Question

Two linear time-invariant systems with transfer functions \[ G_1(s) = \frac{10}{s^2 + s + 1}, \qquad G_2(s) = \frac{10}{s^2 + s\sqrt{10} + 10} \] 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.

Hint:

Overshoot depends only on damping ratio $\zeta$, while damped frequency and settling time depend on $\omega_n$ as well.

Answer 1

The Correct Option is A

The standard second-order form is: \[ G(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}. \] System 1: \[ s^2 + s + 1 \Rightarrow 2\zeta\omega_n = 1,\ \omega_n^2 = 1. \] Thus: \[ \omega_n = 1,\quad \zeta = \frac{1}{2}. \] System 2: \[ s^2 + s\sqrt{10} + 10 \Rightarrow 2\zeta\omega_n = \sqrt{10},\ \omega_n^2 = 10. \] Thus: \[ \omega_n = \sqrt{10}, \quad \zeta = \frac{\sqrt{10}}{2\sqrt{10}} = \frac{1}{2}. \] Both systems have the same damping ratio $\zeta = 0.5$. Percentage overshoot depends ONLY on \(\zeta\), so both systems have identical overshoot. → (A) is true. Steady-state value with unit step: \[ \lim_{s\to 0} sG(s)\frac{1}{s} = G(0) = \frac{10}{1} = 10 \quad \text{and} \quad G_2(0)=1. \] Different. → (B) false. Damped frequency: \[ \omega_d = \omega_n\sqrt{1-\zeta^2}. \] Since \(\omega_n\) differs, they are not equal → (C) false. Settling time: \[ T_s \approx \frac{4}{\zeta\omega_n}. \] Different \(\omega_n\) → different \(T_s\). → (D) false. Final Answer: (A)