1
$\begingroup$

I have a question whether circular convolution and periodically expanded linear convolution corresponds in following case or why it does not?

Think about a signal $x[t]$ and a signal $y[t]$ both of length $N$. We expand the singal $x[t]$ periodically to a signal $x'[t]=[x[t],x[t]]$ with length $2N$. We zeropadd the signal $y[t]$ to a signal $y'[t] = [y[t],0]$ to obtain the same length $2N$.

My question is does the circular convolution result $(x[t] \star y[t] )_{mod_N}$, correspond to the bins $[N...2N]$ in the linear convolution result $x'[t] * y'[t]$?

Best

$\endgroup$
1
$\begingroup$

Yes, they are the same.

Let us take the linear convolution of $x[t]$ and $y[t]$ as 2 portions $H_1$ of length $N$ corresponding to first $N$ samples, and $H_2$ of remaining $N-1$ samples. Circular convolution between $x$ and $y$ causes $H_2$ to overlap over $H_1$ because of time-aliasing. So the first $N-1$ samples of the result is $H_1 + H_2$ with only the last sample being correct (corresponding to linear convolution).

Now, for the linear convolution between $\tilde{x}$ and $\tilde{y}$, the samples from $N$ to $2N-1$ will have additional component of first $N$ samples of $\textrm{conv}(x,y)$. So these samples will correspond circular convolution of $x$ and $y$.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.