$X(z) = 1 / ((z-1)^2(z-3))$ where ROC is $|z|>3$.

I've put the equation into partial fractions then I am not sure on how to proceed further with the first term. Any help is appreciated.

  • $\begingroup$ Confirm it is $$\frac{1}{(z-1)^2(z-3)}$$ $\endgroup$
    – msm
    Dec 18, 2016 at 1:27
  • $\begingroup$ @msm yes that is correct. Updated question. $\endgroup$
    – jump68
    Dec 18, 2016 at 1:31
  • $\begingroup$ Inverse of which transform? $\endgroup$
    – GFauxPas
    Dec 18, 2016 at 1:41
  • 7
    $\begingroup$ Do not delete your question after it has received an answer. That's rude towards the answerer. $\endgroup$ Dec 18, 2016 at 16:30

1 Answer 1




Note that:

  • The inverse of $\frac{z}{z-a}, |z|>a$ is $a^nu[n]$.
  • The inverse transform of $z^{-n_0}X(z)$ is $x[n-n_0]$.
  • $\frac{z}{(z-1)^2}=-z\frac{d}{dz}(\frac{z}{z-1})$.
  • The inverse of $-z\frac{dX(z)}{dz}$ is $nx[n]$.



  • $\begingroup$ any tips on solving the last part $-\frac{1/2}{(z-1)^2}$? $\endgroup$
    – jump68
    Dec 18, 2016 at 1:48

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