Z transform Examples
These examples are solved mostly by using the definition of the Z transform given here, and the common definition of a linear series, also given here:
$ X(z)=\sum_{n=-\infty}^\infty x[n]z^{-n} $
$ \sum_{n=0}^\infty z^n=\frac{1}{1-z} $
Example 1
Find x[n], given that $ X(z)=4z^2+2+3z^{-1} $ for $ |z|\epsilon(0,\infty) $
Solution: Very simple, use the definition, and
$ x[1]=3,x[-2]=4,x[0]=2,else\, x[n]=0 $
Example 2
Given that
$ X(z)=\frac{1}{1+3z} $
Find x[n] with $ |z|<\frac{1}{3} $, and
Solution:
This is done by attempting to make $ X(z) $ fit into the formula given for the z transform, then extracting the $ x[n] $ from it.
We have to first convert this into a proper geometric series, by multiplying top and bottom by $ \frac{1}{3z} $
$ X(z)=\frac{1}{3z}\frac{1}{1+\frac{1}{3z}} $
The geometric series needs to extract a negative term for it like this: $ X(z)=\frac{1}{3z}\frac{1}{1-\frac{-1}{3z}} $
And this is simplified to: $ X(z)=\frac{1}{3z}\sum_{n=0}^\infty \left(\frac{-1}{3z}\right)^n $
Now we need to isolate $ z^{-n} $ so we group like terms:
$ X(z)=\sum_{n=-\infty}^\infty \frac{(-1)^n}{3^{n+1}}z^{-(n+1)}u[n] $
Of course, we want to isolate the $ z^{-n} $ term, so we will substitute $ k=n+1 $
$ X(z)=\sum_{k=-\infty}^\infty \frac{(-1)^{k-1}}{3^k}z^{-k}u[k-1] $
Our answer is clear by looking at the definition, we can now extract $ x[n] $
$ x[n]=\frac{(-1)^{k-1}}{3^k}u[k-1] $
Example 3
Find $ X(z) $, given that $ x[n]=\frac{u[n]}{2} $
Solution: Just use the definition, but note the converging and diverging cases:
$ X(z)=\sum_{-\infty}^\infty 2^{-n}z^{-n}u[n]=\sum_{-\infty}^\infty (2z)^{-n}u[n] $
We can perform the geometric series only when there is convergence, or only in the range of $ \frac{1}{|2z|}<1 $, otherwise, it diverges.
So first lets solve the convergent case: $ \sum_{n=0}^\infty \left(\frac{1}{2z}\right)^n=\frac{1}{1-\frac{1}{2z}}=\frac{2z}{2z-1} $
So the solution is: $ X(z)=\frac{2z}{2z-1} $ with ROC:$ |z|>\frac{1}{2} $
