Question

The transfer function of a system is given by, $\frac{V_0\left(s\right)}{V_i\left(s\right)}=\frac{1-s}{1+s}.$ Let the output of the system be $V_0\left(t\right)=V_{m}sin(\omega t+j)$ for the input, $V_i\left(t\right)=V_{m}sin\left(\omega t\right)$ then the minimum and maximum values of $j$(in radians) are respectively

• A. $\frac{-\pi }{2}and\frac{\pi }{2}$
• B. $\frac{-\pi }{2}and0$
• C. $0and\frac{\pi }{2}$
• D. $-\pi and0$

Answer 1

The correct option is D $-\pi and0$

$\frac{V_0\left(s\right)}{V_i\left(s\right)}=\frac{1-s}{1+s}=H\left(s\right)$

For the minimum and maximum value of $\varphi$.

$H\left(j\omega \right)=\frac{1-j\omega }{1+j\omega }$

$\angle H\left(j\omega \right)=-2ta{n}^{-1}\omega$

$At\omega =0;\angle H\left(j\omega \right)=0^{\circ }$

$At\omega =\infty ;\angle H\left(j\omega \right)=-\pi$