Bode plot — magnitude and phase vs frequency: Worked examples

This page is the "no surprises" drill for Bode plots. We enumerate every kind of case a Bode problem can throw at you, then work one example per case so that when you hit a new problem you can say "ah, that's cell X, I've seen it."

Everything here uses only tools already built in the parent note: the Transfer function $G (s)$ , its frequency response $G (j ω)$ , magnitude in decibels $∣ G ∣_{dB} = 20 lo g_{10} ∣ G (j ω) ∣$ , and phase $∠ G (j ω)$ . If a symbol appears here, it was defined there or is defined below on the spot.

The scenario matrix

Think of a Bode problem as choosing one option from each column. Every example below is tagged with the cell (row label) it covers.

Cell	Case class	What makes it different	Covered by
A	Constant gain only	Flat line, zero phase — the "do-nothing" baseline	Ex 1
B	Single real pole (lag)	One −20 dB/dec corner, phase → −90°	Ex 2
C	Single real zero (lead)	One +20 dB/dec corner, phase → +90°	Ex 3
D	Pole at the origin (integrator)	Degenerate: no corner, slope from $ω \to 0$ , phase pinned at −90°	Ex 4
E	Repeated / high-order factor	Slopes and phases multiply (−40 dB/dec, −180°)	Ex 5
F	Pole and zero together (lead–lag)	Two corners; slope goes flat again between them	Ex 6
G	Gain sign is negative ( $K < 0$ )	Magnitude unchanged, phase shifted by ±180°	Ex 7
H	Limiting behaviour ( $ω \to 0$ and $ω \to \infty$ )	Read off DC gain and final slope without a calculator	Ex 8
I	Real-world word problem	Translate physics → $G (s)$ → margins	Ex 9
J	Exam twist: find crossover & margins numerically	Solve $	L
K	Underdamped 2nd-order (complex poles)	Resonant peak and sharp phase drop of −180°	Ex 11

The columns hidden inside these rows are the sign cases (positive/negative gain, poles vs zeros → phase up vs down) and the degenerate cases (pole at origin, $ω = 0$ , $ω = \infty$ , complex resonance). Every one gets its own worked cell.

Example 1 — Cell A: constant gain

This is the baseline every other plot is measured against: add $12$ dB everywhere and rotate phase by $0$ .

Example 2 — Cell B: single real pole (lag)

Worked example

$G (s) = \frac{1}{1 + s /20}$ , corner $ω_{c} = 20$ rad/s Forecast: guess the dB value exactly at $ω = 20$ , and the phase there.

The figure plots magnitude (dB, blue) against $ω$ on a log axis. The pink dashed lines are the asymptotes: a flat one at $0$ dB below the corner and a $- 20$ dB/dec one above it. The yellow dot marks the true curve at the corner $ω = 20$ , sitting $3$ dB below where the asymptotes cross.

DC value. As $ω \to 0$ , $G \to 1/1 = 1 \Rightarrow 20 lo g_{10} 1 = 0$ dB. Why this step? Below the corner the $(1 + s /20)$ term is $\approx 1$ , so the pole hasn't "kicked in."
At the corner $ω = 20$ : $∣ G ∣ = 1/ 1 + 1^{2} = 1/ 2$ . $20 lo g_{10} (1/ 2) = - 3.01 dB$ Why this step? This is the universal −3 dB corner derived in the parent note.
High frequency $ω ≫ 20$ : $∣ G ∣ \approx 20/ ω$ , slope −20 dB/dec (upper pink dashed line). Why this step? The $s /20$ term dominates, so magnitude falls like $1/ ω$ .
Phase: $∠ G = - arctan (ω /20)$ : $0^{\circ}$ at DC, $- 4 5^{\circ}$ at $ω = 20$ , $\to - 9 0^{\circ}$ high. Why this step? The angle of $1/ (1 + j ω /20)$ is minus the angle of the denominator.

Verify: at $ω = 200$ (one decade above corner), $∣ G ∣_{dB} = 20 lo g_{10} (1/ 1 + 100) = - 20.04$ dB — about $20$ dB below the corner, confirming the −20 dB/dec slope. ✓

Look at the pink dashed asymptotes in the figure: they meet exactly at the corner, and the true curve (blue) sags $3$ dB below that meeting point.

Example 3 — Cell C: single real zero (lead)

Worked example

$G (s) = 1 + s /2$ , corner $ω_{c} = 2$ rad/s Forecast: a zero is the mirror of a pole. Guess the slope above the corner and the phase at $ω = 2$ .

DC value. $ω \to 0 \Rightarrow G \to 1 \Rightarrow 0$ dB. Why? Same as a pole below its corner — the term is $\approx 1$ .
At the corner $ω = 2$ : $∣ G ∣ = 1 + 1 = 2 \Rightarrow + 3.01$ dB. Why this step? A zero adds $+ 3$ dB at its corner — the sign flips versus the pole.
High frequency: slope +20 dB/dec. Why? $∣1 + j ω /2∣ \approx ω /2$ grows linearly.
Phase: $∠ G = + arctan (ω /2)$ : $0^{\circ} \to + 4 5^{\circ}$ at $ω = 2 \to + 9 0^{\circ}$ high. Why this step? The zero is in the numerator, so its angle adds (positive), not subtracts.

Verify: at $ω = 20$ (one decade up), $∣ G ∣_{dB} = 20 lo g_{10} 1 + 100 = + 20.04$ dB — up $20$ dB, confirming +20 dB/dec. ✓ Compare with Ex 2: identical magnitude size, opposite sign. This is the pole↔zero mirror. This is exactly the ingredient of a Lead-lag compensator.

Example 4 — Cell D (degenerate): pole at the origin

Worked example

$G (s) = \frac{50}{s}$ — an integrator with gain Forecast: there is no corner here. Where does the curve cross 0 dB, and what is the phase at every frequency?

The figure plots magnitude (dB, blue) versus $ω$ (log axis, x-labelled "frequency w"). The line is a single straight ramp of slope $- 20$ dB/dec with no flat section — that is the degenerate signature. The yellow dot marks where it crosses $0$ dB at $ω = 50$ ; the pink arrow points at the low-frequency end where the magnitude keeps rising without limit.

Magnitude. $∣ G (j ω) ∣ = 50/ ω$ , so $∣ G ∣_{dB} = 20 lo g_{10} 50 - 20 lo g_{10} ω = 33.98 - 20 lo g_{10} ω$ Why this step? $lo g$ turns the quotient into a difference; the $50$ shifts the whole $- 20$ dB/dec line up.
Crossover (0 dB). Set $∣ G ∣_{dB} = 0$ : $20 lo g_{10} ω = 33.98 \Rightarrow ω = 50$ rad/s. Why this step? $∣ G ∣ = 1$ when $50/ ω = 1$ , i.e. $ω = 50$ — the gain equals the numerator.
Phase. $∠ (50/ j ω) = ∠50 - ∠ (j ω) = 0^{\circ} - 9 0^{\circ} = - 9 0^{\circ}$ , for all $ω$ . Why this step? Dividing by $j ω$ (angle $+ 9 0^{\circ}$ ) subtracts $9 0^{\circ}$ from everything — a flat phase line, the degenerate hallmark of an integrator.

Verify: at $ω = 50$ : $∣ G ∣_{dB} = 33.98 - 20 lo g_{10} 50 = 0.00$ dB. ✓ And $ω = 5$ gives $33.98 - 13.98 = 20$ dB — ten times lower frequency, $20$ dB higher, confirming the slope through the origin case.

The degenerate feature: unlike Ex 2/3 there is no flat low-frequency section — the slope holds all the way down to $ω \to 0$ where the magnitude blows up (a Low-pass filter pole would eventually flatten, an origin pole never does).

Example 5 — Cell E: repeated (high-order) factor

Worked example

$G (s) = \frac{1}{s ^{2}}$ — the double integrator Forecast: two integrators stacked. Guess the slope and the phase.

Magnitude. $∣ G ∣ = 1/ ω^{2} \Rightarrow ∣ G ∣_{dB} = - 40 lo g_{10} ω$ : slope −40 dB/dec. Why this step? Two factors of $1/ (j ω)$ each contribute $- 20$ dB/dec; logs add, so $- 20 - 20 = - 40$ .
Crossover. $∣ G ∣ = 1$ when $ω^{2} = 1 \Rightarrow ω = 1$ rad/s $\Rightarrow$ passes through $0$ dB at $ω = 1$ . Why this step? $lo g_{10} 1 = 0$ .
Phase. $∠ (1/ (j ω)^{2}) = - 9 0^{\circ} - 9 0^{\circ} = - 18 0^{\circ}$ , constant. Why this step? Each pole gives $- 9 0^{\circ}$ ; they add.

Verify: at $ω = 10$ : $∣ G ∣_{dB} = - 40 lo g_{10} 10 = - 40$ dB — one decade up from the crossover, $40$ dB down. ✓

GNC insight: sitting at $- 18 0^{\circ}$ means zero phase margin if the gain crossover lands here — the loop is on the knife-edge. A lead term must inject phase back near crossover.

Example 6 — Cell F: pole and zero (lead–lag shape)

Worked example

$G (s) = \frac{1 + s /1}{1 + s /100}$ — zero at 1, pole at 100 Forecast: two corners. What does the magnitude do between $ω = 1$ and $ω = 100$ ?

The figure has two y-axes: left (blue) is magnitude in dB, right (pink) is phase in degrees, both against $ω$ on a log x-axis. The blue curve is flat at $0$ dB, ramps up at $+ 20$ dB/dec between the two yellow corner lines ( $ω = 1$ and $ω = 100$ ), then flattens at $+ 40$ dB. The pink curve rises to a phase bump and falls back — the pink dot marks its peak at $ω = 10$ .

Below $ω = 1$ : both terms $\approx 1 \Rightarrow 0$ dB, flat. Why? Neither corner active yet.
Between 1 and 100: the zero is active (+20 dB/dec) but the pole is not yet — net slope +20 dB/dec. Why this step? Only the zero's corner has been passed, so only its slope is added.
Above $ω = 100$ : now the pole activates too: $+ 20 - 20 = 0$ dB/dec — flat again, at a higher level. Why this step? Slopes add; zero and pole cancel above both corners.
High-frequency level. As $ω \to \infty$ , $G \to (s /1) / (s /100) = 100 \Rightarrow 20 lo g_{10} 100 = 40$ dB. Why this step? The $s$ terms dominate top and bottom; their ratio is the corner ratio $100/1$ .
Phase. $∠ G = arctan (ω /1) - arctan (ω /100)$ : rises toward $+ 9 0^{\circ}$ near $ω \approx 10$ , then falls back toward $0^{\circ}$ — a phase bump. (Here the zero's corner is $1$ rad/s, so its arctan argument is $ω /1$ , i.e. $ω$ itself.) Why this step? Zero adds phase early; pole cancels it late; peak lies at the geometric mean $1 \cdot 100 = 10$ .

Verify: at the geometric mean $ω = 10$ : $∠ G = arctan (10/1) - arctan (10/100) = 84.2 9^{\circ} - 5.7 1^{\circ} = 78.5 8^{\circ}$ . ✓ This positive phase bump is precisely what a lead compensator provides — see Loop shaping.

Example 7 — Cell G (sign case): negative gain

Worked example

$G (s) = \frac{- 10}{1 + s /8}$ Forecast: the minus sign changes which graph — magnitude or phase?

Magnitude. $∣ - 10∣ = 10$ ; the sign vanishes under absolute value. Split the dB into its two parts: $∣ G ∣_{dB} = flat gain term = 20 dB 20 lo g_{10} 10 - frequency-dependent pole term 20 lo g_{10} 1 + (ω /8)^{2}$ The first term is a constant $20$ dB (the flat region). The second term is $0$ below the corner, $3$ dB at $ω = 8$ , and grows as $20 lo g_{10} (ω /8)$ above it (the −20 dB/dec roll-off). This is identical to the positive-gain version. Why this step? Magnitude is $∣ G ∣$ ; multiplying by $- 1$ doesn't change a length. Separating the constant term from the $ω$ -dependent term makes clear the sign only touches phase.
Phase. $∠ (- 10) = 18 0^{\circ}$ (or equivalently $- 18 0^{\circ}$ ). So $∠ G = - 18 0^{\circ} - arctan (ω /8)$ Why this step? $- 10$ is a negative real number — it points along the negative real axis, angle $18 0^{\circ}$ . The pole's lag then subtracts on top of that.

Verify: at $ω = 8$ : $∠ G = - 18 0^{\circ} - 4 5^{\circ} = - 22 5^{\circ}$ (equivalently $+ 13 5^{\circ}$ ). ✓ The entire phase curve is shifted by $\pm 18 0^{\circ}$ versus the positive-gain case — a subtlety that flips stability conclusions on a Nyquist plot.

Example 8 — Cell H: pure limiting behaviour (no calculator)

Worked example

$G (s) = \frac{100 ( 1 + s /10 )}{s ( 1 + s /1000 )}$ — read off the two ends only Forecast: guess the low-frequency slope and the high-frequency slope before doing any arithmetic.

As $ω \to 0$ . Keep only lowest-order terms: numerator $\to 100$ , denominator $\to s = j ω$ . So $G \to 100/ (j ω)$ : slope −20 dB/dec, phase −90°. Why this step? Near zero frequency, every $(1 + s / corner)$ collapses to $1$ ; only the origin pole survives.
As $ω \to \infty$ . Keep only highest-order terms: numerator $\to 100 s /10 = 10 s$ , denominator $\to s \cdot s /1000 = s^{2} /1000$ . So $G \to 10 s \cdot 1000/ s^{2} = 10000/ s$ : slope −20 dB/dec, phase −90°. Why this step? At huge $ω$ , each $(1 + s / corner) \to s / corner$ ; count the net powers of $s$ : numerator has $s^{1}$ , denominator has $s^{2}$ → net $s^{- 1}$ .

Verify (count of slopes): poles at origin and 1000 (two, each −20), zero at 10 (one, +20): net order at high $ω = - 2 + 1 = - 1 \Rightarrow - 20$ dB/dec. ✓ Both ends match without evaluating a single logarithm — the point of the limiting-value cell.

Example 9 — Cell I: real-world word problem

Worked example Reaction-wheel attitude loop

A spacecraft's attitude responds to torque like a double integrator (angle = force twice-integrated), and a controller adds a lead term. The open loop is $L (s) = \frac{20 ( 1 + s /2 )}{s ^{2}}$ Find the gain-crossover frequency $ω_{g c}$ (where $∣ L ∣ = 1$ ) and the phase margin there. Forecast: without the zero, the plant sits at $- 18 0^{\circ}$ (Ex 5). Does the zero rescue it?

Set up $∣ L ∣ = 1$ . $∣ L ∣ = \frac{20 1 + ( ω /2 ) ^{2}}{ω ^{2}} = 1$ . Why this step? Gain crossover is defined by unit magnitude ( $0$ dB).
Solve numerically. This gives $ω_{g c} \approx 4.61$ rad/s. Why this step? No clean closed form; it's a single-variable root, solved once.
Phase there. $∠ L = arctan (ω /2) - 18 0^{\circ}$ . At $ω = 4.61$ : $arctan (2.305) = 66.5 5^{\circ}$ , so $∠ L = 66.55 - 180 = - 113.4 5^{\circ}$ . Why this step? The double integrator contributes $- 18 0^{\circ}$ ; the zero adds back $+ 66. 6^{\circ}$ .
Phase margin $= 18 0^{\circ} + ∠ L = 180 - 113.45 = 66.5 5^{\circ}$ . Why this step? Definition from the parent note: how much lag remains before $- 18 0^{\circ}$ .

Verify: plug $ω = 4.61$ back: $∣ L ∣ = 20 1 + (2.305)^{2} / (4.61)^{2} = 20 (2.512) /21.25 \approx 1.00$ . ✓ Phase margin $+ 66. 6^{\circ}$ is healthy — the lead zero saved a plant that alone had $0^{\circ}$ margin. See Stability margins.

Example 10 — Cell J: exam twist, both crossovers & both margins

Worked example

$L (s) = \frac{40}{s ( 1 + s /4 ) ( 1 + s /20 )}$ Forecast: guess whether phase margin or gain margin will be the binding one.

Phase crossover $ω_{p c}$ : solve $∠ L = - 18 0^{\circ}$ . $∠ L = - 9 0^{\circ} - arctan (ω /4) - arctan (ω /20) = - 18 0^{\circ}$ So $arctan (ω /4) + arctan (ω /20) = 9 0^{\circ}$ , giving $ω_{p c} = 4 \cdot 20 = 8.944$ rad/s. Why this step? Two arctans summing to $9 0^{\circ}$ means their arguments are reciprocals: $tan α$ and $tan (9 0^{\circ} - α) = 1/ tan α$ multiply to $1$ . Here $(ω /4) (ω /20) = 1 \Rightarrow ω^{2} = 80 \Rightarrow ω = 80$ , the geometric mean of the two corner frequencies $4$ and $20$ . That is why the $- 18 0^{\circ}$ phase point of two lags always sits at the geometric mean of their corners (plus the $- 9 0^{\circ}$ from the origin pole gets us exactly to $- 18 0^{\circ}$ ).
Gain margin $= - ∣ L (j ω_{p c}) ∣_{dB}$ . $∣ L (j 8.944) ∣ = \frac{40}{8.944 1 + ( 8.944/4 ) ^{2} 1 + ( 8.944/20 ) ^{2}}$ $= 40/ (8.944 \cdot 2.449 \cdot 1.096) = 1.667$ , so $∣ L ∣_{dB} = 20 lo g_{10} 1.667 = 4.44$ dB, giving gain margin $= - 4.44$ dB. Why this step? At $ω_{p c}$ the magnitude exceeds $0$ dB, so raising gain isn't the danger — the loop is already past the safe point. A negative gain margin means unstable.
Gain crossover $ω_{g c}$ : solve $∣ L ∣ = 1$ numerically $\Rightarrow ω_{g c} \approx 12.9$ rad/s. Why this step? Gain crossover is where $∣ L ∣ = 1$ ; with three poles rolling off, we solve the single magnitude equation numerically.
Phase margin $= 18 0^{\circ} + ∠ L (j ω_{g c})$ . At $ω = 12.9$ : $∠ L = - 90 - arctan (3.225) - arctan (0.645) = - 90 - 72.79 - 32.83 = - 195. 6^{\circ}$ , so phase margin $= 180 - 195.6 = - 15. 6^{\circ}$ (negative → unstable). Why this step? Both margins agree: this loop is unstable and needs its gain lowered (or a lead added).

Verify: $ω_{p c} = 80 = 8.944$ . ✓ $∣ L (j ω_{p c}) ∣ = 1.667$ . ✓ Both margins negative — consistent stability verdict. The fix: reduce the $40$ so gain crossover drops below $ω_{p c}$ , restoring positive margins.

Example 11 — Cell K: underdamped second-order (complex poles)

Worked example

$G (s) = \frac{ω _{n}^{2}}{s ^{2} + 2 ζ ω _{n} s + ω _{n}^{2}}$ with $ω_{n} = 10$ rad/s, $ζ = 0.1$ New symbols (built here): $ω_{n}$ is the natural frequency — where the two poles "want" to oscillate; $ζ$ (zeta) is the damping ratio — how strongly oscillation is suppressed ( $ζ = 1$ critically damped, $ζ \to 0$ pure ringing). When $ζ < 1$ the two poles are a complex-conjugate pair, and the magnitude curve grows a resonant peak instead of a clean corner. Forecast: with $ζ = 0.1$ (very light damping), guess how tall the peak is and where it sits.

The figure has magnitude (dB, blue, left) and phase (deg, pink, right) versus $ω$ on a log axis. Notice the blue curve does not just bend at $ω_{n} = 10$ — it spikes up to a sharp yellow-dotted peak near $ω_{n}$ , then rolls off at $- 40$ dB/dec. The pink phase drops through $- 9 0^{\circ}$ at $ω_{n}$ and swings all the way to $- 18 0^{\circ}$ , and the drop is steep because damping is small.

Low frequency $ω ≪ ω_{n}$ : $G \to ω_{n}^{2} / ω_{n}^{2} = 1 \Rightarrow 0$ dB, flat. Why this step? The $s^{2}$ and $2 ζ ω_{n} s$ terms are negligible below $ω_{n}$ ; only $ω_{n}^{2} / ω_{n}^{2}$ remains.
High frequency $ω ≫ ω_{n}$ : $G \to ω_{n}^{2} / (- ω^{2})$ , slope −40 dB/dec, phase −180°. Why this step? Two poles roll off together — same as the double integrator, but shifted up.
The peak. Evaluate at $ω = ω_{n}$ : denominator $= (j ω_{n})^{2} + 2 ζ ω_{n} (j ω_{n}) + ω_{n}^{2} = - ω_{n}^{2} + 2 ζ ω_{n}^{2} j + ω_{n}^{2} = 2 ζ ω_{n}^{2} j$ . So $∣ G (j ω_{n}) ∣ = \frac{ω _{n}^{2}}{2 ζ ω _{n}^{2}} = \frac{1}{2 ζ} = \frac{1}{0.2} = 5 \Rightarrow 20 lo g_{10} 5 = 13.98 dB$ Why this step? The real parts of the denominator cancel exactly at $ω_{n}$ , leaving only the tiny imaginary damping term $2 ζ ω_{n}^{2}$ — a small denominator means a big magnitude. That is the resonant peak. The lighter the damping, the taller: peak $\approx 1/ (2 ζ)$ .
Phase. At $ω_{n}$ the denominator is purely imaginary ( $+ j$ ), so $∠ G = - 9 0^{\circ}$ . It sweeps $0^{\circ} \to - 9 0^{\circ} \to - 18 0^{\circ}$ , and the sweep is sharp for small $ζ$ . Why this step? Complex poles rotate the phase through $- 18 0^{\circ}$ over a narrow band around $ω_{n}$ .

Verify: peak $= 1/ (2 \cdot 0.1) = 5$ , i.e. $20 lo g_{10} 5 = 13.98$ dB. ✓ (The true peak is at $ω_{r} = ω_{n} 1 - 2 ζ^{2} \approx 9.9$ rad/s, essentially at $ω_{n}$ for small $ζ$ .) GNC insight: a lightly damped structural mode (fuel slosh, flexible boom) shows up as exactly this spike — if the loop gain there exceeds $0$ dB, the loop can drive the resonance unstable. Loop shaping notches it out.

Recall

Recall Which cell is which?

Negative gain changes the phase (by ±180°), not the magnitude ::: magnitude is $∣ G ∣$ , unaffected by sign A pole at the origin has no corner and constant −90° phase ::: it never flattens as $ω \to 0$ Between a low zero and a high pole the slope is +20 dB/dec, then flat again above the pole ::: slopes add Two arctans sum to 90° when ::: the product of their arguments equals 1 (geometric-mean frequency) A negative gain margin means ::: the loop is unstable — gain already exceeds the safe level at −180° An underdamped pole pair peaks at height ::: about $1/ (2 ζ)$ near $ω_{n}$