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MAE 4182October 7, 2026← Back to Lecture 07

October 7 · Lecture 07 · Through page 13

Optional homework

Voluntary and ungraded. No submission or deadline.

Six questions on what we covered through “Which transient oscillations decay?” Choose any questions you want to practice. Try them on paper before opening the worked solutions.

Questions 01–03

First-order responses

01

A unit impulse

G(s) = 3 divided by s + 3

Apply a unit impulse to this initially resting system.

  1. Find Y(s) and y(t).
  2. Find the pole and time constant τ.
  3. Find y(0+), y(τ), and the long-time value. Does “37% remains” mean y(τ) = 0.37?
Optional hint

A unit impulse has U(s) = 1. Use Y(s) = G(s)U(s) and the pair 1/(s + a) ↔ e⁻ᵃᵗ.

Worked solution
  1. For a unit impulse, U(s) = 1, so
    Y(s) = 3 divided by s + 3 ⇒ y(t) = 3e−3t.
  2. The denominator is zero at p = −3 s−1. Here a = 3, so τ = 1/3 s.
  3. Immediately after the impulse, y(0+) = 3. At one time constant, y(τ) = 3/e ≈ 1.104. As t → ∞, y(t) → 0.

37% is the fraction of the initial value remaining—not the output itself. Here the output at τ is 3 × 0.368 ≈ 1.104, not 0.368.

Review: First-order impulse response
02

The same system, a unit step

G(s) = 3 divided by s + 3

Start a new experiment from rest, now with a unit-step input.

  1. Find Y(s) and y(t). Identify the transient and steady-state terms.
  2. Find y(τ). Has the response finished after one time constant?
  3. Explain why this response approaches 1, while the impulse response in Question 1 approaches 0.

For the last part, think about what each input does after t = 0.

Optional hint

For a unit step, U(s) = 1/s. Write 3/[s(s + 3)] as two simpler fractions.

Worked solution
  1. Change the input, not the system:
    Y(s) = 3 divided by s(s + 3) = 1 divided by s − 1 divided by s + 3.
  2. Invert the two terms:
    y(t) = 1 − e−3t.
    The response starts at 0 and approaches 1. The transient term is −e−3t; the steady-state term is 1.
  3. The system still has τ = 1/3 s. At that time, y(τ) = 1 − e−1 ≈ 0.632: about 63.2% of the final value.
  4. An impulse is a brief hit; afterward the input is zero, and this stable system’s response dies away. A unit step stays at 1, so it continues to drive the system and the output settles at 1.

One time constant does not mean “finished.” About 36.8% of the original gap to the final value remains. Compare with Question 1: the same pole and time constant, but a different input gives a different response.

Review: Step response and time constant
03

Which response is faster?

GA(s) = 2 divided by s + 2,  GB(s) = 5 divided by s + 5

Apply a unit step to each system.

  1. Find both time constants. Which response is faster?
  2. Sketch both responses on the same axes. Mark their initial and final values.
  3. A classmate says “B ends at 5 because its numerator is 5.” Explain the mistake.
Optional hint

Both models have the form a/(s + a). Use τ = 1/a and y(t) = 1 − e⁻ᵃᵗ.

Worked solution
  1. For A, a = 2: yA(t) = 1 − e−2t and τA = 0.5 s. For B, a = 5: yB(t) = 1 − e−5t and τB = 0.2 s.
  2. Both start at 0 and approach 1. B is faster because its time constant is smaller: it reaches 63.2% in 0.2 s rather than 0.5 s.
  3. The numerator 5 does not mean a final value of 5. In this normalized model, GB(0) = 5/5 = 1, and its step response approaches 1.
Two unit-step responses with the same final value and different time constantsBoth curves start at zero and approach 1. A has time constant 0.5 seconds. B has time constant 0.2 seconds and rises faster.00.5100.511.522.5y(t)Time t (s)AB→ 1
B rises faster, but both responses approach 1. A larger a changes the speed, not the final value of a/(s + a).
Review: Compare first-order step responses

Questions 04–06

Second-order model and poles

04

Read the coefficients

G(s) = 16 divided by s2 + 4s + 16
  1. Find ωn, ζ, and α by matching coefficients.
  2. Find the real part of the poles. Will free oscillations decay or grow?
  3. Why is ωn not 16?
Optional hint

Match the constant term first, then the coefficient multiplying s.

Worked solution
  1. Match ωn2 = 16. Taking the positive root gives ωn = 4 rad/s.
  2. Match 2ζωn = 4: 2ζ(4) = 4, so ζ = 0.5.
  3. Then α = ζωn = 2 s−1, and Re(p) = −α = −2 s−1. Free oscillations decay because this real part is negative.

The constant coefficient is ωn2, not ωn. That is why ωn is 4, not 16.

Review: Standard second-order model
05

Natural frequency or oscillation frequency?

ωn = 5 rad/s,  ζ = 0.6
  1. Find α, ω, and both poles.
  2. Does the system oscillate at 5 rad/s? Explain.
  3. Which part of the poles tells you whether free oscillations decay?
Optional hint

Calculate α = ζωₙ and ω = ωₙ√(1 − ζ²) separately, then put them into p = −α ± jω.

Worked solution
  1. α = 0.6 × 5 = 3 s−1.
  2. ω = 5√(1 − 0.62) = 5√0.64 = 4 rad/s.
  3. The poles are p = −3 ± 4j s−1. Their real part, −3, tells us that free oscillations decay. The magnitude of the imaginary part, 4, gives the oscillation frequency.

The actual oscillation frequency is 4 rad/s, not 5 rad/s. For the complex-pole models here, ω equals ωn only when ζ = 0.

Review: Second-order poles and frequency
06

Decay, persist, or grow?

A: p = −1 ± 2j
B: p = ±2j
C: p = +1 ± 2j

These are the pole pairs of three systems (in s−1).

  1. For each system, do free oscillations decay, persist, or grow? Give a one-sentence reason.
  2. Do they have the same oscillation frequency?
  3. Does B’s bounded free oscillation alone prove BIBO stability?
Optional hint

Use the real part for decay or growth, and the magnitude of the imaginary part for oscillation frequency.

Worked solution
  1. A: decay. Re(p) = −1 < 0. The free-oscillation amplitude decreases with time.
  2. B: persist. Re(p) = 0. There is no decay or growth; free oscillations keep their amplitude.
  3. C: grow. Re(p) = +1 > 0. The free-oscillation amplitude increases with time.

All three have the same oscillation frequency, 2 rad/s, because the imaginary parts are ±2j. Their amplitudes behave differently because their real parts differ.

Persistent free oscillations do not prove BIBO stability. BIBO asks whether every bounded input produces a bounded output; an input at the undamped system’s oscillation frequency can make its response grow. No resonance calculation is needed here.

Review: Which transient oscillations decay?