GATE 2020 EE Q.10: solution and answer

GATE 2020 · EEQ.10 · 1 mark · MCQ

Consider a linear time-invariant system whose input $r (t)$ and output $y (t)$ are related by the following differential equation:

\frac{d ^{2} y ( t )}{d t ^{2}} + 4 y (t) = 6 r (t)

The poles of this system are at

(A)
$+ 2 j, - 2 j$
(B)
$+ 2, - 2$
(C)
$+ 4, - 4$
(D)
$+ 4 j, - 4 j$

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Answer and solution

ANSWER

The transfer function is $G (s) = \frac{6}{s ^{2} + 4}$ , so the poles are at $s = + 2 j$ and $s = - 2 j$ . The answer is option (A).

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

Take the Laplace transform with zero initial conditions, since a transfer function is defined that way: $s^{2} Y (s) + 4 Y (s) = 6 R (s)$ .
Transfer function: $G (s) = \frac{Y ( s )}{R ( s )} = \frac{6}{s ^{2} + 4}$ The gain of 6 in the numerator scales the response but has no effect on where the poles are.
The poles are the roots of the denominator, i.e. of the characteristic equation: $s^{2} + 4 = 0 \Rightarrow s^{2} = - 4$ .
$s = \pm - 4 = \pm 2 j$ . Comparing with $s^{2} + 2 ζ ω_{n} s + ω_{n}^{2}$ gives $ω_{n} = 2 rad/s$ and $ζ = 0$ . The system is undamped: it has a pair of simple conjugate poles on the imaginary axis, so it is marginally stable and oscillates steadily at $2 rad/s$ .
The poles are at $+ 2 j$ and $- 2 j$ , which is option (A).

KEY IDEA

Poles come only from the left-hand (output) side of the differential equation: its Laplace transform gives the characteristic polynomial $s^{2} + 4$ . The missing $s$ term means $ζ = 0$ , so the poles sit on the imaginary axis at $\pm j ω_{n}$ with $ω_{n} = 4 = 2 rad/s$ . The input-side gain of 6 plays no part.

WATCH OUT FOR

One slip is solving $s^{2} = 4$ in place of $s^{2} = - 4$ , which gives the real pair $\pm 2$ (option B). Another is reading the coefficient 4 as the pole location, which gives $\pm 4$ or $\pm 4 j$ (options C and D), even though 4 is $ω_{n}^{2}$ , not $ω_{n}$ . A third is letting the right-hand-side gain of 6 affect the poles.

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GATE 2020 EE Q.10: solution and answer

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