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MAE 4182October 14, 2026← Back to the lesson

October 14 · Through “Read frequency and damping from a pole”

Optional homework

01

Derive the second-order poles.

G(s) = ωn2 divided by s2 + 2ζωns + ωn2,  ωn > 0
  1. Derive both poles in terms of ωn and ζ. Show your algebra.
  2. For 0 ≤ ζ < 1, write them as a complex-conjugate pair.
  3. Now use G(s) = 10/(s2 + 2s + 10). Find ωn, ζ, and both poles. Check by completing the square; identify α and ω.
Hint

Set the denominator to zero. Complete the square by adding (ζωₙ)² to both sides, or use the quadratic formula with a = 1, b = 2ζωₙ, c = ωₙ².

Worked solution
  1. The poles are the roots of the denominator:
    s2 + 2ζωns + ωn2 = 0.
    Move the constant to the right and complete the square:
    s2 + 2ζωns + (ζωn)2 = (ζωn)2 − ωn2,
    (s + ζωn)2 = ωn2(ζ2 − 1).
  2. Take both square roots. Because ωn > 0,
    p = −ζωn ± ωn√(ζ2 − 1).
  3. For 0 ≤ ζ < 1, the quantity ζ2 − 1 is negative. Write its square root using j:
    p = −ζωn ± jωn√(1 − ζ2).
    This identifies α = ζωn and ω = ωn√(1 − ζ2). At ζ = 1 the pole repeats; for ζ > 1 both poles are real.
  4. For the numerical example, match ωn2 = 10 and 2ζωn = 2:
    ωn = √10 rad/s,  ζ = 1 divided by √10.
    Then α = 1 s−1, ω = √10√(1 − 1/10) = 3 rad/s, and p = −1 ± 3j.
  5. Check directly by completing the numerical square:
    s2 + 2s + 10 = (s2 + 2s + 1) + 9 = (s + 1)2 + 32.
    Setting this to zero gives (s + 1)2 = −9, so again p = −1 ± 3j.

The remaining constant is 9, not 10: the square already contains 1.

Review: Second-order poles
02

A brief push: find the impulse response.

G(s) = 10 divided by s2 + 2s + 10

Apply a unit impulse to this numerical model, starting from rest before the impulse.

  1. Find Y(s), then use the table above to find y(t).
  2. Which factor makes the response oscillate? Which makes its amplitude decrease?
Hint

A unit impulse has U(s) = 1. The sine table pair needs 3 in its numerator, so write 10 = (10/3) × 3.

Worked solution
  1. A unit impulse gives U(s) = 1. With zero state before the impulse,
    Y(s) = G(s)U(s) = 10 divided by (s + 1)2 + 32.
  2. Match the numerator to the table:
    Y(s) = 10 divided by 3 3 divided by (s + 1)2 + 32.
  3. Read the inverse transform:
    y(t) = 10 divided by 3 e−t sin(3t),  t ≥ 0.

sin(3t) makes the response swing back and forth at 3 rad/s. The multiplier (10/3)e−t gives its shrinking envelope. This is a brief push, not a command that stays on.

Do not write 10e−t sin(3t): that would transform to a numerator of 30, not 10.

Review: Impulse response and the transform table
03

Which swing dies away?

A: e−t sin(3t)
B: sin(3t)
C: et sin(3t)
  1. For each free response, does the oscillation decay, persist, or grow?
  2. Do they have the same oscillation frequency? Explain using the formula.
Hint

sin(3t) stays between −1 and 1. Look only at the factor multiplying it to decide what happens to the amplitude.

Worked solution
  1. A decays: e−t approaches zero, so the oscillation envelope shrinks.
  2. B persists: its envelope stays at 1.
  3. C grows: et increases, so successive swings get larger.

All three oscillate at 3 rad/s because all contain sin(3t). Multiplying by a different exponential changes the envelope, not the sine's frequency.

A response that stays bounded but keeps oscillating does not return to equilibrium. Do not call B an asymptotically stable free response.

Review: Read the exponential and sine factors
04

Change speed or reduce overshoot?

Three standard models G(s) = ωn2/(s2 + 2ζωns + ωn2) receive a unit step from zero initial state.

A: ωn = 2, ζ = 0.2
B: ωn = 4, ζ = 0.2
C: ωn = 4, ζ = 0.6

Natural frequencies are in rad/s. Answer without calculating y(t):

  1. A or B: which oscillates faster?
  2. B or C: which swings less past the target?
  3. Does B have more overshoot than A, or the same percentage overshoot?
Hint

Compare one parameter at a time: A versus B keeps ζ fixed; B versus C keeps ωₙ fixed. No response calculation is needed.

Worked solution
  1. B oscillates faster than A. With the same ζ, doubling ωn also doubles ω = ωn√(1 − ζ2). The response runs through the same shape in half the time.
  2. C has less overshoot than B. In this standard second-order model, increasing ζ from 0.2 to 0.6 reduces the swing past the target.
  3. B does not have more overshoot than A. With ζ unchanged, their percentage overshoot is the same. B reaches its peak earlier.

Compare models with the same input. Here every test is a unit step from zero initial state, with the same final value 1.

Review: Natural frequency and damping ratio
05

Close the loop. What does K change?

Unity negative feedback with a controller gain K and a plant G(s)R enters a summing point with a plus sign. The output Y feeds back directly to its minus input. The error E passes through gain K, producing U, then through G(s), producing Y.REUYKG(s)+−

Unity negative feedback; zero initial state. U is the scaled command.

  1. Starting from the signals in the diagram, find Y/R in terms of K and G.
  2. Use G(s) = 1/[s(s + 1)] and K = 2. Simplify Y/R.
  3. For a fixed error e = 0.1, compare u = Ke at K = 1 and K = 2. What does K do?
Hint

Write E = R − Y, U = KE, and Y = GU. Substitute, then collect every term containing Y on one side.

Worked solution
  1. Follow the arrows:
    E = R − Y,  U = KE,  Y = GU.
    Substitute U and E:
    Y = KG(R − Y) ⇒ (1 + KG)Y = KGR.
    Therefore,
    Y divided by R = KG divided by 1 + KG.
  2. For G(s) = 1/[s(s + 1)],
    Y(s) divided by R(s) = K divided by s2 + s + K = 2 divided by s2 + s + 2  when K = 2.
  3. In the time domain, u = Ke. For the same error e = 0.1, K = 1 gives u = 0.1, while K = 2 gives u = 0.2. The larger K requests twice as much control action for that same error.

K changes the closed-loop denominator as well as the numerator. It is not just a vertical scaling of the same response.

Review: Unity negative feedback and controller gain
06

Read frequency and damping from the poles.

p = −4 ± 3j s−1
  1. Find α, ω, ωn, and ζ.
  2. Is the actual oscillation frequency 3 or 5 rad/s? Does the free oscillation decay or grow?
Hint

First compare p = −4 ± 3j with p = −α ± jω. Then use α² + ω² = ωₙ² and α = ζωₙ.

Worked solution
  1. The poles give α = 4 s−1 and ω = 3 rad/s.
  2. Use the relation from completing the square:
    ωn = √(α2 + ω2) = √(42 + 32) = 5 rad/s.
  3. Then
    ζ = α divided by ωn = 4 divided by 5 = 0.8.

The actual oscillation frequency is 3 rad/s, not 5 rad/s. The negative real part gives an e−4t envelope, so the free oscillation decays.

The imaginary part gives ω. To find ωn, you need both the real and imaginary parts.

Review: The relations from our completed square