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circular conv

Started by harima March 28, 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;



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); _____________________________________
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