Question

The system $900/s(s+1)(s+9)$ is to be such that its gain-crossover frequency becomes same as its uncompensated phase crossover frequency and provides a ${45}^o$ phase margin. To achieve this, one may use

• A. a lag compensator that provies an atenuation of $20dB$ and a phase lag of ${45}^o$ at the frequency of $3\sqrt{3}rad/s$.
• B. a lead compensator that provides an amplification of $20dB$ and a phase lead of ${45}^o$ at the frequency of $3rad/s$.
• C. a lag-lead compensator that provides an amplification of $20dB$ and a phase lag of ${45}^o$ at the frequency at $\sqrt{3}rad/s$
• D. a lag-lead compensator that provides an attenuation of $20dB$ and phase lead of ${45}^o$ at the frequency of $3rad/s$

Answer 1

The correct option is D a lag-lead compensator that provides an attenuation of $20dB$ and phase lead of ${45}^o$ at the frequency of $3rad/s$

Let uncompensated system,
$T\left(s\right)=\frac{900}{s(s+1)(s+9)}$

Phase crossover frequency of uncompensated system $={\omega }_{pc},$ at this frequency phase of $T\left(j\omega \right)is-{180}^o.$

$Puts=j\omega ,$

$T\left(j\omega \right)=\frac{900}{j\omega (j\omega +1)(j\omega +9)}$

$\angle T\left(j\omega \right)=-{90}^o-ta{n}^{-1}\omega -ta{n}^{-1}\left(\frac{\omega }{9}\right)$

$At{\omega }_{p{c}_1},\angle T\left(j\omega \right)=-{180}^o$

$-{180}^o=-{90}^o-ta{n}^{-1}\left({\omega }_{p{c}_1}\right)=ta{n}^{-1}\left(\frac{({\omega }_{pc}{)}_1}{9}\right)$
$-{90}^o=ta{n}^{-1}\frac{({\omega }_{pc}{)}_1+\frac{({\omega }_{pc}{)}_1}{9}}{1-\frac{({\omega }_{pc}{)}_1^2}{9}}$

$\Rightarrow 1-\frac{({\omega }_{pc}{)}_1^2}{9}=0$

$\Rightarrow ({\omega }_{pc}{)}_1=3rad/sec$

Gain cross frequency of compensated system, $({\omega }_{pc}{)}_2=$ phase cross frequency of uncompensated system, $({\omega }_{pc}{)}_1$

$\Rightarrow ({\omega }_{gc}{)}_2=({\omega }_{gc}{)}_1=3rad/sec$

Phase-margin $={180}^o+\angle T\left(j\omega \right){|}_{\omega =({\omega }_{gc}{)}_2}$

$\Rightarrow {45}^o={180}^o+\angle T\left(j\omega \right){|}_{\omega =({\omega }_{pc}{)}_2}$

$At({\omega }_{gc}{)}_2=3rad/sec,$
phase angle of $\angle T\left(j\omega \right)$ is $-{135}^o$, and phase of uncompensated system is $-{180}^o$ at $3rad/sec$.

Let $XdB$ is the gain provided by the compensator, so at gain cros frequency, ${\left|T\right(j\omega \left)\right|}_{com}=1or0dB.$

Gain of uncompensated system ${\left|T\right(j\omega \left)\right|}_{un-com}=\frac{100}{\omega \sqrt{1+{\omega }^2}\sqrt{1+{\left(\frac{\omega }{9}\right)}^2}}$

$|T\left(j\omega \right){|}_{un-com}$ in dB

$=40-20log\omega -20log\sqrt{1+{\omega }^2}-20log\sqrt{1+{\left(\frac{\omega }{9}\right)}^2}$

Gain of compensated system,
$|T\left(j\omega \right){|}_{com}=X+|T(j\omega {)}_{un-com}|$

$|T\left(j\omega \right){|}_{com}$ must be zero at gain cross frequency $({\omega }_{gc}{)}_2$

$|T(j{\omega }_{gm}{)}_2{|}_{com}=X+40-20log({\omega }_{gc}{)}_2$
$-20log\sqrt{1+({\omega }_{gc}{)}_2^2}$
$-20log\sqrt{1+\frac{({\omega }_{gc}{)}^2}{9^2}}=0$
$X+40-20log3-20log\sqrt{1+3^2}$

$X=-20dB$

So, the compensator provides an attenuation of $20dB.$ Hence option (d) is correct.