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ROC of z-transform (derivative)

  1. Oct 16, 2012 #1
    I have a question about the z-transform:

    In the (Oppenheim's Signals and Systems) book it is written:

    If the z-transform of x[n] is X(z) with ROC=R, then the z-transform of nx[n] is -zdX(z)/dz with the same ROC.

    I don't understand why the ROC remains unchanged. Some books say "it follows from the fact that X(z) is analytic", some other books say "note that Laurent z-transform is differentiable within the ROC". But these reasons are not clear to me at all!

    Any help will be appreciated.
     
  2. jcsd
  3. Oct 17, 2012 #2
    I have found some other interpretations about my question:

    1) Multiplication of x[n] by a linearly increasing sequence doesn't change the ROC, since we have z^{-n} factor in the z-transform.

    2) There is no extra pole-zero introduced by differentiation; only their order can change.

    The first one makes sense; however still it is not clear since, for example, if z=e^{jw}, then there is no decaying exponential in the z-transform.

    The second one brings some other questions: Do we always have to think X(z) as a ratio of polynomials? Why don't we have extra pole-zero after differentiation?
     
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