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



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 '16 at 1:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.