Commutativity of Convolution

Convolution is commutative, i.e.,

$$\zbox {x\ast y = y\ast x}$$

Proof:

$$\begin{eqnarray*} (x\ast y)_n &\isdef & \sum_{m=0}^{N-1}x(m) y(n-m) = \sum_{l=n... ...l)\\ &=& \sum_{l=0}^{N-1}y(l) x(n-l) \\ &\isdef & (y \ast x)_n \end{eqnarray*}$$

where in the first step we made the change of summation variable $l\isdeftext n-m$, and in the second step, we made use of the fact that any sum over all $N$ terms is equivalent to a sum from 0 to $N-1$.