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I am having some trouble with this z-transform, and would like a

  1. Apr 29, 2007 #1
    I am having some trouble with this z-transform, and would like a little help.

    Question:
    Find the Z-transform, and sketch the pole/zero plot with the region of convergence.

    [tex] x[n] = \cos( \Omega_0 n) u[n] [/tex]

    Where [tex] u[n] [/tex] is the step function.

    "Answer"

    Directly from a table in my book, [itex] x[n] [/itex] transforms to:

    [tex] X(z) = \frac{1-[\cos \Omega_0]z^{-1}}{1-[2\cos \Omega_0]z^{-1}+z^{-2}} [/tex]

    with a region of convergence: [itex] |z|>1 [/itex]

    For the poles/zeros, a little bit of algebra yields:
    [tex]X(z) = \frac{z(z-\cos \Omega_0)}{z^2-2\cos \Omega_0 z + 1} [/tex]

    Thus: zeros = [itex] \{ 0, \cos \Omega_0 \} [/itex]

    Using the quadratic formula yields:

    poles = [tex] p_{1,2} = \frac{2 \cos \Omega_0 \pm \sqrt{4 \cos^2 \Omega_0 - 4}}{2} = \cos \Omega_0 \pm \sqrt{\cos^2 \Omega_0 -1} [/tex]

    This can be further simplified to: [tex] p_{1,2} = \cos \Omega_0 \pm \sqrt{-\sin^2 \Omega_0} = \cos \Omega_0 \pm j\sin \Omega_0 = \{ e^{j\Omega_0},\,\,\,\, e^{-j\Omega_0} \}[/tex]


    Now if I assume I am right (which I'm not sure about), then:

    ROC = [tex] |z| > 1 [/tex]
    POL = [tex] \{ e^{j\Omega_0}, \,\,\, e^{-j\Omega_0} \}[/tex]
    ZER = [tex] \{ 0, \,\,\, \cos \Omega_0 \}[/tex]

    So how do I plot this?

    Do I say:
    A zero exists at the origin, and the other zero exists from -1 to 1 on the real line.

    A pole exists on the unit circle with the other pole "opposite" it.


    I just need a little help in understanding this. Please let me know, if the in-between steps are necessary. Thanks!
     
  2. jcsd
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