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Controls and Systems · Chapter 8 of 10 · Advanced

Frequency Response and Bode Plots

Drive a linear system with a sine wave and the output is a sine of the same frequency, scaled and shifted. Plotting that scale and shift against frequency is the Bode plot, the engineer's working view of a system.

Readiness check

This chapter works in the frequency domain. Tick only what you can do closed-notes.

Evaluate a transfer function at s = jω.
Find the magnitude and angle of a complex number.
Read a logarithmic frequency axis.
Convert a ratio to decibels with 20 log.
Recall that each pole adds phase lag.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsRevisit complex numbers and transfer functions in Chapter 2.

3 or more weak itemsReview complex arithmetic before continuing.

The core idea

The frequency response is the transfer function evaluated at s = jω. Its magnitude in decibels and its phase, plotted against log frequency, are the Bode plot, built from straight-line asymptotes.

G(jω): gain |G|, phase ∠GdB = 20 log₁₀|G|pole: −20 dB/decade, −90° eventually

Replacing s with jω turns the transfer function into a complex number at each frequency: its magnitude is how much a sine of that frequency is amplified, and its angle is the phase shift. Bode plots show 20 log of the magnitude and the phase against a logarithmic frequency axis, where each first-order factor contributes a simple straight-line shape. A pole bends the magnitude down by 20 dB per decade past its corner frequency and pulls the phase toward −90 degrees; a zero does the opposite. Summing these pieces sketches the whole response, and the frequency where the gain passes unity sets the loop's speed and margin.

The skill works when: you locate the corners, add the asymptote slopes, and read magnitude and phase off the plot.

The skill breaks down when: decibels are taken with 10 log instead of 20, or a corner frequency is misplaced.

The concept. Bode plot of a first-order pole: the magnitude is flat, then breaks down at 20 dB per decade past the corner, while the phase slides from 0 to −90 degrees, passing −45 at the corner.

The skills, taught in order

Five skills define the frequency response, the decibel and phase plots, the asymptote rules, the standard shapes, and the gain crossover.

8.1 The frequency response

Evaluating G(jω) gives a complex number whose magnitude is the steady-state gain to a sine of frequency ω and whose angle is the phase shift. Sweeping ω traces the system's behaviour across all frequencies in one pair of curves.

8.2 Magnitude in decibels and phase

The magnitude is plotted as 20 log₁₀|G| in decibels, which turns products of factors into sums and makes wide ranges legible. The phase is plotted in degrees. Both use a logarithmic frequency axis so first-order factors become straight lines.

8.3 Bode asymptotes and corners

Each factor contributes a simple slope. A pole bends the magnitude down by 20 dB per decade above its corner frequency; a zero bends it up by the same. An integrator gives a constant −20 dB per decade. Phase contributions add the same way.

Factor	Magnitude slope	Phase (final)
Constant gain K	0 dB/dec	0°
Integrator 1/s	−20 dB/dec	−90°
Pole 1/(s + a)	−20 dB/dec above ω = a	−90°
Zero (s + a)	+20 dB/dec above ω = a	+90°

8.4 First- and second-order shapes

A first-order pole is exactly 3 dB down and 45 degrees lagged at its corner. A lightly damped second-order pole pair shows a resonant peak near ω_n whose height grows as the damping ratio falls, and the phase drops through −90 degrees at ω_n.

8.5 Gain crossover

The gain crossover frequency is where the magnitude passes through 0 dB (gain of one). It marks the loop bandwidth and is the frequency at which the phase margin is read, the bridge to the stability margins of the next chapter.

Engineering connection: the gain and phase crossovers found here are exactly where the gain and phase margins of Chapter 9 are measured.

Worked example 1: a first-order Bode point

For G(s) = 10/(s + 10), find the corner frequency, and the magnitude (in decibels) and phase at ω = 10 rad/s.

Figure 1. At the corner frequency the first-order magnitude is exactly 3 dB below its low-frequency value and the phase is −45 degrees, the universal first-order signatures.

ProblemFind the corner frequency, magnitude in dB, and phase at ω = 10 for the system in Figure 1.
Given / findG(s) = 10/(s + 10). Find ω_c, |G| in dB, and ∠G at ω = 10.
AssumptionsLinear system; sinusoidal steady state, so s = jω.
ModelThe corner is at the pole magnitude; evaluate G(j10) for magnitude and phase.
EquationsG(jω) = 10/(jω + 10)|G| = 10/√(ω² + 100), ∠G = −arctan(ω/10)
SolveCorner ω_c = 10 rad/s. |G(j10)| = 10/√(100 + 100) = 10/14.14 = 0.707, which is 20 log(0.707) = −3 dB. Phase = −arctan(10/10) = −45°.
CheckThe low-frequency gain is 10/10 = 1 (0 dB), and at the corner it drops by exactly 3 dB with 45 degrees of lag, the defining first-order behaviour.
ConclusionOne pole gives one corner, a −3 dB point, and a −45 degree phase there. These anchors let you sketch the whole curve by hand.

Result. Corner at 10 rad/s; |G(j10)| = 0.707 (−3 dB), phase −45°.

Worked example 2: the gain crossover frequency

An open-loop transfer function is G(s) = 100/[s(s + 10)]. Find the gain crossover frequency, where the magnitude equals one (0 dB).

Figure 2. The magnitude falls from the integrator until it passes through 0 dB at the gain crossover, where the open-loop gain equals one. Here that is 7.86 rad/s, below the corner at 10.

ProblemFind the gain crossover frequency for the system in Figure 2.
Given / findG(s) = 100/[s(s + 10)]. Find ω_gc where |G(jω)| = 1.
AssumptionsSinusoidal steady state; magnitude condition |G(jω_gc)| = 1.
ModelWrite the magnitude, set it to one, and solve the resulting quadratic in ω².
Equations|G(jω)| = 100/[ω√(ω² + 100)] = 1ω²(ω² + 100) = 10 000
SolveLet u = ω²: u² + 100u − 10 000 = 0, so u = (−100 + √50 000)/2 = 61.8, and ω_gc = √61.8 = 7.86 rad/s.
CheckSubstituting back, 100/[7.86 × √(61.8 + 100)] = 100/[7.86 × 12.72] = 1.00, confirming the crossover. It sits just below the corner at 10, as the slope suggests.
ConclusionThe gain crossover marks the loop bandwidth and the frequency where the phase margin is read in the next chapter.

Result. The gain crossover frequency is ω_gc = 7.86 rad/s.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Using 10 log for amplitude	Decibels off by a factor of two	"Is this a power or an amplitude ratio?"	Amplitude ratios use 20 log₁₀\|G\|.
Misplacing the corner	Slope changes at the wrong frequency	"Where does the pole magnitude sit?"	The corner is at the pole or zero magnitude a.
Forgetting the integrator phase	Phase short by 90 degrees	"Is there a 1/s factor?"	An integrator contributes a constant −90 degrees.
Reading gain crossover off magnitude only	Margin computed at the wrong ω	"Is this the 0 dB crossing?"	Gain crossover is where the magnitude is exactly 0 dB.

Practice ladder

Level 1 · Direct skill

Find the corner frequency and the low-frequency gain of G(s) = 20/(s + 4).

Show answer

Corner at ω = 4 rad/s. Low-frequency gain G(0) = 20/4 = 5, which is 20 log 5 = 14 dB.

Level 2 · Mixed concept

For G(s) = 10/(s + 10), find the magnitude in dB and phase at ω = 100 rad/s.

Show answer

|G| = 10/√(10000 + 100) = 0.0995, which is −20 dB. Phase = −arctan(100/10) = −84°. A decade past the corner the magnitude is down about 20 dB and the phase nears −90 degrees.

Level 3 · Independent problem

For G(s) = 50/[s(s + 5)], estimate the gain crossover using the low-frequency integrator approximation, then refine it.

Show answer

Below the corner, |G| ≈ 50/(5ω) = 10/ω = 1 gives ω ≈ 10, but that is above the corner. Solving ω²(ω² + 25) = 2500 gives ω_gc ≈ 6.4 rad/s. The asymptote estimate must be checked against the corner.

Level 4 · Transfer to real engineering

Take a real system with a known bandwidth (an audio amplifier, a sensor) and sketch its Bode magnitude, identifying the corner and the roll-off.

What good work looks like

A flat region, a corner at the bandwidth, and a −20 dB per decade (or steeper) roll-off, with the gain read in decibels.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I placed the corners and used 20 log for the magnitude."

"Give me three transfer functions; I will find their corner frequencies."

"Draw the Bode plot." Adding the asymptote slopes yourself is the skill.

"What is the crossover frequency?" Solving the magnitude condition is the point.

Portfolio task

Take a real plant, sketch its Bode magnitude and phase from the asymptote rules, and find its gain crossover frequency by solving the magnitude condition.

Must include: corner frequencies, asymptote slopes, and a gain crossover frequency.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. What is the frequency response?

The transfer function evaluated at s = jω: its magnitude is the gain to a sine, its angle the phase shift.

2. How is magnitude plotted?

As 20 log₁₀|G| in decibels, against a logarithmic frequency axis.

3. What does a pole do to the Bode plot?

Bends the magnitude down 20 dB per decade past its corner and the phase toward −90 degrees.

4. What are the first-order signatures at the corner?

−3 dB and −45 degrees.

5. What is the gain crossover frequency?

Where the magnitude passes through 0 dB (gain of one).

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive a first-order Bode point and a crossover from a blank page.

+3 daysSketch the Bode plot of two new systems.

+7 daysCarry the crossovers into stability margins, Chapter 9.

+30 daysConnect the resonant peak to the damping ratio from Chapter 4.

Textbook mapping

Item	Mapping
Primary source	Ogata, Modern Control Engineering, Chapter 7 (Frequency-Response Method, Bode Diagrams)
Cross-reference	Nise, Ch. 10 · Dorf and Bishop, Ch. 8
Core topics	8.1 Frequency response · 8.2 Decibels and phase · 8.3 Asymptotes and corners · 8.4 First/second-order shapes · 8.5 Gain crossover
Engineering connection	The crossovers found here are where the margins are measured.
Read next	Chapter 9: Stability Margins and Nyquist.