# How to go about solving inverse transform of this by hand?

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

• Confirm it is $$\frac{1}{(z-1)^2(z-3)}$$
– msm
Dec 18, 2016 at 1:27
• @msm yes that is correct. Updated question. Dec 18, 2016 at 1:31
• Inverse of which transform? Dec 18, 2016 at 1:41

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

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

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,

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

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