## 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

$x=3,x[-2]=4,x=2,else\, x[n]=0$

### 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:$|z|>\frac{1}{2}$

## Alumni Liaison

Sees the importance of signal filtering in medical imaging Dhruv Lamba, BSEE2010