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Properties of the Z-transform
Prove the following scaling property of the z-transform:
$ z_0^2 x[n] \rightarrow X \left( \frac{z}{z_0}\right) $
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1
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}) $
Answer 2
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) $
Answer 3
$ 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) $
