inverse z-transform

Discussion in 'Homework Help' started by vvkannan, Feb 28, 2009.

  1. vvkannan

    Thread Starter Active Member

    Aug 9, 2008
    138
    11
    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.

    if i solve this by rewriting as

    X(z)/z = z^2 /(z-2)(z-1)^2

    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

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

    i dont know where iam going wrong.
    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?

    Thank you
     
  2. GirishC

    Active Member

    Jan 23, 2009
    58
    0
    Your technique for partial fraction is wrong. carefully look on how to solve repeated root by partial fraction method.
     
  3. vvkannan

    Thread Starter Active Member

    Aug 9, 2008
    138
    11
    Thank you for responding.

    do you mean the second method?

    A and C are solved as usual.
    For B heres how i did it

    differentiating [1/(1 - 2z^-1)] with respect to z^-1 and subtituting
    z^-1 = 1 and i get -2.
    is this method wrong?
    Or is it wrong to find the partial fraction without inverting the z^-1.
    Thank you
     
  4. GirishC

    Active Member

    Jan 23, 2009
    58
    0
    I find A = 4, C = -1

    so I substitue Z = 0 and solve for equation

    Z^2/(Z-2)(Z-1)^2 = A/(Z-2) + B/(Z-2) + c/(Z-1)^2

    I get B = -3

    so which is equal to your first answer :)
     
  5. vvkannan

    Thread Starter Active Member

    Aug 9, 2008
    138
    11
    Yes GirishC i get A=4,B=-3 AND C=-1 when i convert the given function into a rational function in Z

    But my question is why am i not getting this answer when i try to find inverse z-transform keeping the given function as it is.(i.e) without converting it into a function of Z but keeping it in z^-1 form.
    Hope you understand what i mean
     
  6. GirishC

    Active Member

    Jan 23, 2009
    58
    0
    Your denominator has lesser power(Z^-1) than numerator (Z^0).

    Girish
     
  7. vvkannan

    Thread Starter Active Member

    Aug 9, 2008
    138
    11
    Yes the numerator has greater power in Z which means when divided it will result in positive powers of Z.

    positive powers of Z would result in an anticasual signal.

    So to get the casual signal we need lesser powers of z in numerator,correct?

    So on the other hand if it was asked to find an 'anticasual' signal can i do it the second way? that is keeping the powers of Z in numerator greater than denominator?
    Thank you ,your last answer cleared me up a bit
     
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