Question

Consider the standard second-order system of the form: \[ \frac{\omega_n^2}{s^2 + 2\xi\omega_n s + \omega_n^2}, \] with the poles \(p\) and \(p^*\) having negative real parts. The pole locations are also shown in the figure. Now consider two such second-order systems as defined below: System 1: \( \omega_n = 3 \, \text{rad/sec} \) and \( \theta = 60^\circ \). System 2: \( \omega_n = 1 \, \text{rad/sec} \) and \( \theta = 70^\circ \). Which one of the following statements is correct?
\includegraphics[width=0.5\linewidth]{22.png}

• A. Settling time of System 1 is more than that of System 2.
• B. Settling time of System 2 is more than that of System 1.
• C. Settling times of both systems are the same.
• D. Settling time cannot be computed from the given information.

Hint:

Settling time increases with lower damping factors (\( \xi \)) and lower natural frequencies (\( \omega_n \)).

Answer 1

The Correct Option is B

Step 1: Formula for settling time. The settling time (\( t_s \)) for 2\% tolerance is given by: \[ t_s = \frac{4}{\xi \omega_n}, \] where \( \xi = \cos \theta \) is the damping factor. Step 2: Calculating settling time for both systems. For System 1: \[ \omega_n = 3, \, \theta = 60^\circ \quad \Rightarrow \quad \xi = \cos(60^\circ) = 0.5. \] \[ t_s = \frac{4}{0.5 \times 3} = 2.67 \, \text{seconds}. \] For System 2: \[ \omega_n = 1, \, \theta = 70^\circ \quad \Rightarrow \quad \xi = \cos(70^\circ) \approx 0.34. \] \[ t_s = \frac{4}{0.34 \times 1} \approx 11.76 \, \text{seconds}. \] Step 3: Comparing settling times. Since \( t_s \) for System 2 is greater, the correct statement is: \[ \text{Settling time of System 2 is more than that of System 1.} \] Hence, the correct option is (B).