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Find inverse z-transform

  1. Aug 9, 2014 #1
    1. The problem statement, all variables and given/known data
    Find inverse z-transform of
    [itex]X(z) = \frac{z}{(z-0.2)^2(z+0.1)}[/itex]
    2. Relevant equations

    3. The attempt at a solution : partial fraction
    My method :[itex]\frac{X(z)}{z} = \frac{1}{(z-0.2)^2(z+0.1)}[/itex]
    [itex]\frac{X(z)}{z} = \frac{-100/9}{(z-0.2)} + \frac{10/3}{(z-0.2)^2} + \frac{100/9}{z+0.1}[/itex]
    [itex]X(z) = \frac{(-100/9)z}{z-0.2}+\frac{(10/3)z}{(z-0.2)^2}+\frac{(100/9)z}{z+0.1}[/itex]
    [itex]X(z) = \frac{(-100/9)}{1-0.2z^-1}+\frac{(10/3)}{(1-0.2z^-1)^-2}+\frac{(100/9)}{1+0.1z^-1}[/itex]
    → [itex]x[nT]=(-100/9)0.2^n+(50/3)n.0.2^n+(100/9)(-0.1)^n[/itex]

    My friend :
    [itex]Y(z)=\frac{4}{z-0.2} + \frac{6z}{(z-0.2)^2} + \frac{-2}{z+0.1}[/itex]
    [itex]Y(z)=\frac{4z^-1}{1-0.2z^-1} + \frac{6z^-1}{(1-0.2z^-1)^2} + \frac{-2z^-1}{1+0.1z^-1}[/itex]
    → [itex]y[nT]=-4.(0.2)^n.u[n-1]+30n.(0.2)^n.u[n]-2.(-0.1)^nu.[n-1][/itex]

    I don't know which method gives correct result.
     
    Last edited: Aug 9, 2014
  2. jcsd
  3. Aug 9, 2014 #2
    On these face of it, it looks like a fourth order polynomial.
    I type "z=x(z+.1)(z-.2)(z-.2)" into wolframalpha.com and got something entirely different.
    I suspect the course work is expecting a particular strategy.
     
  4. Aug 11, 2014 #3

    rude man

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    Assuming your X(z) partial fraction expansion is correct to this point, the difficulty seems to be to invert the double pole at z = 0.2.

    I rewrote

    [itex]\frac{z}{(z-0.2)^2}[/itex] = z-1/(1 - 2az-1 + a2z-2) with a = 0.2.

    Then I got wolfram alpha to give me x[n] = na(n-1)u[n-1].

    Maybe you can fit this in with the rest which should be conventional inverse transformations.
     
    Last edited: Aug 11, 2014
  5. Aug 18, 2014 #4
    This is the instruction from the book :
    bfadca3eb2740d775078aa479cba59c4.png

    But it does not say if this method can be applied for the fraction which has the order of numerator is equal or larger than denominator.

    Can I use the above method to find z-inverse of following fraction (order of numerator is larger and/or equal)

    [itex]X(z)=\frac{4-z^{-1}}{2-2z^{-1}+z^{-2}}[/itex]
    [itex]X(z)=\frac{2+z^{-1}+z^{-2}}{1+0.4z^{-1}+0.4z^{-2}}[/itex]
    [itex]X(z)=\frac{z^{-2}+z^{-1}}{z^{-3}-z^{-2}+2z^{-1}+1}[/itex]
     
  6. Aug 18, 2014 #5

    rude man

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    That is not the case with your z transform. F(z) = Az/(z-γ)2 has order of denominator > order of numerator.

    You can rewrire F(z) = Az-1/(1 - γz-1)2.

    I see nothing wrong with your instructions. If your table includes that expression then just follow it!
    You seem to have a good z transform table!
     
  7. Aug 19, 2014 #6
    Yes, the instruction works for the problem in my first post. But i am wondering if it works for the function like this :
    [itex]X(z)=\frac{4-z^{-1}}{2-2z^{-1}+z^{-2}}[/itex]

    The numerator's lowest order is (-1), the denominator has (-2) as a lowest order. Can I multiply both numerator and denominator with z2 and do the rest using same method ? In the instruction the order is (-6) and (-5) respectively, so I am not sure if I can follow 8 steps for the above function.
     
  8. Aug 19, 2014 #7

    rude man

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