I'm trying to prove convolution in time domain is same as multiplication in frequency domain but I'm not getting the same answer in matlab.
Here is the code:
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You need to set the length as well in your fft command.
The two signals are of length $5$ and their convolution is of length $5+5-1=9$. So use this instead:
a=[1 2 3 4 5];
b=[6 7 8 9 10];
x=fft(a,9);
y=fft(b,9);
ifft(x.*y)
ans =
Columns 1 through 8
6.0000 19.0000 40.0000 70.0000 110.0000 114.0000 106.0000 85.0000
Column 9
50.0000
conv(a,b)
ans =
6 19 40 70 110 114 106 85 50
-
$\begingroup$ Thanks a lot man I've been struggling a lot for this.No one has even asked any question regarding this. $\endgroup$ – bharath Jul 27 '17 at 7:13
cconv:cconv(a,b)Solution 2 : Try adding cyclic prefix toafor example.aa=[a a]; conv(aa,b);The result isifft(x.*y)from the end of cyclic prefix, in this case since the index 6. $\endgroup$ – AlexTP Jul 27 '17 at 7:06