Question

$\text{For a discrete time signal x(n), the z-tansform is defined as X(z). If x(n) is defined as}x\left(n\right)=\{\begin{array}{cc}\frac{(2{)}^{-n}}{n!};& n\ge 0\\ 0& \text{Otherwise}\end{array}\text{Then the value of X(1) is}$

• A. $2.25$
• B. $3.50$
• C. $1.65$
• D. $4.00$

Answer 1

The correct option is C $1.65$

$x\left(n\right)=\{\begin{array}{cc}\frac{(2{)}^{-n}}{n!};& n\ge 0\\ 0& \text{Otherwise}\end{array}X\left(z\right)={\sum }_{n=-\infty }^{\infty }x\left(n\right)z^{-n}={\sum }_{n=0}^{\infty }\frac{2^{-n}}{n!}{z}^{-n}={\sum }_{n=0}^{\infty }\frac{(2z{)}^{-n}}{n!}={\sum }_{n=0}^{\infty }\frac{{\left(\frac{1}{2z}\right)}^n}{n!}X\left(z\right)=1+\frac{\frac{1}{2z}}{1!}+\frac{{\left(\frac{1}{2z}\right)}^2}{2!}+\frac{{\left(\frac{1}{2z}\right)}^3}{3!}+...X\left(z\right)=e^{1/2z}X\left(1\right)=e^{1/2}=\sqrt{e}=1.648=1.65$