2 X w kw s ; ws T k T
The spectrum of sampled signal is just repetition of actual spectrum at integral multiples of ws .
If w s 2wM , adjacent samples of spectrum overlap, called as aliasing.
Discrete Fourier Transform DFT of n – point sequence is given by: Analysis equation: N1
X k x n .e n 0
j2 kn N
, k = 0, 1, 2…………., N–1
Synthesis equation: x n
1 N1 X k e N K 0
j2 kn N
, n = 0, 1, 2..………., N – 1
Each point of DFT require N complex multiplications and (N – 1) complex additions. Therefore, N point DFT will required N2 complex multiplications and N (N – 1) complex additions. Properties of DFT Sequence x[n] x1[n] x2[n] x[n + N] = x[n] a1 x1 n a2 x2 n
Transform X(k) X1[k] X2[k] X(K+N)= X(k) a1 x1 k a2 x2 k x1 k x2 k
N1
x1 n x2 m nN
n0
Where x2 m n N x2 N m N x n N x N n x n X n e
X(N – k)
N
j
X K e
X K
2 n N
x*(n) x1 n x2 n
j
2k N
N
X*(N – k) 1 X K N x2 k N 1 Circular convolution of 2 DFT sequences
Parseval’s Relation N1
n0
x n
2
2 1 N1 x k N K 0
Fast Fourier Transform (FFT) Algorithms
These are the algorithms for computing DFT when the size N is a power of 2 or when it is a power of 4.
Direct computation of DFT is inefficient because it does not exploit the properties of symmetry and periodicity of the phase factor, WN e
j 2 N
Symmetry property : K+N
WN
2
WNK
Periodicity property : WNK+N WNK
DFT can be expressed as : XR K X I K
N1
xR n cos
n0
N1
xR n sin
n0
2kn 2kn x I n sin N N
2kn 2kn x I n cos N N
No. of operations required for direct computation of DFT 1) 2N2 evolutions of trigonometric functions. 2) 4N2 Real multiplications. 3) 4N (N – 1) real additions. Radix – 2 FFT algorithm There are two types of FFT algorithm: 1) Decimation in time. 2) Decimation in frequency. Radix – 2 algorithm can be implemented over N – point DFT sequence if N = 2n .
We divide the given time sequence by 2 till we get the prime factor. N We split the N–point data sequence into two point data sequences f1 n and f2 n 2 corresponding to even – numbered and numbered samples of x(n). X K
N1
x n WNnk
n0
, K = 0, 1,… ,N – 1.
x nWNnk x nWNnk n even n odd
By substituting n = 2r for n even and n = 2r + 1 for n odd,
X k
N21
r 0
=
2rk
x 2r WN
N21
N21
r 0
rk
x 2r WN 2
r 0
2r 1k x 2r 1WN
2
WN
N21
rk
x 2r 1 WN2
r 0
But WN2 WN . 2
WN2 e
X k
2 j 2
N21
r 0
N
j2
e
WN
2
rk
k
x 2r WN WN 2
WN2H K
G K
=
N2
f1 k
N21
r 0
x 2r 1WNrk
2
, k = 0, 1, …………, N – 1.
f2 k
N 2 N N2 N Now, computation of X[k] requires 2 2 2 2 2
We can further decimate f1 n and f2 n in time. Thus f1 n would result in two
complex multiplications.
point sequences. f2 n decimation by 2 would also result in two
N 4
N point sequence. 4
We can deduce finally that total number of complex multiplication is reduced to N log2 N and the number of complex additions to N log2 N . 2
FFT computation for 8-point DFT
Order of input is bit reversed, due to decimating twice in time domain. N Number of butterflies in per stage. 2 Number of stage is log2 N .
Butterfly computation.
Decimation in frequency We split the DFT formula into two summations, one of which involves the sum over the first data points and the second sum involves the last
N data points. 2
N 2
Thus,
X k
N 1 2
N1
n0
n
X k Since
x n WNKn N x n WNKn N 1 2
x n .WNKn
n0
Nk WN2
Therefore,
e
j
2
N Nk 2 1 W2 x n
2 Nk N 2
x k
n0
N
N Kn W 2 N k
e jk 1
N 1 2
k
x n 1
n0
N x n WNKn 2
Now, divide X(k) into even and odd numbered samples. X 2k
N 1 2
N
x n x n 2 WNKn ,
n0
X 2k 1
k = 0, 1, ……….
2
N1 2
N
n0
x n x n 2 WNn WNKn
, k = 0,1, ………,
2
N log2 N . 2
Total number of complex multiplications are
Total Number of complex addition are Nlog2 N .
Decimation in frequency for N = 8 FFT algorithm
N 1. 2 N 1. 2
Note : 1) Order of input is normal while order of output is bit reversed. 2)
Number of stages is log2 N .
3) Number of butterflies is
N for stage. 2
Butterfly computation for decimation in frequency.