Revision as of 10:37, 19 September 2011

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) $

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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) $

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