Questions: Time-Domain Performance Metrics and Specifications

Question 1

Two second-order systems A and B both have damping ratio ζ = 0.5, but system A has ωₙ = 10 rad/s while system B has ωₙ = 20 rad/s. How do their step responses compare?

• A. System B has lower percentage overshoot because it responds faster
• B. System A has lower percentage overshoot because it responds more slowly and has more time to damp
• C. Both systems have the same percentage overshoot, but system B settles in approximately half the time
• D. The responses are identical since both systems have the same damping ratio and natural frequency determines only the amplitude

Answer: (C)

Percentage overshoot depends only on ζ — the formula %OS = exp(−πζ/√(1−ζ²)) × 100 contains no ωₙ term. With identical ζ = 0.5, both systems have the same overshoot (~16%). But settling time ≈ 4/(ζωₙ): system A settles in 4/(0.5×10) = 0.8 s while system B settles in 4/(0.5×20) = 0.4 s — half the time. Overshoot and response speed are controlled by different parameters and can be adjusted independently (within limits).

Question 2

A control engineer must design a system with both fast rise time and very low overshoot. The current design has unacceptably high overshoot. Increasing the damping ratio ζ will:

• A. Reduce overshoot and also reduce rise time, improving both specifications simultaneously
• B. Reduce overshoot but tend to increase rise time, creating a fundamental design tradeoff
• C. Have no effect on rise time since rise time depends only on ωₙ
• D. Reduce overshoot only at the cost of permanently increased steady-state error

Answer: (B)

Increasing ζ reduces overshoot directly (the %OS formula decreases with increasing ζ). However, higher ζ moves poles toward the real axis — reducing the imaginary component — which slows the oscillatory response and increases rise time. To satisfy both a fast rise time AND low overshoot, the engineer must also increase ωₙ (moving poles further left), not just increase ζ alone. This is the central design tension in second-order system specification.

Question 3

Doubling the natural frequency ωₙ of a second-order system while keeping ζ constant will approximately double the percentage overshoot.

• T. True
• F. False

Answer: False

Percentage overshoot depends only on ζ, not on ωₙ. The formula %OS = exp(−πζ/√(1−ζ²)) × 100 has no ωₙ term. Doubling ωₙ makes the system respond twice as fast (halves rise time and settling time) but leaves overshoot completely unchanged. A common misconception is that faster systems overshoot more — speed and overshoot are controlled by independent parameters.

Question 4

Moving a closed-loop pole further to the left in the s-plane (increasing the magnitude of its real part) reduces both rise time and settling time.

• T. True
• F. False

Answer: True

The real part of the pole is −ζωₙ. Moving poles leftward increases |Re(s)| = ζωₙ, which decreases the time constant τ = 1/(ζωₙ) of the decaying envelope. Settling time ≈ 4τ = 4/(ζωₙ) decreases, and rise time ≈ 1.8/ωₙ also decreases with higher ωₙ. Poles on the far left of the s-plane correspond to rapidly decaying, fast-response systems.

Question 5

A control system has too much overshoot (30%) and too slow a rise time. An engineer proposes increasing ζ to fix the overshoot. What trade-off will they encounter, and what additional design change could address both specifications simultaneously?

Model answer: Increasing ζ reduces overshoot (since %OS decreases as ζ rises) but also slows rise time because poles move toward the real axis, reducing the oscillatory speed. To satisfy both a lower overshoot and a faster rise time, the engineer must also increase ωₙ — moving the poles both further left (faster) and at a steeper angle (higher ζ, less overshoot). In the s-plane, the target is a pole location with larger |Re(s)| and a steeper angle from the negative real axis, corresponding to simultaneously higher ωₙ and higher ζ.

This is why controller design begins with converting performance specs into desired pole locations in the s-plane. A 5% overshoot requirement maps to ζ ≈ 0.69; a 50 ms settling time maps to ζωₙ ≥ 4/0.05 = 80. These two requirements together specify a region in the s-plane (left of −80, at an angle corresponding to ζ = 0.69) where the poles must lie. Root locus and lead-lag design are then tools for placing poles in this target region.