# How to prove that this system is an invertible system or not?

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 ?

• 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 May 24 at 12:07
• Are you asking if it's invertible ? May 24 at 12:10
• Yes i am asking if this an invertible system or not. May 24 at 12:47

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.

• So $x(t)=\dot y(t)+y(t)$ and that means it is invertible. May 24 at 14:15
• 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? May 24 at 19:20
• @TimWescott Our definition: The system has an inverse if there is a function $h_{i}(t)$ such that $h(t)*h_{i}(t)=\delta(t)$ May 26 at 18:16
• Are you allowed to use $\frac{d}{dt} \delta(t)$ in $h_i$? May 26 at 21:16
• 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 May 27 at 16:35