Questions: Transient Response Damping and Oscillation

Question 1

Two second-order systems both have damping ratio ζ = 0.3, but different natural frequencies: System A has ωₙ = 5 rad/s and System B has ωₙ = 20 rad/s. Which statement is correct?

• A. System B has larger percent overshoot because it oscillates faster and reaches higher peaks
• B. Both systems have the same percent overshoot, but System B settles to steady state faster
• C. System A has larger percent overshoot because its slower oscillations persist longer before decaying
• D. Percent overshoot cannot be compared without knowing the magnitude of the input step

Answer: (B)

Percent overshoot depends only on ζ — the formula M_p = e^(−ζπ/√(1−ζ²)) × 100% contains no ωₙ term. Both systems have the same ζ = 0.3, so both have approximately 37% overshoot. However, System B's higher natural frequency means its time-to-peak and settling time are shorter by a factor of four. A common error is assuming that faster oscillation implies higher peaks; in fact, the faster decay exactly compensates, leaving overshoot unchanged.

Question 2

A control engineer wants to design a system with no more than 5% percent overshoot. Which is the correct approach?

• A. Set ωₙ high enough that oscillations complete before significant overshoot accumulates
• B. Set ζ ≥ 0.69, because overshoot depends only on the damping ratio — increasing ωₙ alone cannot reduce it
• C. Set ζ = 0.3, which gives approximately 5% overshoot for standard second-order systems
• D. Require ζωₙ > 10 so the decay envelope collapses before the first oscillation peak

Answer: (B)

Because M_p depends only on ζ, the engineer must specify a minimum ζ. For ≤5% overshoot, ζ ≥ 0.69 is required. Increasing ωₙ changes how fast the system responds but not how much it overshoots — overshoot is a property of the damping alone. Note that ζ = 0.3 gives approximately 37% overshoot, not 5%.

Question 3

For an underdamped second-order system, the actual oscillation frequency (damped natural frequency ωd) is always less than the undamped natural frequency ωₙ.

• T. True
• F. False

Answer: True

ωd = ωₙ√(1 − ζ²). Since 0 < ζ < 1 for underdamped systems, √(1 − ζ²) < 1, so ωd < ωₙ. Damping slows the oscillation. At the extreme ζ → 1, ωd → 0 (oscillation disappears at critical damping). At ζ = 0, ωd = ωₙ (undamped system oscillates at its natural frequency). The measured oscillation period is always longer than 2π/ωₙ.

Question 4

Increasing the natural frequency ωₙ of an underdamped control system will reduce percent overshoot because the system responds faster and reaches its setpoint before significant overshoot can occur.

• T. True
• F. False

Answer: False

Percent overshoot M_p = e^(−ζπ/√(1−ζ²)) × 100% does not contain ωₙ. Increasing ωₙ makes the system reach its peak faster, but it also makes the decay envelope shrink faster by exactly the same factor — the two effects cancel, leaving overshoot unchanged. Only ζ controls overshoot. To reduce overshoot, the engineer must increase ζ (add more damping), not increase ωₙ.

Question 5

Explain why the percent overshoot formula M_p = e^(−ζπ/√(1−ζ²)) depends only on ζ and not on ωₙ.

Model answer: Overshoot is determined by how much the decay envelope e^(−ζωₙt) has shrunk by the time the system reaches its first peak. The time to first peak is t_peak = π/ωd = π/(ωₙ√(1−ζ²)). Substituting into the decay envelope: the exponent is −ζωₙ × π/(ωₙ√(1−ζ²)) = −ζπ/√(1−ζ²). The ωₙ cancels exactly. Physically, a higher ωₙ causes the system to reach its peak faster, but it also causes the envelope to decay faster by the same proportion — so the envelope value at the peak is unchanged. Only ζ sets the ratio of damping strength to oscillation speed.

This cancellation is why ζ is the single most important design parameter for overshoot. Engineers targeting a specific overshoot specification can translate it directly into a ζ constraint, independent of speed requirements (which are governed by ωₙ independently).