Why is this convolution true?

Question

I am a little puzzled by how the following summation has been written as a convolution, with one of the inputs reversed in time. Consider the following sum on the LHS, and the convolution on the RHS. I am trying to ascertain why they are equal.

$$ \sum_{i=-\infty}^{\infty} h[i-l] \ x[i] = h[-l] * x[l] $$

How/Why was the $h[-l]$ ascertained here? I have read and re-read the wiki on convolution and other links, but I cannot seem to figure out how you get it. Can someone please explain to me step by step what is going on here? Thanks!

EDIT:

Case1: I am told to assume $g[n] = h[-n]$. This seems to work for this case, but not for another case. In this case, I get:

$$ \sum_{i=-\infty}^{\infty} h[i-l] \ x[i] = \sum_{i=-\infty}^{\infty} g[-(i-l)] \ x[i] = \sum_{i=-\infty}^{\infty} g[l-i] \ x[i] = g[l] * x[l] = h[-l]*x[l] $$

This seems to work.

Case 2: However, if I try it on a different scenario, it does not work. Suppose we wanted to calculate the convolution sum of $h[k] * h^*[-k]$. The right answer is that:

$$ h[k] * h^*[-k] = \sum_{n=-\infty}^{\infty} h[n] \ h*[n+k] $$

But using this method, I would get:

$$ h[k] * h^*[-k] = h[k] * g^*[k] = \sum_{n=-\infty}^{\infty} h[n] \ g^*[k-n] = \sum_{n=-\infty}^{\infty} h[n] \ h^*[-(k-n)] = \sum_{n=-\infty}^{\infty} h[n] \ h^*[n-k] $$

...which is not the same as the right answer.

Answer 1

Yes, let $g$ be $h$ reversed, and rewrite as $\sum g[l-i] x[i]$, and that already shows you that your sum is $(g \ast x)[l]$. The quick memory trick is that in the sum above, if you sum the indices, $l-i+i = l$ tells you which entry of the convolution is being computed.

Note $g[x] = h[-x]$.

So I feel the notation is slightly confusing in what you wrote. I would have written that $(g \ast x)[l] = \sum_i g[l-i]x[i] = \sum_i h[i-l]x[i]$, just so the indices don't get confusing.

---

In your linked document, let us translate the notation to something more reasonable:

• Fourth line (first convolution appearance): $(h \ast R_{xx})[m+l]$.
• Fifth line: $ (h^\ast_r \ast ( h \ast R_{xx} ))[m] $ where $h^\ast_r$ is the reversed sequence of $h^\ast$.

Why not just stop here? Now by associativity and commutativity of convolutions, you can jump to the bottom already... And again, that notation does leave something to complain about.

---

Notation:

$(g\ast h)[k] = \sum_i g[i] h[k-i]$

So your sum $\sum h[n] h^\ast[-k-n] = (h\ast h^\ast)[-k]$. (sum the indices to get which index of the convolution you solve for)

on the other hand $\sum h[n] h^\ast[-(k-n)] = (h\ast h^\ast[-(\cdot)])[k]$

Hope this clarifies somewhat what I am getting at.

Answer 2

what is going on here?

An error, or an arguably bad notation. Actually, neither the more usual $h[n] * x[n]$ is very correct.

The convolution of the signals $x[n]$ and $h[n]$ is another signal $g[n] = \sum_{i=-\infty}^{\infty} h[i-n] \ x[i]$; this is totally correct. Now, if we denote this operation by $x[n] * h[n]$, we shoud write $g[n] = x[n] * h[n]$, and we could conclude that, say, $g[8] = x[8] * h[8]$... which, of course is nonsense. The correct form should be $(h * x)[n]$, as the Wikipedia article uses - but one often writes $x[n] * h[n]$ (abuse of notation) to mean that.