Question

The magnitude Bode plot of a network is shown in the figure

The maximum phase angle ${\varphi }_m$ and the corresponding gain $G_m$ respectively, are

• A. $-{30}^{\circ }and1.73dB$
• B. $-{30}^{\circ }and4.77dB$
• C. $+{30}^{\circ }and4.77dB$
• D. $+{30}^{\circ }and1.73dB$

Answer 1

The correct option is C $+{30}^{\circ }and4.77dB$

From the given Bode plot, it is clear that corner frequencies,
${\omega }_{c_1}=\frac{1}{3}and{\omega }_{c_2}=1$
$\therefore$ Transfer function of given system is given by
$T\left(s\right)=\frac{C\left(s\right)}{R\left(s\right)}=\left[\frac{1+3s}{1+s}\right]$

$=\left[\frac{1+Ts}{1+\alpha Ts}\right]$

$Here,T=3and\alpha T=1or\alpha =\frac{1}{3}$

As $\alpha <1$, therefore the transfer function T(s)
represents lead compensator having $\alpha =\frac{1}{3}$

$\therefore$ Maximum phase shift,
${\varphi }_m=ta{n}_{-1}\left[\frac{1-\alpha }{2\sqrt{\alpha }}\right]=ta{n}^1\left[\frac{1-\frac{1}{3}}{\frac{2}{\sqrt{3}}}\right]$

$=ta{n}^1[\frac{2}{3}\times \frac{\sqrt{3}}{2}]=ta{n}^{-1}\left[\frac{1}{\sqrt{3}}\right]={30}^{\circ }$

$\therefore {\varphi }_m={30}^{\circ }$

$AlsoT\left(j\omega \right)=\frac{1+j3\omega }{1+j\omega }$

$\therefore \left|T\left(j\omega \right)\right|=\frac{\sqrt{1+9{\omega }^2}}{\sqrt{1+{\omega }^2}}$

$\therefore Gain{G}_{m}indB$ = $20log\left|T\right(j\omega \left)\right|$

$=20lo{g}_{10}\left[\frac{\sqrt{1+9{\omega }_m^2}}{\sqrt{1+{\omega }_m^2}}\right]$

$Now,{\omega }_m=\frac{1}{T\sqrt{\alpha }}=\frac{1}{\sqrt{3}}$

$\therefore G_m=20lo{g}_{10}\left[\sqrt{\frac{1+9\times {\left(\frac{1}{\sqrt{3}}\right)}^2}{1+{\left(\frac{1}{\sqrt{3}}\right)}^2}}\right]$

$=20lo{g}_{10}\left[\sqrt{\frac{1+3}{1+\frac{1}{3}}}\right]=20lo{g}_{10}\sqrt{3}$

$or{G}_m=4.77dB$