Question

$\text{Consider the z-transform}X\left(z\right)=\frac{10-8z^{-1}}{2-5z^{-1}+2z^{-2}}\text{if ROC includes unit circle then the}\text{value of x(1) is}$

• A. $1.5$
• B. $0.5$
• C. $1$
• D. $2$

Answer 1

The correct option is B $0.5$

$\text{Given}X\left(z\right)=\frac{10-8z^{-1}}{2-5z^{-1}+2z^{-2}}=\frac{2}{(2-z^{-1})}+\frac{4}{(1-2z^{-1})}X\left(z\right)=\frac{2z}{2z-1}+\frac{4z}{z-2}X\left(z\right)=\frac{z}{(z-\frac{1}{2})}+\frac{4z}{(z-2)}\text{Since, ROC includes unit circle,}\therefore \text{ROC of X(z) is}\frac{1}{2}<|z|<2x\left[n\right]={\left(\frac{1}{2}\right)}^{n}u\left[n\right]-4\left(2^n\right)u[-n-1]\therefore x\left(1\right)=\frac{1}{2}=0.5$