GATE 2020 IN Q.14: solution and answer

GATE 2020 · INQ.14 · 1 mark · MCQ

The closed loop transfer function of a control system is given by $\frac{C ( s )}{R ( s )} = \frac{1}{s + 1}$ . For the input $r (t) = sin t$ , the steady state response $c (t)$ is ____

(A)
$1$
(B)
$\frac{1}{2} cos t$
(C)
$\frac{1}{2} sin (t + \frac{π}{4})$
(D)
$\frac{1}{2} sin (t - \frac{π}{4})$

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Answer and solution

ANSWER

The steady-state response is $c (t) = \frac{1}{2} sin (t - \frac{π}{4})$ , which is option (D).

Official answer: DSource: Official answer key, GATE 2020 IN (IIT Delhi)

The only pole of $G (s) = \frac{1}{s + 1}$ is at $s = - 1$ , in the left half-plane, so the system is stable and its transient term $e^{- t}$ dies out. For a stable system with input $r (t) = A sin ω t$ , the steady-state output is a sinusoid at the same frequency: $c_{ss} (t) = A ∣ G (j ω) ∣ sin (ω t + ∠ G (j ω))$
The input $sin t$ has $A = 1$ and $ω = 1 rad/s$ . Substituting $s = j ω = j 1$ gives $G (j 1) = \frac{1}{1 + j}$ .
Magnitude: $∣ G (j 1) ∣ = \frac{1}{1 ^{2} + 1 ^{2}} = \frac{1}{2}$ .
Phase: $∠ G (j 1) = 0 - tan^{- 1} (\frac{1}{1}) = - 4 5^{\circ} = - \frac{π}{4}$ . The pole introduces a phase lag, so the output lags the input.
Therefore $c_{ss} (t) = \frac{1}{2} sin (t - \frac{π}{4})$ which is option (D).

KEY IDEA

For a stable LTI system, a sinusoid at $ω$ produces a steady-state sinusoid at the same $ω$ . Its amplitude is multiplied by $∣ G (j ω) ∣$ and its phase is shifted by $∠ G (j ω)$ . Here $G (j 1) = \frac{1}{1 + j}$ , so the amplitude becomes $\frac{1}{2}$ and the phase shifts by $- \frac{π}{4}$ .

WATCH OUT FOR

Getting the sign of the phase wrong and picking (C). A pole in the denominator makes the phase negative: $∠ \frac{1}{1 + j} = - tan^{- 1} 1 = - \frac{π}{4}$ , so the output lags the input. Picking (A) is another mistake: it uses the DC gain $G (0) = 1$ , which only applies to a step input, not to a sinusoid.

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GATE 2020 IN Q.14: solution and answer

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