# Does convolution of delayed sequences produces a delayed output?

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?

## 1 Answer

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!