A second-order control system has ωₙ = 4 rad/s and ζ = 0.3. A designer wants to approximately halve the settling time without changing the percent overshoot. What adjustment achieves this?
ADouble ζ to 0.6 while keeping ωₙ = 4 rad/s
BDouble ωₙ to 8 rad/s while keeping ζ = 0.3
CHalve ζ to 0.15 while keeping ωₙ = 4 rad/s
DHalve both ωₙ to 2 rad/s and ζ to 0.15
Settling time is approximately 4/(ζωₙ) — the decay time constant is τ = 1/(ζωₙ). To halve settling time, you need to double the decay rate ζωₙ. Keeping ζ constant and doubling ωₙ doubles the product ζωₙ, halving settling time. Since percent overshoot depends only on ζ (not ωₙ), overshoot is unchanged. Option A would also speed settling but changes the overshoot by altering ζ. This highlights that ωₙ and ζ can be tuned somewhat independently.
Question 2 Multiple Choice
The closed-loop poles of a second-order system are located at s = −2 ± j2. What are the natural frequency ωₙ and damping ratio ζ?
Aωₙ = 2√2 rad/s, ζ = cos(45°) ≈ 0.707
Bωₙ = 2 rad/s, ζ = 0.5
Cωₙ = 4 rad/s, ζ = 0.707
Dωₙ = 2 rad/s, ζ = 1
The poles lie at distance ωₙ from the origin: ωₙ = √(2² + 2²) = √8 = 2√2 ≈ 2.83 rad/s. The angle from the negative real axis is θ = arctan(2/2) = 45°, and ζ = cos(45°) ≈ 0.707. This geometric reading — ωₙ is the pole's distance from the origin, ζ is the cosine of its angle from the negative real axis — is the key s-plane insight that lets you infer time-domain performance directly from the pole plot.
Question 3 True / False
For a standard second-order system, increasing the natural frequency ωₙ while holding the damping ratio ζ constant will increase the percent overshoot.
TTrue
FFalse
Answer: False
Percent overshoot is determined solely by ζ: %OS = exp(−πζ/√(1−ζ²)) × 100. The natural frequency ωₙ controls the speed of the response (rise time, settling time) but does not affect the overshoot. Changing ωₙ moves the poles radially in the s-plane — closer to or farther from the origin — without changing the angle, so ζ = cos(θ) is unchanged and overshoot is unchanged.
Question 4 True / False
In the complex s-plane, the poles of a standard second-order transfer function lie on a circle of radius ωₙ centered at the origin, and the damping ratio equals the cosine of the angle the pole makes with the negative real axis.
TTrue
FFalse
Answer: True
The poles are at s = −ζωₙ ± jωₙ√(1−ζ²). Their distance from the origin is √((ζωₙ)² + (ωₙ√(1−ζ²))²) = ωₙ√(ζ²+1−ζ²) = ωₙ. The angle from the negative real axis satisfies cos(θ) = ζωₙ/ωₙ = ζ. This geometric picture is why root-locus design is powerful: you can read ωₙ and ζ — and therefore all time-domain specs — directly from where the poles sit in the s-plane.
Question 5 Short Answer
Explain the geometric interpretation of ωₙ and ζ in the s-plane, and describe how a control engineer can read approximate time-domain performance from a pole location without solving the differential equation.
Think about your answer, then reveal below.
Model answer: Poles of a standard second-order system lie on a circle of radius ωₙ centered at the origin. The angle θ from the negative real axis satisfies ζ = cos(θ). From these two geometric quantities, all key time-domain metrics follow: settling time ≈ 4/(ζωₙ) = 4/(real part magnitude); damped oscillation frequency ω_d = ωₙ√(1−ζ²) = imaginary part; percent overshoot = exp(−πζ/√(1−ζ²)) × 100, a function of ζ alone. A pole further from the origin (larger ωₙ) responds faster; a pole closer to the negative real axis (smaller angle, larger ζ) overshoots less.
This geometric view is the foundation of root-locus design: the designer specifies desired time-domain performance, converts it to a target region in the s-plane (e.g., a sector defined by minimum ζ and a vertical line defined by minimum decay rate), and then uses gain or compensator design to place poles in that region. The power of the ωₙ–ζ parameterization is that it makes the mapping between frequency-domain pole locations and time-domain behavior transparent and intuitive.