Topic: Properties of z-transform

## Question

Prove the following property of the z-transform:

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

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

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

Now if we look at that last expression, we see that it is just the expressing for the z-transform, $X(z) =\sum_{n=-\infty}^{\infty} x[n]z^{-n}$, but with $z$ replaced by $\frac{z}{z_0}$

Good!

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