can a square z-transform function be expressed as the convolution of itself in time-domain

Question

Given $H(z) = (1+z^{-1}+z^{-2})^2$

I want to make use of the property of z-transform that

$$x_1(n)\ast x_2(n) \rightarrow X_1(z)X_2(z)$$

so if $x_1(n) = x_2(n)$ then the $X_1(z)X_2(z) = X_1(z)^2 = X_2(z)^2$

But I don't know that if a function can have a convolution with itself or not. If it a available then

Apply it to $H(z)$ say $H(z) = X(z)^2$ and $X(z) = 1+z^{-1}+z^{-2}$

Since $x(n)$ is finite, the convolution is easy to calculate though being long

Answer 1

Yes of course can a function have a convolution with itself. For instance the convolution of the $rect$ function with itself is a triangle.

In your example you have $x(n) = (1,1,1)$ for n = 0,1,2 respectively.

The convolution is defined as

$y(n) = \sum_{k=-\infty}^\infty x(k)h(n-k) = (x*h)(n)$.

In particular (for your kausal signals)

$h(n) = \sum_{k=0}^2 x(k)x(n-k) = (1,2,3,2,1 )$, for $n = 0...4$