Let $h(n)$ denote the \emph{impulse response} of systems $H$ to the input signal $x(n)$. For example, let $x(n)=\delta (n) + 2\delta (n-1)$ then $y(n) = h(n) + 2h(n-1)$, such that 
$$ x(n) = \delta(n) + 2\delta(n-1) \to \fbox{H} \to h(n) + 2h(n-1) = y(n)$$

This is because the system is defined to be time invariant. An arbitrary signal can then be decomposed into linear combinations of shifted impulses:

$$ x(n) =\cdots + x(-1)\delta(n+1) + x(0)\delta(n) + x(1)\delta(n-1) + \cdots $$
$$ y(n) =\cdots + x(-1)h(n+1) + x(0)h(n) + x(1)h(n-1) + \cdots $$

Since we are talking about DT signals, this brings us to a summation formula. The \emph{convolution sum} is denoted as 

$$ y(n) = \sum \limits _{m=-\infty}^\infty x(m)h(n-m) = (x * h)(n)$$

The convolution of the input signal with the impulse response gives us the output of a linear time-invariant system. But what about the convolution of the impulse response with the input signal? Let $k = n - m$, then we write the formula as follows:

$$ y(n) = \sum \limits _{k=\infty}^{-\infty} x(n-k)h(k) = \sum \limits _{k=-\infty}^{\infty} h(k)x(n-k) = (h * x)(n)$$

Notice that the limits did switch, but since it is a summation it does not affect the value. This demonstrates an important property of convolution sums, \emph{commutativity}:
$$ x * h = h * n $$

The roles of input response can be change or reversed:

$$ x \to \fbox{h} \to y $$
$$\mbox{  or  }$$
$$ h \to \fbox{x} \to y$$

Both produce the same output signal of the LTI. Note, if a system has an impulse response $h$, this usually implies that it is an LTI system.

Linear Time-Invariant Systems