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:
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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