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?

  • 2
    $\begingroup$ Anywhere you have an $n$, you want to replace it with a $-n$. $\endgroup$
    – Baddioes
    Commented May 8 at 5:13
  • 4
    $\begingroup$ Why don't you write down the convolution sum, so you can be sure what's going on? $\endgroup$
    – Matt L.
    Commented May 8 at 9:18
  • $\begingroup$ 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? $\endgroup$ Commented May 12 at 11:49
  • $\begingroup$ 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. $\endgroup$ Commented May 12 at 11:55

1 Answer 1

  1. The operation of $n \to -n$ is basically reflecting the signal.
  2. The convolution with shifted Kronecker Delta is the shifting operator.

So basically the composition of those operations is a shifted reflected signal.


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