Practice Question on "Digital Signal Processing"
Topic: Properties of z-transform
Question
Prove the following property of the z-transform:
$ z_0^n 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
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!
