GATE 2017 EE (EE2) – Question 17

Control Systems · Lag, Lead and Lead-Lag compensators; P, PI and PID controllers · 1 mark · Multiple choice

A compensator has
$$C(s)=\frac{(1+s/0.1)(1+s/100)}{(1+s)(1+s/10)}.$$
In which frequency range does its phase lead reach a maximum?

$0.1<\omega<1$
$1<\omega<10$
$10<\omega<100$
$\omega>100$

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Show answer and explanation

Correct answer: (A) $0.1<\omega<1$

Explanation

The zero at 0.1 and pole at 1 form a lead pair, whose isolated maximum occurs near $\sqrt{0.1\cdot1}=0.316$ rad/s. The pole at 10 and zero at 100 form a lag pair, contributing only a small negative phase at that low frequency.

The total phase is $\tan^{-1}(10\omega)+\tan^{-1}(\omega/100)-\tan^{-1}\omega-\tan^{-1}(\omega/10)$. Its positive maximum lies near 0.30 rad/s, within **0.1<ω<1, A**.

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