Stem time convolution using conv, filter, cconv and multiplication in frequency domain

1 vue (au cours des 30 derniers jours)
It is asked to graph time convolution using conv, filter, cconv and multiplication in frequency domain. All the answers seem to agree except the multiplication in frequency domain. What is wrong with it? Any suggestions?
u=@(n)1.0.*(n>=2);
y=@(n)abs(2.*n+1).*(u(n+1)-u(n-5));
n_x=1:10;
n_y=-2:20;
y_n=u(n_y);
x_n=y(n_x);
c=conv(x_n,y_n);
n_c= n_x(1)+n_y(1):n_x(end)+n_y(end);
figure()
%
subplot(4,1,1)
stem(n_c,c)
xlabel('n')
xlim([-3,15])
title('u(n) * |2n+1|(u(n+1)-u(n-5)) using conv')
%
cf=filter(x_n,1,y_n);
cf=[cf,zeros(1,length(n_c)-length(cf))];
subplot(4,1,2)
stem(n_c,cf)
xlabel('n')
xlim([-3,15])
title('u(n) * |2n+1|(u(n+1)-u(n-5)) using filter')
%
cc=cconv(y_n,x_n);
subplot(4,1,3)
stem(n_c,cc)
xlabel('n')
xlim([-3,15])
title('u(n) * |2n+1|(u(n+1)-u(n-5)) using cconv')
subplot(4,1,4)
z=ifft(fft(x_n).*fft(y_n));
stem(n_ ,z);
xlabel('n')
xlim([-3,15])
title('u(n) * |2n+1|(u(n+1)-u(n-5)) multiplying in frequency domain')
Thanks

Réponses (1)

Nalini Vishnoi
Nalini Vishnoi le 20 Mai 2015
Hi,
Your last section when executed gave error because you are multiplying two vectors (element by element) when they are of different lengths. The correct way to use 'fft' to perform 'convolution' is the following:
% ensures the vectors being multiplied are of the same size/length
z=ifft(fft(x_n, numel(n_c)).*fft(y_n, numel(n_c)));
stem(n_c,z);
I hope this helps.
Nalini

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