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

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?

• It doesn't. You can find the response of a linear time-varying system by convolution; you just need to deal with the fact that the system's impulse response will be time-varying. Commented Mar 18, 2023 at 7:03
• It does need to be time-invariant in order to be characterized by its response to $\delta(t)$. For time-varying systems we need to know their responses to all possible shifts $\delta(t-\tau)$, $\forall\tau$. Commented Mar 18, 2023 at 10:45

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.
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 $$S(\delta[.-k])$$.