Question

The z-transform of the discrete-time signal $x\left[n\right]=n{2}^{n}sin\left(\frac{\pi }{2}n\right)u\left[n\right]$ is

• A. $z(z^2-4)$
• B. $\frac{2z(z^2+4)}{(z^2-4{)}^2}$
• C. $\frac{2(z^2-4)}{(z^2+4{)}^2}$
• D. $\frac{2z(z^2-4)}{(z^2+4{)}^2}$

Answer 1

The correct option is D $\frac{2z(z^2-4)}{(z^2+4{)}^2}$

Given discrete-time signal

$x\left[n\right]=n{2}^{n}sin\left(\frac{n}{2}n\right)u\left[n\right]$

we know that,

$z\left[sin\left(\frac{\pi }{2}n\right)u\right[n\left]\right]=\frac{2sin\left(\frac{\pi }{2}\right)}{z^2-2zcos\left(\frac{\pi }{2}\right)+1}=\frac{z}{z^2+1}$

Using the multiplication by exponential property,

we have

$z\left[2^{n}sin\left(\frac{\pi }{2}n\right)u\right[n\left]\right]=z\left[sin\left(\frac{\pi }{2}n\right)u\right[n\left]\right]{\mid }_{z\to \left(\frac{z}{2}\right)}\left[\begin{array}{c}x\left[n\right]\leftrightarrow x\left(z\right)\\ a^{n}x\left[n\right]\leftrightarrow X\left(\frac{z}{a}\right)\end{array}\right]$


$=\frac{z}{z^2+1}{\mid }_{z\to \frac{z}{2}}=\frac{2z}{x^2+4}$

Using differentiation in z-domain property

$Z\left[n{2}^{n}sin\left(\frac{\pi }{2}u\right[n\left]\right)\right]=-z\frac{d}{dz}\left\{Z\left[n{2}^{n}sin\left(\frac{\pi }{2}n\right)u\right[n\left]\right]\right\}\left[\begin{array}{c}x\left[n\right]\leftrightarrow X\left(z\right)\\ nx\left[n\right]\leftrightarrow -z\frac{d}{dz}X\left(z\right)\end{array}\right]$

$=-z\frac{d}{dz}\left(\frac{2z}{z^2+4}\right)$

$=-z\left[\frac{(z^2+4)\left(2\right)-2z\left(2z\right)}{(z^2+4{)}^2}\right]=-z\left[\frac{-2x^2+8}{(z^2+4{)}^2}\right]$

$Z\left[n{2}^{n}sin\left(\frac{\pi }{2}n\right)u\right[n\left]\right]=\frac{2z(z^2-4)}{(z^2+4{)}^2}$