Gain margin, phase margin — stability margins: Worked examples

This page is the drill hall for stability margins. The parent note built the two rulers — gain margin (GM) and phase margin (PM) — from the critical point $- 1$ . Here we throw every kind of case at those rulers so that no exam question, no sign, no degenerate loop can surprise you.

Before we start, three reminders in plain words so nobody is lost from line one:

The scenario matrix

Every question this topic can ask is one of these cells. Each worked example below is tagged with the cell(s) it covers.

Cell	What makes it distinct	Example
A. Clean GM	Loop has a real phase crossover; compute GM	Ex 1
B. Clean PM	Loop crosses $	L
C. Both margins together	Same loop, find GM and PM	Ex 3
D. Infinite GM (degenerate)	Phase never reaches $- 18 0^{\circ}$ ⟹ GM $= \infty$	Ex 4
E. Negative margin (unstable)	Gain pushed too high ⟹ GM $< 0$ dB, PM $< 0$	Ex 5
F. Time-delay twist	Delay eats phase margin; find $τ_{ma x}$	Ex 6
G. Design / word problem	Choose $K$ to hit a target PM (link to overshoot)	Ex 7
H. Exam trap	Conditionally stable / sign confusion	Ex 8

Worked example Ex 1 — Cell A: clean gain margin

Statement. For $L (s) = \frac{4}{s ( s + 1 ) ( s + 2 )}$ , find the gain margin in dB.

Forecast: Guess — is GM here bigger or smaller than Worked Example 1 in the parent (where $K = 1$ gave GM $= 6$ )? Write your guess before reading on.

Find $ω_{p c}$ (where $∠ L = - 18 0^{\circ}$ ). $∠ L = - 9 0^{\circ} - arctan ω - arctan (ω /2)$ . Setting $= - 18 0^{\circ}$ gives $arctan ω + arctan (ω /2) = 9 0^{\circ}$ . Why this step? GM must be read exactly where the phase already sits at the danger angle — only extra size is then needed to reach $- 1$ . The tangent of a sum blows up (to $\infty$ ) when the two arctans add to $9 0^{\circ}$ ; using $tan (a + b) = \frac{t a n a + t a n b}{1 - t a n a t a n b}$ , the denominator $1 - ω \cdot \frac{ω}{2} = 0$ , so $ω_{p c} = 2$ .
Size there. $∣ L ∣ = \frac{4}{ω ω ^{2} + 1 ω ^{2} + 4}$ . At $ω = 2$ : denominator $= 2 \cdot 3 \cdot 6 = 6$ , so $∣ L ∣ = \frac{4}{6} = \frac{2}{3}$ . Why this step? $∣ L ∣$ at $ω_{p c}$ is the one number GM depends on.
GM $= \frac{1}{∣ L ∣} = \frac{3}{2} = 1.5$ , i.e. $20 lo g_{10} 1.5 \approx 3.52$ dB. Why this step? We invert $∣ L ∣$ because GM is the factor you can multiply the gain by before $∣ L ∣$ reaches $1$ at the danger phase: if $∣ L ∣ = \frac{2}{3}$ , multiplying by $\frac{3}{2}$ lands it exactly on $1$ (i.e. on $- 1$ ). Bigger $∣ L ∣$ ⟹ closer to disaster ⟹ smaller GM, which is exactly what dividing does.

Verify: The gain went up $4 \times$ vs the parent's $K = 1$ (GM $6$ ). Multiplying $K$ scales $∣ L ∣$ up, so GM must divide by $4$ : $6/4 = 1.5$ . ✓ Matches — and it is smaller, as bigger gain shrinks GM.

Worked example Ex 2 — Cell B: clean phase margin

Statement. For $L (s) = \frac{1}{s ( s + 1 )}$ , find the phase margin (repeat of parent Ex 2 but we verify by figure).

Forecast: The parent got PM $\approx 51. 8^{\circ}$ . Is that "healthy"? Guess a yes/no.

Gain crossover. $∣ L ∣ = \frac{1}{ω ω ^{2} + 1} = 1 \Rightarrow ω^{4} + ω^{2} - 1 = 0$ . Let $u = ω^{2}$ : $u = \frac{- 1 + 5}{2} \approx 0.618$ , so $ω_{g c} = 0.618 \approx 0.786$ . Why this step? PM is read where the size is exactly $1$ ; only extra phase lag is then needed to reach $- 18 0^{\circ}$ .
Phase there. $∠ L = - 9 0^{\circ} - arctan (0.786) = - 9 0^{\circ} - 38. 2^{\circ} = - 128. 2^{\circ}$ .
PM $= 18 0^{\circ} + (- 128. 2^{\circ}) = 51. 8^{\circ}$ .

Look at the figure: the loop curve crosses the unit circle (size $= 1$ ) at the amber dot; the amber wedge from that ray back to the $- 18 0^{\circ}$ axis is the phase margin.

Verify: $tan (38. 2^{\circ}) = 0.787 \approx ω_{g c}$ . ✓ And $3 0^{\circ}$ – $6 0^{\circ}$ is healthy, so yes. ✓

Worked example Ex 3 — Cell C: both margins on one loop

Statement. For $L (s) = \frac{2}{s ( s + 1 ) ( s + 2 )}$ , find both GM (dB) and PM.

Forecast: With $K = 2$ (double the parent's $K = 1$ ), guess whether both margins are still positive.

GM. Same $ω_{p c} = 2$ (phase does not depend on $K$ ). $∣ L ∣ = \frac{2}{6} = \frac{1}{3}$ , so GM $= 3 = 20 lo g_{10} 3 \approx 9.54$ dB. Why this step? $K$ only slides the size curve; the danger frequency is unchanged.
PM — find $ω_{g c}$ . Solve $∣ L ∣ = \frac{2}{ω ω ^{2} + 1 ω ^{2} + 4} = 1$ . Numerically $ω_{g c} \approx 0.749$ . Why this step? Size $= 1$ is where PM lives.
Phase there. $∠ L = - 9 0^{\circ} - arctan (0.749) - arctan (0.749/2) = - 9 0^{\circ} - 36.8 3^{\circ} - 20.5 3^{\circ} = - 147.3 6^{\circ}$ .
PM $= 18 0^{\circ} - 147.3 6^{\circ} = 32. 6^{\circ}$ .

Verify: Both positive ⟹ stable, consistent with GM $= 3 > 1$ . Sanity on step 2: plug $ω = 0.749$ into $∣ L ∣$ ⟹ $\approx 1.00$ . ✓ (See Bode plots to read these two numbers straight off the two stacked graphs.)

Worked example Ex 4 — Cell D: infinite gain margin (degenerate)

Statement. For the first-order loop $L (s) = \frac{5}{s + 1}$ , find GM and PM.

Forecast: How much extra gain before instability? Guess "some finite factor" vs "unlimited."

Phase range. $∠ L = - arctan ω$ . As $ω : 0 \to \infty$ , $∠ L$ goes from $0^{\circ}$ down toward $- 9 0^{\circ}$ but never reaches $- 18 0^{\circ}$ . Why this step? GM lives at $- 18 0^{\circ}$ — if the phase never gets there, there is no $ω_{p c}$ .
GM $= \infty$ (or " $\infty$ dB"): no finite gain multiple can ever push a curve that stops at $- 9 0^{\circ}$ onto the negative-real axis $- 1$ .
PM. Size $∣ L ∣ = \frac{5}{ω ^{2} + 1} = 1 \Rightarrow ω^{2} + 1 = 25 \Rightarrow ω_{g c} = 24 \approx 4.899$ . Phase $= - arctan (4.899) = - 78.4 6^{\circ}$ . PM $= 18 0^{\circ} - 78.4 6^{\circ} = 101. 5^{\circ}$ .

Verify: A single pole can never make a simple loop unstable by gain alone — classic result, GM $= \infty$ . ✓ PM huge ( $> 9 0^{\circ}$ ) ⟹ extremely damped, matches Root locus staying in the left half-plane for all $K$ . ✓

Worked example Ex 5 — Cell E: negative margin (already unstable)

Statement. Take Ex 1's plant but crank the gain: $L (s) = \frac{12}{s ( s + 1 ) ( s + 2 )}$ . Find GM and PM; is the loop stable?

Forecast: We know $K = 6$ is the exact edge (parent Ex 1: GM $= 6$ at $K = 1$ ). We chose $K = 12$ . Guess the sign of GM.

GM. $ω_{p c} = 2$ unchanged. $∣ L ∣ = \frac{12}{6} = 2$ , so GM $= \frac{1}{2} = 0.5 = 20 lo g_{10} 0.5 \approx - 6.02$ dB. Why this step? GM $< 1$ (negative dB) is the algebraic signature of instability: the curve has already swallowed $- 1$ .
PM. Solve $∣ L ∣ = 1$ : $\frac{12}{ω ω ^{2} + 1 ω ^{2} + 4} = 1 \Rightarrow ω_{g c} \approx 1.96$ . Phase $= - 9 0^{\circ} - arctan (1.96) - arctan (0.98) = - 9 0^{\circ} - 62.9 7^{\circ} - 44.4 2^{\circ} = - 197. 4^{\circ}$ . PM $= 18 0^{\circ} - 197. 4^{\circ} \approx - 17. 4^{\circ}$ . Why this step? Negative PM confirms the phase has already crossed the danger angle before the gain drops to $1$ .
Verdict. GM $< 0$ dB and PM $< 0$ ⟹ unstable.

Verify: Critical gain is $K = 6$ (GM $= 1$ ). We used $12 = 2 \times 6$ , so GM must be $1/2$ . ✓ Both margins negative ⟹ the Nyquist stability criterion would show $- 1$ encircled. ✓

Worked example Ex 6 — Cell F: the time-delay twist

Statement. The stable loop $L (s) = \frac{1}{s ( s + 1 )}$ (Ex 2: PM $= 51. 8^{\circ}$ , $ω_{g c} = 0.786$ ) gets a transport delay $τ$ (e.g. a communication lag). What is the largest $τ$ that keeps it stable?

Forecast: Delay adds only phase lag, no size change. Guess whether $τ_{ma x}$ is a fraction of a second or several seconds.

What a delay does. A delay $e^{- s τ}$ has size $1$ (never touches the magnitude curve) but adds phase $- ω τ$ radians. Why this step? Because size is untouched, $ω_{g c}$ stays $0.786$ — the extra lag simply eats into PM there.
Eat all the margin. Instability at the moment the added lag equals the whole PM: $τ_{ma x} = \frac{PM in radians}{ω _{g c}} .$ PM $= 51. 8^{\circ} = 51.8 \times \frac{π}{180} = 0.9040$ rad.
Compute. $τ_{ma x} = \frac{0.9040}{0.786} = 1.150$ s.

The figure shows the amber PM wedge being nibbled away by the delay's growing lag until it vanishes.

Verify: Units: $\frac{rad}{rad/s} = s$ . ✓ Numeric: $0.786 \times 1.150 = 0.904$ rad $= 51. 8^{\circ}$ of extra lag = exactly the PM. ✓ See Time delay and Padé approximation for approximating $e^{- s τ}$ as a rational function.

Worked example Ex 7 — Cell G: design / word problem (choose the gain)

Statement. A satellite reaction-wheel loop is modelled $L (s) = \frac{K}{s ( s + 2 )}$ . Mission spec: PM $= 4 5^{\circ}$ (for a well-damped, low-overshoot response). Find $K$ .

Forecast: There is a handy rule (justified below) that PM in degrees divided by $100$ estimates the damping ratio $ζ$ , so PM $= 4 5^{\circ}$ targets $ζ \approx 0.45$ . Guess whether the required $K$ is above or below $4$ .

Impose PM $= 4 5^{\circ}$ . PM $= 18 0^{\circ} + ∠ L (j ω_{g c}) = 4 5^{\circ} \Rightarrow ∠ L (j ω_{g c}) = - 13 5^{\circ}$ . $∠ L = - 9 0^{\circ} - arctan (ω /2) = - 13 5^{\circ} \Rightarrow arctan (ω /2) = 4 5^{\circ} \Rightarrow ω_{g c} = 2$ . Why this step? Phase does not depend on $K$ , so the target angle fixes $ω_{g c}$ first.
Impose size $= 1$ at that $ω_{g c}$ . $∣ L ∣ = \frac{K}{ω ω ^{2} + 4} = 1$ at $ω = 2$ : $\frac{K}{2 \cdot 8} = 1 \Rightarrow K = 28 = 42 \approx 5.657$ . Why this step? Now use $K$ (the only free knob) to slide the size curve so unity gain lands exactly at $ω_{g c} = 2$ .

Verify: Check PM at $K = 5.657$ : $ω_{g c} = 2$ , phase $= - 9 0^{\circ} - arctan (1) = - 13 5^{\circ}$ , PM $= 4 5^{\circ}$ . ✓ Expected overshoot from $ζ \approx 0.45$ is roughly $20%$ (see Damping ratio and overshoot); to trim overshoot without losing speed, reshape with PID controller tuning or Loop shaping. $K \approx 5.66 > 4$ , matching the forecast direction.

Worked example Ex 8 — Cell H: the exam trap (sign / "which frequency" confusion)

Statement. A student writes: "For $L (s) = \frac{10}{s ( s + 1 ) ( s + 2 )}$ , $∠ L (j ω_{g c}) = - 15 0^{\circ}$ , therefore PM $= - 15 0^{\circ}$ and the system is unstable." Find the true PM and expose the error.

Forecast: Guess whether $- 15 0^{\circ}$ of phase means a positive or negative PM.

Spot the trap. PM is not the phase itself; it is the gap to $- 18 0^{\circ}$ : $PM = 18 0^{\circ} + ∠ L (j ω_{g c})$ . Why this step? The most common mistake (parent's Mistake #4) is reading $∠ L$ as PM.
Correct value. PM $= 18 0^{\circ} + (- 15 0^{\circ}) = + 3 0^{\circ}$ . Positive ⟹ stable (barely; ringing likely).
Second trap — which frequency? If the student had used $ω_{p c}$ (where $∠ = - 18 0^{\circ}$ ) to read PM, they'd get $0^{\circ}$ wrongly. Remember the mnemonic "GaP": margin is measured where the other quantity hits its critical value.

Verify: $180 + (- 150) = 30 > 0$ ⟹ stable, contradicting the student. ✓ A $- 15 0^{\circ}$ phase is less lag than the $- 18 0^{\circ}$ danger point, so there must still be positive margin — geometry confirms the algebra.

Recall Rapid self-test

Which margin do you read at the phase crossover? ::: Gain margin (GM $= 1/∣ L (j ω_{p c}) ∣$ ). A first-order loop $\frac{K}{s + 1}$ has what gain margin? ::: Infinite — phase never reaches $- 18 0^{\circ}$ . A delay adds size or phase? ::: Only phase, $- ω τ$ ; size stays $1$ . If PM $\approx - 17. 4^{\circ}$ and GM $= - 6$ dB, is the loop stable? ::: No — both negative means already unstable. To hit PM $= 4 5^{\circ}$ you first fix which quantity? ::: The gain-crossover frequency, from the target phase (phase is independent of $K$ ). Student says PM $= ∠ L (j ω_{g c})$ . Fix? ::: PM $= 18 0^{\circ} + ∠ L (j ω_{g c})$ — the gap, not the angle.