Difference between revisions of "Practice prove modulation property z transform" - Rhea

Latest revision as of 18:05, 3 March 2015

Question

Prove the following property of the z-transform:

$ z_0^n x[n] \rightarrow X \left( \frac{z}{z_0}\right) $

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Answer 1

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

$ 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} $

Now if we look at that last expression, we see that it is just the expressing for the z-transform, $ X(z) =\sum_{n=-\infty}^{\infty} x[n]z^{-n} $, but with $ z $ replaced by $ \frac{z}{z_0} $

Good!

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) $

Short and sweet!

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