# why z-transform differentiation is needed?

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

$$x(n)=|n|\left(\frac{1}{2}\right)^{|n|}$$

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

And,

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

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?

• 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. – GKH Jan 5 at 8:34
• If $x(n) \overset{ZT}\longleftrightarrow X(z)$, then according to the differentiation in frequency property of Z transform, $$nx(n) \overset {ZT}\longleftrightarrow -z \dfrac{d}{dz}X(z)$$ – Shehin Jan 5 at 10:06

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):
$$\mathcal{Z}\big\{nx[n]\big\}=-z\frac{dX(z)}{dz}\tag{1}$$