convolution product calculater impulse response
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amjad hammad
on 17 Feb 2022
Commented: amjad hammad
on 19 Feb 2022
can anyone tell me where is the problem
LDKI
x(n)=[1,3,-2,5,7]
h(n)= [1,5,8,-5,13]
x=[1,3,-2,5,7]
h= [1,5,8,-5,13]
dx=length(x)
dh=length(h)
dy=dx+dh-1
for i=1:dy
y1(i)=0;
for j=1:dx
y1(i)=y1(i)+x(j)*h(i-j+1)
end
end
check1=conv(x1,h1)
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Accepted Answer
David Goodmanson
on 19 Feb 2022
Edited: David Goodmanson
on 19 Feb 2022
Hi amjad,
You have to keep the indices within the bounds set by dx and dh. For the j loop, you can use the line
for j=max(1,i-dh+1):min(i,dx)
Also, in case you run the script more than once with different x and h, it pays to initialize y1 with
y1 = zeros(1,dy);
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