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Residue theorem

  1. Jan 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider the following Z tranform
    [itex]X(z)=\frac{z^{-1}}{1-2z^{-1}+2z^{-2}}[/itex]

    Calculate the poles and resides by hand.

    2. Relevant equations
    To find residue at pole p1

    [itex](z-p_{1})X(z)z^{k-1}[/itex] Evaluate at [itex]z=p_{1}[/itex]


    3. The attempt at a solution


    OK so ive taken the X(z) and converted to positive power of z by multiplying by z^2/z^2

    I get [itex]X(z)=\frac{z}{z^{2}+2z+2}[/itex]
    I know there is 1 zero at z=o.

    For the poles ive used the quadratic equation on the denominator. I come up with
    $$p1=\frac{-2\pm\sqrt{-4}}{2} p2=\frac{-2 \pm\ 2j}{2}$$

    simplifys to
    $$p1=-1-1j$$
    $$p2=-1+1j$$


    Ok so im stuck on finding the residues.

    The way I see using the above formula I end up with a 0 in the denominator. I know this incorrect. Am I missing some sort of simplification before i evaluate at z=p?

    If I take [itex]z^{2}+2z+2[/itex] and substitute [itex]z=-1-1j[/itex] I get 0.

    When I use the equation

    [itex](z-p_{1})X(z)z^{k-1}[/itex] Evaluate at [itex]z=p_{1}[/itex]

    and substitute z with p1 that first term in the parenthesis = o. (p1-p1)=0.

    Should something cancel out before i evaluate z=p1so I dont get the 0 in the numerator or denominator? If so I dont see it. Ive played with the numbers and moved them around for a couple of days now and I juts dont see it. It doesnt look complicated to me but it never comes out not 0.

    any help is appreciated! Thanks!
     
  2. jcsd
  3. Jan 11, 2012 #2

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    Hey Evo8! :smile:

    The denominator of X(z) can and should be factorized into (z-p1)(z-p2).
    Then by multiplying X(z) with (z-p1) that factor cancels after which you can calculate the residu.
     
  4. Jan 11, 2012 #3
    Hi I like Serena!

    I guess I dont see how it factors out to that then. This is what I would have assumed should be done. I was just under the impression that the denominator does not factor out that way.

    When I take (z-p1)(z-p2) and expand this is what I get
    $$z^{2}=p_{2}z-p_{1}+p_{1}p_{2}$$

    My expanded set of terms is
    $$z^{2}+2z+2$$ or $$1-2z^{-1}+2z^{-2}$$

    I will google basic factoring and see if I cant figure out what Im missing here. P.s. If you thought my algebra was already sub par, factoring was my LEAST favorite part of algebra ha!
     
  5. Jan 11, 2012 #4

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    You solved for the denominator being zero and found p1 and p2.
    This means that the denominator factorizes into (z-p1)(z-p2).
    The typical verification is by expanding it again and check you get your original denominator.

    I'm afraid you did not expand it correctly.
    The proper expansion of (z-p1)(z-p2) is ##z^2 -p_1 z -p_2 z + p_1 p_2##.
    When you work that out, you should get ##z^{2}+2z+2##.
    (Do you?)

    Anyway, if you multiply (z-p1) with ##X(z) = {z \over (z-p_1)(z-p_2)}##, you should get ##z \over (z-p_2)##, which is what you need to evaluate.
     
  6. Jan 11, 2012 #5
    I must have typed the expansion incorrectly. When I look at what I wrote down I have what you put. I dont even know how I got that = sign up there:confused:. Anyway thanks for pointing that out.

    This makes a little more sense now. So I can literally plug my two poles into the (z-p1)(z-p2)? Now where getting somewhere. I knew it had to be something small that I was forgetting.

    Thanks for the help! Hopefully I will be able to continue without issue!
     
  7. Jan 11, 2012 #6

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    Oh, the = sign is on the same key as the +, and it's next to the -.
    It's a pretty standard typo, it even looks a little bit like a minus sign.
    I didn't really think you put the = sign there intentionally. ;)
     
  8. Jan 11, 2012 #7
    Ok so after evaluating at expression im left with this for the residue at p1.
    $$
    \frac{(-1-1j)(-1-1j)^{k-1}}{(-1-1j)-(-1+1j)}
    $$

    What is the k? In my book all it says is that there is a dependence on k which is clear here. The example given also uses variable and constants for everything so the result is in terms of k.

    Thanks for any hints/help
     
  9. Jan 12, 2012 #8

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    Looks good!

    I think you have the inverse z-tranform there, that is, ##x[k]=\mathcal{Z}^{-1}\{X(z)\}##.
     
    Last edited: Jan 12, 2012
  10. Jan 19, 2012 #9
    sorry for the delayed response! Thanks so much for your help on this one I like Serena!

    Much appreciated as always
     
  11. Jan 21, 2012 #10

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    I've looked it up.

    The "residue" of a simple pole p1 is (z-p1)X(z) evaluated at z=p1.

    The inverse z-transform is the sum of both (z-p1)X(z)z^{-k} evaluated at z=p1 and (z-p2)X(z)z^{-k} evaluated at z=p2, yielding x[k].
     
  12. Jan 21, 2012 #11
    That looks right to me. I sometimes end up with repeated poles so my p1 and p2 are the same. So for the first term i get a 0 which brings everything down to 0. I always forget to manipulate and cancel out though. This is where I got lost. This example I was able to complete I believe.

    Thanks for posting that up!
     
  13. Jan 21, 2012 #12

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    :)

    In this example you only have simple poles.
    Simple poles (typically) only have one factor (z-p1) in the denominator of X(z).

    If you have for instance (z-p1)^2 in the denominator, which is what you seem to refer to, you need a more advanced formula.
     
  14. Jan 21, 2012 #13
    Indeed. That is exactly the form that where I used the mixed poles method and got hung up. :)
     
  15. Jan 21, 2012 #14

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    Okay. ;)


    As you can see in the wikipedia article, a simple pole has residue:
    969f2ce4cb6868464466832c12d79cc6.png
    This is what you've been working with.


    A pole of order n (in your example you would have n=2) has residue:
    fc16ae96cdbc7aea1023462e07d19753.png


    More specifically, if you have (z-p1)^2 in the denominator, your residue is:
    ##{d \over dz}((z-p_1)^2 X(z))## evaluated at z=p1.
     
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