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GATE 2025 EE – Question 65

Control Systems · Lag, Lead and Lead-Lag compensators; P, PI and PID controllers · 2 marks · Numerical answer

A controller $D(s)$ of the form $(1+K_Ds)$ is to be designed for the plant $G(s)=\dfrac{1000\sqrt2}{s(s+10)^2}$ as shown in the figure. The value of $K_D$ that yields a phase margin of 45° at the gain cross-over frequency of 10 rad/sec is ________________ (round off to one decimal place).

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Correct answer: 0.09 to 0.11

Explanation

At $\omega=10$: $|G(j10)|=\dfrac{1000\sqrt2}{10\times200}=\dfrac{\sqrt2}{2}$, and the phase is $-90^\circ-2\tan^{-1}(1)=-180^\circ$. Gain crossover needs $\dfrac{\sqrt2}{2}\sqrt{1+100K_D^2}=1$, so $1+100K_D^2=2$ and $K_D=0.1$. The controller adds $\tan^{-1}(10K_D)=45^\circ$, giving a phase margin of $45^\circ$ ✓.