Second-order step response characteristics: For the linear time-invariant system with transfer function G(s) = 1 / (s^2 + 2s + 3), which statement best describes the unit step response?

Difficulty: Easy

Correct Answer: It exhibits a damped oscillatory response (stable underdamped).

Explanation:


Introduction / Context:
Recognising the qualitative time response of a given transfer function is a core skill in control engineering. Here, the denominator s^2 + 2s + 3 defines a second-order system. From its coefficients we can infer damping ratio, natural frequency, stability, and whether the unit step response oscillates or not. The goal is to select the statement that best matches the true response behaviour.


Given Data / Assumptions:

  • Transfer function: G(s) = 1 / (s^2 + 2s + 3).
  • Input: unit step r(t) with Laplace transform 1/s.
  • System starts from rest (zero initial conditions) and is linear time-invariant.


Concept / Approach:
Compare the standard second-order form: G(s) = ω_n^2 / (s^2 + 2ζω_n s + ω_n^2). For s^2 + 2s + 3 we have ω_n^2 = 3 ⇒ ω_n = √3, and 2ζω_n = 2 ⇒ ζ = 2/(2√3) = 1/√3 ≈ 0.577. Since 0 < ζ < 1, the system is underdamped and stable, which means the step response will overshoot and oscillate with exponentially decaying envelope.


Step-by-Step Solution:

Identify ω_n: ω_n = √3 ≈ 1.732.Compute damping ratio: ζ = 1/√3 ≈ 0.577 (between 0 and 1).Conclude stability: all poles have negative real parts (−1 ± j√2).Infer response: stable, damped oscillations with finite overshoot.Check initial slope using the initial value theorem on ẏ(t): ẏ(0⁺) = lim_{s→∞} s[sY(s)] = 0 → slope at t = 0⁺ is zero, not non-zero.


Verification / Alternative check:
Poles at −1 ± j√2 confirm an exponentially decaying sinusoid in the homogeneous response. The particular solution reaches a finite steady value, so the total step response is underdamped and stable.


Why Other Options Are Wrong:

Non-zero initial slope — false; here the initial slope is zero for this strictly proper form.Overdamped — would require ζ ≥ 1, not satisfied.Unstable — poles have negative real parts; system is stable.Zero-overshoot/zero-oscillation — contradicts ζ ≈ 0.577.


Common Pitfalls:
Mixing up damping categories or assuming any second-order denominator implies overdamping. Always compute ζ and inspect pole locations.


Final Answer:
It exhibits a damped oscillatory response (stable underdamped).

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