PYQs / GATE EC / 2022 / Set 1 / Q50 GATE EC 2022 Set 1 — Question 50 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MSQ +2 / -0 Medium First & Second-Order Response Time-Domain Analysis Control Systems Transient Response Specs
Control Systems → Time-Domain Analysis → First & Second-Order Response
Last updated 5 September 2026
Question Two linear time-invariant systems with transfer functions
G 1 ( s ) = 10 s 2 + s + 1 and G 2 ( s ) = 10 s 2 + s 10 + 10 G_1(s) = \frac{10}{s^2 + s + 1} \text{ and } G_2(s) = \frac{10}{s^2 + s\sqrt{10} + 10} G 1 ( s ) = s 2 + s + 1 10 and G 2 ( s ) = s 2 + s 10 + 10 10 have unit step responses
y 1 ( t ) y_1(t) y 1 ( t ) and
y 2 ( t ) y_2(t) y 2 ( t ) , respectively. Which of the following statements is/are true?
Correct answer (A) y₁(t) and y₂(t) have the same percentage peak overshoot.
Solution For a standard second-order system
G ( s ) = K ω n 2 s 2 + 2 ζ ω n s + ω n 2 G(s) = \frac{K\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2} G ( s ) = s 2 + 2 ζ ω n s + ω n 2 K ω n 2 :
1. The percentage peak overshoot is given by M p = e − π ζ 1 − ζ 2 × 100 % M_p = e^{-\frac{\pi\zeta}{\sqrt{1-\zeta^2}}} \times 100\% M p = e − 1 − ζ 2 π ζ × 100% ζ \zeta ζ 2. The steady-state value for a unit step input is y ( ∞ ) = G ( 0 ) = K y(\infty) = G(0) = K y ( ∞ ) = G ( 0 ) = K 3. The damped frequency of oscillation is ω d = ω n 1 − ζ 2 \omega_d = \omega_n \sqrt{1-\zeta^2} ω d = ω n 1 − ζ 2 4. The 2% settling time is t s = 4 ζ ω n t_s = \frac{4}{\zeta\omega_n} t s = ζ ω n 4
For
G 1 ( s ) = 10 s 2 + s + 1 G_1(s) = \frac{10}{s^2 + s + 1} G 1 ( s ) = s 2 + s + 1 10 :
ω n 1 2 = 1 ⇒ ω n 1 = 1 \omega_{n1}^2 = 1 \Rightarrow \omega_{n1} = 1 ω n 1 2 = 1 ⇒ ω n 1 = 1 2 ζ 1 ω n 1 = 1 ⇒ 2 ζ 1 ( 1 ) = 1 ⇒ ζ 1 = 0.5 2\zeta_1\omega_{n1} = 1 \Rightarrow 2\zeta_1(1) = 1 \Rightarrow \zeta_1 = 0.5 2 ζ 1 ω n 1 = 1 ⇒ 2 ζ 1 ( 1 ) = 1 ⇒ ζ 1 = 0.5 y 1 ( ∞ ) = G 1 ( 0 ) = 10 y_1(\infty) = G_1(0) = 10 y 1 ( ∞ ) = G 1 ( 0 ) = 10 ω d 1 = 1 1 − 0.5 2 = 0.75 \omega_{d1} = 1\sqrt{1-0.5^2} = \sqrt{0.75} ω d 1 = 1 1 − 0. 5 2 = 0.75 rad/s
t s 1 = 4 0.5 × 1 = 8 t_{s1} = \frac{4}{0.5 \times 1} = 8 t s 1 = 0.5 × 1 4 = 8 s
For
G 2 ( s ) = 10 s 2 + s 10 + 10 G_2(s) = \frac{10}{s^2 + s\sqrt{10} + 10} G 2 ( s ) = s 2 + s 10 + 10 10 :
ω n 2 2 = 10 ⇒ ω n 2 = 10 \omega_{n2}^2 = 10 \Rightarrow \omega_{n2} = \sqrt{10} ω n 2 2 = 10 ⇒ ω n 2 = 10 2 ζ 2 ω n 2 = 10 ⇒ 2 ζ 2 ( 10 ) = 10 ⇒ ζ 2 = 0.5 2\zeta_2\omega_{n2} = \sqrt{10} \Rightarrow 2\zeta_2(\sqrt{10}) = \sqrt{10} \Rightarrow \zeta_2 = 0.5 2 ζ 2 ω n 2 = 10 ⇒ 2 ζ 2 ( 10 ) = 10 ⇒ ζ 2 = 0.5 y 2 ( ∞ ) = G 2 ( 0 ) = 10 10 = 1 y_2(\infty) = G_2(0) = \frac{10}{10} = 1 y 2 ( ∞ ) = G 2 ( 0 ) = 10 10 = 1 ω d 2 = 10 1 − 0.5 2 = 7.5 \omega_{d2} = \sqrt{10}\sqrt{1-0.5^2} = \sqrt{7.5} ω d 2 = 10 1 − 0. 5 2 = 7.5 rad/s
t s 2 = 4 0.5 × 10 = 8 10 t_{s2} = \frac{4}{0.5 \times \sqrt{10}} = \frac{8}{\sqrt{10}} t s 2 = 0.5 × 10 4 = 10 8 s
Comparing the two systems:
Both have ζ = 0.5 \zeta = 0.5 ζ = 0.5 Steady-state values are 10 and 1. Statement (B) is false. Damped frequencies are 0.75 \sqrt{0.75} 0.75 7.5 \sqrt{7.5} 7.5 Settling times are 8 s and 8 / 10 8/\sqrt{10} 8/ 10 Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker. Review answer and solution without JavaScript Interactive answer checking needs JavaScript. The published solution is available below.
Correct answer (A) y₁(t) and y₂(t) have the same percentage peak overshoot.
Solution For a standard second-order system
G ( s ) = K ω n 2 s 2 + 2 ζ ω n s + ω n 2 G(s) = \frac{K\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2} G ( s ) = s 2 + 2 ζ ω n s + ω n 2 K ω n 2 :
1. The percentage peak overshoot is given by M p = e − π ζ 1 − ζ 2 × 100 % M_p = e^{-\frac{\pi\zeta}{\sqrt{1-\zeta^2}}} \times 100\% M p = e − 1 − ζ 2 π ζ × 100% ζ \zeta ζ 2. The steady-state value for a unit step input is y ( ∞ ) = G ( 0 ) = K y(\infty) = G(0) = K y ( ∞ ) = G ( 0 ) = K 3. The damped frequency of oscillation is ω d = ω n 1 − ζ 2 \omega_d = \omega_n \sqrt{1-\zeta^2} ω d = ω n 1 − ζ 2 4. The 2% settling time is t s = 4 ζ ω n t_s = \frac{4}{\zeta\omega_n} t s = ζ ω n 4
For
G 1 ( s ) = 10 s 2 + s + 1 G_1(s) = \frac{10}{s^2 + s + 1} G 1 ( s ) = s 2 + s + 1 10 :
ω n 1 2 = 1 ⇒ ω n 1 = 1 \omega_{n1}^2 = 1 \Rightarrow \omega_{n1} = 1 ω n 1 2 = 1 ⇒ ω n 1 = 1 2 ζ 1 ω n 1 = 1 ⇒ 2 ζ 1 ( 1 ) = 1 ⇒ ζ 1 = 0.5 2\zeta_1\omega_{n1} = 1 \Rightarrow 2\zeta_1(1) = 1 \Rightarrow \zeta_1 = 0.5 2 ζ 1 ω n 1 = 1 ⇒ 2 ζ 1 ( 1 ) = 1 ⇒ ζ 1 = 0.5 y 1 ( ∞ ) = G 1 ( 0 ) = 10 y_1(\infty) = G_1(0) = 10 y 1 ( ∞ ) = G 1 ( 0 ) = 10 ω d 1 = 1 1 − 0.5 2 = 0.75 \omega_{d1} = 1\sqrt{1-0.5^2} = \sqrt{0.75} ω d 1 = 1 1 − 0. 5 2 = 0.75 rad/s
t s 1 = 4 0.5 × 1 = 8 t_{s1} = \frac{4}{0.5 \times 1} = 8 t s 1 = 0.5 × 1 4 = 8 s
For
G 2 ( s ) = 10 s 2 + s 10 + 10 G_2(s) = \frac{10}{s^2 + s\sqrt{10} + 10} G 2 ( s ) = s 2 + s 10 + 10 10 :
ω n 2 2 = 10 ⇒ ω n 2 = 10 \omega_{n2}^2 = 10 \Rightarrow \omega_{n2} = \sqrt{10} ω n 2 2 = 10 ⇒ ω n 2 = 10 2 ζ 2 ω n 2 = 10 ⇒ 2 ζ 2 ( 10 ) = 10 ⇒ ζ 2 = 0.5 2\zeta_2\omega_{n2} = \sqrt{10} \Rightarrow 2\zeta_2(\sqrt{10}) = \sqrt{10} \Rightarrow \zeta_2 = 0.5 2 ζ 2 ω n 2 = 10 ⇒ 2 ζ 2 ( 10 ) = 10 ⇒ ζ 2 = 0.5 y 2 ( ∞ ) = G 2 ( 0 ) = 10 10 = 1 y_2(\infty) = G_2(0) = \frac{10}{10} = 1 y 2 ( ∞ ) = G 2 ( 0 ) = 10 10 = 1 ω d 2 = 10 1 − 0.5 2 = 7.5 \omega_{d2} = \sqrt{10}\sqrt{1-0.5^2} = \sqrt{7.5} ω d 2 = 10 1 − 0. 5 2 = 7.5 rad/s
t s 2 = 4 0.5 × 10 = 8 10 t_{s2} = \frac{4}{0.5 \times \sqrt{10}} = \frac{8}{\sqrt{10}} t s 2 = 0.5 × 10 4 = 10 8 s
Comparing the two systems:
Both have ζ = 0.5 \zeta = 0.5 ζ = 0.5 Steady-state values are 10 and 1. Statement (B) is false. Damped frequencies are 0.75 \sqrt{0.75} 0.75 7.5 \sqrt{7.5} 7.5 Settling times are 8 s and 8 / 10 8/\sqrt{10} 8/ 10 Understand the concept, then try another question Revisit Control Systems with concept notes, common mistakes and an original worked example before your next attempt.
More questions on Time-Domain Analysis ← Q49 Full paper Q51 → 