Understanding Convolution Summation in Discrete time signals

Question

General definition of convolution states: $$ u(n)*s(n) = \sum_k u(k)s(n-k) $$

However, unable to grasp the fundamental over here, I am wondering what summation would the following represent: $u(n-1)*s(n)$?

Answer 1

Let's say the impulse response of the system is $y[n]=s[n]$. Also, we can write any discrete signal as the sum of impulse functions such as

$$u[n] = \sum_{k=0}^\infty u[k] \delta[n-k]$$

Now using linearity and time invariance, we can see that

$$y[n] = \sum_{k=0}^\infty u[k] s[n-k]$$

This is so fundamental so it has a name.

Answer 2

If you define the shifted version of $u[n]$ as another signal $v[n]\triangleq u[n-1], n \in \mathbb{Z}$, it follows from the convolutions formula that $$ v[n]*s[n]=\sum_k v[k] s[n-k] $$

Now, by replacing $v[n]$ with $u[n-1]$ in the above formula, we have

$$ u[n-1]*s[n]=\sum_k u[k-1] s[n-k] $$