User:M1ss1ontomars2k4/sandbox

From the definition of convolution, we know:

$$ \begin{align} (x* h)(n)& = \sum_{m=-\infty}^\infty{x[m]h[n-m]} \\

\end{align} $$

where $$h[n]$$ is the shifted delta, $$\delta[n-n_0]$$. Let us substitute this in to obtain:

$$ \sum_{m=-\infty}^\infty{x[m]\delta[(n-m)-n_0]} $$

Let $$k=n-m-n_0$$. Then we have:

$$ \begin{align} (x * h)(n)& = \sum_{m=-\infty}^\infty{x[m]\delta[(n-m)-n_0]} \\ & = \sum_{m=-\infty}^\infty{x[(n-n_0)-k]\delta[k]} \\ & = \sum_{k=-\infty}^\infty{x[(n-n_0)-k]\delta[k]} \\ & = (x*\delta)(n-n_0) \end{align} $$

i.e., the result of $$(x*h)$$ where $$h[n]$$ is the shifted delta is simply the convolution of $$x$$ and $$\delta$$, shifted by the same amount.