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WCS\whw\DSP_2008_9-FFT.ppt\Feb2008

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The Hong Kong Polytechnic UniversityDepartment of Electronic and Information Engineering

Prof. W.C. Siu Subject: Digital Signal Processing February 2008

Digital Signal Processing:6. The Fast Fourier Transform (FFT)

Recall the DFT equation again,

for k = 0, 1,…, N-1

If N = 8, we have

for k = 0, 1,…,7

1N

0n

nkW)n(x)k(X

18

0n

nkW)n(x)k(X

k3k2k0 W)3(xW)2(xW)1(xW)0(x

k7k6k5k4 W)7(xW)6(xW)5(xW)4(x


2



or00000000 W)7(xW)6(xW)5(xW)4(xW)3(xW)2(xW)1(xW)0(x)0(X

76543210 W)7(xW)6(xW)5(xW)4(xW)3(xW)2(xW)1(xW)0(x)1(X







2


3



In matrix form

Hence there are a lot of duplicated calculations,great simplification is possible.

)7(x

)6(x

)5(x

)4(x

)3(x

)2(x

)1(x

)0(x

WWWWWWWW

WWWWWWWW

WWWWWWWW

WWWWWWWW

WWWWWWWW

WWWWWWWW

WWWWWWWW

WWWWWWWW

)7(X

)6(X

)5(X

)4(X

)3(X

)2(X

)1(X

)0(X

12345670

24602460

36147250

40404040

52741630

64206420

76543210

00000000


4



Note that

1. the number of operations required for the computation of a DFT of a sequence with N elements,

Multiplications: M = N2, the order of N2

Additions: A = N(N-1), also the order of N2

and

2. the Circulant Property of the DFT is important.

X(rN+k) = X(k)

N 2N-N 0t

• • •• • •

X(k)

3


5



The Fast Fourier TransformDecimation-in-Time Decomposition:

Recall the DFT of N-point

where

Define

x1(n) = x(2n)

x2(n) = x(2n+1)

for

where

the length of x1(n) and x2(n) is N1

1N

0n

nkW)n(x)k(X

N

2jN eW,1W

r2N

)1N(,,1,0n 1

2N

1N


6



Define the DFT of {x1(n)} and {x2(n)}

where

1N

0n

nk111

1

W)n(x)k(X

1N

0n

nk122

1

W)n(x)k(X

2N

2j

1 WeW 1

4


7



Since

and

for any integer r.

i.e. X1(k) and X2(k) are periodic function of

for finding

we only need to find

)rNk(X)k(X 122

2

NN1

)1N(X,,)1(X),0(X 111

1)2

N(X,(1),X(0),X 111

)rNk(X)k(X 111


8



Now

= x(0)W0 + x(2)W2k + x(4)W4k + …+ x(N-2)W(N-2)k

+ x(1)Wk + x(3)W3k + x(5)W5k + …+ x(N-1)W(N-1)k

1N

0n

nkW)n(x)k(X

5


9



Now

= x(0)W0 + x(2)W2k + x(4)W4k + …+ x(N-2)W(N-2)k

+ x(1)Wk + x(3)W3k + x(5)W5k + …+ x(N-1)W(N-1)k

[x(1)W0 + x(3)W2k + x(5)W4k + …+ x(N-1)W(N-2)k]Wk

1N

0n

nkW)n(x)k(X


10



Now

= x(0)W0 + x(2)W2k + x(4)W4k + …+ x(N-2)W(N-2)k

+ x(1)Wk + x(3)W3k + x(5)W5k + …+ x(N-1)W(N-1)k

[x(1)W0 + x(3)W2k + x(5)W4k + …+ x(N-1)W(N-2)k]Wk

for k = 0,1,2, …, N-1

1N

0n

nkW)n(x)k(X

k21 (k)WX(k)XX(k)

6


11



)7(x

)5(x

)3(x

)1(x

WWWW

WWWW

WWWW

WWWW

W000

0W00

00W0

000W

)6(x

)4(x

)2(x

)0(x

WWWW

WWWW

WWWW

WWWW

)3(X

)2(X

)1(X

)0(X

91

61

31

01

61

41

21

01

31

21

11

01

01

01

01

01

3

2

1

0

91

61

31

01

61

41

21

01

31

21

11

01

01

01

01

01

Note the equation for k = 0,1,2, …, N-1

can be written as

)7(x

)5(x

)3(x

)1(x

WWWW

WWWW

WWWW

WWWW

W000

0W00

00W0

000W

)6(x

)4(x

)2(x

)0(x

WWWW

WWWW

WWWW

WWWW

)7(X

)6(X

)5(X

)4(X

91

61

31

01

61

41

21

01

31

21

11

01

01

01

01

01

7

6

5

4

91

61

31

01

61

41

21

01

31

21

11

01

01

01

01

01

k21 (k)WX(k)XX(k)


12



Note very carefully that we have converted an N-length DFT into two length-½N DFT with some additional operations.

That is N-length DFT 2 length-½N DFTs

+ N complex multiplications+ N complex additions

Let us check the number of multiplications at this stage, say for example:

complex multiplications for X1(k)

complex multiplications for X2(k)

N complex multiplications for WkX2(k)

The total number of complex multiplications is

2

2

N

2

2

N

2

NNN

2

N

2

N 222

N

2

11N

7


13



An example

N = 8 = 23

X(k) = x(0) + x(2)W2k + x(4)W4k + x(6)W6k

+ [x(1) + x(3)W2k + x(5)W4k + x(7)W6k ] Wk

where k = 0, 1,…,7

X(k) = X1(k) + X2(k)Wk

where X1(k) = x(0) + x(2)W2k + x(4)W4k + x(6)W6k

X2(k) = x(1) + x(3)W2k + x(5)W4k + x(7)W6k


14



Wm will repeat itself at a distance of m = 8

W2m will repeat itself at a distance of m = 4

since W1 = W 2

we have,

X1(k) = X1(k + 4)

X2(k) = X2(k + 4)

8


15



For X1(k) = x(0) + x(2)W2k + x(4)W4k + x(6)W6k

where k = 0, 1,…,7

X1(0) = x(0)W0 + x(2)W0 + x(4)W0 + x(6)W0

X1(1) = x(0)W0 + x(2)W2 + x(4)W4 + x(6)W6

X1(2) = x(0)W0 + x(2)W4 + x(4)W8 + x(6)W12

X1(3) = x(0)W0 + x(2)W6 + x(4)W12 + x(6)W18


16



X1(4) = x(0)W0 + x(2)W8 + x(4)W16 + x(6)W24

= x(0)W0 + x(2)W0 + x(4)W0 + x(6)W0 = X1(0)

X1(5) = x(0)W0 + x(2)W10 + x(4)W20 + x(6)W30

= x(0)W0 + x(2)W2 + x(4)W4 + x(6)W6 = X1(1)

Similarly

X1(6) = X1(2)

X1(7) = X1(3)

9


17



We just have to compute

X1(k), for k = 0, 1, 2, 3

i.e

for k = 0, 1, 2, 3

nk111 W)n(x)k(X


18



=1= X1(k)

X1(k+4) = X1(k)

only X1(k) for need to be computed.

1N

0n

)4k(n111

1

W)n(x)4k(X,Or

1N

0n

n41

nk11

1

WW)n(x

2

N,,1,0k

10


19



for k = 0, 1, 2, 3

Similarly we may decompose {x(0), x(2), x(4), x(6)}

into two sets:

{x(0), x(4)} and {x(2), x(6)}

2 length-2 sequences

k6k4k21 W)6(xW)4(xW)2(x)0(x)k(X


20



Again define

where W2 = W12= W4

for k = 0, 1, 2, 3

We have

X11(0) = x(0) + x(4)W0

X11(1) = x(0) + x(4)W4

X11(2) = x(0) + x(4)W8 = X11(0)

X11(3) = x(0) + x(4)W12 = X11(1)

we only have to find X11(k) for k = 0, 1

k4k211 W)4(x)0(xW)4(x)0(x)k(X

11


21



Similarly,

for k = 0, 1

Also

k212 W)6(x)2(x)k(X

k212111 W)k(X)k(X)k(X


22



Similarly

for k = 0, 1

for k = 0, 1, 2, 3

k421 W)5(x)1(x)k(X

k422 W)7(x)3(x)k(X

k222212 W)k(X)k(X)k(X

12


23



the splitting of N is shown below

24

28

24

2

• For each stage we need to compute N numbers

• Each operation is computed from 2 numbers obtained in the previous stage


24



Sub-Summary:

(i) for the 1st stage

X(k) = X1(k) +Wk X2(k)

for k = 0, 1,…,7

For each k, we need one complex multiplication

N complex multiplications

13


25



(ii) for the 2nd stage

complex multiplications


for k = 0, 1, 2, 3

)k(XW)k(X)k(X 12k2

111

2

N

)k(XW)k(X)k(X 22k2

212

2

N

N complex multiplications


26



(iii) for the 3rd stage

for k = 0, 1

W n Twiddle Factor

8 complex multiplications

k411 W)4(x)0(x)k(X

k412 W)6(x)2(x)k(X

k421 W)5(x)1(x)k(X

k422 W)7(x)3(x)k(X

14


27



Hence, the total number of complex multiplications for the whole process is

M = N log2 N

For the present example

M = 8 log2 8 = 24


28



Equations:(Decimation in Time Decomposition)

3rd stage X11(0) = x(0) + x(4)W0 X21(0) = x(1) + x(5)W0

X11(1) = x(0) + x(4)W4 X21(1) = x(1) + x(5)W4

X12(0) = x(2) + x(6)W0 X22(0) = x(3) + x(7)W0

X12(1) = x(2) + x(6)W4 X22(1) = x(3) + x(7)W4

2nd stage X1(0) = X11(0) + X12(0)W0 X2(0) = X21(0) + X22(0)W0

X1(1) = X11(1) + X12(1)W2 X2(1) = X21(1) + X22(1)W2

X1(2) = X11(0) + X12(1)W4 X2(2) = X21(0) + X22(0)W4

X1(3) = X11(1) + X12(1)W6 X2(3) = X21(1) + X22(1)W6

1st stage X(0) = X1(0) + X2(0)W0 X(4) = X1(0) + X2(0)W4

X(1) = X1(1) + X2(1)W1 X(5) = X1(1) + X2(1)W5

X(2) = X1(2) + X2(2)W2 X(6) = X1(2) + X2(2)W6

X(3) = X1(3) + X2(3)W3 X(7) = X1(3) + X2(3)W7

Equations

15


29



Decimation-in-time decomposition of an eight-pointDFT computation

x(0)

x(1)

x(2)

x(3)

x(4)

x(5)

x(6)

x(7)

X11(0)

X21(0)

X12(0)

X22(0)

X11(1)

X21(1)

X12(1)

X22(1)

W4

W4

W4

W4

X1(0)

X2(0)

X1(2)

X2(2)

X1(1)

X2(1)

X1(3)

X2(3)

W4

W4

W2

W6

W2

W6

X(0)

X(4)

X(2)

X(6)

W4

W6

W2

X(1)W

1

W5

W3

W7

X(5)

X(3)

X(7)


30



W3

W7

W2

W6

W0

W4

Decimation-in-time decomposition of an eight-point

DFT computation

16


31



Flow graph of basic butterfly computation (DIT)

A

B

A’

B’

Wr

2

Nr

W

BWA'B 2

Nr

BWA'A r

BWA r


32



Flow graph of simplified butterfly computation

requiring only one complex multiplication

-1

A

B

A’

B’Wr

17


33



This suggests a means of reducing the number of complex multiplications by a factor of 2

the whole FFT process requires only approximately


and the total number of complex additions in the FFT is

Complex additions

NlogN2

12

NlogN 2


34



If one complex multiplication requires

4 real multiplications

2 real additions

the FFT requires totally

2N log2N real multiplications and

2N log2N real additions

Brief Reasons

18


35



e.g. N = Sequence lengthM’ = number of multiplication per pointA’ = number of additions per point

N = 1024 (210) complex data

* refers to the case that 1 complex multiplication is equivalent to 4 real multiplications and 2 real additions

+ refers to the case that 1 complex multiplication is equivalent to 3 real multiplications and 3 real additions

1520101024A’

152051024M’

FFT+

(real)orFFT*

(real)FFT

(complex)

DFT(complex)


36



For an optimized arrangement, we can have

N = 1024

M’ 10

A’ 30

.)etc,jW,1W( 4

N0

19


37



Decimation in Frequency decomposition:~~ a second form of the FFT ~~

(i) Different from the Decimation-in-Time (DIT) FFT, the Decimation-in-Frequency FFT tries to decimate (to extract even and odd terms) the frequency domain results to form the fast algorithm.

(ii) Furthermore we will start by simply splitting the N-point input sequence {x(n)} into two -point sequences: {x(n)} and , corresponding respectively to the first samples and the last samples of {x(n)}.2

N)}n(x{ 2

N2N

2N


38



i.e.

for k=0,1,…, N-1

Then consider the even and odd-numbered values of X(k) separately.

1N

0n

nkW)n(x)k(X

nk

12

N

0n

k2

N

W)2

Nn(xW)n(x

12

N

0n

12

N

0n

k)2

Nn(nk W)

2

Nn(xW)n(x

1

2

N

0n

12

N

0n

nkk2

Nnk W)

2

Nn(xWW)n(x

20


39



For k even, replacing k by 2k, we obtain

for

Replacing k by 2k+1 for k = odd, we have

nk2

12

N

0n

W)2

Nn(x)n(x)k2(X

12

N,,1,0k

nk2n

12

N

0n

WW)2

Nn(x)n(x

12

N,,1,0kfor

)1k2(n

12

N

0n

)1k2(W)

2

Nn(xW)n(x)1k2(X 2

N


40



)7(x)3(x

)6(x)2(x

)5(x)1(x

)4(x)0(x

wwww

wwww

wwww

wwww

)6(X

)4(X

)2(X

)0(X

91

61

31

01

61

41

21

01

31

21

11

01

01

01

01

01

)7(x)3(x

)6(x)2(x

)5(x)1(x

)4(x)0(x

w000

0w00

00w0

000w

wwww

wwww

wwww

wwww

)7(X

)5(X

)3(X

)1(X

31

21

11

01

91

61

31

01

61

41

21

01

31

21

11

01

01

01

01

01

nk2

12

N

0n

W)2

Nn(x)n(x)k2(X

nk2n

12

N

0n

WW)2

Nn(x)n(x)1k2(X

21


41



Decimation in frequency signal flow graph, N = 8

x(0)

x(1)

x(2)

x(3)

x(4)

x(5)

x(6)

x(7)

-1

-W2 W

2

-W1

W1

W3

-W3

-1

-1

-1

-1

-1

W2

-W2

W2

-W2

X(0)

X(4)

X(2)

X(6)

X(1)

X(5)

X(3)

X(7)


42



Flow graph of a typical butterfly computation (DIF)

-1

A

B

A’

B’Wr

22


43



Comments on the FFT:

Advantages:(i) Reasonably good saving in terms of computation, i.e. it

is efficient(ii) Short Program required(iii) Easy to understand

Disadvantages:(i) A relatively large number of operations still required,

especially multiplications(ii) Almost the same number of multiplications for both real

and complex input(iii) Problem of the generation of Wn’s for n=0,1,2,…N-1.


44



A Generalized Approach for the FFT

All FFT algorithms can be derived

from successive applications of a single

operation, and the representation of a one-

dimensional string of inputs as a two-

dimensional array.

1N

0n

nkW)n(x)k(X

23


45



To simplify its computation, let us consider

the General Criteria for mapping n and k

indices

n L1n1 + L2n2

k K1k1 + K2k2

Hence)kKkK)(nLnL(nk 22112211WW

2222121221211111 knKLknKLknKLknKLW 2222121221211111 knKLknKLknKLknKL WWWW


46



In order to get rid of one twiddle factors,

We can let either

L1K2 = N, hence

L2K1 = N, hence

1WW 212121 kNnknKL

1WW 121212 kNnknKL

24


47



To transform the DFT equation into 2-dimensional form,

assume that, N = N2N1 and

let n n2 + N2n1

k N1k2 + k1

(i.e. L2 = 1, L1 = N2, K2 = N1 , K1 = 1

hence L1K2 = N)

)kkN)(nNn(nk 121122WW 11212221 knNknknN WWW

22112

2

2

1

1

112 knNkn1N

0n

1N

0n

knN1212 WWW)n,n(x)k,k(X


48



Let us further assume that N2 = 2,

Hence N = 2N1 and N1 =

let n n2 + 2n1

k k2 + k1

(i.e. L2 = 1, L1 = 2, K2 = , K1 = 1 hence L1K2 = N)

)k2/Nk)(n2n(nk 1212WW

111222 kn2kn2/kNn WWW

2/kNnkn12

0n

12/N

0n

kn21212

2212

2 1

11 WWW)n,n(x)k,k(X

2N

2N

2N

Appendix B:Mapping

25


49



This becomes the Equation for

Decimation-In-Time decomposition (DIT) algorithm.

We have transformed a length-N DFT into 2 length-½NDFTs at the expenses of two Twiddle Factor multiplications and two additions.

12/N

0n

kn21

12/N

0n

kkn211

1

11

1

111 W)n,1(xWW)n,0(x)k,0(X

12/N

0n

kn21

12/N

0n

kkn211

1

11

1

111 W)n,1(xWW)n,0(x)k,1(X


50



Recall again:We have transformed a length-N DFT into 2 length-DFTs at the expenses of N Twiddle Factor multiplications and N additions. This is considered as a one stage decomposition.

For the second stage, the same procedure can be done, hence a length- DFT can also be decomposed into 2 length- DFTs at the expenses of Twiddle Factor multiplications and additions. However, there are 2 length- DFTs to be computed; hence the total numbers of multiplications and addition are respectively also N.

This is the essence of the FFT.

2N

2N

4N

2N

2N

26


51



Example: For N=8 with DIT decomposition,

assume N = N2N1 = 2 x 4 and

Hence n n2 + 2n1

k 4k2 + k1

(i.e. L2 = 1, L1 = 2, K2 = 4, K1 = 1 hence L1K2 = 8 = N)

)kk4)(n2n(nk 1212WW

111222 kn2knkn4 WWW

2212

2 1

11 kn4kn12

0n

14

0n

kn21212 WWW)n,n(x)k,k(X


52



This result is the same as before:

the decimation in time decomposition algorithm.

14

0n

kn21

14

0n

kkn211

1

11

1

111 W)n,1(xWW)n,0(x)k,0(X

14

0n

kn21

14

0n

kkn211

1

11

1

111 W)n,1(xWW)n,0(x)k,1(X

27


53



Diagram showing the decomposition:

The total number of multiplications = multiplications per stage x the Number of stages= N x log2N

Similarly, the total number of adds = N x log2N

2N

4N

2N

N

4N

4N

4N


54



Decimation-in-frequency Algorithm (DIF)Similarly, we can define the following mappings for the DIF algorithm.

n 4n2 + n1

k k2 + 2k1

(i.e. L2 = 4, L1 = 1, K2 = 1, K1 = 2, hence L2K2 = 8 = N)

112122 kn2knkn4nk WWWW

2221

2 1

11 kn4kn12

0n

14

0n

kn21212 WWW)n,n(x)k,k(X

Appendix B:Mapping

28


55



and

Equation for decimation in frequency decomposition algorithm

14

0n

kn2111

1

11W)n,1(x)n,0(x)k,0(X

14

0n

kn2n111

1

111WW))n,1(x)n,0(x()k,1(X


56



1. We have recall the properties of DFT, in particular

(i) DFT Kernel,

(ii) Circulant Property

X(-k) = X(N-k)

X(rN+k) = X(k), for r = integer

N

2j

eW

,1W0

,1WrN

,1W 2N

and,jW 4N

jW 4N3

Chapter Summary (Chapter 9)

29


57



2. Decimation-In-Time FFT

(i) To decimate the time-domain signal representation of the sequence,

x1(n) = {x(0), x(2), x(4), ….}

x2(n) = {x(1), x(3), x(5), ….}

(ii) This gives a resultant formulation as shown below,

X(k) = X1(k) + Wk X2(k)

for and

1N

0n

nk111

1

W)n(x)k(X

1N

0n

nk122

1

W)n(x)k(X


58



3. Decimation-In-Frequency FFT

(i) To decimate the frequency-domain signal representation of the sequence,

X1(k) = X (2k), for k = 0, 1, 2, …(N/2-1)

X2(k) = X(2k+1), for k = 0, 1, 2, …(N/2-1)

(ii) This gives a resultant formulation as shown below,

nk2

12

N

0n

W)2

Nn(x)n(x)k2(X

nk2n

12

N

0n

WW)2

Nn(x)n(x)1k2(X

30


59



4. Number of operations

(i) For DFT:

Multiplications: M = N2, the order of N2

Additions: A = N(N-1), also the order of N2

(complex operations)

(ii) For FFT

Multiplications: ½ Nlog2N, the order of log2N

Additions: N log2N, also the order of log2N

(complex operations)


60



5. Realisation using Butterflies

(i) Butterflies for DIT-FFT

(ii) Butterflies for DIF-FFT

-1

A

B

A’

B’Wr

-1

A

B

A’

B’Wr

31


61



6. Generalized formulation for the FFT

(i) For DIT-FFT:

let n n2 + 2n1

k ½ N k2 + k1

1

0n

kn21

k1

0n

kn211

2N

1

111

2N

1

11 W)n,1(xWW)n,0(x)k,0(X

1

0n

kn21

k1

0n

kn211

2N

1

111

2N

1

11 W)n,1(xWW)n,0(x)k,1(X


62



6. Generalized formulation for the FFT (con’t)

(ii) For DIF-FFT

n ½Nn2 + n1

k k2 + 2k1

and

1

0n

kn2111

2N

1

11W)n,1(x)n,0(x)k,0(X

1

0n

kn2n111

2N

1

111WW)n,1(x)n,0(x)k,1(X

32


63



The Endof Chapter 6


64



Appendix: Complex number multiplication

Consider

x = x0 + jx1

w = w0 + jw1

Approach 1 (Direct Approach)

y = y0 + jy1 = (x0 + jx1)(w0 + jw1)

= (x0w0 - x1w1) +j(x0w1 + x1w0)

4 multiplications and 2 additions

or y0 = x0w0 - x1w1

y1 = x0w1 + x1w0

33


65



Approach 2

Let a0 = x0 + x1

b1 = w0 + w1 Pre-multiplication

b2 = w1 – w0 additions

m0 = a0w0

m1 = x1b1

m2 = x0b2

hence

y0 = m0 – m1 (= x0w0 – x1w1)

y1 = m0 + m2 (= x0w1 + x1w0)


66



Hence it involves


(if additions for bi’s are counted)


(if additions for bi’s are not counted)

34


67



The Endof Chapter 6


68



For example if N =8, and let also N = N2N1 = 2 x 4,

n n2 + 2n1

k 4k2 + k1for n1, k1 = 0,1,..3

n2, k2 = 0,1Or (n2 ,n1) n

(0,0) 0(0,1) 2(0,2) 4(0,3) 6(1,0) 1(1,1) 3(1,2) 5

(1,3) 7 How about (k2,k1) ?

This is actually the Radix-2 algorithm.

Note: x(0) = x(0,0)x(1)= x(1,0)x(2) = ?..

35


69



For example let also N = N2N1 = 2 x 4,

n n2 + 2n1

k 4k2 + k1for n1, k1 = 0,1,..3

n2, k2 = 0,1Or (n2 ,n1) n

(0,0) 0(0,1) 2(0,2) 4(0,3) 6(1,0) 1(1,1) 3(1,2) 5(1,3) 7 How about (k2,k1) ?

k 4k2 + k1 Or (k2 ,k1) k

(0,0) 0(0,1) 1(0,2) 2(0,3) 3(1,0) 4(1,1) 5(1,2) 6(1,3) 7

Return


70



Find the mapping for N = N2N1 = 2 x8,

n 8n2 + n1

k k2 + 2k1

for n1, k1 = 0,1,..7n2, k2 = 0,1

This is actually the Radix-2 algorithm,

Decimation-in-Frequency Algorithm.

Or (n2 ,n1) n(0,0) 0

Or (k2 ,k1) n

(0,0) 0

(0,1) 2

36


71



Find the mapping for N = N2N1 = 2 x8,

n 8n2 + n1

k k2 + 2k1 for n1, k1 = 0,1,..7n2, k2 = 0,1

Or (n2 ,n1) n(0,0) 0(0,1) 1(0,2) 2(0,3) 3(0,4) 4(0,5) 5(0,6) 6(0,7) 7(1,0) 8(1,1) 9(1,2) 10(1,3) 11(1,4) 12(1,5) 13(1,6) 14(1,7) 15

k k2 + 2k1Or (k2 ,k1) n

(0,0) 0

(0,1) 2

(0,2) 4(0,3) 6

(0,4) 8

(0,5) 10

(0,6) 12

(0,7) 14

(1,0) 1

(1,1) 3

(1,2) 5

(1,3) 7

(1,4) 9(1,5) 11

(1,6) 13

(1,7) 15

Return-DIT

Return-DIF


72



The Endof Chapter 6

37


73



Number of Complex Multiplications is

M = ½ N log2NNow each complex multiplication requires

4 real multiplications and 2 real additions

For this complex multiplications, we requireMreal = (½ N log2N).4 real multiplications

= 2N log2N real multiplications

Areal = (½ N log2N).2 real adds= N log2N real adds

Number of Complex Adds required is

A = N log2NObviously, each complex add requires

2 real additionHence the number of real adds for these complex adds becomes

Areal = (N log2N).2 real adds = 2N log2N real adds

Hence the total number of Adds becomesA = N log2N + 2N log2N

= 3N log2N

Return


74



The Endof Chapter 6

38


75



(i) for the 1st stage

X(k) = X1(k) +Wk X2(k)

for k = 0,1,…,7 W n = e2/N = e2/8

(ii) for the 2nd stage

for k = 0,1,2,3 (W1)n = (W2)n =e2.2/N = e2/4

(iii) for the 3rd stage

for k = 0, 1 and (W2)

n = (W12)n =e4.2/N = e2/2

k411 W)4(x)0(x)k(X

k412 W)6(x)2(x)k(X

k421 W)5(x)1(x)k(X

k422 W)7(x)3(x)k(X

3 Set of Equations: (p.21-24) Wn = Twiddle factor

)k(XW)k(X)k(X 12k2

111 )k(XW)k(X)k(X 22

k2212

Return, p.28


76



The Endof Chapter 6

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