Convolution example: inputA[64],InputB[64] and Output[128]

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AndyK
AndyK 2022-12-12
回答： Paul 2022-12-12
Hi all,
I need help to reproduce a convolution example presented here:
https://www.keil.com/pack/doc/CMSIS/DSP/html/group__ConvolutionExample.html
A[k] = FFT(a[n],N)
B[k] = FFT(b[n],N)
conv(a[n], b[n]) = IFFT(A[k] * B[k], N)
/* ----------------------------------------------------------------------
* Test input data for Floating point Convolution example for 32-blockSize
* Generated by the MATLAB randn() function
* ------------------------------------------------------------------- */
float32_t Ak[MAX_BLOCKSIZE]; /* Input A */
float32_t Bk[MAX_BLOCKSIZE]; /* Input B */
float32_t AxB[MAX_BLOCKSIZE * 2]; /* Output */
  1 个评论
Bruno Luong
Bruno Luong 2022-12-12
编辑：Bruno Luong 2022-12-12
None of the standard (MATLAB) shape produce output of length nA+nB where nA and nB are the lenghs of a and b respectively (64 in the thread subject) , but
  • 'full': nA+nB-1
  • 'same': nA
  • 'valid': nA-nB+1

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回答（1 个）

Paul
Paul 2022-12-12
Ran in:
Hi AndyK,
I'm going to assume that N = na + nb - 1, where na and nb are the lengths of a[n] and b[n] respectively. Under this assumption, the code would look like this. If we make N > na + nb - 1, like to the nextpower of 2, then ci should have N - (na + nb - 1) trailing zeros (to within rounding error)
rng(100)
na = 5;
nb = 5;
a = rand(1,na);
b = rand(1,nb);
c = conv(a,b);
N = na + nb - 1;
A = fft(a,N);
B = fft(b,N);
C = A.*B;
ci = ifft(C);
c
c = 1×9
0.0661 0.3983 0.6871 0.6916 1.2684 0.9189 0.3635 0.4865 0.0027
ci
ci = 1×9
0.0661 0.3983 0.6871 0.6916 1.2684 0.9189 0.3635 0.4865 0.0027

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