Design Of Control Systems – GATE EE Control SystemsPractice Questions & PYQs

$T(s)=\frac{1+3Ts}{1+Ts}$ Frequency at which $\angle T(j\omega )$ is maximum, $(\omega _m)=\frac{1}{T\sqrt{\alpha }}$ $\alpha =\frac{1}{1/3}=3$ $\omega _m=\frac{1}{T\sqrt{3}}=\sqrt{\frac{1}{3T^2}}$

Pole zero diagram of compensator transfer function is shown below. Maximum phase lead is between 0.1 and 1. $0.1 \lt \omega \lt 1$

Assuming

$G_c(s)=\frac{s+a}{s+b}$ $G_c(j\omega )=\frac{j\omega +a}{j\omega +b}$ $\angle G_c(j\omega )=tan^{-1}\frac{\omega }{a}-tan^{-1}\frac{\omega }{b}$ $\;\;\;= tan^{-1}\left ( \frac{\frac{\omega }{a}-\frac{\omega }{b}}{1+\frac{\omega ^2 }{ab}} \right )$ For $G_c(s)$ to be a lead compensator $\angle G_c(j\omega ) \gt 0$ $\frac{\omega }{a} \gt \frac{\omega }{b}$ $\Rightarrow b \gt a$ Option (A) Satisfies the above equation.

$\omega _m=\sqrt{\omega _{c1} \times \omega _{c2}}$ $\;\;=\sqrt{1 \times 2}=\sqrt{2}$ rad/sec.

$C_1=\frac{10(s+1)}{(s+10)}$ Zero at s=-1 Pole at s=-10 As zero is closer origin, zero dominates pole. Hence $C_1$ is lead compensator. $C_2=\frac{s+10}{10(s+1)}$ Zero at s=-10 Pole at s=-1 As pole is closer to origin, pole dominates zero. Hence $C_2$ is lag compensator.

Let uncompensated syatem, $T(s)=\frac{900}{s(s+1)(s+9)}$ Phase crossover frequency of uncompensated system $=(\omega _{pc})$ , at this frequency phase of $T(j \omega )$ is $-180^{\circ}$ Put $s=j\omega \text{ \in } T(s)$

Gain cross frequency of compensated system, $(\omega _{gc})_2$ = phase cross frequency of uncompensated system, $(\omega _{pc})_1$ $\Rightarrow \;\;(\omega _{gc})_2=(\omega _{gc})_1=\; 3$ rad/sec Phase-margin $=180^{\circ}+\angle T(j\omega )|_{\omega =(\omega _{gc})_2}$ $\Rightarrow \; \; \; 45^{\circ}=180^{\circ}+\angle T(j\omega )|_{\omega =(\omega _{pc})_2}$ At $(\omega _{gc})_2=3$ rad/sec, phase angle of $\angle T(j\omega )$ is $-135^{\circ}$ and phase of uncompensated systen is $-180^{\circ}$ at 3 rad/sec. Therefore, the compensator provides phase lead of $45^{\circ}$ at the frequency of 3 rad/sec. Let X dB is the gain provided by the compensator, so at gain cross frequency, $|T(j\omega )|_{com}$ =1 or 0 dB. Gain of uncompensated system $=|T(j\omega )|_{un-com} =\frac{100}{\omega \sqrt{1+\omega ^2}\sqrt{1+\left ( \frac{\omega }{9} \right )^2}}$ $|T(j\omega )|_{un-com}$ in dB $\;=40-20 \log \omega -20 \log \sqrt{1+\omega ^2} -20 \log \sqrt{1+\left ( \frac{\omega }{9} \right )^2}$ Gain of compensated system, $|T(j\omega )|_{com}=X+|T(j\omega )_{un-com}|$ $|T(j\omega )|_{com}$ must be zero at gain cross frequency $(\omega _{gc})_2$ $|T(j\omega_{gc} )_2|_{com}=X+40-20 \log (\omega _{gc})_2 -20 \log \sqrt{1+(\omega _{gc})^2_2}-20 \log \sqrt{1+\frac{(\omega _{gc})^2}{9^2}}=0$ $X+40-20 \log 3-20 \log \sqrt{1+3^2}-20 \log \sqrt{1+\left ( \frac{3}{9} \right )^2}=0$ $X=-20dB$ So, the compensator provides an attenuation od 20dB. Hence option (D) is correct.

Transfer function $=\frac{k\left ( 1+\frac{s}{a} \right )}{\left ( 1+\frac{s}{b} \right )}$ Zero of TF=-a Pole of TF=-b For a lead-compensator, the zero is nearer to origin as compared to pole, hence the effect of zero is dominant, therefore, the lead-compensator when introduced in series with forward path of the trnasfer function the phase shift is increased. So, from pole-zero configuration of the compensator $a \lt b$ .