GATE 2023 EE Q.38: solution and answer

GATE 2023 · EEQ.38 · 2 marks · MCQ

Consider a lead compensator of the form

K (s) = \frac{1 + \frac{s}{a}}{1 + \frac{s}{β a}}, β > 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 $a, β$ , respectively, are

(A)
1, 16
(B)
2, 4
(C)
3, 5
(D)
2.66, 2.25

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Answer and solution

ANSWER

(B): $a = 2$ , $β = 4$ . Maximum phase lead occurs at $ω_{m} = a β = 4 rad/s$ , and the gain there is $10 lo g_{10} β = 6 dB$ , which gives $β = 4$ and so $a = 2$ .

Official answer: BSource: Official answer key, GATE 2023 EE (IIT Kanpur)

Corner frequencies: the zero is at $ω = a$ and the pole at $ω = β a$ . Because $β > 1$ , the zero comes before the pole, so this is a lead network.
Maximum phase lead occurs at the geometric mean of the two corner frequencies: $ω_{m} = a \cdot β a = a β = 4 rad/s$
Asymptotic Bode magnitude: it is 0 dB up to $ω = a$ , then rises at $+ 20 dB/decade$ until $ω = β a$ . Since $1 < β < β$ , $ω_{m}$ lies on this rising segment, so $∣ K (j ω_{m}) ∣_{dB} = 20 lo g_{10} \frac{ω _{m}}{a} = 20 lo g_{10} β = 10 lo g_{10} β$
Set this equal to 6 dB: $10 lo g_{10} β = 6 \Rightarrow β = 1 0^{0.6} \approx 3.98 \approx 4$ (the 6 dB in the question is the usual rounding of $20 lo g_{10} 2 = 6.02 dB$ , i.e. a gain of 2).
Substitute into the $ω_{m}$ condition: $a = \frac{4}{β} = \frac{4}{2} = 2$
Check each option against both conditions, $a β = 4 rad/s$ and $10 lo g_{10} β = 6 dB$ : (A) $a β = 1 \times 4 = 4$ , but $10 lo g_{10} 16 \approx 12 dB$ . Fails on gain. (B) $a β = 2 \times 2 = 4$ and $10 lo g_{10} 4 \approx 6 dB$ . Meets both. (C) $a β = 35 \approx 6.7 rad/s$ . Fails on frequency (and $10 lo g_{10} 5 \approx 7 dB$ ). (D) $a β = 2.66 \times 1.5 \approx 4$ , but $10 lo g_{10} 2.25 \approx 3.5 dB$ . Fails on gain. Only (B), $a = 2, β = 4$ , satisfies both.

KEY IDEA

For a lead network $\frac{1 + s / a}{1 + s / ( β a )}$ , the maximum phase lead occurs at the geometric mean of the corners, $ω_{m} = a β$ . The gain there is $β$ , which is $10 lo g_{10} β$ dB, or half the high-frequency gain of $20 lo g_{10} β$ dB. The asymptotic and exact values agree at this frequency: $∣ K (j ω_{m}) ∣ = 1 + β / 1 + 1/ β = β$ . One condition fixes $β$ , and the other then fixes $a$ .

WATCH OUT FOR

Using $20 lo g_{10} β$ (the high-frequency gain, reached only above $ω = β a$ ) as the gain at $ω_{m}$ . Setting $20 lo g_{10} β = 6$ gives $β \approx 2$ , and $a β = 4$ then puts the answer near (D). Option (A) catches students who check only $a β = 4$ and ignore the gain condition. Another slip is placing $ω_{m}$ at the arithmetic mean $(a + β a) /2$ of the corners instead of the geometric mean.

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GATE 2023 EE Q.38: solution and answer

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