A second-order mechanical system has poles at s = −0.05 ± 20j rad/s. A sinusoidal force is applied at several frequencies. What do the pole locations predict about the system's frequency response?
AThe poles are far from the imaginary axis, indicating heavy damping and no resonance peak
BThe poles are close to the imaginary axis (small real part relative to imaginary part), indicating light damping and a sharp resonance peak near ω = 20 rad/s
CThe imaginary part of the poles indicates the system is marginally stable and will oscillate indefinitely
DPole locations cannot predict the frequency response — only the time-domain step response
The real part of the poles (−0.05) represents the decay rate; the imaginary part (20) represents the damped oscillation frequency. A small real part relative to the imaginary part means the poles sit close to the imaginary axis — indicating very light damping (ζ ≈ 0.05/20 = 0.0025). Poles near the imaginary axis produce a sharp, tall resonance peak in the frequency response near ω = 20 rad/s. In the Tacoma Narrows analogy: light structural damping meant the poles were nearly on the imaginary axis, and wind driving near the natural frequency produced catastrophic amplitude buildup.
Question 2 Multiple Choice
Two second-order systems share the same natural frequency ω_n. System A has ζ = 0.05 and System B has ζ = 0.7. Both are driven by a sinusoidal input at ω_n. What does each system do?
ABoth respond identically because they share the same natural frequency
BSystem A exhibits a very large resonance peak; System B has little or no amplitude peak near ω_n
CSystem B has a larger peak because higher ζ means more energy is stored per cycle
DNeither system peaks exactly at ω_n; they both peak at lower frequencies
Damping ratio ζ controls peak magnitude: the resonance peak is approximately 1/(2ζ) for small ζ. System A (ζ = 0.05) has a peak of ~10 times the DC gain — a large, sharp resonance. System B (ζ = 0.7 > 1/√2 ≈ 0.707) is just barely overdamped in frequency response terms — near or below the Butterworth threshold — producing little or no peak. Resonance requires that energy input accumulates faster than the system dissipates it; high damping bleeds energy too quickly for significant amplitude buildup.
Question 3 True / False
Increasing the damping ratio ζ from 0.1 to 0.8 reduces both the overshoot in the step response and the height of the resonance peak in the frequency response.
TTrue
FFalse
Answer: True
This is the fundamental unity of time-domain and frequency-domain descriptions: they are two views of the same physics. In the time domain, low ζ causes underdamped ringing and overshoot after a step; higher ζ dampens this. In the frequency domain, low ζ produces a tall resonance peak; higher ζ flattens it. Both behaviors trace back to the same pole locations: poles close to the imaginary axis produce both time-domain oscillation and frequency-domain resonance. Moving poles leftward (increasing ζ) simultaneously reduces overshoot and peak magnitude.
Question 4 True / False
The resonant frequency of an underdamped second-order system is generally equal to its natural frequency ω_n.
TTrue
FFalse
Answer: False
The resonant frequency (where the frequency response magnitude peaks) is ω_r = ω_n√(1 − 2ζ²), which equals ω_n only when ζ = 0 (undamped). For any real underdamped system (0 < ζ < 1/√2), the resonant frequency is slightly below ω_n. The difference is small for light damping (ζ = 0.1 gives ω_r ≈ 0.99ω_n) but becomes significant as ζ increases. Above ζ = 1/√2, there is no resonance peak at all. The natural frequency ω_n is a system property; the resonant frequency ω_r depends on both ω_n and the damping ratio.
Question 5 Short Answer
Explain why a second-order system's pole locations in the complex s-plane simultaneously determine its transient step response and its steady-state resonance behavior.
Think about your answer, then reveal below.
Model answer: The poles of H(s) are the values of s where the transfer function's denominator is zero: s = −ζω_n ± jω_n√(1−ζ²). In the time domain, poles correspond to the natural modes of the system — each pole contributes a term e^(pole·t) to the impulse response. The real part (−ζω_n) sets the exponential decay rate; the imaginary part (ω_n√(1−ζ²)) sets the oscillation frequency. In the frequency domain, evaluating H(jω) as ω varies sweeps along the imaginary axis; the magnitude peaks when ω is closest to the imaginary part of the poles. Poles near the imaginary axis (small real part, low damping) produce both slow-decaying oscillatory transients and sharp resonance peaks — the same proximity to the imaginary axis controls both phenomena.
This unification is the central insight of Laplace analysis: the s-domain representation encodes both transient and steady-state behavior in a single mathematical object. Engineers use this to design for both: a filter specification in the frequency domain (flat passband, steep rolloff) translates directly into pole placement requirements in the s-plane, which then predicts the transient ringing behavior of the same filter when exposed to sudden inputs.