Question

The open loop transfer function of a unity feedback control system is given as $G\left(s\right)=\frac{as+1}{s^2}.$ The value of 'a' to give a phase margin of ${45}^{\circ }$ is equal to.

• A. 0.141
• B. 0.441
• C. 0.841
• D. 1.141

Answer 1

The correct option is C 0.841


$G\left(s\right)=\frac{as+1}{s^2}$
${\omega }_g$=gain crossover frequency, at which open gain is loop .
$Phasemargin={180}^{\circ }+\angle G\left(j\omega \right)$
$={45}^{\circ }={180}^{\circ }+\angle G\left(j{\omega }_g\right)$
$\angle G\left(j{\omega }_g\right)=-{135}^{\circ }$
$\angle G\left(j{\omega }_g\right)=\frac{ja\omega {}_g+1}{(j{\omega }_g{)}^2}$
$=-2\times {90}^{\circ }+ta{n}^{-1}a{\omega }_g$
$\angle G\left(j{\omega }_g\right)=-{180}^{\circ }+ta{n}^{-1}a{\omega }_g=-{135}^{\circ }$
$ta{n}^{-1}a{\omega }_g={45}^{\circ }$
$a{\omega }_g=1$
${\omega }_g=\frac{1}{a}$
$Gain=|G\left(j{\omega }_g\right)|=1$
$\Rightarrow \frac{\sqrt{1+a^2{\omega }_g^2}}{{\omega }_g^2}=1$
$\Rightarrow \frac{\sqrt{1+a^2\times \frac{1}{a^2}}}{1/a^2}=1$
$\frac{1}{a^2}=\sqrt{2}$
$a=\frac{1}{2^{1/4}}=0.841$