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Discussion Groups | Matlab DSP | circular conv

Technical discussion about Matlab and issues related to Digital Signal Processing.

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circular conv - harima - Mar 28 8:50:00 2003

Hi there, Ive a doubt in finding circular convolution of two unequal sequences in MATLAB. Should the resultant length be equal to the highest of the given two vectors. Ive written a code for it but not sure about its result. Hope there would b some to help me. If neone have still optimised code, do post it. Thanks, Hari. x = [1 1 1 1 2 1]; h = [1 1 2 1]; Nx = length(x); Nh = length(h); N = max(Nx,Nh); x = [x zeros(1,N-Nx)]; h = [h zeros(1,N-Nh)]; m = 0:1:N-1; M = mod(-m,N) h = h(M+1) for n = 1:1:N m = n-1; p = 0:1:N-1; q = mod(p-m,N) hm = h(q+1) H(n,:) = hm end Y = x*H' k = 0:N-1; subplot(4,4,1); stem(k,Y); subplot(4,4,2); C = conv(x,h); K = 0:Nx+Nh-4;

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Re: circular conv - Claudio - Mar 31 17:11:00 2003

That's right, in circular convolution the resultant length is the size of the longest vector. You can test your results by using the circular convolution matrix, this link may help you: http://cnx.rice.edu/content/m10459/latest/ Regards, Claudio U. V. ----- Original Message ----- From: harima To: matlab Sent: Sunday, March 30, 2003 2:28 AM Subject: [matlab] circular conv Hi there, Ive a doubt in finding circular convolution of two unequal sequences in MATLAB. Should the resultant length be equal to the highest of the given two vectors. Ive written a code for it but not sure about its result. Hope there would b some to help me. If neone have still optimised code, do post it. Thanks, Hari. x = [1 1 1 1 2 1]; h = [1 1 2 1]; Nx = length(x); Nh = length(h); N = max(Nx,Nh); x = [x zeros(1,N-Nx)]; h = [h zeros(1,N-Nh)]; m = 0:1:N-1; M = mod(-m,N) h = h(M+1) for n = 1:1:N m = n-1; p = 0:1:N-1; q = mod(p-m,N) hm = h(q+1) H(n,:) = hm end Y = x*H' k = 0:N-1; subplot(4,4,1); stem(k,Y); subplot(4,4,2); C = conv(x,h); Yahoo! Groups Sponsor ADVERTISEMENT _____________________________________ /groups.php3 Your use of Yahoo! Groups is subject to the Yahoo! Terms of Service.