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

3 visualizzazioni (ultimi 30 giorni)
AndyK
AndyK il 12 Dic 2022
Risposto: Paul il 12 Dic 2022
Hi all,
I need help to reproduce a convolution example presented here:
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 Commento
Bruno Luong
Bruno Luong il 12 Dic 2022
Modificato: Bruno Luong il 12 Dic 2022
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

Accedi per commentare.

Risposte (1)

Paul
Paul il 12 Dic 2022
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

Prodotti


Release

R2022b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by