System invertability

Question

Can someone please explain how to determine if a system is invertible or not. The particular systems that I am looking at are:

$y(t) = \cos \left( x(t) \right)$ - I believe this is non-invertable because of the periodic function

$y[n] = nx[n]$ - I think this is invertable and the inverse is $x[n]/n$

$y[n] = x[4n + 1]$ - Is the inverse $x[1/4n - 1]$?

$y[n] = x[n - 2] - 2x[n - 8]$ - I do not know even how to begin

I am really confused with systems... did not understand a single thing in the lecture today, and now reading the book ("Signals and Systems", Oppenheim) I am getting frustrated. Is there an algebraic procedure for determining the inverse of a system?

Please help.

Answer 1

$$ y[n] = x[n-2] - 2 x[x-8] $$

that's actually a linear and time-invariant system. you invert it by computing the Z transform

$$ Y(z) = z^{-2}X(z) - 2 z^{-8}X(z) $$

solve for $X(z)$, then inverse Z transform. you might find out that, to invert this system, you will have to look at future samples, so the inverse would not be causal.

Answer 2

Let's look at each of the systems:

1. This is generally not invertible, simply because you can't invert the cosine function unless you have some additional information about its argument. E.g., if you knew that $0\le x(t)\le \pi$ were satisfied for all $t$, then you could uniquely reconstruct $x(t)$ from $y(t)$, but otherwise you can't.

2. In this case you have a problem with $n=0$. For $n=0$ you have $y(0)=0$ regardless of the value of $x(n)$. So the system can't be inverted.

3. The output is simply $\ldots, x[-7], x[-3], x[1], x[5], \ldots$, so you only take one sample out of four and throw the others away. No way you can reconstruct these values from $y[n]$.

4. Here Robert already gave you a good hint in his answer. You can also look at it directly in the time domain. Exchanging the roles of $x[n]$ and $y[n]$ you get

$$x[n]=y[n-2]-2y[n-8]$$

Rearranging gives

$$y[n-2]=x[n]+2y[n-8]$$

or, equivalently,

$$y[n]=x[n+2]+2y[n-6]$$

This is a recursive system which (theoretically) inverts the given system. However, as pointed out by Robert, in order to compute $y[n]$ you need the value of the input two samples further in the future, i.e. this system is not causal. In general you would like a system that is causal and stable, but in this case there is no such system which inverts the given system.