Convolution is commutative; find y[n]

LTI system

An LTI system is one that is both linear (superposition holds) and time-invariant (a delay in the input gives the same delay in the output). It is completely described by its impulse response $h[n]$, and the output for any input is $y[n]=x[n]*h[n]=\sum_k x[k]h[n-k]$.

Commutative property

To show $x[n]*h[n]=h[n]*x[n]$:

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

Put $m=n-k$, so $k=n-m$. As $k$ goes from $-\infty$ to $\infty$, $m$ also covers $-\infty$ to $\infty$:

$$x[n]*h[n]=\sum_{m=-\infty}^{\infty}x[n-m]\,h[m]=\sum_{m=-\infty}^{\infty}h[m]\,x[n-m]=h[n]*x[n]$$

So a system with impulse response $h$ driven by $x$ gives the same output as a system with impulse response $x$ driven by $h$.

Computing y[n]

$x[n]=\{1,2,3,4\}$ with $x[0]=1$ (so $n=0..3$); $h[n]=\{2,1,2\}$ with $h[0]=1$ (so $h[-1]=2,\ h[0]=1,\ h[1]=2$).

Start index $=0+(-1)=-1$, length $=4+3-1=6$, so $y[n]$ runs from $n=-1$ to $4$.

For example, $y[1]=x[0]h[1]+x[1]h[0]+x[2]h[-1]=2+2+6=10$. Sum check: $\sum y=50=10\times5$.

Answer: $y[n]=\{2,\ 5,\ 10,\ 15,\ 10,\ 8\}$ for $n=-1,\dots,4$, i.e. $y[0]=5$.

Computing $h*x$ (each $h$ value times $x$) gives the same numbers, which verifies the commutative property.