# Convolution with Kronecker delta

I know that convolution with delta shifts a signal. As for example, $$x \!\left[ n \right] * \delta \!\left[ n - 2 \right] = x \!\left[ n - 2 \right]$$. How to do convolution with $$x \!\left[ -n \right]$$?

Is $$x \!\left[ -n \right] * \delta \!\left[ n - 2 \right] = x \!\left[ -n + 2 \right]$$ or $$x \!\left[ -n - 2 \right]$$?

I think $$\delta \!\left[ n - 2 \right] = \delta \!\left[ 2 - n \right]$$, then $$x \!\left[ -n \right] * \delta \!\left[ 2 - n \right] = x \!\left[ 2 - n \right]$$. Is this approach correct?

• Anywhere you have an $n$, you want to replace it with a $-n$. Commented May 8 at 5:13
• Why don't you write down the convolution sum, so you can be sure what's going on? Commented May 8 at 9:18
• expanding on Matt’s reply: $$\delta[n] * x[-n] = \sum_{k = -\infty}^{\infty} \delta[k] x[(-n)-k]$$ - which values of k aren’t 0 in the sum? Commented May 12 at 11:49
• it might be better to define $g[n] = x[-n]$. now we know $$x[-n] * \delta[n] = (g*\delta)[n] = \sum_{k=-\infty}^{\infty} g[k] \delta [n-k] = \sum_{k=-\infty}^{\infty} x[-k] \delta [n-k]$$ - now think about which value of k isn’t zeroed out by the delta. Commented May 12 at 11:55

1. The operation of $$n \to -n$$ is basically reflecting the signal.