Find the z-transform and sketch pole-zero plot and ROC for


Right now, i can get the following,

$$x(n)=\begin{cases} n(\frac{1}{2})^n, & n \geq 0 \\ -n(\frac{1}{2})^{-n}, & n \lt 0 \end{cases}$$

Thus, $$n\left(\frac{1}{2} \right)^n u(n) -n\left(\frac{1}{2} \right)^{-n}u(-n-1)$$


$$\left(\frac{1}{2} \right)^n u(n) \underleftrightarrow{z} \frac{1}{1-\frac{1}{2} z^{-1}} ,|z|>\frac{1}{2}$$ $$\left(-2 \right)^n u(-n-1) \underleftrightarrow{z} \frac{1}{1-2 z^{-1}}, |z|<2$$

From here, i know my poles is $\frac{1}{2}$ and $2$. But, the solution that i have still continue working on to differentiate the equtions in z-domain and obtained the following,

$$\frac{\frac{1}{2} z^{-1}}{\left( 1-\frac{1}{2} z^{-1} \right)^{2}}+\frac{2z^{-1}}{\left( 1-2z^{-1} \right)^2} ,\frac{1}{2}<|z|<2$$

Does the differentitaion important and why does it need so?

  • $\begingroup$ Hi! Well, you have found the Z-Transform for, let's say, parts of your signals. Clearly what you have transformed does not correspond to the signal you have at the beginning. $\endgroup$
    – GKH
    Jan 5, 2020 at 8:34
  • $\begingroup$ If $x(n) \overset{ZT}\longleftrightarrow X(z)$, then according to the differentiation in frequency property of Z transform, $$ nx(n) \overset {ZT}\longleftrightarrow -z \dfrac{d}{dz}X(z)$$ $\endgroup$
    – Shehin
    Jan 5, 2020 at 10:06

1 Answer 1


You just wrote down the $\mathcal{Z}$-transform of $\left(\frac12\right)^nu[n]$, but you need the $\mathcal{Z}$-transform of $n\left(\frac12\right)^nu[n]$ (note the factor $n$).

In order to find that $\mathcal{Z}$-transform you can use the differentiation property (see this table):



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