Line 34: Line 34:
  
 
=== Answer 3===
 
=== Answer 3===
Write it here.
+
<math>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) </math>
  
 
----
 
----
 
[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]]
 
[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]]

Revision as of 11:37, 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) $


Share your answers below

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) $


Back to ECE438 Fall 2011 Prof. Boutin

Alumni Liaison

To all math majors: "Mathematics is a wonderfully rich subject."

Dr. Paul Garrett