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178 Chapter 5 THE DISCRETE FOURIER TRANSFORM
% Check for length of x2
if length(x2) > N
error(’N must be >= the length of x2’)
end
x1=[x1 zeros(1,N-length(x1))];
x2=[x2 zeros(1,N-length(x2))];
m = [0:1:N-1]; x2 = x2(mod(-m,N)+1); H = zeros(N,N);
for n = 1:1:N
H(n,:) = cirshftt(x2,n-1,N);
end
y = x1*conj(H’);
Problems P5.24 and P5.25 explore an approach to eliminate the for...
end loop in the circonvt function. The third approach would be to implement
the frequency-domain operation (5.40) using the dft function.
This is explored in Problem P5.26.
□ EXAMPLE 5.14 Let us use MATLAB to perform the circular convolution in Example 5.13.
Solution The sequences are x 1(n) ={1, 2, 2} and x 2(n) ={1, 2, 3, 4}.
MATLAB script:
>> x1 = [1,2,2]; x2 = [1,2,3,4]; y = circonvt(x1, x2, 4)
y =
15 12 9 14
Hence
x 1(n) 4○ x 2(n) ={15, 12, 9, 14}
as before.
□
□ EXAMPLE 5.15 In this example we will study the effect of N on the circular convolution. Obviously,
N ≥ 4; otherwise there will be a time-domain aliasing for x 2(n). We will
use the same two sequences from Example 5.13.
a. Compute x 1(n) 5○ x 2(n).
b. Compute x 1(n) 6○ x 2(n).
c. Comment on the results.
Solution
The sequences are x 1(n) ={1, 2, 2} and x 2(n) ={1, 2, 3, 4}. Even though the
sequences are the same as in Example 5.14, we should expect different results
for different values of N. This is not the case with the linear convolution, which
is unique, given two sequences.
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