Question:medium

A unity feedback second order system has \( G(s) = \frac{100{S^2+10S+100} \). The 2% settling time is in second.

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Always remember the difference between the two common settling time error criteria used in exams:
• For a 2% error band, use: \(t_s = \frac{4}{\zeta\omega_n}\)
• For a 5% error band, use: \(t_s = \frac{3}{\zeta\omega_n}\)
Updated On: Jul 4, 2026
  • 0.4
  • 0.8
  • 1
  • 2
Show Solution

The Correct Option is B

Solution and Explanation

Understanding the Concept: A standard prototype second-order closed-loop control system function is conventionally modeled as: \[ T(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2} \] Where \(\omega_n\) represents the undamped natural frequency and \(\zeta\) is the damping ratio parameter. The settling time (\(t_s\)) represents the total time required for the system's response transient output to sink and stabilize within a specified error band around its steady-state value. For a standard 2% error tolerance band criterion, the settling time formula is given by: \[ t_s = \frac{4}{\zeta\omega_n} \] Where the term \(\zeta\omega_n\) is also known as the attenuation factor (\(\alpha\)).

Step 1: Identify system equations.

The problem states that the system has an open-loop transfer function \(G(s)\) with unity feedback (\(H(s)=1\)). However, let us examine the provided expression layout: \(G(s) = \frac{100}{s^2+10s+100}\). Typically, when a second-order system formula is written directly in this layout with a quadratic denominator, it represents the overall closed-loop transfer function \(T(s)\). Let us perform a comparison against the standard form: \[ T(s) = \frac{100}{s^2 + 10s + 100} \]

Step 2: Extract \(\omega_n\) and \(\zeta\).

Comparing the denominator coefficients with \(s^2 + 2\zeta\omega_n s + \omega_n^2 = 0\): From the constant term: \[ \omega_n^2 = 100 \implies \omega_n = \sqrt{100} = 10 \text{ rad/s} \] From the coefficient of \(s\): \[ 2\zeta\omega_n = 10 \] We can directly observe that the complete term \(\zeta\omega_n\) (which represents the real part of the system poles) is: \[ \zeta\omega_n = \frac{10}{2} = 5 \]

Step 3: Compute the 2% settling time.

Using the standard 2% error performance criteria formula: \[ t_s = \frac{4}{\zeta\omega_n} \] Substitute the calculated factor value \(\zeta\omega_n = 5\) directly into the denominator: \[ t_s = \frac{4}{5} = 0.8 \text{ seconds} \] This perfectly matches option (B).
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