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System Dynamics · Module 10 of 10

Frequency Response and System Analysis

Drive a system with a sinusoid and it answers with the same frequency, scaled and shifted. Sweeping the frequency traces the system's signature: a corner where first-order systems roll off, a peak where second-order systems resonate.

01

Readiness check

This closing module reads the response across frequency. Tick only what you can do closed-notes.

  • Evaluate a transfer function at s = jω.
  • Find the magnitude and angle of a complex number.
  • Recall a decibel as 20 log10 of a ratio.
  • Recall the first-order corner frequency 1/τ.
  • Recall ωn and ζ for a second-order system.
0 or 1 weak itemsContinue with this module.
2 weak itemsRevisit phasors and complex numbers in Mathematics, Module 18.
3 or more weak itemsReview second-order response in Module 8.
02

The core idea

The frequency response is the transfer function evaluated at s = jω. Its magnitude scales the input sinusoid and its angle shifts it. A first-order system has a corner where it begins to roll off; a second-order one can peak at resonance.

G(jω) = magnitude ∠ phasefirst-order corner: ω = 1/τresonant peak Mr = 1/(2ζ√(1−ζ²))

Drive a stable linear system with a sinusoid and, once the transient dies, the output is a sinusoid of the same frequency, scaled in amplitude and shifted in phase. Both are read directly from the transfer function evaluated at s = jω: the magnitude |G(jω)| is the amplitude ratio, and the angle is the phase shift. Sweeping ω and plotting magnitude (usually in decibels) and phase against log frequency is the Bode view, the standard signature of a system. A first-order system is flat at low frequency and rolls off beyond its corner ω = 1/τ at 20 decibels per decade, with 45 degrees of lag at the corner. A lightly damped second-order system instead peaks near its natural frequency, the resonance, with a peak height Mr = 1/(2ζ√(1 − ζ2)) that grows as damping falls.

The skill works when: you evaluate G(jω), read magnitude and phase, and locate the corner or resonance.
The skill breaks down when: magnitude and phase are mixed up, or resonance is expected from a well-damped system.
The concept. The frequency response of a first-order system: flat until the corner, then rolling off at 20 dB per decade. A second-order system would instead show a peak near its natural frequency.
03

The skills, taught in order

Five skills build the frequency response function, the Bode view, and the first and second-order signatures.

10.1 The frequency response function

Setting s = jω in the transfer function gives G(jω), a complex number at each frequency. Its magnitude is the steady amplitude ratio between output and input sinusoids; its angle is the phase shift between them. This is the steady-state response to a sinusoid.

10.2 Magnitude and phase

At each frequency, |G(jω)| scales the input amplitude and ∠G(jω) shifts its phase. Plotting both against frequency separates how much the system passes from how much it delays, the two halves of the frequency response.

10.3 The Bode view

Plotting magnitude in decibels (20 log10|G|) and phase in degrees against log frequency gives the Bode diagram. Its straight-line asymptotes make the corner, the roll-off slope, and the phase shift easy to read and to sketch.

SystemLow frequencySignatureRoll-off
First orderflat (DC gain)corner at 1/τ−20 dB/decade
Second order, light dampingflatresonant peak near ωn−40 dB/decade

The two canonical frequency signatures. The corner and the peak are what identify a system by ear or by instrument.

10.4 The first-order corner frequency

A first-order system G(jω) = K/(1 + jωτ) is flat at K until the corner ω = 1/τ, where the magnitude has dropped to 1/√2 (−3 dB) and the phase reached −45 degrees. Beyond it the magnitude falls at 20 decibels per decade toward zero.

10.5 The second-order resonant peak

A lightly damped second-order system peaks near ωn, at the resonant frequency ωr = ωn√(1 − 2ζ2), with a peak magnitude Mr = 1/(2ζ√(1 − ζ2)). Lower damping gives a taller, sharper peak; for ζ above about 0.707 no peak appears at all.

Engineering connection: the frequency response is the foundation of Bode and Nyquist stability analysis in Control Systems, the recommended next course.

04

Worked example 1: a first-order frequency response

A first-order system has G(s) = 2/(0.1s + 1). Find the DC gain, the corner frequency, and the magnitude and phase at the corner.

Figure 1. The response is flat at the DC gain of 2 (6.02 dB) up to the corner at 10 rad/s, where it has fallen 3 dB and lags by 45 degrees, then rolls off at 20 dB per decade.
  1. ProblemFind the DC gain, corner frequency, and the magnitude and phase at the corner for the system in Figure 1.
  2. Given / findG(s) = 2/(0.1s + 1), so K = 2, τ = 0.1 s. Find the DC gain, ωc, |G(jωc)|, ∠G(jωc).
  3. AssumptionsStable first-order system in steady sinusoidal state.
  4. ModelG(jω) = K/(1 + jωτ); corner at ω = 1/τ; at the corner |G| = K/√2 and phase = −45 degrees.
  5. Equationsωc = 1/τ|G(jωc)| = K/√2∠G(jωc) = −45°
  6. SolveDC gain = K = 2 (20 log 2 = 6.02 dB). ωc = 1/0.1 = 10 rad/s. |G(jωc)| = 2/√2 = 1.41 (3.01 dB below the DC level). Phase = −45°.
  7. CheckAt the corner the magnitude is always 1/√2 of the passband and the phase exactly −45 degrees, the defining marks of a first-order corner, independent of K and τ.
  8. ConclusionThe system passes slow signals at a gain of 2 and begins attenuating above 10 rad/s. The corner sets the usable bandwidth.
Result. DC gain 2, corner 10 rad/s, |G| = 1.41 (−3 dB) and −45° at the corner.
05

Worked example 2: a second-order resonant peak

A second-order system has ωn = 20 rad/s and ζ = 0.25. Find the resonant peak magnitude and the resonant frequency.

Figure 2. With ζ = 0.25 the response rises to a peak just over twice the DC level, near but slightly below the natural frequency. Lighter damping would make the peak taller and sharper.
  1. ProblemFind Mr and ωr for the system in Figure 2.
  2. Given / findωn = 20 rad/s, ζ = 0.25. Find the resonant peak Mr and frequency ωr.
  3. AssumptionsUnderdamped second-order system with ζ < 0.707, so a peak exists.
  4. ModelMr = 1/(2ζ√(1 − ζ2)); ωr = ωn√(1 − 2ζ2).
  5. EquationsMr = 1/(2ζ√(1 − ζ2))ωr = ωn√(1 − 2ζ2)
  6. SolveMr = 1/(2 × 0.25 × √(1 − 0.0625)) = 1/(0.5 × 0.968) = 1/0.484 = 2.07. ωr = 20√(1 − 0.125) = 20√0.875 = 20 × 0.935 = 18.7 rad/s.
  7. Checkζ = 0.25 is below 0.707, so a peak must exist, and ωr sits just below ωn, as expected. A smaller ζ would push Mr higher and ωr closer to ωn.
  8. ConclusionThe system amplifies inputs near 18.7 rad/s by a factor of about 2.1, the resonance. This peak is what excessive vibration, or a sharp filter, exploits.
Result. Mr = 2.07, ωr = 18.7 rad/s.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Magnitude and phase confusedPhase used as an amplitude ratio"Which is |G| and which is ∠G?"Magnitude scales; angle shifts.
Peak from a damped systemResonance expected at ζ = 1"Is ζ below 0.707?"No peak appears for ζ ≥ 0.707.
Corner without the 2π or 1/τCorner frequency wrong"Is the corner at 1/τ?"The first-order corner is ω = 1/τ.
Resonance at exactly ωnωr taken equal to ωn"Did I reduce by √(1−2ζ²)?"ωr = ωn√(1 − 2ζ2), below ωn.
07

Practice ladder

Level 1 · Direct skill

A first-order system has τ = 0.05 s. Find the corner frequency.

Show answer

ωc = 1/τ = 1/0.05 = 20 rad/s.

Level 2 · Mixed concept

A second-order system has ζ = 0.5 and ωn = 10 rad/s. Find the resonant peak magnitude.

Show answer

Mr = 1/(2 × 0.5 × √(1 − 0.25)) = 1/(1 × 0.866) = 1.15. A modest peak, since ζ is near the limit.

Level 3 · Independent problem

For ζ = 0.1 and ωn = 50 rad/s, find the resonant peak and resonant frequency.

Show answer

Mr = 1/(2 × 0.1 × √0.99) = 1/(0.2 × 0.995) = 5.03. ωr = 50√(1 − 0.02) = 50 × 0.99 = 49.5 rad/s.

Transfer task | Real engineering

A machine vibrates badly when driven near a particular frequency. Explain, using the resonant peak, what its damping ratio tells you and how to reduce the vibration.

What good work looks like

A tall peak means a small ζ, so the machine strongly amplifies inputs near ωn. Adding damping lowers and broadens the peak, or shifting ωn away from the driving frequency avoids it; both follow directly from Mr and ωr.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I evaluated G at s = jω, not s alone."
"Give me three systems; I will find each corner or resonant peak."
"Plot the Bode diagram for me." Reading the magnitude and phase yourself is the skill.
"Does it resonate?" Reasoning from the damping ratio is the point.

Portfolio task

Take a real system, compute its frequency response at several frequencies, identify its corner or resonant peak, and relate that to its time-domain behaviour.

Must include: magnitude and phase at several frequencies, a corner or peak, and a time-domain link.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. What is the frequency response function?

The transfer function evaluated at s = jω, giving amplitude ratio and phase shift.

2. What happens at a first-order corner?

The magnitude drops to 1/√2 (−3 dB) and the phase reaches −45 degrees, at ω = 1/τ.

3. Write the resonant peak magnitude.

Mr = 1/(2ζ√(1 − ζ2)).

4. When does no resonant peak appear?

For ζ ≥ 0.707, the response has no peak.

5. What is the Bode view?

Magnitude in decibels and phase in degrees plotted against log frequency.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive the first-order corner and second-order peak from a blank page.
+3 daysFind the frequency signature of three new systems.
+7 daysCombine modeling, response, and frequency analysis into one study.
+30 daysCarry the frequency response into Control Systems, the recommended next course, and revisit the whole course through the System Dynamics hub.
10

Textbook mapping

This module follows Karnopp, Margolis, and Rosenberg, System Dynamics: Modeling, Simulation, and Control of Mechatronic Systems, 5th edition. Use these references to read further.

Topic in this moduleWhere to read more
Frequency response and the Bode viewKarnopp, Margolis & Rosenberg, Chapter 9
First-order corner frequencyKarnopp, Margolis & Rosenberg, Chapter 9
Second-order resonanceKarnopp, Margolis & Rosenberg, Chapter 9

Chapter numbers refer to the 5th edition. The frequency-response methods are standard, so any recent edition will align closely.