Question

The transfer funciton of a system is given by $\frac{Y\left(s\right)}{X\left(s\right)}=\frac{e^{-0.1s}}{1+s}.$ If x(t) is $0.5sint,$ then the phase angle between the output and the input will be

• A. $-{39.27}^o$
• B. $-{45}^o$
• C. $-{50.73}^o$
• D. ${90}^o$

Answer 1

The correct option is C $-{50.73}^o$

$Phaseangle=-ta{n}^{-1}\omega -\frac{0.100\times {180}^o}{\pi }$
$For\omega =1$
$\therefore x\left(t\right)=0.5sint,\omega =1$
$\angle Y\left(j\omega \right)-\angle X\left(j\omega \right)=-ta{n}^{-1}\left(1\right)-\frac{0.1\times {180}^o}{\pi }$

$=-{50.73}^o$