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ROC in Sign function Z-transform

  1. Sep 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Calculate the Z-transform of the function x[n] = u[n]-u[-n-1].

    2. Relevant equations
    [itex]X(z) = ZT\{x[n]\} = \sum_{n=-\infty}^{\infty}x[n]z^{-n}[/itex]

    [itex]ZT\{u[n]\} = \displaystyle\frac{1}{1-z^{-1}}[/itex], ROC: |z| > 1.
    [itex]ZT\{-u[-n-1]\} = \displaystyle\frac{1}{1-z^{-1}}[/itex], ROC: |z| < 1.

    [itex]ZT\{x[n]\} = X(z)[/itex], ROC: R1
    [itex]ZT\{y[n]\} = Y(z)[/itex], ROC: R2
    [itex]ZT\{ax[n]+by[n]\} = aX(z)+bY(z)[/itex], ROC: at least [itex]R1\cap R2[/itex]

    3. The attempt at a solution
    Using formulas in section 2. it is obvious that [itex]X(z) = ZT\{x[n]\} = \displaystyle\frac{2}{1-z^{-1}}[/itex], but which is the ROC? The intersection between |z| > 1 and |z|< 1 is null.

    Thank you.
  2. jcsd
  3. Sep 9, 2011 #2
    Isn't it that if there the intersection of the ROC is null then you can't use the z-transform because it diverges?
  4. Sep 9, 2011 #3
    I thought that, but if you take z = e^(j*Omega), you obtain the correct Fourier transform.

    In addition (without rigorousness), this z-transform was asked in an exam and I'm sure it exists. What is happening?
  5. Sep 9, 2011 #4
    So taking the z-transform and evaluating at z = e^(j*Omega) is the definition of the DTFT, right? Isn't the DTFT more prone to instability than the z-transform?

    Hmm.. I will think about this more. Maybe someone else is better suited to answer this question as my DSP class begins covering the z-transform this upcoming Monday (good time to start reading up on it though!).

    Is this for a graduate level DSP course? Just curious.
    Last edited: Sep 9, 2011
  6. Sep 10, 2011 #5
    Yes, DTFT only exists if the unit circle |z| = 1 is in the ROC.

    So I don't know if I'm making mistakes and this ROC is miscalculated or if there is another explanation to this behavior.

    Yes. Electrical engineering, more exactly.
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