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, $\backslash delta[n-n\_0]$. 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 $\backslash delta$, shifted by the same amount.