GATE 2020 EE Q.41: solution and answer

GATE 2020 · EEQ.41 · 2 marks · MCQ

A stable real linear time-invariant system with single pole at $p$ , has a transfer function $H (s) = \frac{s ^{2} + 100}{s - p}$ with a dc gain of 5. The smallest positive frequency, in rad/s, at unity gain is closest to:

(A)
8.84
(B)
11.08
(C)
78.13
(D)
122.87

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Answer and solution

ANSWER

The dc gain gives $p = - 20$ . Setting $∣ H (j ω) ∣ = 1$ gives $ω^{4} - 201 ω^{2} + 9600 = 0$ , so $ω^{2} = 78.13$ or $122.87$ . The smallest positive unity-gain frequency is $ω = 78.13 \approx 8.84 rad/s$ , option (A).

Official answer: ASource: Official answer key, GATE 2020 EE (IIT Delhi)

DC gain fixes the pole. Put $s = 0$ : $H (0) = \frac{100}{- p} = 5 \Rightarrow p = - 20$ . The pole is in the left half-plane, which agrees with the system being stable. So $H (s) = \frac{s ^{2} + 100}{s + 20}$ .
Magnitude on the $j ω$ axis. With $s = j ω$ , the numerator is $100 - ω^{2}$ (real) and the denominator is $20 + j ω$ , so $∣ H (j ω) ∣ = \frac{∣100 - ω ^{2} ∣}{ω ^{2} + 400}$
Unity-gain condition $∣ H (j ω) ∣ = 1$ . Square both sides: $(100 - ω^{2})^{2} = ω^{2} + 400$
Substitute $x = ω^{2}$ : $x^{2} - 200 x + 10000 = x + 400 \Rightarrow x^{2} - 201 x + 9600 = 0$ .
Quadratic formula: $x = \frac{201 \pm 20 1 ^{2} - 4 ( 9600 )}{2} = \frac{201 \pm 40401 - 38400}{2} = \frac{201 \pm 2001}{2} = \frac{201 \pm 44.73}{2}$ which gives $x = 78.13$ or $x = 122.87$ .
Take square roots, since these are values of $ω^{2}$ : $ω = 78.13 = 8.84 rad/s$ or $ω = 122.87 = 11.08 rad/s$ . Check at $ω = 8.84$ : $∣100 - 78.13∣ = 21.87$ and $78.13 + 400 = 21.87$ , so the gain is indeed 1.
Pick the smaller root. The gain starts at 5 at $ω = 0$ , falls to zero at the notch $ω = 10 rad/s$ (the zeros $s = \pm j 10$ ), then rises without bound. It therefore crosses unity once below the notch and once above it. The smallest positive unity-gain frequency is $8.84 rad/s$ , option (A).

KEY IDEA

Use the dc gain $H (0)$ to find the unknown pole. Then set $∣ H (j ω) ∣^{2} = 1$ , which is a quadratic in $ω^{2}$ . The zeros at $\pm j 10$ make a notch at $ω = 10$ , so the gain falls from 5 to 0 and then rises again, and it crosses unity once on each side of the notch. The smaller crossing is the answer.

WATCH OUT FOR

Picking 78.13 (option C) or 122.87 (option D). Those are the roots for $ω^{2}$ , not $ω$ , and the options are set up to catch anyone who forgets the square root. Another trap is a sign slip in $H (0) = 100/ (- p)$ that gives $p = + 20$ . That pole is unstable and contradicts the problem statement, though it gives the same magnitude, so check the sign against stability.

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GATE 2020 EE Q.41: solution and answer

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