Inverse system of $y(t)=\int_{-\infty}^te^{-(t-\tau)}x(\tau)d\tau$

Question

How to find the inverse system of $$y(t)=\int_{-\infty}^te^{-(t-\tau)}x(\tau)d\tau$$ with input $x$ and output $y$?

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The solution for this is $y(t) = x(t) + \dot x(t)$. I am confused about how to get this answer. Thanks!

Answer 1

We have,

$$y(t) = e^{-t}\int_{-\infty}^te^\tau x(\tau)d\tau $$

The impulse response is,

$$h(t) = e^{-t}\int_{-\infty}^t e^{\tau} \delta(\tau)d\tau = e^{-t}(1)=e^{-t} \Rightarrow H(s) = \mathcal L \{e^{-t}\} = \frac1{s+1}$$

Now to determine the inverse transfer function, $$\begin{align}&H(s)\cdot H^{-1}(s) = 1 \\ \Rightarrow~&H^{-1}(s) =\frac{Y(s)}{X(s)}= s+1\\ \Rightarrow~& Y(s) = sX(s)+X(s)\\ \Rightarrow ~&y(t) = \frac{dx(t)}{dt}+x(t) ~~~~~~(\text{assuming } x(0) = 0 \text{ in the inverse system})\end{align}$$