General first-order unit-step response: for a stable, positive-gain first-order transfer function, describe how the output approaches its final steady state (qualitative trajectory shape).

Difficulty: Easy

Correct Answer: A monotonic increase

Explanation:


Introduction / Context:
First-order systems appear everywhere: tanks, thermal masses, RC circuits. Their hallmark is an exponential approach to a new steady state following a step input, with trajectory governed by the time constant τ and steady-state gain K.

Given Data / Assumptions:

  • Single real pole at s = −1/τ, τ > 0.
  • Positive steady-state gain K > 0.
  • Stable system (pole in left half plane).


Concept / Approach:
The standard step response is y(t) = K * (1 − e^(−t/τ)) for t ≥ 0 (assuming zero initial bias). This rises smoothly from 0 toward K without overshoot or oscillation. Its slope is maximum at t = 0 and decays exponentially; the output never decreases at any time when K > 0.

Step-by-Step Solution:

Recognize first-order canonical form: τ dy/dt + y = K u.Apply step input u = 1: solution is y(t) = K (1 − e^(−t/τ)).Note y′(t) = (K/τ) e^(−t/τ) ≥ 0 → strictly increasing to the asymptote.


Verification / Alternative check:
At t = τ, the response reaches about 63.2% of final; at t ≈ 5τ, it is essentially at steady state, consistent with a monotonic rise.


Why Other Options Are Wrong:

Monotonic decrease: would require negative gain or a negative step.Increase then decrease, or oscillatory patterns: those imply higher-order dynamics or zeros causing non-minimum-phase behavior, not present here.


Common Pitfalls:
Confusing first-order behavior with second-order underdamped responses that can overshoot.


Final Answer:
A monotonic increase

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