A first-order system has a time constant τ = 4 s and DC gain K = 10. Immediately after a unit step input is applied at t = 0, an engineer claims the system has 'basically settled' at t = 4 s. What fraction of the final value has the output actually reached at that moment?
A100% — the system is fully settled at t = τ
B86% — it has completed two time constants worth of response
C63% — it has completed exactly one time constant
D50% — the half-life of an exponential
At t = τ, the step response is K(1 − e⁻¹) ≈ 0.632K, exactly 63% of final value. The system is not settled — engineering convention defines settling at t = 5τ (within 1% of final value). Confusing 'one time constant elapsed' with 'settled' is the classic error.
Question 2 Multiple Choice
A first-order system has time constant τ = 0.01 s. A designer doubles the time constant to τ = 0.02 s. What happens to the system's bandwidth?
ABandwidth doubles, since the system now has more time to respond
BBandwidth is unchanged, since it depends only on DC gain
CBandwidth halves, since ω_b = 1/τ and τ has doubled
DBandwidth increases by √2, following the −3 dB rule
The break frequency (bandwidth) is ω_b = 1/τ. Doubling τ halves the bandwidth — a slower system passes fewer high-frequency signals. This is the key time-frequency duality: larger τ means slower settling AND narrower bandwidth. They are two descriptions of the same constraint.
Question 3 True / False
A first-order system is within 1% of its final steady-state value at t = 5τ.
TTrue
FFalse
Answer: True
At t = 5τ, the step response is K(1 − e⁻⁵) ≈ K(1 − 0.0067) = 99.3% of final value. The engineering convention that '5 time constants = settled' follows directly from this calculation.
Question 4 True / False
A first-order system with a larger time constant is faster because it takes larger steps toward the final value each second.
TTrue
FFalse
Answer: False
A larger time constant means a SLOWER system — τ is the ratio of energy storage to dissipation, so larger τ means the system stores more energy relative to how quickly it can dissipate it. The step response rises as 1 − e^(−t/τ); larger τ stretches the exponential out over a longer time window, taking more time to reach steady state.
Question 5 Short Answer
A first-order transfer function has a pole at s = −50. What is the system's time constant, and what does the pole location tell you about the system's settling speed?
Think about your answer, then reveal below.
Model answer: τ = 1/50 = 0.02 s; the system settles in about 5τ = 0.1 s
The pole sits at s = −1/τ, so τ = 1/|pole| = 1/50 = 0.02 s. Poles further left in the s-plane (more negative real part) correspond to smaller time constants and faster settling. This is why pole placement is the core of control design: moving poles leftward speeds the response.