Question 2016.S2.21: For the network shown in the figure below, the frequency (in......

From the given circuit.

\frac{V_{ o }(s)}{V_{ in }}=\frac{R_2+\frac{1}{s} C}{R_1+R_2+1 / s C}=\frac{1+R_2 C s}{1+\left\lgroup \frac{R_1+R_2}{R_2} \right\rgroup R_2 C s}

Let R_2 C=\tau \text { and } \frac{R_1+R_2}{R_2}=a

Then equation for lag compensation is

\begin{gathered}\frac{V_{ o }(s)}{V_{\text {in }}(s)}=\frac{1+\tau s}{(1+a \tau s)} \\\tau=R_2 C=1 \times 1=1 s \\a=\frac{1+9}{1}=10 \\\omega_{ n }=\frac{1}{\tau \sqrt{a}}=\frac{1}{\sqrt{10}}=0.316   rad / s\end{gathered}