How to determine if the system is invertible

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: $$ y^{-1}(t)=\frac{1}{e^{t}}\frac{d}{dt}\left(e^{t}x(t)\right) $$

Answer

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).