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GATE 2023 EE – Question 38

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

Consider a lead compensator of the form

$$K(s)=\frac{1+\frac sa}{1+\frac s{\beta a}},\quad\beta>1,\ a>0$$

The frequency at which this compensator produces maximum phase lead is 4 rad/s. At this frequency, the gain amplification provided by the controller, assuming asymptotic Bode-magnitude plot of $K(s)$, is 6 dB. The values of $a$, $\beta$, respectively, are

  1. 1, 16
  2. 2, 4
  3. 3, 5
  4. 2.66, 2.25

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Correct answer: (B) 2, 4

Explanation

The maximum phase lead occurs at $\omega_m=a\sqrt\beta=4$. On the asymptotic plot the gain at $\omega_m$ is $20\log\dfrac{\omega_m}{a}=20\log\sqrt\beta=6$ dB, so $\sqrt\beta=2$ and $\beta=4$. Then $a=4/2=2$.