Topic: Properties of z-transform

## Question

Prove the following scaling property of the z-transform:

$z_0^2 x[n] \rightarrow X \left( \frac{z}{z_0}\right)$

I think there is a mistake, it should be $z_0^n$ instead of $z_0^2$.

proof:

$x'[n]=z_0^n x[n]$

$Z[x'[n]]=\sum_{n=-\infty}^{\infty}x'[n]z^{-n}=\sum_{n=-\infty}^{\infty}z_0^n x[n]z^{-n}=\sum_{n=-\infty}^{\infty}x[n](\frac{z}{z_0})^{-n}$

$let k=\frac{z}{z_0}$

$Z[z_0^n x[n]]=\sum_{n=-\infty}^{\infty}x[n]k^{-n}=X(k)=X(\frac{z}{z_0})$

Instructor's comment: It is a bit confusing to use k as a complex variable. Usually, k represents an integer. -pm

I agreed with above, it should be $z_0^n$ not $z_0^2$, otherwise $z_0^2$ is just a constant and the transform will just be $z_0^2 { X \left( z \right)}$

TA's comments: Good catch and reasoning.

$Z \left( z_0^n x[n] \right) =\sum_{n=-\infty}^{\infty} z_0^n x[n]z^{-n} =\sum_{n=-\infty}^{\infty} x[n]\left({\frac{z}{z_0}}\right)^{-n} = X \left( \frac{z}{z_0}\right)$

$Z \left( z_0^n x[n] \right) =\sum_{n=-\infty}^{\infty} z_0^n x[n]z^{-n} =\sum_{n=-\infty}^{\infty} x[n]\left({\frac{z}{z_0}}\right)^{-n} = X \left( \frac{z}{z_0}\right)$