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consider a unity feedback control system for an open loop transfer function \(G(s)=\frac{5}{s(s+1)}\). Arrange the time constant in ascending order at different value of damping ratio of

A. \(\xi=3\)

B. \(\xi=7\)

C. \(\xi=1\)

D. \(\xi=10\)

E. \(\xi=5\)

Choose the correct answer from the options given below :

This question was previously asked in
UGC NET 2023 Electronic Science Question Paper (13-Dec-2023) (Shift 1)
The correct answer is

\(D \lt B \lt E \lt A \lt C\)

 What the time constant depends on. For a standard second-order system

\(T(s)=\dfrac{\omega_n^{2}}{s^{2}+2\xi\omega_n s+\omega_n^{2}}\)

the transient decays as \(e^{-\xi\omega_n t}\), so the decay rate is the real part of the poles, \(\sigma=\xi\omega_n\), and the time constant is

\(\tau=\dfrac{1}{\xi\omega_n}\)

What is fixed here. With \(G(s)=\dfrac{5}{s(s+1)}\) under unity feedback, the characteristic equation is \(s^{2}+s+5=0\), so the natural frequency \(\omega_n=\sqrt5\) is set by the loop and is the same for every case. The only variable is ξ, giving

\(\tau \propto \dfrac{1}{\xi}\)

So larger damping ⇒ shorter time constant. Rank the given damping ratios in descending order and the time constants come out in ascending order:

ItemξRelative τ ∝ 1/ξ
D100.10 — smallest
B70.14
E50.20
A30.33
C11.00 — largest

Ascending order of time constant:

\(D \lt B \lt E \lt A \lt C\)

Physical reading. All the quoted values satisfy ξ ≥ 1, so the system is critically damped (ξ = 1) or over-damped (ξ > 1) in every case — no oscillation, just an exponential approach to the final value. Heavier damping pushes the dominant pole further from the origin, so the exponential envelope \(e^{-t/\tau}\) collapses faster. Watch the direction of the question: it asks for the time constants in ascending order, which is the reverse of the ξ ordering.

Hence, the ascending order is \(D \lt B \lt E \lt A \lt C\).

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