GATE 2018 EC Q.41: solution and answer

GATE 2018 · ECQ.41 · 2 marks · NAT

For a unity feedback control system with the forward path transfer function

G (s) = \frac{K}{s ( s + 2 )}

The peak resonant magnitude $M_{r}$ of the closed-loop frequency response is 2. The corresponding value of the gain $K$ (correct to two decimal places) is ________.

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Answer and solution

ANSWER

$K = 8 + 43 \approx 14.93$ .

Official answer: 14 to 17Source: Official answer key, GATE 2018 EC (IIT Guwahati)

Closed-loop transfer function (unity feedback): $T (s) = \frac{G ( s )}{1 + G ( s )} = \frac{K}{s ^{2} + 2 s + K}$
Compare with the standard second-order form $\frac{ω _{n}^{2}}{s ^{2} + 2 ζ ω _{n} s + ω _{n}^{2}}$ : $ω_{n}^{2} = K$ gives $ω_{n} = K$ , and $2 ζ ω_{n} = 2$ gives $ζ = \frac{1}{K}$ , so $K = \frac{1}{ζ ^{2}}$ .
Resonant peak of a second-order system (a peak exists only for $ζ < \frac{1}{2} \approx 0.707$ ): $M_{r} = \frac{1}{2 ζ 1 - ζ ^{2}} = 2 \Rightarrow ζ 1 - ζ^{2} = \frac{1}{4} \Rightarrow ζ^{2} (1 - ζ^{2}) = \frac{1}{16}$
Solve the quadratic in $ζ^{2}$ : $ζ^{4} - ζ^{2} + \frac{1}{16} = 0 \Rightarrow ζ^{2} = \frac{1 \pm 1 - \frac{1}{4}}{2} = \frac{1 \pm \frac{3}{2}}{2}$ which gives $ζ^{2} = 0.0670$ or $ζ^{2} = 0.9330$ .
Reject $ζ^{2} = 0.9330$ : then $ζ = 0.966 > 0.707$ , the closed-loop magnitude falls steadily from 1 and there is no resonant peak, so the $M_{r}$ formula does not apply. Keep $ζ^{2} = \frac{2 - 3}{4} = 0.06699$ , so $ζ = 0.2588$ .
Gain: $K = \frac{1}{ζ ^{2}} = \frac{4}{2 - 3} = \frac{4 ( 2 + 3 )}{( 2 - 3 ) ( 2 + 3 )} = 4 (2 + 3) = 8 + 43 \approx 14.93$
Check: the resonant frequency is $ω_{r} = ω_{n} 1 - 2 ζ^{2} = K - 2 \approx 3.596 rad/s$ , which is real, so the peak exists. Evaluating $∣ T (j ω) ∣ = \frac{K}{∣ K - ω ^{2} + j 2 ω ∣}$ at that frequency gives exactly 2.

KEY IDEA

For $G (s) = \frac{K}{s ( s + a )}$ with unity feedback, the closed loop is a standard second-order system with $ω_{n} = K$ and $ζ = \frac{a}{2 K}$ . Setting $M_{r} = \frac{1}{2 ζ 1 - ζ ^{2}}$ gives a quadratic in $ζ^{2}$ . Take only the root with $ζ < 0.707$ , because a resonant peak exists only in that range. Then convert $ζ$ back to $K$ .

WATCH OUT FOR

Taking the larger root $ζ^{2} = 0.933$ , which gives $K \approx 1.07$ . That value of $ζ$ is above 0.707, where the closed-loop response has no resonant peak, so it cannot have $M_{r} = 2$ . A second slip is to apply $M_{r}$ to the open-loop $G (j ω)$ instead of the closed-loop $T (j ω)$ . A third is to write $ζ = 1/ K$ instead of $ζ = 1/ K$ when comparing with $2 ζ ω_{n} = 2$ .

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