Question

Let $X\left(z\right)=\frac{z}{1-z^{-5}}$ be the z - transform of a casual signal x[n]. Then the value of $x\left[0\right]+x[-1]+x\left[2\right]+x\left[3\right]+x\left[4\right]$ is

• A. 2
• B. 4
• C. 0
• D. 1

Answer 1

The correct option is A 2

$X\left(z\right)=\frac{z^6}{z^5-1}=z+z^{-4}+z^{-9}+$ .....

From z-transform definition,

$X\left(z\right)={\sum }_{n=-\infty }^{\infty }x\left[n\right]z^{-n}=x\left(0\right)+x\left(1\right)z^{-1}+x\left(2\right)z^{-2}+...$

Comparing :

$x\left(0\right)=0,x\left(2\right)=x\left(3\right)=0$

$x(-1)=1$

$x\left(4\right)=1$

$\therefore x\left[0\right]+x[-1]+x\left[2\right]+x\left[3\right]+x\left[4\right]=2$