Is the following system invertible or not?

$$y[n]=x[n] * x[n^2]$$ where $*$ stands for the aperiodic convolution operator. I have not been able to find a mathematically sufficient argument for it...


The system is nonlinear (bilinear in $x$), with a nonlinear law (square) on indices. Odds are the system is not invertible. One can try to prove it in its full generally, or try to find a counterexample. Let us try the lazy way, using properties at hand.

The system is bilinear. Hence, the output for $k x[n]$, $k\in \mathbb{R} $, will be $k^2 y[n]$. Resultingly, since $(-k)^2 = k^2$, $k x[n]$ and $-k x[n]$ have the same output. If they are nonzero, they differ and yield the same result. Verdict: non-invertible.

This is a first way. Another is possible, along the same line, using the indices $n$ and $n^2$. Fat32 just did it, using prime delays. With more generality: take some signal $x[n]$, assuming that:

  • there exists at least one non-square index $n_+$, for which $x[n_+] \neq 0$: for instance $x[-2] =1$ (many negative numbers are not perfect squares),
  • $x$ is zero at all square indices: $x[0] = x[1] = x[4] = x[9]=\cdots = 0$.

Then $x[n^2]$ is identically zero, while $x[n]$ is not and the convolution is zero.


This system is not invertible and a single counter-example is sufficient to prove it.

First express the signal $x[n^2]$ as $w[n]$ and then the output is $$y[n]= x[n] \star w[n]$$

Then let an example input be $x[n] = \delta[n-3]$. Then you can verify easily that $w[n] = 0$ for all $n$. Hence the output $y[n]$ for the input $x[n]=\delta[n-3]$ is $$ y[n] = x[n] \star w[n] = 0$$ for all $n$.

Hence you can conlcude that from the computed output $y[n] = 0$ you cannot re-construct back $x[n] = \delta[n-3]$ therefore the system is not invertible.

Furthermore, you can also verify that the output $y[n]$ will be identically zero for many distinct inputs such as $x[n] = \delta[n-5] + \delta[n-7]$ or $x[n] = \delta[n-11] + \delta[n-19]$. Therefore you can further conclude that distinct inputs producing same outputs would violate the principle of invertibility.


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