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How could i go throw proving that this system $y(t)=\int_{-\infty}^{t}e^{-(t-\tau)}x(\tau)d\tau$ is invertible system or not ?

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    $\begingroup$ Question is not very clear. Do you want to check if $x(t)$ can be recovered from $y(t)$? i.e. Are you looking if deconvolution is possible $\endgroup$
    – AnVij
    May 24 at 12:07
  • $\begingroup$ Are you asking if it's invertible ? $\endgroup$
    – Hilmar
    May 24 at 12:10
  • $\begingroup$ Yes i am asking if this an invertible system or not. $\endgroup$ May 24 at 12:47
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To invert your system you take the derivative in $t$

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

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

The problem is that the integral reduces to $y(t)$ as well.

If you look at the response, the transfer function is $1/(s+1)$ the inverse response would be $s + 1$, a system with more zeros than poles. Does it mean it is not invertible in the context you are asking it.

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    $\begingroup$ So $x(t)=\dot y(t)+y(t)$ and that means it is invertible. $\endgroup$ May 24 at 14:15
  • $\begingroup$ Typically, a system $x(t) = a y(t) + b \dot y(t)\, \forall a, b \ne 0$ is considered non-causal, because of the "naked" differentiation. Some authors would consider your system to be non-invertible because of this non-causality. Some would consider it OK. So whether that system is invertible depends on the ground rules under which you are working. If it's for a class -- what does the prof and/or book say on this point? $\endgroup$
    – TimWescott
    May 24 at 19:20
  • $\begingroup$ @TimWescott Our definition: The system has an inverse if there is a function $h_{i}(t)$ such that $h(t)*h_{i}(t)=\delta(t)$ $\endgroup$ May 26 at 18:16
  • $\begingroup$ Are you allowed to use $\frac{d}{dt} \delta(t)$ in $h_i$? $\endgroup$
    – TimWescott
    May 26 at 21:16
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    $\begingroup$ I think yes but I am not certain i could ask the prof about that, but why this specific function could be a problem.$$Thank\space you\space for\space helping\space appreciate\space that$$@TimWescott $\endgroup$ May 27 at 16:35

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