Questions: Second-Order System Time Response

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

A second-order system has ωₙ = 10 rad/s and ζ = 0.6. A step input is applied. At what frequency does the step response oscillate?

A10 rad/s — that is the natural frequency of the system
B6 rad/s — the product ζ × ωₙ gives the oscillation frequency
C8 rad/s — the damped natural frequency ωd = ωₙ√(1 − ζ²) = 10√(1 − 0.36) ≈ 8 rad/s
DThe response does not oscillate because ζ = 0.6 is nearly critically damped
Question 2 Multiple Choice

A control system specification requires settling within 2 seconds and less than 5% overshoot. A designer is choosing between ζ = 1.0 (critically damped) and ζ = 0.707 (underdamped). Which choice is likely better?

Aζ = 1.0 is always the best choice — it never overshoots and thus guarantees the <5% overshoot specification without any risk
Bζ = 0.707 is likely better — it produces only ~4.3% overshoot (within spec) and typically settles faster than the critically damped case at the same ωₙ, meaning it can meet both specs simultaneously
Cζ = 1.0 settles faster than ζ = 0.707 because it doesn't waste time in oscillation, so it better satisfies the settling time requirement
Dζ = 0.707 cannot be used because any underdamped system violates an overshoot specification by definition
Question 3 True / False

The natural frequency ωₙ is the frequency at which an underdamped second-order system actually oscillates in its step response.

TTrue
FFalse
Question 4 True / False

For a second-order system, increasing the damping ratio ζ while holding the natural frequency ωₙ constant will reduce overshoot but will generally increase the settling time.

TTrue
FFalse
Question 5 Short Answer

Explain why ζ ≈ 0.707 is often preferred over ζ = 1.0 (critically damped) in practical control system design, even though critical damping guarantees zero overshoot.

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