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 ${v_0}\left( t \right) = {V_m}\sin \left( {\omega t + \phi } \right)$ for the input, ${v_i}\left( t \right) = {V_m}\sin \left( {\omega t} \right)$. Then the lower and upper limiting values of the continuous unwrapped phase as ω ranges from 0 to infinity of ϕ (in radians) are respectively

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 ${v_0}\left( t \right) = {V_m}\sin \left( {\omega t + \phi } \right)$ for the input, ${v_i}\left( t \right) = {V_m}\sin \left( {\omega t} \right)$. Then the lower and upper limiting values of the continuous unwrapped phase as ω ranges from 0 to infinity of ϕ (in radians) are respectively

Prepp Team · Accepted Answer

The magnitude is unity and $\phi(\omega)=\arg(1-j\omega)-\arg(1+j\omega)=-2\tan^{-1}\omega$. It tends from 0 at zero frequency to −π as frequency tends to infinity. Thus the lower and upper limiting phases are −π and 0.

Prepp Team · Answer

$- \frac{\pi }{2}\;and\frac{\pi }{2}$

Prepp Team · Answer

$- \frac{\pi }{2}\;and\;0$

Prepp Team · Answer

$0\;and\frac{\pi }{2}$

Important Questions from Closed Loop Control Systems

Start Your Preparation with Prepp Mobile App