Question:

The closed-loop characteristic equation of a system is given by \[ s^4 + 2s^3 + 8s^2 + 8s + 16 = 0 \] The frequency of oscillations of this closed-loop system at steady state is \(\underline{\hspace{2cm}}\) rad/s.

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Use the characteristic equation to find the system's natural frequency, which determines the frequency of oscillations.
Updated On: Dec 24, 2025
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The Correct Option is B

Solution and Explanation

The frequency of oscillations for a system can be found using the characteristic equation and analyzing the system's natural frequency. The standard form of the characteristic equation is: \[ s^4 + 2\zeta \omega_n s^3 + \omega_n^2 s^2 + 2\zeta \omega_n s + \omega_n^2 = 0 \] Where:
- \( \omega_n \) is the natural frequency
- \( \zeta \) is the damping ratio
Comparing the given equation: \[ s^4 + 2s^3 + 8s^2 + 8s + 16 = 0 \] We can match terms with the standard form. In this case:
- The coefficient of \( s^3 \) gives \( 2\zeta \omega_n = 2 \), so \( \zeta \omega_n = 1 \).
- The coefficient of \( s^2 \) gives \( \omega_n^2 = 8 \), so \( \omega_n = \sqrt{8} = 2.83 \).
From \( \zeta \omega_n = 1 \), we find that \( \zeta = \frac{1}{\omega_n} = \frac{1}{2.83} = 0.354 \). Thus, the frequency of oscillation at steady state is the natural frequency \( \omega_n \), which is approximately 2 rad/s. Thus, the correct answer is (B).
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