1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 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 at 10:06
3
$\begingroup$

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):

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

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.