GATE 2024 IN Q.45: solution and answer

GATE 2024 · INQ.45 · 2 marks · MCQ

The transfer function of a system is
$G (s) = \frac{ω _{n}^{2}}{s ^{2} + 2 ξ ω _{n} s + ω _{n}^{2}}$
Choose the range of $ξ$ and $ω_{n}$ (in rad/s) from the following options such that the poles lie on the shaded region of the $s$ -plane as shown in the figure.

GATE 2024 IN question 45: An s-plane with a vertical jω axis and a horizontal σ axis, both drawn with arrows, and the label 's-plane' in the first… — Figure from the GATE 2024 IN paper. It is part of the question.

(A)
$ξ \geq \frac{1}{2}$ and $ω_{n} \geq 2$
(B)
$ξ \geq \frac{1}{4}$ and $ω_{n} \geq 2$
(C)
$ξ \geq \frac{1}{2}$ and $ω_{n} \geq 3$
(D)
$ξ \geq \frac{1}{4}$ and $ω_{n} \geq 3$

Answer and solution

ANSWER

Option (A): $ξ \geq \frac{1}{2}$ and $ω_{n} \geq 2 rad/s$ .

Official answer: ASource: Official answer key, GATE 2024 IN (IISc Bengaluru)

Pole geometry. For $0 < ξ < 1$ the poles of $G (s)$ are $s = - ξ ω_{n} \pm j ω_{n} 1 - ξ^{2}$ Their distance from the origin is $∣ s ∣ = ξ^{2} ω_{n}^{2} + ω_{n}^{2} (1 - ξ^{2}) = ω_{n}$ If $θ$ is the angle each pole makes with the negative real axis, then $cos θ = \frac{ξ ω _{n}}{ω _{n}} = ξ$ . So on the $s$ -plane $ω_{n}$ sets the radius and $ξ$ sets the angle. At $ξ = 1$ both poles sit at $- ω_{n}$ with $θ = 0$ , which fits the same picture. The figure's rays and arc are drawn in exactly these radius and angle terms.
Read the boundary rays. The rays are at $12 0^{\circ}$ from the positive $σ$ axis. Measured from the negative real axis, that is $θ = 18 0^{\circ} - 12 0^{\circ} = 6 0^{\circ}$ , above and below the axis.
Damping condition. The shaded region lies between the two rays and contains the negative real axis, so every pole in it has $θ \leq 6 0^{\circ}$ . Since $cos θ$ decreases on $0^{\circ} \leq θ \leq 9 0^{\circ}$ , $ξ = cos θ \geq cos 6 0^{\circ} = \frac{1}{2}$
Natural-frequency condition. The shaded region lies outside the arc of radius 2, so $∣ s ∣ \geq 2$ . Because $∣ s ∣ = ω_{n}$ , this gives $ω_{n} \geq 2 rad/s$ .
Combine the conditions and check the other options. Together the conditions are $ξ \geq \frac{1}{2}$ and $ω_{n} \geq 2$ , which is option (A). Option (B) allows $ξ = \frac{1}{4}$ , which gives $θ = cos^{- 1} \frac{1}{4} \approx 75. 5^{\circ} > 6 0^{\circ}$ , so the poles would fall outside the wedge. Options (C) and (D) allow values such as $ω_{n} = 1.8$ , between $3 \approx 1.73$ and $2$ , which puts the poles inside the arc, in the unshaded region.
Where $3$ comes from. The corner where the upper ray meets the arc is $2∠12 0^{\circ} = 2 cos 12 0^{\circ} + j 2 sin 12 0^{\circ} = - 1 + j 3$ . The number $3$ is that point's height above the real axis, not its distance from the origin. The distance, which is what $ω_{n}$ measures, is $2$ .

KEY IDEA

For a complex pole pair of the standard second-order system, $∣ s ∣ = ω_{n}$ , so circles centred at the origin are lines of constant $ω_{n}$ . The angle from the negative real axis is $cos^{- 1} ξ$ , so rays from the origin are lines of constant $ξ$ . A wedge of half-angle $θ$ around the negative real axis therefore means $ξ \geq cos θ$ , and the region outside a circle of radius $r$ means $ω_{n} \geq r$ .

WATCH OUT FOR

One mistake is to use the marked $12 0^{\circ}$ directly, which gives $cos 12 0^{\circ} = - \frac{1}{2}$ . The damping angle has to be measured from the NEGATIVE real axis, which makes it $6 0^{\circ}$ and gives $ξ \geq \frac{1}{2}$ . The other mistake is to take the corner point's imaginary part $3$ , from $- 1 + j 3$ , as the bound on $ω_{n}$ (options C and D). The bound is the radius, 2, not the height. The value $ξ \geq \frac{1}{4}$ in option B matches no boundary in the figure: $cos^{- 1} \frac{1}{4} \approx 75. 5^{\circ}$ , which lies outside the wedge.

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GATE 2024 IN Q.45: solution and answer

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