GATE 2023 IN Q.21: solution and answer

GATE 2023 · INQ.21 · 1 mark · MCQ

A system has the transfer-function

\frac{Y ( s )}{X ( s )} = \frac{s - π}{s + π}

Let $u (t)$ be the unit-step function. The input $x (t)$ that results in a steady-state output $y (t) = sin π t$ is ______.

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

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Answer and solution

ANSWER

The answer is (C): $x (t) = sin (π t - \frac{π}{2}) u (t)$ . At $ω = π$ we get $H (j π) = j$ , so the system passes the signal at unit gain and adds a $+ π /2$ phase lead. An input lagging by $π /2$ therefore comes out as $sin π t$ .

Official answer: CSource: Official answer key, GATE 2023 IN (IIT Kanpur)

The only pole is at $s = - π$ , in the left half-plane, so the system is stable and a sinusoidal steady state exists. The transient from the step switch-on decays as $e^{- π t}$ . Sinusoidal steady-state rule: an input $sin (ω t + θ) u (t)$ gives the steady-state output $∣ H (j ω) ∣ sin (ω t + θ + ∠ H (j ω))$ .
The output frequency is $ω = π rad/s$ . An LTI system does not change frequency, so the input must also be at $ω = π$ . Every option is, so the answer comes down to the phase (and gain).
Put $s = j π$ : $H (j π) = \frac{j π - π}{j π + π} = \frac{- 1 + j}{1 + j} = \frac{( - 1 + j ) ( 1 - j )}{( 1 + j ) ( 1 - j )} = \frac{- 1 + 2 j + 1}{2} = \frac{2 j}{2} = j$
So $∣ H (j π) ∣ = 1$ , as expected for an all-pass network (the zero at $+ π$ mirrors the pole at $- π$ ), and $∠ H (j π) = + \frac{π}{2}$ . Checking with angles: $∠ (- π + j π) - ∠ (π + j π) = 13 5^{\circ} - 4 5^{\circ} = 9 0^{\circ}$ .
At this frequency the system adds $+ π /2$ of phase and leaves the amplitude unchanged. For the output to be $sin (π t)$ , the input must therefore have phase $- π /2$ : $x (t) = sin (π t - \frac{π}{2}) u (t) \Rightarrow y_{ss} (t) = 1 \cdot sin (π t - \frac{π}{2} + \frac{π}{2}) = sin π t$
Ruling out the others: (A) gives $sin (π t + π /2)$ . (B) gives $sin (π t + π) = - sin π t$ . (D) is $cos (π t + π /4) = sin (π t + 3 π /4)$ , which gives $sin (π t + 5 π /4)$ . None of these is $sin π t$ , so the answer is (C).

KEY IDEA

In sinusoidal steady state a stable LTI system scales the amplitude by $∣ H (j ω) ∣$ and shifts the phase by $∠ H (j ω)$ , and it never changes the frequency. To find the input from the output, you undo that shift. Here $(s - π) / (s + π)$ is all-pass: the gain is 1 at every frequency, and at $ω = π$ the phase is exactly $+ 9 0^{\circ}$ . So the input has to lag the output by $π /2$ .

WATCH OUT FOR

The usual slip is getting the sign of the phase wrong. The numerator $j π - π$ sits in the second quadrant, at $13 5^{\circ}$ and not $4 5^{\circ}$ . People who treat $(s - π) / (s + π)$ as a lag, or read the numerator's angle as $- 4 5^{\circ}$ , conclude that the system shifts the phase by $- π /2$ . That leads them to pick (B), which actually produces $- sin π t$ . A second mistake is to pick (A) on the grounds that $∣ H ∣ = 1$ , so the input must equal the output. That forgets that an all-pass network still shifts the phase.

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GATE 2023 IN Q.21: solution and answer

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