For a second-order control system whose transient response is underdamped, which statements about reaching steady state and overshoot relative to the input are correct?

Difficulty: Easy

Correct Answer: Both a and b

Explanation:


Introduction / Context:
Underdamped transient behavior is common in second-order systems such as mass–spring–damper models or closed-loop control loops with moderate damping. This question checks recognition of key underdamped traits: relatively fast response with overshoot.


Given Data / Assumptions:

  • System is underdamped (damping ratio zeta between 0 and 1).
  • Input is a step command used to observe transient behavior.
  • Comparisons are made to an overdamped system (zeta > 1).


Concept / Approach:

Underdamped systems respond quickly but oscillate about the target, producing overshoot before settling. Overdamped systems are slower but monotonic. Zero overshoot characterizes critically damped or overdamped cases, not underdamped ones.


Step-by-Step Solution:

1) Underdamped implies oscillatory decay toward steady state, hence nonzero overshoot for a typical step input.2) Relative to overdamped, the rise time and settling time can be shorter for certain zeta in (0,1), so steady state is often reached faster.3) Because of the oscillatory nature, the output exceeds the final value at least once (overshoot > 0).4) Therefore: (a) is true, (b) is true, and (c) is false; the combined correct choice is 'Both a and b'.


Verification / Alternative check:

Standard second-order step response shows maximum overshoot Mp = exp(-pi*zeta / sqrt(1 - zeta^2)) for 0 < zeta < 1, which is strictly greater than 0, confirming overshoot occurs.


Why Other Options Are Wrong:

Option c is wrong because percent overshoot is not zero when underdamped. 'None of the above' is wrong because (a) and (b) hold for underdamped behavior.


Common Pitfalls:

Confusing critically damped (fast and no overshoot) with underdamped (fast but with overshoot). Also, forgetting that settling time depends on both zeta and natural frequency.


Final Answer:

Both a and b

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