The open loop transfer function of a unity feedback control system is given by \(G(s)=\frac{25}{s(s+5)}\). The natural frequency of oscillator is fixed. Arrange the damped frequency of oscillation for the following damping ratio in ascending order A. \(\xi = 0.5\) B. \(\xi = 0.1\) C. \(\xi = 0.3\) D. \(\xi = 0.25\) E. \(\xi = 0.4\) Choose the correct answer from the options given below :
\(A \lt E \lt C \lt D \lt B\)
Step 1 — find the closed-loop form. With unity feedback and \(G(s)=\dfrac{25}{s(s+5)}\):
\(T(s)=\dfrac{G}{1+G}=\dfrac{25}{s^{2}+5s+25}\)
Comparing with the standard second-order denominator \(s^{2}+2\xi\omega_n s+\omega_n^{2}\) gives
\(\omega_n=\sqrt{25}=5\ \text{rad/s}\)
and the question explicitly holds this natural frequency fixed for all five cases.
Step 2 — the quantity being ranked. The damped frequency of oscillation is the imaginary part of the closed-loop poles \(s=-\xi\omega_n \pm j\omega_n\sqrt{1-\xi^{2}}\):
\(\omega_d=\omega_n\sqrt{1-\xi^{2}}\)
With ωn constant, ωd decreases monotonically as ξ increases — heavier damping means slower ringing.
Step 3 — evaluate for each damping ratio (ωn = 5 rad/s):
| Item | ξ | \(\omega_d=5\sqrt{1-\xi^{2}}\) |
|---|---|---|
| A | 0.50 | 4.33 rad/s — smallest |
| E | 0.40 | 4.58 rad/s |
| C | 0.30 | 4.77 rad/s |
| D | 0.25 | 4.84 rad/s |
| B | 0.10 | 4.97 rad/s — largest |
Step 4 — read off the ascending order.
\(A \lt E \lt C \lt D \lt B\)
i.e. the ascending order of ωd is exactly the descending order of ξ.
Points worth noting. All five values satisfy ξ < 1, so every case is under-damped and genuinely oscillatory. Notice how weakly ωd depends on ξ for small ξ (at ξ = 0.1 it is still 99.5 % of ωn) — the square root flattens the dependence. What ξ does change strongly is the overshoot, \(M_p=e^{-\pi\xi/\sqrt{1-\xi^{2}}}\), and the decay rate ξωn.
Hence, the ascending order of damped frequency is \(A \lt E \lt C \lt D \lt B\).
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