# Inverse z transform - contour integration

#### xxxyyyba

Joined Aug 7, 2012
289
Hi!
Find inverse z transform of $$X(z)=\frac{1}{2-3z}$$, if $$|z|>\frac{2}{3}$$ using definition formula.
I found that $$x(n)$$ is $$\frac{1}{3}(\frac{2}{3})^{n-1}u(n-1)$$ (using other method). But how can I find it using definition formula, $$x(n)=\frac{1}{2\pi j}\oint_{C}^{ } X(z)z^{n-1}dz$$?

#### MrAl

Joined Jun 17, 2014
8,497
Hello there,

It's been over 20 years since i had to do this so i only have a limited amount of information, but one way to evaluate this integral is by the use of the residue theorem. You could look that up for more information.

Side Note:
The integral itself is derived by taking the 'normal' inverse transform and multiplying both sides by z^(k-1) and then integrating with a contour integral for which the path encloses the origin and is inside the region of convergence. Since we already have the right integral we dont have to do this though.

A quick example is:
X(z)=1/(1-a*z^-1), z>a

so inside the contour integral we end up with:
z^(n-1)/((1-a*z^-1)

which simplified a little leads to:
z^n/(z-a)

For a>=0 the contour of integration only encloses one pole at z=0 so for n>=0 we get:
x(n)=a^n

You would also have to evaluate this for n<0, where there is more than one pole to consider.

Sorry i dont have more information on hand at the moment, but if you look up the residue theorem you should be able to find more information about this. This is a topic that is covered more extensively in digital signal processing.

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