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# The FFT Algorithm

Module by: Robert Nowak. E-mail the author

Summary: This module introduces CTFT, DTFT, DFT and frequency domain.

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## The Fast Fourier Transform FFT

An efficient computational algorithm for computing the DFT.

DFT can be expensive to compute directly k,0kN1:Xk=n=0N1xne(i2πkNn) k 0 k N 1 X k n 0 N 1 x n 2 k N n

For each k, we must execute N complex multiplies and N-1 complex adds. The total cost of direct computation of an N-point DFT is N2 N 2 complex mults and N(N1) N N 1 complex adds. How many adds and mults of real #'s?

This " N2 N 2 " computation rapidly gets out of hand, as N gets large:

Table 1
NN 1 10 100 1000 106 106
N2 N2 1 100 10,000 106 106 1012 1012

The FFT is much more efficient for computing the DFT. FFT requires only " ONlogN O N N " computations to compute the N-point DFT.

Table 2
NN 10 100 1000 106 106
N2 N2 100 10,000 106 106 1012 1012
NlogN N 10 N 10 200 3000 6×106 6 106

How long is 1012 10 12 μμsec? More than 10 days! How long is 6×106 6 106 μμsec?

The FFT and digital computers revolutionized DSP (1960 - 1980).

## How does the FFT work?

• The FFT exploits the symmetries of the complex exponentials WN kn=e(i2πNkn) WN k n 2 N k n
• WN kn WN k n are called "twiddle factors"

### Symmetry 1: Complex Conjugate Symmetry

WN k(Nn)= WN (kn)= WN kn¯ WN k N n WN k n WN k n e(i2πkN(Nn))=ei2πkNn=e(i2πkNn)¯ 2 k N N n 2 k N n 2 k N n

### Symmetry 2: Periodicity in n and k

WN kn= WN k(N+n)= WN (k+N)n WN k n WN k N n WN k N n e(i2πNkn)=e(i2πNk(N+n))=e(i2πN(k+N)n) 2 N k n 2 N k N n 2 N k N n WN =e(i2πN) WN 2 N

## Decimation in Time FFT

• Just one of many different FFT algorithms
• IDEA: Build a DFT out of smaller and smaller DFTs by decomposing xn x n into smaller and smaller subsequences
• Assume N=2m N 2 m (a power of 2)

### Derivation

N is even, so we can complete Xk X k by separating xn x n into two subsequences each of length N2 N 2 . xn{N2  if  n=evenN2  if  n=odd x n N 2 n even N 2 n odd k,0kN1:Xk=n=0N1xn WN kn k 0 k N 1 X k n 0 N 1 x n WN k n Xk=n=2rxn WN kn+n=2r+1xn WN kn X k n 2 r x n WN k n n 2 r 1 x n WN k n where 0rN21 0 r N 2 1 , So Xk=r=0N21x2r WN 2kr+r=0N21x2r+1 WN (2r+1)k=r=0N21x2r WN 2kr+ WN kr=0N21x2r+1 WN 2kr X k r 0 N 2 1 x 2 r WN 2 k r r 0 N 2 1 x 2 r 1 WN 2 r 1 k r 0 N 2 1 x 2 r WN 2 k r WN k r 0 N 2 1 x 2 r 1 WN 2 k r where WN 2=e(i2πN2)=e(i2πNN2)= W N 2 WN 2 2 N 2 2 N N 2 W N 2 , So Xk=r=0N21x2r W N 2 kr+ WN kr=0N21x2r+1 W N 2 kr X k r 0 N 2 1 x 2 r W N 2 k r WN k r 0 N 2 1 x 2 r 1 W N 2 k r where r=0N21x2r W N 2 kr r 0 N 2 1 x 2 r W N 2 k r is N2 N 2 -point DFT of even samples Gk G k and r=0N21x2r+1 W N 2 kr r 0 N 2 1 x 2 r 1 W N 2 k r is N2 N 2 -point DFT of odd samples Hk H k . k,0kN1:Xk=Gk+ WN kHk k 0 k N 1 X k G k WN k H k Decomposition of an N-point DFT as a sum of 2 N2 N 2 -point DFTs.

Why would we want to do this? Because it is more efficient!

#### Recall:

Cost to compute an N-point DFT is approximately N2 N 2 complex mults and adds.
But decomposition into 2 N2 N 2 -point DFTs + combination requires only N22+N22+N=N22+N N 2 2 N 2 2 N N 2 2 N where the first part is the number of complex mults and adds for N2 N 2 -point DFT Gk G k . The second part is the number of complex mults and adds for N2 N 2 -point DFT Hk H k . The third part is the number of complex mults and adds for combination. And the total is N22+N N 2 2 N complex mults and adds.

#### Example 1: Savings

For N=1000 N 1000 , N2=106 N 2 10 6 N22+N=1062+1000 N 2 2 N 10 6 2 1000 Because 1000 is small compare to 500,000, N22+N1062 N 2 2 N 10 6 2

So why stop here?! Keep decomposing. Break each of the N2 N 2 -point DFTs into two N4 N 4 -point DFTs, etc, ....

We can keep decomposing: N2N4N8N2m1N2m=1 N 2 N 4 N 8 N 2 m 1 N 2 m 1 where m=log2N m 2 N times.

Computational cost: N N-pt DFT  two N2 N 2 -pt DFTs. The cost is N22N22+N N 2 2 N 2 2 N . So replacing each N2 N 2 -pt DFT with two N4 N 4 -pt DFTs will reduce cost to 2(2N42+N2)+N=4N42+2N=N222+2N=N22p+pN 2 2 N 4 2 N 2 N 4 N 4 2 2 N N 2 2 2 2 N N 2 2 p p N As we keep going p=34m p 3 4 m , where m=log2N m 2 N . We get the cost N22log2N+Nlog2N=N2N+Nlog2N=N+Nlog2N N 2 2 2 N N 2 N N 2 N N 2 N N N 2 N N+Nlog2N N N 2 N is the total number of complex adds and mults.

For large N, costNlog2N cost N 2 N or " ONlog2N O N 2 N ", since Nlog2NN N 2 N N for large N.

### Note:

Weird order of time samples

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