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Say, I have $x[n]$ and $y[n]$ be periodic sequences of period $N$. Let us define $z[n] =x[n] \ast y[n]$ where $\ast$ denote the circular convolution operation.

What is $x[n+\delta] \ast y[n+\delta]?$

My answer:

Taking FFT: $Z[k] =X[k]Y[k]$. Now, FFT of $x[n+\delta] =e^{j2\pi\delta n/N}X[k]$ and $y[n+\delta] =e^{j2\pi\delta n/N}Y[k]$. Since we know circular convolution translates to multiplication of FFTs, we have the FFT of $x[n+\delta] \ast y[n+\delta]$ equal to $e^{j2\pi 2\times \delta n/N}X[k]Y[k] = e^{j2\pi 2\times \delta n/N}Z[k].$

Now, note that $e^{j2\pi 2\times \delta n/N}Z[k]$ has an IFFT of $z[n+2\delta]$.

So, $x[n+\delta] \ast y[n+\delta] = z[n+2\delta]$.

Am I correct?

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First of all FFT (Fast Fourier Transform) is just a fast implementation of the DFT (Discrete Fourier Transform) so the operation in your attempt is the DFT.

Next let's consider the expression: $x[n+\Delta] \ast y[n+\Delta]$ (lowercase delta, $\delta$, is used for the Dirac delta function so I changed the variable to uppercase delta, $\Delta$) in the context of regular linear convolution.

Remember that convolution is commutative, distributive and associative. With that in mind we can rewrite the expression as:

\begin{align*} x[n+\Delta] \ast y[n+\Delta] &= \overbrace{\left(x[n] \ast \delta[n+\Delta] \right)}^{x[n+\Delta]} \ast \overbrace{\left(y[n] \ast \delta[n+\Delta] \right)}^{y[n+\Delta]}\\ & = \overbrace{(x[n]\ast y[n])}^{z[n]} \ast \overbrace{(\delta[n+\Delta] \ast \delta[n+\Delta])}^{\delta[n+2\Delta]}\\ & = z[n+2\Delta] \end{align*}

Now I did not write this in the context of circular convolution because shift, $\Delta$, is a circular shift in that case. Meaning whatever falls off on one side of a finite sequence comes back up on the other side. So you have to take that into consideration!

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