Skip to the questions
MAE 4182October 20, 2026← Back to the lesson

October 20 · Root locus and frequency response

After-class practice

01

Predict the tracking motion

Consider the stable LTI system

G(s) = 2 divided by s + 2,  r(t) = 3 sin(2t).

Find the steady-state output amplitude, phase, and yss(t).

Hint

Identify R and ω. Evaluate G(jω), then use Y = R|G(jω)| and φ = ∠G(jω).

Worked solution
  1. The input amplitude is R = 3 and its frequency is ω = 2 rad/s. Substitute s = j2:
    G(j2) = 2 divided by 2 + j2.
  2. Find the magnitude and phase:
    |G(j2)| = 2 divided by √(2² + 2²) = 1 divided by √2,
    φ = 0 − tan⁻¹(2/2) = −45° = −π/4.
    The denominator has phase +45°; taking its reciprocal changes the sign.
  3. Multiply the input amplitude by the gain. Keep the same frequency:
    Y = 3 divided by √2 ≈ 2.12,
    yss(t) = 3 divided by √2 sin(2t − π/4).
    The output is smaller and lags by 45°. Its frequency is still 2 rad/s.

|G(jω)| is the amplitude ratio, not the output amplitude. Multiply it by R. Use radians inside the sine.

Try a faster input

Keep G(s) unchanged, but use r(t) = 3 sin(8t). Find the new steady-state response. Compared with the first input, is the amplitude larger or smaller? Is the phase lag greater or smaller?

Worked solution
G(j8) = 2 divided by 2 + j8,  |G(j8)| = 2 divided by √68.
Y = 6 divided by √68 ≈ 0.728,  φ = −tan⁻¹(8/2) ≈ −75.96°.
yss(t) = 6 divided by √68 sin(8t − arctan 4).

The output amplitude decreases from about 2.12 to 0.728. The phase lag increases from 45° to about 75.96°. The output frequency remains equal to its input frequency, now 8 rad/s.

These are steady-state responses: the startup transient has decayed.

Review the classroom example ↗
02

Turn up the input frequency

Try ω = 0.5, then 8. What changes? What stays the same?

G(s) = 1s + 2 r(t) = √2 sin(ωt)
Input r(t)
Input sinusoidTime from zero to eight seconds in a steady-state window. Shared vertical scale from minus 1.6 to 1.6. Frequency 2.00 radians per second. Amplitude 1.414.-10102468Time (s)
Output yss(t)
Steady-state output sinusoidTime from zero to eight seconds in a steady-state window. Shared vertical scale from minus 1.6 to 1.6. Frequency 2.00 radians per second. Amplitude 0.500. Phase -45.0 degrees relative to input. The startup transient is not plotted.-10102468Time (s)

Steady state: after the startup transient has died out. Both plots use the same scales.

Same frequency 2.0 rad/sAmplitude ratio |G| 0.354Phase φ -45.0°
Compare your observations

Both signals repeat at the input frequency. In this model, higher frequency makes the output smaller and more delayed in phase. The system has not changed—only the input frequency has.

03

Read gain and phase

G(s) = 1 divided by s + 2,  r(t) = 2 sin(2t).

Use the marked frequency. What are the output amplitude, frequency, and phase?

Gain 20 log₁₀|G(jω)| · dB
Bode magnitude plot for G(s) = 1/(s+2)Logarithmic frequency axis from 0.1 to 100 radians per second. Marker at input frequency 2 radians per second. Gain is minus 9.03 decibels, equivalent to amplitude ratio 1 over square root of 8.0-10-20-30-400.1110100−9.03 dBω (rad/s) · log scale
Phase ∠G(jω) · degrees
Bode phase plot for G(s) = 1/(s+2)Logarithmic frequency axis from 0.1 to 100 radians per second. Marker at input frequency 2 radians per second. Phase is minus 45 degrees.0-45-900.1110100−45°ω (rad/s) · log scale
Worked solution
  1. The input has amplitude R = 2 and frequency ω = 2 rad/s. At this frequency, read −9.03 dB from the gain plot and −45° from the phase plot.
  2. Convert decibels to an amplitude ratio:
    |G(j2)| = 10−9.03/20 ≈ 0.354.
  3. Multiply by the input amplitude. Keep the same input frequency:
    Y ≈ 2(0.354) = 0.707,  ω = 2 rad/s,  φ = −45°.
    The negative phase means that the steady-state output lags behind the input.
  4. For this model, the exact amplitude ratio is 1/√8, so
    yss(t) = 1 divided by √2 sin(2t − π/4).
    The plotted dB value is rounded; the exact formula gives an output amplitude of 1/√2.

The dB reading is not the output amplitude. Convert it to |G|, then multiply by R. Use radians inside the sine.

04

Compare two gains

For unity negative feedback with forward path K L(s), use the characteristic equation below:

L(s) = s + 1 divided by s2,  s2 + Ks + K = 0.

Find the closed-loop poles for K = 2 and K = 5. For each gain, is the system stable? Do its natural modes oscillate?

Hint

Substitute each K and use the quadratic formula. The real part tells you whether a mode decays; a nonzero imaginary part gives oscillation.

Worked solution
  1. For K = 2:
    s2 + 2s + 2 = 0  →  s = −1 ± j.
    Both poles have real part −1, so the system is stable. The natural modes contain e−t cos t and e−t sin t: they oscillate while decaying.
  2. For K = 5:
    s2 + 5s + 5 = 0  →  s = −5 ± √5 divided by 2 ≈ −1.382, −3.618.
    Both poles are real and negative, so the system is stable. Its natural modes are decaying exponentials with no oscillation.

Both gains give stable poles. Stability alone does not tell you whether the natural modes oscillate.

Review the root-locus examples ↗
05

Read two basic factors

GA(s) = 10 divided by s,  GB(s) = 0.1s2.

For each factor, identify g and n in G(s) = gsn. Find the magnitude ratio and dB value at ω = 1 rad/s, the magnitude slope in dB/decade, and the phase for ω > 0.

Why does changing a positive gain g shift the magnitude level without changing the phase?

Hint

Use |G(jω)| = gωn, M = 20 log₁₀g + 20n log₁₀ω, and φ = n × 90°.

Worked solution
  1. For GA(s) = 10/s, g = 10 and n = −1:
    |GA(j1)| = 10,  M(1) = 20 log₁₀10 = +20 dB.
    The magnitude slope is −20 dB/decade, and the phase is −90°.
  2. For GB(s) = 0.1s2, g = 0.1 and n = 2:
    |GB(j1)| = 0.1,  M(1) = 20 log₁₀0.1 = −20 dB.
    The magnitude slope is +40 dB/decade, and the phase is +180° (equivalently −180°).
  3. A positive real gain scales the magnitude and contributes zero phase. In dB, changing g adds the constant 20 log₁₀g; the slope and phase still depend on n.
Review gain and powers of s ↗
06

Check the corner frequency

G(s) = 1 + 2s.
  1. Find the corner frequency, then the exact G(jω), magnitude ratio, dB value, and phase at that frequency.
  2. Give the low-frequency magnitude level and the high-frequency slope.
  3. The two magnitude asymptotes meet at 0 dB at the corner. Why is this not the exact value?
Hint

Match 1 + 2s to 1 + τs. At the corner, ωτ = 1. At low or high frequency, decide which term dominates.

Worked solution
  1. Here τ = 2 s, so
    ωc = 1 divided by τ = 0.5 rad/s,  G(j0.5) = 1 + j.
    |G(j0.5)| = √2,  M = 20 log₁₀√2 ≈ +3.01 dB,  φ = +45°.
  2. At low frequency, 2ω ≪ 1, so G(jω) ≈ 1 and the magnitude is approximately 0 dB. At high frequency, 2ω ≫ 1, so G(jω) ≈ j2ω and the magnitude rises at +20 dB/decade.
  3. At the corner, 1 and j have equal magnitude; neither can be neglected. The straight-line approximations each keep only one term and give magnitude 1, or 0 dB. The exact magnitude is |1 + j| = √2, so the exact curve is +3.01 dB there.
Review the first-order zero ↗