Question

If the compensated system shown in the figure has a phase margin of ${60}^{\circ }$ at the crossover frequency of 1 rad /sec, then value of the gain k is

• A. 0.366
• B. 0.732
• C. 1.366
• D. 2.738

Answer 1

The correct option is C 1.366

$G\left(s\right)=\frac{k+0.366s}{s(s+1)}$ and $H\left(s\right)=1$

$OLTF=G\left(s\right)H\left(s\right)=\frac{k+0.366s}{s(s+1)}$

$Phasemargin={60}^{\circ }at{\omega }_{gc}$
(gain crossover frequency)

$\Rightarrow {180}^{\circ }+\angle G\left(j{\omega }_{gc}\right)H\left(j{\omega }_{gc}\right)={60}^{\circ }$
$\angle G\left(j{\omega }_{gc}\right)H\left(j{\omega }_{gc}\right)=-{120}^{\circ }$

$-{90}^{\circ }-ta{n}^{-1}{\omega }_{gc}+ta{n}^{-1}\left(\frac{0.366{\omega }_{gc}}{k}\right)=-{120}^{\circ }$

${\omega }_{gc}=1rad/sec$

$-{90}^{\circ }-ta{n}^{-1}\left(1\right)+ta{n}^{-1}\left(\frac{0.366\times 1}{k}\right)=-{120}^{circ}$

$=-{90}^{\circ }-{45}^{\circ }+ta{n}^{-1}\left(\frac{0.366}{k}\right)=-{120}^{\circ }$

$\Rightarrow ta{n}^{-1}\left(\frac{0.366}{k}\right)={15}^{\circ }$

$\Rightarrow k=\frac{0.366}{tan{15}^{\circ }}=1.366$