Inverse z transform - contour integration

Discussion in 'Homework Help' started by xxxyyyba, Apr 30, 2015.

  1. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    Hi!
    Here is my task:
    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?
    Thanks in advance
     
  2. MrAl

    Well-Known Member

    Jun 17, 2014
    2,438
    492
    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.
     
    Last edited: Apr 30, 2015
    xxxyyyba likes this.
  3. xxxyyyba

    Thread Starter Member

    Aug 7, 2012
    249
    2
    Thanks for reply. I must study complex analysis in detail :)
     
Loading...