# 2.11 Fast convolution (ticket #5015)

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Efficient computation of convolution using FFTs.

## Fast circular convolution

Since, $\sum_{m=0}^{N-1} x(m)h(n-m)\mod N=y(n)\mathrm{is equivalent to}Y(k)=X(k)H(k)$ $y(n)$ can be computed as $y(n)=\mathrm{IDFT}(\mathrm{DFT}(x(n))\mathrm{DFT}(h(n)))$

• ## Direct

• $N^{2}$ complex multiplies.
• $N(N-1)$ complex adds.
• ## Via ffts

• 3 FFTs + $N$ multipies.
• $N+\frac{3N}{2}\log_{2}N$ complex multiplies.
• $3N\log_{2}N$ complex adds.
If $H(k)$ can be precomputed, cost is only 2 FFts + $N$ multiplies.

## Fast linear convolution

DFT produces cicular convolution. For linear convolution, we must zero-pad sequences so that circular wrap-around alwayswraps over zeros.

To achieve linear convolution using fast circular convolution, we must use zero-padded DFTs of length $N\ge L+M-1$

Choose shortest convenient $N()$ (usually smallest power-of-two greater than or equal to $L+M-1$ ) $y(n)={\mathrm{IDFT}}_{N}({\mathrm{DFT}}_{N}(x(n)){\mathrm{DFT}}_{N}(h(n)))$

There is some inefficiency when compared to circular convolution due to longer zero-padded DFTs . Still, $O(\frac{N}{\log_{2}N})$ savings over direct computation.

## Running convolution

Suppose $L$ , as in a real time filter application, or $(L, M)$ . There are efficient block methods for computing fast convolution.

## Overlap-save (ols) method

Note that if a length- $M$ filter $h(n)$ is circularly convulved with a length- $N$ segment of a signal $x(n)$ ,

the first $M-1$ samples are wrapped around and thus is incorrect . However, for $M-1\le n\le N-1$ ,the convolution is linear convolution, so these samples are correct. Thus $N-M+1$ good outputs are produced for each length- $N$ circular convolution.

The Overlap-Save Method: Break long signal into successive blocks of $N$ samples, each block overlapping the previous block by $M-1$ samples. Perform circular convolution of each block with filter $h(m)$ . Discard first $M-1$ points in each output block, and concatenate the remaining points to create $y(n)$ .

Computation cost for a length- $N$ equals $2^{n}$ FFT per output sample is (assuming precomputed $H(k)$ ) 2 FFTs and $N$ multiplies $\frac{2\frac{N}{2}\log_{2}N+N}{N-M+1}=\frac{N(\log_{2}N+1)}{N-M+1}\mathrm{complex multiplies}$ $\frac{2N\log_{2}N}{N-M+1}=\frac{2N\log_{2}N}{N-M+1}\mathrm{complex adds}$

Compare to $M$ mults, $M-1$ adds per output point for direct method. For a given $M$ , optimal $N$ can be determined by finding $N$ minimizing operation counts. Usualy, optimal $N$ is $4M\le {N}_{\mathrm{opt}}\le 8M$ .

Zero-pad length- $L$ blocks by $M-1$ samples.

Add successive blocks, overlapped by $M-1$ samples, so that the tails sum to produce the complete linear convolution.

Computational Cost: Two length $N=L+M-1$ FFTs and $M$ mults and $M-1$ adds per $L$ output points; essentially the sames as OLS method.

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