Exercises — Gain margin, phase margin — stability margins

WHAT we do: plug straight into the definition. GM is the factor you can multiply the loop gain by before $|L|$ reaches $1$ at the dangerous phase, so it is the reciprocal of the magnitude there. $\text{GM}=\frac{1}{0.25}=4.$ In dB: the decibel is just $20{\log }_{10}$ of a magnitude ratio (a compressed ruler so huge and tiny numbers fit on one axis — the same ruler Bode plots use). ${\text{GM}}_{\text{dB}}=20{\log }_{10}(4)\approx 12.04\text{dB}.$ Sanity: GM $>1$ (positive dB) ⟹ stable margin. You could quadruple the gain before touching $-1$.

WHAT & WHY: PM is the gap from the current phase up to the danger angle $-180^{\circ }$. It is not the angle itself (classic trap — see below). $\text{PM}=180^{\circ }+(-120^{\circ })=60^{\circ }.$ Positive ⟹ stable, and $60^{\circ }$ sits at the top of the healthy band $30^{\circ }$–$60^{\circ }$.

False. Gain margin lives at the phase crossover ($\angle L=-180^{\circ }$), because only there is the loop already pointing at $-1$'s direction, so a pure gain boost is all that's left to reach it. (PM is the one read at $|L|=1$.)

WHY $K$ drops out: a constant gain $K>0$ scales magnitude but adds zero phase. Phase crossover is defined purely by the phase equation, so $K$ is invisible here. The phase: each factor contributes an angle. A pure integrator $1/s$ gives $-90^{\circ }$; each $(s+a)$ in the denominator gives $-\arctan (\omega /a)$. $\angle L=-90^{\circ }-\arctan \omega -\arctan \frac{\omega }{2}=-180^{\circ }.$ So $\arctan \omega +\arctan \frac{\omega }{2}=90^{\circ }$. When two arctangents sum to $90^{\circ }$, their tangents are reciprocals, i.e. $\omega \cdot \frac{\omega }{2}=1$: $\frac{{\omega }^2}{2}=1\Rightarrow {\omega }_{pc}=\sqrt{2}\approx 1.414\text{rad/s}.$ See the phase curve in Figure s01 crossing $-180^{\circ }$ at exactly $\sqrt{2}$.

WHAT: magnitude at ${\omega }_{pc}=\sqrt{2}$. Magnitude multiplies factor-by-factor: $|L|=\frac{K}{\omega \sqrt{{\omega }^2+1}\sqrt{{\omega }^2+4}}.$ At ${\omega }^2=2$: $\sqrt{{\omega }^2+1}=\sqrt{3},\sqrt{{\omega }^2+4}=\sqrt{6},\omega =\sqrt{2}$. $|L|=\frac{4}{\sqrt{2}\cdot \sqrt{3}\cdot \sqrt{6}}=\frac{4}{\sqrt{36}}=\frac{4}{6}=\frac{2}{3}.$ $\text{GM}=\frac{1}{2/3}=\frac{3}{2}=1.5,{\text{GM}}_{\text{dB}}=20{\log }_{10}(1.5)\approx 3.52\text{dB}.$ Contrast with the parent note (where $K=1$ gave GM $=6$, i.e. $15.6$ dB): quadrupling $K$ shrank GM by exactly a factor $4$ ($15.6-12.04\approx 3.5$ dB). Gain and margin trade off directly.

Rule of thumb: $\zeta \approx \frac{{\text{PM}}^{\circ }}{100}$ for PM up to $\sim 60^{\circ }$. $\zeta \approx \frac{51.8}{100}\approx \mathrm{0.52.}$ A damping ratio near $0.5$ means moderate, well-controlled overshoot — see Damping ratio and overshoot. For a standard second-order system, $\zeta =0.5$ gives roughly $16\%$ overshoot, a fast but not wild response.

WHAT: marginal instability means the curve exactly touches $-1$, so $|L(j{\omega }_{pc})|=1$ with $K=K_{\max }$. From Problem 2.2, at ${\omega }_{pc}=\sqrt{2}$, $|L|=K/6$. Setting this to $1$: $\frac{K_{\max }}{6}=1\Rightarrow K_{\max }=6.$ Interpretation: at $K=6$ the loop sustains a steady oscillation at $\omega =\sqrt{2}$ rad/s. Any $K>6$ ⟹ GM $<1$ ⟹ unstable. This is exactly the gain the Root locus branches cross the imaginary axis, and the Nyquist stability criterion would show the curve passing through $-1$.

WHY a delay is dangerous: a pure delay $e^{-j\omega \tau }$ has magnitude $1$ (it never changes gain) but adds phase $-\omega \tau$ — a phase lag that grows with frequency. It eats directly into the phase margin without warning on the magnitude plot. This is the Time delay and Padé approximation story. Convert PM to radians: $40^{\circ }\times \frac{\pi }{180}=0.698$ rad. Max delay: the delay's phase lag at the gain crossover must not exceed the margin: ${\omega }_{gc}{\tau }_{\max }=\text{PM (rad)}\Rightarrow {\tau }_{\max }=\frac{0.698}{2}\approx 0.349\text{s}.$ Beyond $0.349$ s the extra lag pushes the phase past $-180^{\circ }$ at unity gain ⟹ PM goes negative ⟹ unstable.

Loop B is far more robust. Gain margin only measures safety along one direction (pure gain scaling toward $-1$). A loop can have a comfortable GM yet pass close to $-1$ from a different angle — that closeness is what PM captures. Loop A's PM $=15^{\circ }$ ⟹ $\zeta \approx 0.15$ ⟹ heavy ringing and large overshoot; the Nyquist stability criterion plot skims $-1$. Loop B's PM $=50^{\circ }$ ⟹ $\zeta \approx 0.5$ ⟹ smooth response. Moral (parent mistake #2 made concrete): you need both margins healthy; either alone can lie.

Set up the phase condition first (it fixes ${\omega }_{gc}$ independent of $K$'s value once we demand a target angle): At gain crossover we need $\angle L(j{\omega }_{gc})=-180^{\circ }+45^{\circ }=-135^{\circ }$. $\angle L=-90^{\circ }-\arctan \frac{\omega }{2}=-135^{\circ }\Rightarrow \arctan \frac{\omega }{2}=45^{\circ }\Rightarrow \frac{\omega }{2}=1.$ So ${\omega }_{gc}=2$ rad/s. Now enforce $|L|=1$ at that frequency to pin down $K$: $|L|=\frac{K}{\omega \sqrt{{\omega }^2+4}}=\frac{K}{2\sqrt{8}}=\frac{K}{2\cdot 2.828}=1.$ $K=2\sqrt{8}=4\sqrt{2}\approx \mathrm{5.657.}$ Check: $\zeta \approx \text{PM}/100=0.45$ — nicely damped, small overshoot. This is a mini PID controller tuning step: we picked a pure gain to hit a phase-margin spec.

Find ${\omega }_{pc}$: $\angle L=-90^{\circ }-\arctan (\omega /2)=-180^{\circ }$ demands $\arctan (\omega /2)=90^{\circ }$, i.e. $\omega \to \infty$. Interpret: the phase only approaches $-180^{\circ }$ as $\omega \to \infty$; it never actually reaches it for a finite frequency. As $\omega \to \infty$, $|L|\to 0$. $\text{GM}=\frac{1}{|L(j\infty )|}=\frac{1}{0}=\infty .$ Meaning: infinite gain margin — you can multiply $K$ by any factor and this two-pole loop never becomes unstable (its phase can't quite reach $-180^{\circ }$). All the design freedom here lives in the phase margin. See Figure s02 for the two Nyquist shapes side by side.

Gain-margin constraint. From Problem 2.2's pattern, $|L(j\sqrt{2})|=K/6$, so $\text{GM}=6/K$. Require $\text{GM}\ge 2$: $6/K\ge 2\Rightarrow K\le 3.$ Check PM at $K=2$. First find ${\omega }_{gc}$: solve $|L|=1$: $\frac{2}{\omega \sqrt{{\omega }^2+1}\sqrt{{\omega }^2+4}}=1.$ Numerically ${\omega }_{gc}\approx 0.749$ rad/s (verified in VERIFY). Phase there: $\angle L=-90^{\circ }-\arctan (0.749)-\arctan (0.749/2).$ $\arctan (0.749)=36.83^{\circ },\arctan (0.3745)=20.53^{\circ }.$ $\angle L=-90-36.83-20.53=-147.36^{\circ },\text{PM}=180-147.36=32.6^{\circ }.$ PM $=32.6^{\circ }<40^{\circ }$ fails! Even though $K=2$ passes the gain-margin test (GM $=3=9.5$ dB), it violates the phase-margin spec. Conclusion: the phase margin binds first. We must lower $K$ further. This is a textbook Loop shaping insight: gain and phase specs can pull in opposite directions, and you honour the tighter one.

WHAT: we scan $K$ downward until PM rises to $40^{\circ }$. Lowering $K$ pulls the magnitude curve down, moving ${\omega }_{gc}$ to a lower frequency where the phase is less negative ⟹ larger PM. Numerically, PM $=40^{\circ }$ occurs near $K\approx 1.54$ (VERIFY confirms PM $\approx 40^{\circ }$ there). Final range (both specs, GM $\le 3$ and PM $\ge 40^{\circ }$): $0<K\lesssim \mathrm{1.54.}$ The gain-margin ceiling $K\le 3$ is never the active limit — the phase margin caps us at roughly $1.54$. A robust practical pick like $K=1$ (PM $\approx 45^{\circ }$, GM $=6$) sits comfortably inside.

1.1 GM $=4=12.04$ dB · 1.2 PM $=60^{\circ }$ · 1.3 phase crossover. 2.1 ${\omega }_{pc}=\sqrt{2}$ · 2.2 GM $=1.5=3.52$ dB · 2.3 $\zeta \approx 0.52$. 3.1 $K_{\max }=6$ · 3.2 ${\tau }_{\max }\approx 0.349$ s · 3.3 Loop B robust. 4.1 $K=4\sqrt{2}\approx 5.657,{\omega }_{gc}=2$ · 4.2 GM $=\infty$. 5.1 GM binds at $K\le 3$ but PM fails at $K=2$ ($32.6^{\circ }$) · 5.2 usable $K\lesssim 1.54$.