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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) $


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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)} $

$ 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

Write it here.


Back to ECE438 Fall 2011 Prof. Boutin

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang