Question

The asymptotic Bode magnitude plot of lead network with its pole and zero on the left half of the s-plane is shown in the adjoining figure. The frequency at which the phase angle of the network in maximum (in rad/s) is

• A. $\frac{3}{\sqrt{10}}$
• B. $\frac{1}{\sqrt{20}}$
• C. $\frac{1}{20}$
• D. $\frac{1}{30}$

Answer 1

The correct option is B $\frac{1}{\sqrt{20}}$

System bode plot starts with 0 dB/decade, thus it is a type zero system with transfer fuction of
$\frac{k(1+\frac{s}{0.1})}{(1+\frac{s}{0.5})}=\frac{k[1+10s]}{(1+\frac{10s}{5})}=\frac{k(1+10s)}{(1+2s)}$

$20logk=0$

$\therefore logk=0$

$\therefore k=1$

$\therefore$Transfer function of system
$=\frac{(1+10s)}{(1+2s)}$

$\varphi =-ta{n}^{-1}\left(2\omega \right)+ta{n}^{-1}\left(10\omega \right)$

for $\varphi$ to be maximum,
$\frac{d\varphi }{d\omega }=0$

$-\frac{2}{1+(2\omega {)}^2}+\frac{10}{1+(10\omega {)}^2}=0$

$-\frac{2}{1+4{\omega }^2}+\frac{10}{1+100{\omega }^2}=0$

$\frac{10}{1+100{\omega }^2}=\frac{2}{1+4{\omega }^2}$

$5(1+4{\omega }^2)=1+100{\omega }^2$

$5+20{\omega }^2=1+100{\omega }^2$

$4=80{\omega }^2$

$\therefore {\omega }^2=\frac{1}{20}$

$\omega =\frac{1}{\sqrt{20}}rad/sec$