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$X(z) = 1 / ((z-1)^2(z-3))$ 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.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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$$X(z)=\frac{1}{(z-1)^2(z-3)}=\frac{1/4}{(z-3)}-\frac{1/4}{z-1}-\frac{1/2}{(z-1)^2},\,|z|>3$$

$$X(z)=0.25z^{-1}\frac{z}{z-3}-0.25z^{-1}\frac{z}{z-1}-0.5z^{-1}\left(\frac{z}{(z-1)^2}\right)$$

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 $nx[n]$.

Hence,

$$x[n]=(0.25)3^{n-1}u[n-1]-(0.25)u[n-1]-(0.5)(n-1)u[n-1]$$

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  • $\begingroup$ 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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