$X(z) = 1 / ((z-1)^2(z-3))$ 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 $$\frac{1}{(z-1)^2(z-3)}$$ $\endgroup$– msmDec 18, 2016 at 1:27
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$\begingroup$ @msm yes that is correct. Updated question. $\endgroup$– jump68Dec 18, 2016 at 1:31
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$\begingroup$ Inverse of which transform? $\endgroup$– GFauxPasDec 18, 2016 at 1:41
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7$\begingroup$ Do not delete your question after it has received an answer. That's rude towards the answerer. $\endgroup$– Daniel FischerDec 18, 2016 at 16:30
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1 Answer
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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 $-\frac{1/2}{(z-1)^2}$? $\endgroup$– jump68Dec 18, 2016 at 1:48