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Compute the Z-inverse transform

  1. Jul 22, 2006 #1
    Hello i would need some help to compute the Z-inverse transform of:

    [tex] \frac{1}{z^{2}-1} [/tex] and [tex] \frac{1}{(z-1)(z^{2}-1)} [/tex]

    i don,t know if they can be computed using the residue theorem as the inverse in general has the form:

    [tex] 2\pi i a(n)=\oint f(z)z^{n-1}dz [/tex] for a curve on complex plane...:grumpy: :grumpy:
  2. jcsd
  3. Jul 22, 2006 #2


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    Why would you say that? It's not at all difficult to calculate the residues of [itex]\frac{z^{n-1}}{z^2- 1}[/itex] at z= 1 and -1 where the function has poles of order 1. Similarly for the residues of [itex]\frac{z^{n-1}}{(z-1)(z^2-1)}= \frac{z^{n-1}}{(z-1)^2(z+1)}[/itex\ at z= 1 and -1 with a pole of order 1 at z= -1 and a pole of order 2 at z= 1. Looks pretty easy to me.
  4. Jul 24, 2006 #3
    Thanks..that was precisely the info i needed, if Z-inverse transform could be calculated by "residue theorem" if so i will try the functions above ..
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