GATE 2022 EC Q.50: solution and answer

GATE 2022 · ECQ.50 · 2 marks · MSQ

Two linear time-invariant systems with transfer functions

G_{1} (s) = \frac{10}{s ^{2} + s + 1} and G_{2} (s) = \frac{10}{s ^{2} + s 10 + 10}

have unit step responses $y_{1} (t)$ and $y_{2} (t)$ , respectively. Which of the following statements is/are true?

(A)
$y_{1} (t)$ and $y_{2} (t)$ have the same percentage peak overshoot.
(B)
$y_{1} (t)$ and $y_{2} (t)$ have the same steady-state value.
(C)
$y_{1} (t)$ and $y_{2} (t)$ have the same damped frequency of oscillation.
(D)
$y_{1} (t)$ and $y_{2} (t)$ have the same $2%$ settling time.

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Answer and solution

ANSWER

Only (A) is true. Both systems have $ζ = 0.5$ , so they share a percentage peak overshoot of about $16.3%$ . Their steady-state values (10 and 1), damped frequencies (0.866 and 2.74 rad/s) and 2% settling times (8 s and about 2.53 s) are all different.

Official answer: ASource: Official answer key, GATE 2022 EC (IIT Kharagpur)

Write each denominator in the standard form $s^{2} + 2 ζ ω_{n} s + ω_{n}^{2}$ . For $G_{1} (s) = \frac{10}{s ^{2} + s + 1}$ : $ω_{n}^{2} = 1 \Rightarrow ω_{n} = 1 rad/s$ , and $2 ζ ω_{n} = 1 \Rightarrow ζ = 0.5$ .
For $G_{2} (s) = \frac{10}{s ^{2} + 10 s + 10}$ : $ω_{n}^{2} = 10 \Rightarrow ω_{n} = 10 rad/s$ , and $2 ζ 10 = 10 \Rightarrow ζ = 0.5$ . Both systems are underdamped with the same damping ratio.
Percentage peak overshoot is measured relative to the steady-state value, so it does not depend on the gain, and it depends only on $ζ$ : $M_{p} = e^{- π ζ / 1 - ζ^{2}} = e^{- π / 3} \approx 0.163$ That is about $16.3%$ for both. So (A) is true.
Steady-state value of a unit-step response (final value theorem) equals the DC gain $G (0)$ : $G_{1} (0) = \frac{10}{1} = 10$ , while $G_{2} (0) = \frac{10}{10} = 1$ . They differ, so (B) is false.
Damped frequency $ω_{d} = ω_{n} 1 - ζ^{2}$ : $ω_{d 1} = 1 \cdot \frac{3}{2} \approx 0.866 rad/s$ , while $ω_{d 2} = 10 \cdot \frac{3}{2} \approx 2.74 rad/s$ . They differ, so (C) is false.
$2%$ settling time $t_{s} \approx \frac{4}{ζ ω _{n}}$ : $t_{s 1} = \frac{4}{0.5 \times 1} = 8 s$ , while $t_{s 2} = \frac{4}{0.5 10} \approx 2.53 s$ . The conclusion does not depend on the approximation: with the same $ζ$ , the second response is the first one time-compressed by the factor $10$ , so every time measure (exact settling time included) is $10$ times smaller. They differ, so (D) is false.
Only (A) holds. The two systems share $ζ$ but have different $ω_{n}$ and different DC gains.

KEY IDEA

For a standard second-order system, percentage overshoot depends only on $ζ$ . The time scale ( $ω_{d}$ , $t_{s}$ , $t_{p}$ ) is set by $ω_{n}$ , and the steady-state value is set by the DC gain $G (0)$ . Two systems with the same $ζ$ but different $ω_{n}$ and gain therefore share only the percentage overshoot.

WATCH OUT FOR

Seeing the same numerator (10) in both transfer functions and assuming the steady-state values match, so marking (B). The steady-state value is $G (0)$ , numerator divided by the constant term of the denominator: $10/1 = 10$ against $10/10 = 1$ . A second slip is thinking a different gain changes the percentage overshoot. It does not, because overshoot is measured relative to the final value.

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GATE 2022 EC Q.50: solution and answer

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