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MAE 4182October 19, 2026← Back to the lesson

October 19 · Poles, damping and gain

Optional homework

01

Calculate the second-order poles.

G(s) = 9 divided by s2 + 3s + 9
  1. Calculate both poles. Show the algebra.
  2. Find the natural frequency ωn and damping ratio ζ.
Hint

Set the denominator equal to zero. Use the quadratic formula or complete the square. Match the denominator to s² + 2ζωₙs + ωₙ².

Worked solution
  1. The poles satisfy
    s2 + 3s + 9 = 0.
    The quadratic formula gives
    p₁,₂ = −3 ± √(9 − 36) divided by 2 = −3 divided by 2 ± j3√3 divided by 2.
    The negative discriminant, −27, produces a complex-conjugate pair.
  2. Compare with the standard denominator:
    s2 + 2ζωns + ωn2.
    Its constant term gives ωn2 = 9, so ωn = 3 rad/s. Its s coefficient gives
    2ζ(3) = 3  ⇒  ζ = 1 divided by 2.
  3. Check using the pole formula:
    p = −ζωn ± jωn√(1 − ζ2) = −1.5 ± j3√3 divided by 2.
    Both methods give the same poles.

ωn is 3 rad/s, not 9 rad/s. The constant coefficient is ωn2.

Review: Second-order poles
02

Read frequency and damping from a pole pair.

p = −3 ± j4 s−1
  1. Find ωn, ζ, and the actual oscillation frequency ω.
  2. A classmate says “ωn = 4 because the imaginary part is 4.” Explain the mistake.
Hint

Identify α and ω from p = −α ± jω. Then use ωₙ² = α² + ω² and α = ζωₙ.

Worked solution
  1. Compare the given pair with p = −α ± jω:
    α = 3 s−1,   ω = 4 rad/s.
  2. Use both coordinates to find the natural frequency:
    ωn = √(α2 + ω2) = √(32 + 42) = 5 rad/s.
  3. Then
    ζ = α divided by ωn = 3 divided by 5 = 0.6.

The imaginary coordinate gives the actual oscillation frequency, 4 rad/s. The natural frequency is the distance from the origin to either pole, 5 rad/s.

Do not identify the imaginary part alone as ωn. The real part also contributes to the distance.

Review: Poles, frequency and damping
03

Does the oscillation decay, grow, or persist?

A: p = −1 ± j2
B: p = +1 ± j2
C: p = ±j2

Consider a nonzero free response associated with each pair.

  1. Classify each oscillation: decay, growth, or constant amplitude.
  2. Write the exponential factor that explains each answer. Do the three oscillation frequencies differ?
Hint

A pair σ ± jω gives free-response terms eσtcos(ωt) and eσtsin(ωt). Compare e−t, et and e0t.

Worked solution
  1. A decays. Its real part is −1, giving terms
    e−tcos(2t),   e−tsin(2t).
    The envelope shrinks to zero because e−t → 0.
  2. B grows. Its real part is +1, giving terms
    etcos(2t),   etsin(2t).
    The envelope grows because et increases.
  3. C persists. Its real part is zero, so e0t = 1. Its terms are
    cos(2t),   sin(2t).
    The oscillation amplitude does not decay.

All three pairs have imaginary parts ±2, so they have the same oscillation frequency, 2 rad/s. The real part changes the envelope.

C does not decay to equilibrium. Persistent oscillation is not an asymptotically stable free response.

Review: Poles and motion
04

Close the loop and match the coefficients.

Unity negative feedback with controller gain K and plant G(s)R enters the positive input of a summing point. The output Y feeds back to its negative input. The error E passes through K, producing U, then through G(s), producing Y.REUYKG(s)+−
G(s) = 1 divided by s(s + 2),   K > 0

Unity negative feedback; zero initial state.

  1. Starting from the diagram, derive T(s) = Y(s)/R(s).
  2. Express ωn and ζ in terms of K.
  3. Find ωn and ζ when K = 4.
Hint

Start from E = R − Y, U = KE, and Y = GU. Collect the Y terms. Then compare the closed-loop denominator with s² + 2ζωₙs + ωₙ².

Worked solution
  1. Write the signal equations and substitute:
    E = R − Y,   U = KE,   Y = GU,
    Y = KG(R − Y)  ⇒  (1 + KG)Y = KGR.
    Therefore,
    T(s) = Y(s) divided by R(s) = KG(s) divided by 1 + KG(s).
  2. Substitute the given plant:
    T(s) = K divided by s(s + 2) + K = K divided by s2 + 2s + K.
  3. Match the constant and s coefficients:
    ωn2 = K,   2ζωn = 2.
    Since K > 0,
    ωn = √K,   ζ = 1 divided by √K.
  4. At K = 4,
    T(s) = 4 divided by s2 + 2s + 4,   ωn = 2 rad/s,   ζ = 1 divided by 2.
    One gain determines both parameters.

This plant has s(s + 2), not s(s + 1). Re-match the coefficients instead of copying ζ = 1/(2√K) from the classroom model.

Review: Feedback gain and the second-order model
05

Find all gains that meet both requirements.

T(s) = K divided by s2 + s + K,   K > 1 divided by 4

Requirements:

ωn ≥ 1.5 rad/s,   ζ ≥ 1 divided by 4
  1. Find every K that meets both requirements.
  2. Choose one valid K and check both requirements numerically.
Hint

Match the denominator to the standard second-order form. Turn each requirement into a bound on K, then take the overlap. K is positive, so √K is positive.

Worked solution
  1. Match the coefficients of the given closed-loop model:
    ωn2 = K,   2ζωn = 1.
    Thus,
    ωn = √K,   ζ = 1 divided by 2√K.
  2. The natural-frequency requirement gives a lower bound:
    √K ≥ 3 divided by 2  ⇒  K ≥ 9 divided by 4.
  3. The damping requirement gives an upper bound:
    1 divided by 2√K ≥ 1 divided by 4  ⇒  4 ≥ 2√K  ⇒  K ≤ 4.
    Multiplication by the positive quantity 8√K does not reverse the inequality.
  4. Both requirements hold exactly when
    9 divided by 4 ≤ K ≤ 4.
    Both endpoints are included; this interval also satisfies K > ¼.
  5. For example, choose K = 3 and check:
    ωn = √3 ≈ 1.732 rad/s ≥ 1.5 rad/s,
    ζ = 1 divided by 2√3 ≈ 0.289 ≥ 0.25.
    K = 3 works, but it is not the only answer.

Checking one gain does not find all valid gains. First find the overlapping interval, then choose a value within it.

Review: Choose a gain