Second-order system – phase lag in frequency response For a standard underdamped second-order system excited by a sinusoidal input, how does the phase lag of the output behave as frequency increases?

Introduction / Context:
Frequency response characterises how a system scales and shifts sinusoids at different frequencies. For a standard underdamped second-order system, the phase lag evolves from near 0° at low frequency to a larger negative value at high frequency. Knowing the asymptotic limit is essential for Bode plot interpretation and controller design.

Given Data / Assumptions:

• Canonical transfer: G(s) = ω_n^2 / (s^2 + 2 ζ ω_n s + ω_n^2) with 0 < ζ < 1.
• Input is sinusoidal; steady-state response is considered.
• System is stable.

Concept / Approach:
The phase of G(jω) is φ(ω) = −tan^{-1}(2 ζ ω/ω_n / (1 − (ω/ω_n)^2)). As ω → 0, numerator term is small and denominator near 1, so φ → 0°. As ω increases past resonance, the argument of tan^{-1} grows in magnitude, and as ω → ∞, the 1 − (ω/ω_n)^2 term is large negative, driving φ toward −180°. Thus the phase lag approaches 180° asymptotically from 0° as frequency sweeps from low to high.

Step-by-Step Solution:

Verification / Alternative check:
Bode phase plots for second-order systems show an S-shaped transition from 0° to −180°, with the midpoint around ω/ω_n ≈ 1 and slope depending on ζ.

Why Other Options Are Wrong:

• 30° or 120°: Specific points, not asymptotic behaviour.
• 90° at most: Incorrect; total lag can approach 180° for second-order systems.

Common Pitfalls:
Confusing first-order (−90° limit) and second-order (−180° limit) behaviours; damping ratio shifts the transition frequency band but not the asymptotic limit.

Final Answer:
Approaches 180° asymptotically