Question

For the network shown in the figure below the frequency (in rad/s) at which the maximum phase lag occurs is ____.

• A. 0.316

Answer 1

The correct option is A 0.316
Assuming,
$R_1=9\Omega$
$R_1=1\Omega$
We can write,
$\frac{V_o\left(s\right)}{V_{in}\left(s\right)}=\frac{R_2+\frac{1}{sC}}{R_1+R_2+\frac{1}{sC}}$
$=\frac{1+R_2.Cs}{1+(R_1+R_2)Cs}$

Let,
$R_{2}C=T$
$\frac{R_1+R_2}{R_2}=\beta$

Hence,
$\frac{V_o\left(s\right)}{V_{in}\left(s\right)}=\frac{1+Ts}{(1+\beta Ts)}$

Which represent a lag compensator.

$\because$ Here, $T=R_{2}C=1.1=1sec$
$\beta =\frac{1+9}{1}=10$

Maximum phase lag occurs at frequency
${\omega }_n=\frac{1}{T\sqrt{\beta }}=\frac{1}{1\sqrt{10}}$
$=0.316$ rad/sec