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

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

답변 (1개)

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

카테고리

질문:

DDD
2015년 5월 19일

답변:

2015년 5월 20일

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