Why time invariant system in order to know any output for any input using the impulse response?

Question

Suppose we have a system $S$.We give it the input $\delta(t)$ and get a response $S(\delta(t))$.If the system is linear and time invariant we can calculate every output $y(t)$ from any input $x(t)$ by finding the convolution of $x(t)$ and $S(\delta(t))$.But why does the system need to be time invariant?

Answer 1

If the system is time-varying, its response to an impulse at $t=0$ might be very different from its response to an impulse at any other time instant. Hence, knowing only its response to $\delta(t)$ (i.e., an impulse at $t=0$) is not sufficient to completely characterize a time-varying system.

Answer 2

Let us discretize the system to better capture the analogy. Now, $\delta[n]$ denotes the so-called discrete Kronecker delta. Take length-2 signal $x[n]$ defined as:

$ x[n]=\left\{ \begin{array}{ll} \alpha\neq0\;\mathrm{if }\;n=0\\ \beta\neq0\;\mathrm{if }\;n=1\\ 0\;\mathrm{elsewhere }.\\ \end{array} \right. $

Then, we can rewrite $x [n] = \alpha \delta[n]+ \beta\delta[n-1] $. By the linearity-additive property, you can compute the system output to $x$ by knowning that of $\alpha \delta[n]$ and $\beta\delta[n-1]$. By the linearity-scaling property, you can get the system output to $\alpha \delta[n]$ from that of $\delta[n]$ (by multiplying it by $\alpha$), and the system output to $\beta \delta[n-1]$ from that of $\delta[n-1]$ (by multiplying it by $\beta$). If now the system is time-invariant, knowing the system output to $\delta[n]$ offers you the output to $\delta[n-1]$ (by a time-shift).

Therefore, being linear and time-invariant, you can characterize all of $S$ by a convolution of any input signal by a "single" convolution with $S(\delta[.])$. If not, you still need the knowledge of all delayed Kronecker deltas $S(\delta[.-k])$.

About the same goes for continuous systems, the discrete setting sounds more intuitive to me.