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Math Help - Z-transform to solve difference equation

  1. #1
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    Z-transform to solve difference equation

    Hello all hopefully you can be of help
    I have to solve the following difference equation using z transforms

    Xq+2 - 5Xq+1 +6Xq = 2

    conditions Xo=0 and X1=1

    Not to sure how to do it

    so far i have took the z transforms and got

    (z^2-z) -5zx + 6x = 2z/z-1

    I'm not sure if this is correct but even if it is i have no idea of what to do next
    Any help would be greatly appreciated
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  2. #2
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    Just to clarify:

    You're trying to solve

    x_{q+2}-5x_{q+1}+6x_{q}=2,

    subject to x_{0}=0, and x_{1}=1.

    Is this correct?
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  3. #3
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    Yes thats correct using z-transforms
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  4. #4
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    There are different definitions of the z-transform: bilateral, unilateral, and geophysical. Which one are you using? I'm guessing unilateral?
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  5. #5
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    yeah i believe its unilateral

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  6. #6
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    So, when I take the z-transform of the difference equation, x_{q+2}-5x_{q+1}+6x_{q}=2, I get the following:

    z^{2}X(z)-5zX(z)+6X(z)=2z/(z-1).

    As with most transform techniques, the next step is to solve the problem in the new domain (i.e., solve for X(z)), and then perform the inverse z-transform. You might need partial fraction decomposition.
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  7. #7
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    Okay im still not 100% sure

    But so far i've got it to this form

    2z/(z-1)(z-3)(z-2)

    Then using partial fractions got it to,

    -4/z-2 +1/z-1 +3/z-3

    You mention inverse z-transform, have i missed a stage? or am i heading in the right direction?

    (sorry being vague very new to z-transforms)




    Many thanks figured it out.
    Last edited by Rainey77; June 9th 2010 at 01:28 PM. Reason: Problem solved
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  8. #8
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    Hmm. You can do it that way. Looking ahead, it's a bit easier to work with terms that look like 1/(1-az^{-1}). Try a partial fraction decomposition using that form. In other words, take your

    \frac{2z}{(z-1)(z-2)(z-3)}=\frac{2z^{-2}}{(1-z^{-1})(1-2z^{-1})(1-3z^{-1})}, and expand it out that way.
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  9. #9
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    If you're having difficulty expanding it out with z^{-1}'s in there, just substitute y=z^{-1}, expand it out in partial fractions, and then substitute back in for z^{-1}.
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  10. #10
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    think i may have solved it
    ended up with

    -4(2^t) + 1 + 3(-3^t)
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  11. #11
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    Hmm. That doesn't jibe with what RSolve gives me in Mathematica. Also, you want to be careful with your parentheses. What's being exponentiated?

    I got the partial fraction expansion of

    \frac{1}{1-z^{-1}}-\frac{2}{1-2z^{-1}}+\frac{1}{1-3z^{-1}}.

    The inverse z-transform of this is

    1-C_{1}2^{q+1}+C_{2}3^{q}.

    Plugging in the initial conditions yields two simultaneous equations:

    0=1-2C_{1}+C_{2}
    1=1-4C_{1}+3C_{2}.

    Solving gives C_{1}=\frac{3}{2}, and C_{2}=2. Plugging this into the expression above yields

    x_{q}=1-3\cdot 2^{q}+2\cdot 3^{q}, which is what RSolve also yields.
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