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} $

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010