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| − | = Properties of | + | <center><font size= 4> |
| + | '''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]''' | ||
| + | </font size> | ||
| + | |||
| + | Topic: Properties of z-transform | ||
| + | |||
| + | </center> | ||
| + | ---- | ||
| + | ==Question== | ||
Prove the following scaling property of the z-transform: | Prove the following scaling property of the z-transform: | ||
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<math>Z[z_0^n x[n]]=\sum_{n=-\infty}^{\infty}x[n]k^{-n}=X(k)=X(\frac{z}{z_0})</math> | <math>Z[z_0^n x[n]]=\sum_{n=-\infty}^{\infty}x[n]k^{-n}=X(k)=X(\frac{z}{z_0})</math> | ||
| − | + | :<span style="color:orange">Instructor's comment: It is a bit confusing to use k as a complex variable. Usually, k represents an integer. -pm </span> | |
=== Answer 2=== | === Answer 2=== | ||
I agreed with above, it should be <math>z_0^n</math> not <math>z_0^2</math>, otherwise <math>z_0^2</math> is just a constant and the transform will just be <math>z_0^2 { X \left( z \right)} </math> | I agreed with above, it should be <math>z_0^n</math> not <math>z_0^2</math>, otherwise <math>z_0^2</math> is just a constant and the transform will just be <math>z_0^2 { X \left( z \right)} </math> | ||
| + | |||
| + | :<span style="color:green">TA's comments: Good catch and reasoning.</span> | ||
<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> | <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> | ||
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=== Answer 3=== | === Answer 3=== | ||
| − | + | <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]] | ||
Latest revision as of 11:45, 26 November 2013
Practice Question on "Digital Signal Processing"
Topic: Properties of z-transform
Question
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}) $
- Instructor's comment: It is a bit confusing to use k as a complex variable. Usually, k represents an integer. -pm
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) $
