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Inverse z-transform

  1. Mar 19, 2006 #1
    I need to find the inverse z-transform for this function:

    [tex] \frac {Z^2-Zr Cos W0} {Z^2-r^2 Sin^2 W0} [/tex]
     
    Last edited: Mar 20, 2006
  2. jcsd
  3. Mar 19, 2006 #2

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    Since [tex]\omega_0[/tex] is a constant, what you really want is the inverse transform of [tex]\frac{z^2-a}{z^2-b^2} = \frac{Az}{z-b} + \frac{Bz}{z+b}[/tex].
    Find A and B by expanding the fuction in partial fractions. The inverse transform is easy to calculate.
     
  4. Mar 20, 2006 #3
    thanks but I see that you didn't pay attention to the first order term (zr cos w0), I wonder whether the partial funtions expanded should be changed or it might be the same?!
     
  5. Mar 20, 2006 #4

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    You are right. It should be:
    [tex]\frac{z^2-a}{z^2-b^2} = \frac{Az+B}{z-b} + \frac{Cz+D}{z+b}[/tex].
     
  6. Mar 21, 2006 #5
    I tried to find constants:A,B,C,D but useless, i can't so i invented fiction ways ....but there's one thing which we didn't pay attention on at all, we need to divide P(z) by Q(z) manually till order of P(z) becomes less than Q(z) then find IZT of the function.

    [tex] \frac {z^2-az} {z^2-b^2}=1 + \frac {-az+b^2}{z^2-b^2} [/tex]

    so now we can expand the resulted function in partial fractionds and find the constansts

    [tex]=\frac {-az+b^2}{z^2-b^2}=\frac{A}{z-b}+\frac {B}{z+b} [/tex]

    A=(b-a)/2 & B=(-b-a)/2

    so IZT function is:
    [tex] x(n)= \delta(n)+A b^{n-1} u(n-1)+B (-b)^{n-1} u(n-1) [/tex]
     
    Last edited: Mar 21, 2006
  7. Mar 21, 2006 #6
  8. Mar 21, 2006 #7

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