Question

For a discrete time signal x(n), the z-transform is defined as X(z), If $x\left(n\right)=\{\begin{array}{cc}\frac{(2{)}^{-n}}{n!};& n\ge 0\\ 0;& Otherwise\end{array}$

Then the value of X(1) is _______

• A. 1.65

Answer 1

The correct option is A 1.65
$X\left(z\right)={\sum }_{n=-\infty }^{\infty }x\left(n\right)z^{-n}={\sum }_{n=0}^{\infty }\frac{2^{-n}}{n!}{z}^{-n}$

$={\sum }_{n=0}^{\infty }\frac{(2z{)}^{-n}}{n!}={\sum }_{n=0}^{\infty }\frac{{\left(\frac{1}{2z}\right)}^n}{n!}$

$X\left(z\right)=1+\frac{1}{\frac{2z}{1!}}+\frac{{\left(\frac{1}{2z}\right)}^2}{2!}+\frac{\left(\frac{1}{2z}\right)}{3!}+...$

$X\left(z\right)=e^{1/2z}$

$X\left(1\right)=e^{1/2}=\sqrt{e}=1.648=1.65$