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Is there any systematic way to determine if the system is invertible?

My general approach is first trying to find the inverse system by using mathematical method; that is, solving for the output in terms of input. If this is difficult, my guess would be that the system is non-invertible, and the goal is then to sit and think of the two inputs $x_{1}(t)$ and $x_{2}(t)$ that generate the same output $y(t)$.

However, I was just wondering if this method is 100% correct:

  • If we can find the input $x(t)$ in terms of output $y(t)$ (by solving mathematically), does that always mean the system is invertible?
  • Is there any possibility that can lead to the failure in reasoning?

For example, determine if the following system is invertible: $$ y(t)=\int_{-\infty}^{t}e^{-(t-\tau)}x(\tau)d\tau $$

Firstly \begin{align} y(t)&=e^{-t}\int_{-\infty}^{t}e^{\tau}x(\tau)d\tau\\ e^{t}y(t)&=\int_{-\infty}^{t}e^{\tau}x(\tau)d\tau\\ \frac{d}{dt}\left(e^{t}y(t)\right) &= e^{t}x(t)\\ x(t)&=\frac{1}{e^{t}}\frac{d}{dt}\left(e^{t}y(t)\right) \end{align} So the inverse system is: $$ y^{-1}(t)=\frac{1}{e^{t}}\frac{d}{dt}\left(e^{t}x(t)\right) $$

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  • $\begingroup$ might wanna apply the product rule to your result. (either that or recognize that the first equation is a convolution integral and derive the impulse response from it and then use Laplace Transform.) $\endgroup$ – robert bristow-johnson Sep 7 '16 at 17:56
  • $\begingroup$ Thanks, I might have to study more about the Laplace transform, since this is just the first week of my course $\endgroup$ – IgNite Sep 7 '16 at 17:59
  • $\begingroup$ well, you will undoubtably get to Laplace Transform in this class. do apply the product rule of differentiation to your answer. it will show that your result is also LTI. $\endgroup$ – robert bristow-johnson Sep 7 '16 at 18:03
  • $\begingroup$ Looks like this is correct. I would like to give one more system. Consider the first difference system( backward or forward). Is it invertible? $\endgroup$ – Vinayak Jul 18 '17 at 5:47
  • $\begingroup$ Sir a problem in gate 17 y(n)=1+(-1)^n x(n) Please suggest me solutions and related topics to understand concept. Please comment $\endgroup$ – Shrey Chandra Feb 6 '18 at 11:58
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It depends on what exactly you mean by "invertible". In system theory, what is often meant is if there is a causal and stable system that can invert a given system, because otherwise there might be an inverse system but you can't implement it.

For linear time-invariant systems there is a straightforward method, as mentioned in the comments by Robert Bristow-Johnson. First, note that the input-output relationship of an LTI system is given by the convolution integral

$$y(t)=\int_{-\infty}^{\infty}h(t-\tau)x(\tau)d\tau\tag{1}$$

Comparing $(1)$ to the system given in your question shows that the system is LTI with impulse response

$$h(t)=e^{-t}u(t)\tag{2}$$

where $u(t)$ is the unit step function. The corresponding transfer function is

$$H(s)=\frac{1}{1+s}\tag{3}$$

This system is causal and stable. However, its inverse system

$$G(s)=\frac{1}{H(s)}=1+s\tag{4}$$

is not stable. It has a pole at infinity. The input-output relation of the inverse system is

$$y(t)=x(t)+x'(t)\tag{5}$$

where $x'(t)$ is the derivative of $x(t)$. So in order to implement the inverse system you need a differentiator, which is not stable in the bounded-input-bounded-output (BIBO) sense. I.e., bounded input signals can lead to unbounded output signals (just use a rectangular input function as an example).

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