signal processing fundamentals – part i spectrum analysis ...enpklun/eie327/fft2.pdf · 14 signal...

1

6.6. Fast Fourier TransformFast Fourier Transform

Signal Processing Fundamentals – Part ISpectrum Analysis and Filtering

6. Fast Fourier Transform

-1

x[0]

x[4]x[2]

x[6]

x[1]

x[5]x[3]

x[7]

X[0]

X[1]X[2]

X[3]

X[4]

X[5]X[6]

X[7]WN3WN2WN1WN0

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1-1WN1

WN0

WN1WN0

2

A Historical Perspective

• The Cooley and Tukey Fast Fourier Transform (FFT) algorithm is a turning point to the computation of DFT

• Before that, DFT was never practical except running by some very expensive computers

• FFT gives nearly a factor of 1000 improvement in the computation speed over the traditional approach in most computing environments



3



• Principal discoveries of efficient methods of computing the DFT are listed as follows:

Researcher(s) Date Feature

Cooley & Tukey 1965 Any 2n composite integer

I.J. Good 1958 Any integer with relatively prime factors

L.H. Thomas 1948 Any integer with relatively prime factors

Danielson 1942 2n

K. Stumpff 1939 2nk, 3nk

C. Runge 1903 2nk

J.D. Everett 1860 12

A. Smith 1846 4, 8, 16, 32

F. Carlini 1828 12

C.F. Gauss 1805 Any composite integer

4



• Basic principle of FFT

Divide and Conquer

)(cost)(cost)(cost

problemoriginaloverheadssubproblem

<

+∑• Need to satisfy the following criterion

⇒ To break down a big problem to a number of smaller problems and tackle them individually

5



)(cost)(cost)(cost


<

+∑

Big Problem SmallerProblem

SmallerProblem

SmallerProblems

OverheadApply Divide and Conquer iteratively

6

Operation of FFT



Length-N/4 DFTs

For each round, the size of the problem is divided by 2

Length-N DFT Length-N/2DFT

Length-N/2DFT

Overhead

7



Mathematical Derivation• Recall the DFT of x[n], for n = 0, 1,…,N-1

∑−

==

1

0][][

N

n

nkNWnxkX

• Separate the summation into two groups

∑∑ +=nodd

nkN

neven

nkN WnxWnxkX ][][][

for k = 0, 1,…,N-1

for k = 0, 1,…,N-1

NnkjnkN eWwhere

/2π−=

8



Example• If DFT of x[n] is

kkkk WxWxWxWxkX .34.2

4.1

4.0

4 ]3[]2[]1[]0[][ +++=

• We separate it into two groups with odd and even n

{ } { }∑∑ +=

+++=

nodd

nk

neven

nk

kkkk

WnxWnx

WxWxWxWxkX

44

.34

.14

.24

.04

][][

]3[]1[]2[]0[][

9



∑∑

∑∑−

=

+−

=++=

+=

12/

0

)12(12/

0

2 ]12[]2[

][][][

N

r

krN

N

r

rkN

nodd

nkN

neven

nkN

WrxWrx

WnxWnxkX

for k = 0, 1,…,N-1

∑∑−

=

−

=++=

12/

0

212/

0

2 ]12[]2[N

r

rkN

kN

N

r

rkN WrxWWrx

10



Length-N/2 DFT of data with even n

Overhead

][][

]12[]2[][12/

02/

12/

02/

kHWkG

WrxWWrxkX

kN

N

r

rkN

kN

N

r

rkN

+=

++= ∑∑−

=

−

=

• Since

rkN

NrkjNrkjrkN

W

eeW

2/

)2//(2/222

=

== −− ππ

for k = 0, 1,…,N-1

Length-N/2 DFT of data with odd n

11



Example: N = 8

N/2 = 4 point DFT

x[0]

x[2]

x[4]

x[6]

N/2 = 4 point DFT

x[1]

x[3]x[5]

x[7]

G[0]

G[1]

G[2]G[3]

H[0]

H[1]

H[2]H[3]

X[0]

X[1]X[2]

X[3]

X[4]

X[5]X[6]

X[7]W86W85W84

W83W82W81W80

W87

12



How much it saves?• Assume x[n] has N data and each data is a

complex number

∑−

==

1

0][][

N

n

nkNWnxkX for k = 0, 1,…,N-1

• For each k, it needs N complex multiplications and N-1 complex additions

• In overall, it needs N2 complex multiplications and N(N-1) complex additions

13



• FFT converts an N-point DFT to 2 N/2-points DFTs

][][][ kHWkGkX kN+= for k = 0, 1,…,N-1

(N/2)2 complex multiplicationsN/2(N/2-1) complex additions

N complex multiplications

14



262,144x/261,632+ 131,584x/131,072+

N DFT 1-stage FFT4 16x/12+ 12x/8+

8 64x/56+ 40x/32+

16 256x/240+ 144x/128+

32 1,024x/992+ 544x/512+

64 4,096x/4,032+ 2,112x/2,048+

128 16,384x/16,256+ 8,320x/8,192+

256 65,536x/65,280+ 33,024x/32,768+

1024 1,048,576x/1,047,552+ 525,312x/524,288+

512

15

)(cost)(cost)(cost


<

+∑



• Obviously

• We can re-apply the same approach to the decomposed N/2 problems

16



Example: Apply to G[k] and H[k]x[0]

x[4]

x[2]

x[6]

x[1]

x[5]

x[3]

x[7]

X[0]

X[1]X[2]

X[3]

X[4]

X[5]X[6]

X[7]W87W86W85W84

W83W82W81W80

N/4 = 2 point DFT

N/4 = 2 point DFT

N/4 = 2 point DFT

N/4 = 2 point DFT

W43

W42

W41W40

W43

W42W41W40

17



• The complexity is further reduced

• The above decomposition is recursively performed until a length-2 DFT is computed

• A length-2 DFT can be implemented as a butterflyas follows:

12

02

02

02

2

1

0

]1[]0[]1[

]1[]0[]0[

][][

WxWxX

WxWxX

WnxkX nkn

+=

+=⇒

= ∑=

x[0]

x[1]

X[0]

X[1]

W20 1=

12/212 −==− πjeW

18



x[0]

x[4]x[2]

x[6]

x[1]

x[5]x[3]

x[7]

X[0]

X[1]X[2]

X[3]

X[4]

X[5]X[6]

X[7]W87W86W85W84

W83W82W81W80

W43

W42

W41W40

W43

W42W41W40

-1

-1

-1

-1

19



• The complexity can be further reduced. For every butterfly, it can be converted as follows:

rNW

)2/( NrNW

+ rNW

1

-1

rNW

rNW−

rN

jrN

NN

rN

NrN WeWWWW −===

−+ π2/)2/(Since

20



-1

x[0]

x[4]x[2]

x[6]

x[1]

x[5]x[3]

x[7]

X[0]

X[1]X[2]

X[3]

X[4]

X[5]X[6]

X[7]W83W82W81W80

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1-1W41

W40

W41W40

FFTTotal complex multiplication: 5Total complex addition: 24

DFTTotal complex multiplication: 64Total complex addition: 56

W20

W20

W20

W20

21



Remarks:• FFT of the form above is called decimation-in-time

(DIT) FFT (or Cooley and Tukey FFT)• In general, the complexity of DIT FFT is

(N/2)log2N complex multiplications (W0 is considered as multiplication in this expression)

Nlog2N complex additions

• DIT FFT requires N to be a power of 2• If N is not a power of 2, need zero padding to let N

be a power of 2 before FFT • More than a hundred of FFT algorithms after DIT

FFT for different applications

22



Exercise1. Show why the complexity of DIT FFT is given

by

(N/2)log2N complex multiplications (W0 is considered as multiplication in this expression)

Nlog2N complex additions

2. If N = 256, what are the complexities of DIT FFT and DFT?

23



Exercise (cont)

3. If x[n] has the values {8 7 6 5 4 3 2 1}, use FFT to calculate its spectrum.

24



Solution

1. For a length-N DFT, there will be log2(N) stages. For each stage, there will be N/2 butterflies. For each butterfly, there will 1 complex multiplication and 2 complex additions. Hence the total no. of complex multiplications is N/2log2N. And the number of additions is Nlog2N

25



2. If N = 256, DFT requires N2 = 2562 = 65536 complex multiplications and N(N - 1) = 256(255) = 65280 complex additions.

By using FFT, it requires N/2log2N = 128*8 = 1024 complex multiplications and Nlog2N = 256*8 = 2048 complex additions

26



3. If x[n] has the values {8 7 6 5 4 3 2 1}, its spectrum can be calculated using FFT as follows:

After 1st stage, the output is

{12 4 8 4 10 4 6 4}

After 2nd stage, the output is

{20 4-j4 4 4+j4 16 4-j4 4 4+j4}

After 3rd stage, the output is

{36 4-j9.657 4-j4 4-j1.657

4 4+j1.657 4+j4 4+j9.657}

signal processing fundamentals – part i spectrum analysis ...enpklun/eie327/fft2.pdf · 14 signal...

Documents