1.12 The fft algorithm

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The FFT, an efficient way to compute the DFT, is introduced and derived throughout this module.

FFT
(Fast Fourier Transform) An efficient computational algorithm for computing the DFT .

The fast fourier transform fft

DFT can be expensive to compute directly $\forall k, 0\le k\le N-1\colon X(k)=\sum_{n=0}^{N-1} x(n)e^{-(i\times 2\pi \frac{k}{N}n)}$

For each $k$ , we must execute:

• $N$ complex multiplies
• $N-1$ complex adds
The total cost of direct computation of an $N$ -point DFT is
• $N^{2}$ complex multiplies
• $N(N-1)$ complex adds
How many adds and mults of real numbers are required?

This " $O(N^{2})$ " computation rapidly gets out of hand, as $N$ gets large:

 $N$ 1 10 100 1000 $10^{6}$ $N^{2}$ 1 100 10,000 $10^{6}$ $10^{12}$

The FFT provides us with a much more efficient way of computing the DFT. The FFT requires only " $O(N\lg N)$ " computations to compute the $N$ -point DFT.

 $N$ 10 100 1000 $10^{6}$ $N^{2}$ 100 10,000 $10^{6}$ $10^{12}$ $N\lg N$ 10 200 3000 $6E6$

How long is $10^{12}\mathrm{sec}$ ? More than 10 days! How long is $6E6\mathrm{sec}$ ?

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

How does the fft work?

• The FFT exploits the symmetries of the complex exponentials ${W}_{N}^{(kn)}=e^{-(i\frac{2\pi }{N}kn)}$
• ${W}_{N}^{(kn)}$ are called " twiddle factors "

Complex conjugate symmetry

${W}_{N}^{(k(N-n))}={W}_{N}^{-(kn)}=\overline{{W}_{N}^{(kn)}}$ $e^{-(i\times 2\pi \frac{k}{N}(N-n))}=e^{i\times 2\pi \frac{k}{N}n}=\overline{e^{-(i\times 2\pi \frac{k}{N}n)}}$

Periodicity in n and k

${W}_{N}^{(kn)}={W}_{N}^{(k(N+n))}={W}_{N}^{((k+N)n)}$ $e^{-(i\frac{2\pi }{N}kn)}=e^{-(i\frac{2\pi }{N}k(N+n))}=e^{-(i\frac{2\pi }{N}(k+N)n)}$ ${W}_{N}=e^{-(i\frac{2\pi }{N})}$

Decimation in time fft

• Just one of many different FFT algorithms
• The idea is to build a DFT out of smaller and smaller DFTs by decomposing $x(n)$ into smaller and smaller subsequences.
• Assume $N=2^{m}$ (a power of 2)

Derivation

$N$ is even , so we can complete $X(k)$ by separating $x(n)$ into two subsequences each of length $\frac{N}{2}$ . $x(n)\to \begin{cases}\frac{N}{2} & \text{if n=\mathrm{even}}\\ \frac{N}{2} & \text{if n=\mathrm{odd}}\end{cases}$ $\forall k, 0\le k\le N-1\colon X(k)=\sum_{n=0}^{N-1} x(n){W}_{N}^{(kn)}$ $X(k)=\sum_{n=2r} x(n){W}_{N}^{(kn)}+\sum_{n=2r+1} x(n){W}_{N}^{(kn)}$ where $0\le r\le \frac{N}{2}-1$ . So

$X(k)=\sum_{r=0}^{\frac{N}{2}-1} x(2r){W}_{N}^{(2kr)}+\sum_{r=0}^{\frac{N}{2}-1} x(2r+1){W}_{N}^{((2r+1)k)}=\sum_{r=0}^{\frac{N}{2}-1} x(2r){W}_{N}^{2}^{(kr)}+{W}_{N}^{k}\sum_{r=0}^{\frac{N}{2}-1} x(2r+1){W}_{N}^{2}^{(kr)}$
where ${W}_{N}^{2}=e^{-(i\frac{2\pi }{N}\times 2)}=e^{-(i\frac{2\pi }{\frac{N}{2}})}={W}_{\frac{N}{2}}$ . So $X(k)=\sum_{r=0}^{\frac{N}{2}-1} x(2r){W}_{\frac{N}{2}}^{(kr)}+{W}_{N}^{k}\sum_{r=0}^{\frac{N}{2}-1} x(2r+1){W}_{\frac{N}{2}}^{(kr)}$ where $\sum_{r=0}^{\frac{N}{2}-1} x(2r){W}_{\frac{N}{2}}^{(kr)}$ is $\frac{N}{2}$ -point DFT of even samples ( $G(k)$ ) and $\sum_{r=0}^{\frac{N}{2}-1} x(2r+1){W}_{\frac{N}{2}}^{(kr)}$ is $\frac{N}{2}$ -point DFT of odd samples ( $H(k)$ ). $\forall k, 0\le k\le N-1\colon X(k)=G(k)+{W}_{N}^{k}H(k)$ Decomposition of an $N$ -point DFT as a sum of 2 $\frac{N}{2}$ -point DFTs.

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

Cost to compute an $N$ -point DFT is approximately $N^{2}$ complex mults and adds.
But decomposition into 2 $\frac{N}{2}$ -point DFTs + combination requires only $\left(\frac{N}{2}\right)^{2}+\left(\frac{N}{2}\right)^{2}+N=\frac{N^{2}}{2}+N$ where the first part is the number of complex mults and adds for $\frac{N}{2}$ -point DFT, $G(k)$ . The second part is the number of complex mults and adds for $\frac{N}{2}$ -point DFT, $H(k)$ . The third part is the number of complex mults and adds for combination. And the total is $\frac{N^{2}}{2}+N$ complex mults and adds.

Savings

For $N=1000$ , $N^{2}=10^{6}$ $\frac{N^{2}}{2}+N=\frac{10^{6}}{2}+1000$ Because 1000 is small compared to 500,000, $\frac{N^{2}}{2}+N\approx \frac{10^{6}}{2}$

So why stop here?! Keep decomposing. Break each of the $\frac{N}{2}$ -point DFTs into two $\frac{N}{4}$ -point DFTs, etc. , ....

We can keep decomposing: $\frac{N}{2^{1}}=\{\frac{N}{2}, \frac{N}{4}, \frac{N}{8}, , \frac{N}{2^{(m-1)}}, \frac{N}{2^{m}}\}=1$ where $m=\log_{2}N=\text{times}$

Computational cost: $N$ -pt DFTtwo $\frac{N}{2}$ -pt DFTs. The cost is $N^{2}\to 2\left(\frac{N}{2}\right)^{2}+N$ . So replacing each $\frac{N}{2}$ -pt DFT with two $\frac{N}{4}$ -pt DFTs will reduce cost to $2(2\left(\frac{N}{4}\right)^{2}+\frac{N}{2})+N=4\left(\frac{N}{4}\right)^{2}+2N=\frac{N^{2}}{2^{2}}+2N=\frac{N^{2}}{2^{p}}+pN$ As we keep going $p=\{3, 4, , m\}$ , where $m=\log_{2}N$ . We get the cost $\frac{N^{2}}{2^{\log_{2}N}}+N\log_{2}N=\frac{N^{2}}{N}+N\log_{2}N=N+N\log_{2}N$ $N+N\log_{2}N$ is the total number of complex adds and mults.

For large $N$ , $\mathrm{cost}\approx N\log_{2}N$ or " $O(N\log_{2}N)$ ", since $(N\log_{2}N, N)$ for large $N$ .

Weird order of time samples

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research.net
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sciencedirect big data base
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or in general
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in general
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On having this app for quite a bit time, Haven't realised there's a chat room in it.
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how did you get the value of 2000N.What calculations are needed to arrive at it
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