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Inversee z -transform

  1. Mar 1, 2009 #1

    Hello all,

    I have a question on z-transforms.The question states

    "determine the casual signal x(n) having the z-transform
    X(z) = 1/(1-2z^-1)(1 - z^-1)^2".

    This is how i tried

    If i solve this by rewriting as

    X(z)/z = z^2 /(z-2)(z-1)^2 and using partial fractions

    i get x(n) as 4(2^n) u(n) - 3u(n) - nu(n).

    But if i try to find the value by writing the partial fractions for the given function as it is
    (i.e) A/(1 - 2z^-1) + B/(1 - z^-1) + C /(1 - z^-1)^2
    i get A,B,C as 4,-2,-1
    and hence x(n) as 4(2^n) - 2u(n) - n u(n).As seen i get the coefficient as -2 instead of -3 for the second term.

    So is it right on my part to take partial fractions as i have done in the second case and solve the given function in z-inverse as it is or it should be solved after converting to a function in z instead of z^-1?
    Thank you
  2. jcsd
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