Part 7 · Chapter 29

Nonlinear Control Systems

Every method in this book so far has rested on one assumption — that doubling the input doubles the output — and every real actuator violates it; this chapter asks what survives when superposition goes, and answers with three tools that need no transfer function at all: the phase plane, which draws the motion; the describing function, which recovers a Nyquist test for the oscillations that linear systems cannot have; and Lyapunov's direct method, which settles stability from an energy argument alone.

Control Systems Prof. Mithun Mondal Reading time ≈ 50 min
i What you'll learn
  • The four behaviours a nonlinear system shows and a linear one cannot: amplitude-dependent response, multiple equilibria, limit cycles, and jump resonance.
  • The standard hard nonlinearities — saturation, dead zone, backlash, relay — and which of them carry memory.
  • How linearisation settles local stability through the Jacobian, and the two cases where it says nothing at all.
  • Phase-plane analysis: trajectories, the six types of singular point, and reading a limit cycle off the portrait.
  • The describing function \(N(A)\) as an amplitude-dependent gain, derived for saturation and for a relay.
  • How the intersection of \(G(j\omega)\) with the \(-1/N(A)\) locus predicts a limit cycle's amplitude and frequency — and decides whether it is stable.
  • Lyapunov's direct method, the Lyapunov equation \(A^{T}P+PA=-Q\), and why failing to find a \(V\) proves nothing.
Section 29-1

Where Linear Theory Runs Out

Twenty-eight chapters have been built on one property. Chapter 3 defined the transfer function by assuming that the response to a sum of inputs is the sum of the responses; Chapter 14 defined the frequency response by assuming that a sinusoid in produces a sinusoid out at the same frequency; the root locus, the Nyquist criterion and the state transition matrix all inherit that assumption. Superposition is not one tool among many — it is the foundation the entire edifice stands on.

No physical system obeys it. An amplifier clips. A valve cannot open beyond its stop. A gear train has lash. Static friction resists motion until it suddenly does not. These are not modelling refinements to be added later; they are the reason a controller that looks perfect on paper hunts, sticks or oscillates on the bench. When superposition fails, four behaviours become possible that no linear model can produce.

BehaviourWhat it looks likeWhy linear systems cannot do it
Amplitude-dependent responseA loop is well damped for small commands and rings badly for large onesA linear system's poles do not depend on signal size
Multiple equilibriaThe system rests at several different states depending on where it started\(A\mathbf{x}=0\) has one solution unless \(A\) is singular, and then a whole subspace
Limit cyclesA self-sustained oscillation of fixed amplitude and frequency, reached from any nearby startLinear oscillation requires poles exactly on the axis, and its amplitude is set by the initial condition
Jump resonance and subharmonicsThe frequency-response magnitude jumps discontinuously as \(\omega\) is swept; output contains \(\omega/2\), \(\omega/3\)A linear system's steady-state output contains only the input frequency

The limit cycle deserves particular attention, because it is the phenomenon this chapter spends most of its effort on. In a linear system a sustained oscillation requires a pole pair sitting exactly on the imaginary axis — an infinitely improbable condition — and even then the amplitude is whatever the initial condition made it. A limit cycle is different in kind: it is an isolated closed trajectory that nearby motions converge to. Its amplitude and frequency are properties of the system, not of how it was started. That is why a relay-controlled thermostat always hunts with the same swing, and why a servo with backlash always buzzes at the same frequency however it is disturbed.

Nonlinearity is not always the enemy. A relay is the cheapest actuator there is, and bang-bang control is provably time-optimal for a double integrator — no smooth controller can beat it. Deliberate dead zones suppress chatter around a set point; deliberate saturation protects hardware; deliberate nonlinear damping gives a fast approach with a gentle arrival. The distinction worth making is between incidental nonlinearities, which the plant forces on you, and intentional ones, which you introduce because they do something linear elements cannot.
Section 29-2

The Common Nonlinearities

In practice a small catalogue covers most of what is met. Each is described by a static input–output characteristic \(y=f(x)\), and the single most useful thing to notice about any of them is whether that characteristic is single-valued or multi-valued. A single-valued characteristic — saturation, dead zone, an ideal relay — assigns one output to each input, and the corresponding describing function of Section 29-5 turns out to be a real number: pure amplitude-dependent gain, no phase shift. A multi-valued one — backlash, a relay with hysteresis — depends on the direction of travel, which is to say it has memory, and its describing function is complex: it contributes phase lag, and phase lag is what destabilises loops.

x y S −S slope K
Saturation — single-valued
x y δ −δ slope K
Dead zone — single-valued
x y h −h +M −M
Relay with hysteresis — multi-valued
NonlinearityWhere it comes fromEffect on the loop
SaturationAmplifier rails, valve stops, torque limitsEffective gain falls with amplitude; large-signal response slows; integral windup
Dead zoneValve overlap, stiction, deliberate insensitivityEffective gain falls with small amplitude; leaves a residual steady-state error
BacklashGear lash, mechanical playMulti-valued: contributes up to \(90^\circ\) of phase lag and readily sustains a limit cycle
Relay (on–off)Contactors, thermostats, thruster valvesInfinite small-signal gain; almost always produces a limit cycle
Relay with dead zone or hysteresisContact chatter suppression, Schmitt triggersDead zone reduces the effective gain; hysteresis adds phase lag and lowers the hunting frequency
Coulomb frictionDry bearings, brushes, sealsBehaves as a dead zone at the input and as a constant disturbance at the output
QuantisationFinite word length in the digital controller of Chapter 28Small-amplitude limit cycles in the least significant bits
Square-lawAerodynamic drag, orifice flow, radiationGain rises with amplitude — the opposite of saturation

Almost all of these can be pushed into a single standard arrangement: a static nonlinearity \(f(\cdot)\) in the forward path, with everything else — plant, sensor, linear compensator — lumped into one linear transfer function \(G(s)\) around it. That configuration is called the Lur'e problem, and every analytical method in the rest of this chapter except the phase plane is built for it.

Section 29-3

Linearisation and Its Limits

The first move is always the cheapest one: pretend the nonlinearity is not there, and see how far that carries. Chapter 2 did exactly this for physical models, expanding each nonlinear relation about an operating point and keeping the first term; Chapter 23 rewrote the procedure in matrix form. It is worth restating precisely, because knowing what it proves matters more than knowing how to do it.

Given \(\dot{\mathbf{x}}=\mathbf{f}(\mathbf{x})\) with an equilibrium \(\mathbf{x}^{*}\) satisfying \(\mathbf{f}(\mathbf{x}^{*})=\mathbf{0}\), write \(\mathbf{x}=\mathbf{x}^{*}+\delta\mathbf{x}\) and expand:

Linearisation about an equilibrium
\[ \dot{\delta\mathbf{x}}=\underbrace{\left.\frac{\partial \mathbf{f}}{\partial \mathbf{x}}\right|_{\mathbf{x}^{*}}}_{\textstyle A}\delta\mathbf{x}+\mathcal{O}\!\left(\|\delta\mathbf{x}\|^{2}\right) \]
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Lyapunov's indirect (first) method
If every eigenvalue of \(A\) has negative real part, \(\mathbf{x}^{*}\) is locally asymptotically stable. If any eigenvalue has positive real part, \(\mathbf{x}^{*}\) is unstable. If eigenvalues sit on the imaginary axis and none is in the right half-plane, the test is inconclusive.

Note the word locally. The theorem promises stability in some neighbourhood of \(\mathbf{x}^{*}\), and says nothing whatever about how large that neighbourhood is.

Three limitations follow, and each one motivates a later section. First, the result is local: a system can be beautifully stable for small disturbances and diverge for larger ones, which is precisely the amplitude-dependence of Section 29-1. Estimating the region of attraction needs Lyapunov's direct method, Section 29-7. Second, the borderline case genuinely occurs — Example 5 gives a system whose linearisation is a pure centre, undecidable by eigenvalues, yet which is globally asymptotically stable. Third, and most importantly for control engineering, the hard nonlinearities of Section 29-2 are not differentiable at the origin at all. A relay has no Jacobian; neither does backlash. For those the linearised model does not merely mislead, it does not exist. That is what the phase plane and the describing function are for.

Section 29-4

Phase-Plane Analysis

For a second-order system there is an exact method that needs no linearity at all: draw the motion. Take the state variables \(x_1=x\) and \(x_2=\dot x\) of Chapter 23 and plot \(x_2\) against \(x_1\) as time runs. The resulting curve is a trajectory, and the family of trajectories from all initial conditions is the phase portrait. Everything about the system's behaviour is visible in it.

Time is eliminated by dividing one state equation by the other. For \(\ddot x=f(x,\dot x)\),

The trajectory slope
\[ \frac{dx_2}{dx_1}=\frac{dx_2/dt}{dx_1/dt}=\frac{f(x_1,x_2)}{x_2} \]

Two consequences are immediate and worth having at your fingertips. Where \(x_2>0\) the trajectory moves to the right, and where \(x_2<0\) it moves to the left — the upper half of the plane is always traversed rightward. And wherever a trajectory crosses the \(x_1\) axis with \(f\ne0\), the slope is infinite, so it crosses perpendicularly. Sketching a portrait usually starts from those two facts, refined by the isocline method: setting the slope to a constant \(m\) gives the curve \(f(x_1,x_2)=mx_2\), along which every trajectory has slope \(m\); drawing a handful of isoclines with short slope marks lets the trajectories be threaded in by eye.

The points where both derivatives vanish are the singular points — the equilibria. Near each of them the linearisation of Section 29-3 applies, and its eigenvalues fix the local shape. There are six possibilities, and recognising them from a sketch is a standard examination question.

Eigenvalues of \(A\)Singular pointLocal trajectory shapeStable?
Real, both negativeStable nodeApproach the point directly, without spirallingYes (overdamped)
Real, both positiveUnstable nodeRecede directlyNo
Real, opposite signsSaddle pointApproach along one direction, leave along anotherNever
Complex, \(\operatorname{Re}\lt0\)Stable focusSpiral inwardYes (underdamped)
Complex, \(\operatorname{Re}\gt0\)Unstable focusSpiral outwardNo
Pure imaginaryCentreClosed orbits around the pointLinearisation inconclusive

The saddle point is the one with no linear-systems intuition behind it, because a linear system has only one equilibrium and a saddle needs neighbours. Its two incoming directions form a curve — the separatrix — that divides the plane into regions whose trajectories end up in entirely different places. When a nonlinear system has several equilibria, it is the saddles between them that carve the state space into basins of attraction.

A limit cycle appears in the portrait as an isolated closed curve. If nearby trajectories spiral onto it from both sides it is stable, and the system will settle into that oscillation from any nearby start; if they spiral away it is unstable, and it marks the boundary of a region of attraction. The figure below shows the portrait of a relay-controlled inertia, where the closed curve is not isolated at all — a whole family of them fills the plane — and shows what a rate-feedback term does to it.

x₁ = x x₂ = ẋ u = −M sgn(x): closed orbits, never settles s = x₁ + λx₂ = 0 start reaches the line then slides to the origin
On–off control of an inertia: closed orbits alone, sliding motion with rate feedback
The phase plane is exact but two-dimensional. Nothing has been approximated in the figure above — those parabolas are the true trajectories. That exactness is why the method remains the reference against which every approximate technique is checked. Its cost is dimension: three states need a phase space that cannot be drawn, and four states cannot be visualised at all. Beyond second order the phase plane gives way to the describing function, which sacrifices exactness for the ability to handle any order.
Section 29-5

The Describing Function

The frequency-response methods of Chapters 14 to 18 are the most powerful design tools in the book, and they are lost the moment a nonlinearity enters, because a sinusoid into a nonlinearity does not come out a sinusoid. The describing function recovers them by an approximation of considerable audacity: throw the harmonics away.

Drive the nonlinearity with \(x(t)=A\sin\omega t\). Its output \(y(t)\) is periodic with the same period, so it has a Fourier series \(y(t)=Y_0+Y_1\sin(\omega t+\phi_1)+Y_2\sin(2\omega t+\phi_2)+\dots\). Assume the characteristic is odd-symmetric, which kills \(Y_0\), and then simply discard every term beyond the first. What remains is a sinusoid at the input frequency, with an amplitude and a phase that depend on \(A\). Define their ratio as a gain.

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The describing function
\[ N(A)=\frac{Y_1}{A}\,\angle\,\phi_1 \qquad\text{where}\qquad Y_1\angle\phi_1 \ \text{is the fundamental of the output for } x=A\sin\omega t \]

\(N(A)\) is an amplitude-dependent complex gain. For a memoryless odd characteristic it is real, and \(N\) is simply the ratio by which the nonlinearity scales a sinusoid of size \(A\). For a nonlinearity with memory it is complex, and its angle is genuine phase lag.

The approximation stands or falls on one assumption, which must be checked in every application: the linear part of the loop must be low-pass, so that the harmonics generated by the nonlinearity are filtered out before they return to its input. Since most plants roll off at \(-40\) dB/decade or steeper, the assumption is usually reasonable — but it fails for resonant or lightly damped plants, and it fails when the nonlinearity generates a strong third harmonic that lands near a plant resonance.

Deriving a describing function is a Fourier coefficient calculation. Take the ideal relay of output \(\pm M\): the output is a square wave of amplitude \(M\) in phase with the input, whatever \(A\) is. Its fundamental coefficient is

Worked derivation — the ideal relay
\[ Y_1=\frac{2}{\pi}\int_0^{\pi}M\sin\theta\,d\theta=\frac{4M}{\pi}\qquad\Longrightarrow\qquad N(A)=\frac{4M}{\pi A} \]

The relay's describing function is real and falls off as \(1/A\): it presents infinite gain to a vanishingly small signal and negligible gain to a large one. Saturation takes more work but follows the same route. With slope \(K\) up to \(\pm S\) and constant beyond, and writing \(\theta_1=\sin^{-1}(S/A)\) for the angle at which the input reaches the break point,

Worked derivation — saturation, A > S
\[ Y_1=\frac{4}{\pi}\left[\int_0^{\theta_1}\!KA\sin^{2}\theta\,d\theta+\int_{\theta_1}^{\pi/2}\!KS\sin\theta\,d\theta\right]=\frac{4}{\pi}\left[\frac{KA\theta_1}{2}-\frac{KS}{2}\cos\theta_1+KS\cos\theta_1\right] \]
\[ N(A)=\frac{Y_1}{A}=\frac{2K}{\pi}\left[\sin^{-1}\!\frac{S}{A}+\frac{S}{A}\sqrt{1-\left(\frac{S}{A}\right)^{2}}\;\right] \]

The middle step uses \(\sin2\theta_1=2(S/A)\cos\theta_1\) to collapse the first integral. Check the limits: as \(A\to S\) the bracket tends to \(\pi/2\) and \(N\to K\), the unsaturated gain, as it must; as \(A\to\infty\) the bracket tends to zero and \(N\to0\). So saturation is a gain that falls monotonically from \(K\) to zero as the signal grows.

Nonlinearity\(N(A)\)Range of \(N\)
Ideal relay, \(\pm M\)\(\dfrac{4M}{\pi A}\)\(0\) to \(\infty\), real
Relay with dead zone \(\pm\delta\)\(\dfrac{4M}{\pi A}\sqrt{1-\left(\dfrac{\delta}{A}\right)^{2}},\;A\ge\delta\)\(0\) to \(2M/(\pi\delta)\), real; zero for \(A\lt\delta\)
Relay with hysteresis \(\pm h\)\(\dfrac{4M}{\pi A}\,\angle-\sin^{-1}\dfrac{h}{A}\)complex; phase lag up to \(90^\circ\)
Saturation, slope \(K\), limit \(\pm S\)\(\dfrac{2K}{\pi}\left[\sin^{-1}\dfrac{S}{A}+\dfrac{S}{A}\sqrt{1-\left(\dfrac{S}{A}\right)^{2}}\right]\)\(K\) down to \(0\), real
Dead zone \(\pm\delta\), slope \(K\)\(K-\;\)(the saturation expression with \(S=\delta\))\(0\) up to \(K\), real
Backlash, width \(b\)magnitude and phase both functions of \(b/A\)complex; phase lag approaching \(90^\circ\)
Dead zone and saturation are complements. Put a dead zone of width \(\delta\) and a saturation with break point \(S=\delta\) in parallel, both of slope \(K\), and the two outputs add back to the plain gain \(Kx\). The describing functions must therefore add to \(K\) — which is exactly what the table says. That is why the two entries share one formula, and why the two nonlinearities have opposite effects on a loop: saturation reduces the effective gain as the signal grows, dead zone reduces it as the signal shrinks.
Section 29-6

Predicting Limit Cycles

With \(N(A)\) in hand, the loop of Section 29-2 — nonlinearity \(f\) followed by linear \(G(s)\), closed with negative feedback — has become an ordinary linear loop whose forward gain happens to depend on the signal amplitude. Its characteristic equation is therefore the familiar one, and a sustained oscillation is exactly the condition that it be satisfied on the imaginary axis:

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The limit-cycle condition
\[ 1+N(A)\,G(j\omega)=0 \qquad\Longleftrightarrow\qquad G(j\omega)=-\frac{1}{N(A)} \]

Two real equations — the real and imaginary parts — in two unknowns, the amplitude \(A\) and the frequency \(\omega\). A solution predicts a limit cycle of that amplitude at that frequency; no solution predicts none.

The equation is best solved graphically, and the picture is a direct extension of the Nyquist diagram of Chapter 15. Plot \(G(j\omega)\) as usual. On the same axes plot the locus of \(-1/N(A)\) as \(A\) runs from zero to infinity. In the linear case that locus degenerates to the single critical point \(-1\); here it is a curve, and the question "does the Nyquist plot pass through the critical point" becomes "does it cross the critical locus". Each intersection is a predicted limit cycle: \(\omega\) is read from the \(G\) curve and \(A\) from the \(-1/N\) curve.

The shape of the critical locus is characteristic of the nonlinearity. For an ideal relay \(N=4M/(\pi A)\) is real and positive, so \(-1/N=-\pi A/(4M)\) runs along the negative real axis from the origin outward — a relay loop therefore has a limit cycle whenever \(G\) crosses the negative real axis at all, which is to say almost always. For saturation \(N\) falls from \(K\) to \(0\), so \(-1/N\) runs along the negative real axis from \(-1/K\) outward, and a limit cycle exists only if \(G\) crosses the real axis to the left of \(-1/K\) — that is, only if the small-signal linear loop would have been unstable. For a relay with hysteresis the locus is a horizontal line below the real axis at \(\operatorname{Im}=-\pi h/(4M)\), as Example 4 derives.

Re Im G(jω) = 6/[jω(jω+1)(jω+2)] ω increasing → −1/N = −πA/4M (ideal relay), A increasing ← −1/N for a relay with hysteresis: Im = −πh/4M ω = 1.414, A = 1.27 ω = 1.242, A = 1.64 0
Each crossing of the critical locus is a predicted limit cycle

An intersection tells you a limit cycle exists; it does not by itself tell you whether the system will settle into it. That requires a perturbation argument, and the argument is a direct reading of the Nyquist criterion of Chapter 16. Suppose the oscillation is nudged to a slightly larger amplitude. The operating point moves along the \(-1/N\) locus in the direction of increasing \(A\). If that direction takes the point into the region that \(G(j\omega)\) encircles — the region a Nyquist analysis would call unstable — the oscillation grows further and the limit cycle is not sustained at that amplitude. If it takes the point out of the encircled region, the oscillation decays back and the limit cycle is stable.

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Stability of a predicted limit cycle
Follow the \(-1/N(A)\) locus in the direction of increasing \(A\). If it passes from the region enclosed by the \(G(j\omega)\) plot into the region outside it, the limit cycle is stable; if from outside to inside, it is unstable.

A stable limit cycle is a persistent hunt the machine will exhibit; an unstable one is invisible in practice but marks a threshold — disturbances smaller than \(A\) die away, larger ones diverge.

The prediction is approximate, and it can be badly wrong. Everything above rests on discarding the harmonics. When the linear part filters well the predicted amplitude is typically within a few percent and the frequency closer still; when it does not, the describing function can predict a limit cycle that does not exist, or miss one that does. Two safeguards are standard: check that \(|G(j n\omega)|\ll|G(j\omega)|\) for \(n=3,5\) at the predicted frequency, and confirm any consequential result by simulation. The describing function is a design instrument, not a proof — which is exactly why Section 29-7 exists.
Section 29-7

Lyapunov's Direct Method

Chapter 10 defined stability through the location of poles, which presupposes a transfer function. Lyapunov's contribution, made in 1892 and largely ignored by engineers for sixty years, is a definition and a test that never mention poles, linearity, or even the solution of the differential equation.

The idea comes from mechanics. If the total energy of a system is positive whenever the system is away from rest, and if that energy is strictly decreasing along every motion, then the motion must eventually run out of energy and come to rest. Nothing in that argument requires the equations to be linear; it requires only an energy-like function. Lyapunov's insight was that any function with those two properties will do, whether or not it is a physical energy.

Let \(\dot{\mathbf{x}}=\mathbf{f}(\mathbf{x})\) with \(\mathbf{f}(\mathbf{0})=\mathbf{0}\). A scalar function \(V(\mathbf{x})\) is positive definite in a region if \(V(\mathbf{0})=0\) and \(V(\mathbf{x})>0\) for every other \(\mathbf{x}\) there. Its rate of change along the trajectories is obtained by the chain rule without solving anything:

The derivative along a trajectory
\[ \dot V(\mathbf{x})=\frac{\partial V}{\partial \mathbf{x}}\,\dot{\mathbf{x}}=\nabla V(\mathbf{x})\cdot\mathbf{f}(\mathbf{x}) \]
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Lyapunov's direct (second) method
If there exists \(V(\mathbf{x})\) positive definite with \(\dot V(\mathbf{x})\) negative semi-definite, the origin is stable. If \(\dot V\) is negative definite, the origin is asymptotically stable. If in addition \(V\) is positive definite everywhere and \(V(\mathbf{x})\to\infty\) as \(\|\mathbf{x}\|\to\infty\) (radially unbounded), the origin is globally asymptotically stable.

The conditions are sufficient, not necessary. Failing to find a suitable \(V\) proves nothing whatever about the system — only about the search.

That last sentence is the method's defining characteristic, and the source of both its power and its frustration. There is no algorithm that produces \(V\) for a general nonlinear system. What exists is a set of constructions that work for particular classes: the variable-gradient method, which posits \(\nabla V\) with unknown coefficients and forces \(\dot V\) negative; Krasovskii's method, which tries \(V=\mathbf{f}^{T}\mathbf{f}\); and for mechanical systems, the physical energy itself, which is very often the right answer.

For a linear system the search is completely solved. Take the quadratic form \(V=\mathbf{x}^{T}P\mathbf{x}\) with \(P\) symmetric positive definite, and evaluate its derivative along \(\dot{\mathbf{x}}=A\mathbf{x}\):

Worked derivation — the Lyapunov equation
\[ \dot V=\dot{\mathbf{x}}^{T}P\mathbf{x}+\mathbf{x}^{T}P\dot{\mathbf{x}}=\mathbf{x}^{T}\!\left(A^{T}P+PA\right)\!\mathbf{x}\;\equiv\;-\mathbf{x}^{T}Q\mathbf{x} \]
\[ \boxed{\,A^{T}P+PA=-Q\,} \]

The logic then runs backwards, which is what makes it useful. Choose any positive definite \(Q\) — \(Q=I\) is the standard choice — and solve the linear equations for \(P\). The theorem states that \(A\) has all its eigenvalues in the left half-plane if and only if the resulting \(P\) is positive definite. For linear systems, therefore, Lyapunov's method is not merely sufficient but exact, and it provides a stability test that requires no characteristic polynomial at all — only the solution of \(n(n+1)/2\) linear equations and a positive-definiteness check by Sylvester's criterion. Example 6 works one through.

The same machinery estimates a region of attraction, which is something nothing else in this book can do. If \(\dot V<0\) only inside some bounded region \(\Omega\), then the largest level set \(\{V(\mathbf{x})\le c\}\) that fits entirely inside \(\Omega\) is guaranteed to be attracted to the origin. That level set is a rigorous, if conservative, answer to the question the linearisation of Section 29-3 refused to answer: how large a disturbance can this system survive?

Section 29-8

Absolute Stability and Nonlinear Design

The describing function predicts; Lyapunov proves but demands ingenuity. Between them sits a class of results that prove stability for an entire family of nonlinearities at once, using only a frequency-domain test on the linear part. They address the Lur'e problem of Section 29-2 with the nonlinearity confined to a sector: \(k_1x^2\le x f(x)\le k_2x^2\), which says only that the characteristic stays between two straight lines through the origin. Nothing else about \(f\) is assumed — not continuity, not differentiability, not even that it is the same function from one instant to the next.

CriterionSectorTest on \(G(j\omega)\)
Circle criterion\([k_1,k_2]\), possibly time-varying \(f\)The Nyquist plot must not enter or encircle the disc whose real-axis diameter runs from \(-1/k_1\) to \(-1/k_2\)
Popov criterion\([0,k]\), \(f\) time-invariantThere must exist \(q\ge0\) with \(\operatorname{Re}\left[(1+j\omega q)G(j\omega)\right]+\dfrac{1}{k}\gt0\) for all \(\omega\)

The circle criterion is the exact generalisation of Nyquist: shrink the sector to a single gain \(k_1=k_2=k\) and the disc collapses to the point \(-1/k\), recovering Chapter 16 word for word. The Popov criterion is the sharper of the two because it exploits time-invariance, and it has a graphical form as neat as a Bode plot — plot \(\operatorname{Re}G(j\omega)\) against \(\omega\operatorname{Im}G(j\omega)\), and the condition is that this Popov plot lie entirely to the right of a straight line of slope \(1/q\) through the point \(-1/k\). Both criteria are sufficient only, and both are conservative: a system can fail them and still be perfectly stable. Their value is that they are guarantees, and a guarantee for a whole sector covers all the modelling uncertainty in the nonlinearity at once.

It is worth recording why these results were needed. Aizerman conjectured that a system whose nonlinearity lies in a sector for which every constant gain would be stabilising must itself be stable. The conjecture is intuitive, widely believed for years, and false — counterexamples exhibit limit cycles. The sector criteria are what replaced it.

Design methods that deliberately exploit nonlinearity fall into a few families, each of which the reader will meet again in more depth elsewhere.

MethodIdeaCost
Gain schedulingLinearise at several operating points, design a linear controller for each, interpolate the gainsNo global guarantee; scheduling variable must move slowly
Feedback linearisationChoose \(u\) to cancel the nonlinearity exactly, leaving a linear system to control by Chapter 26's methodsNeeds an exact model; cancellation fails outside it
Sliding-mode controlDrive the state onto a surface \(s(\mathbf{x})=0\) with a switching law, then let it slide along the surface to the originChattering at the switching frequency; needs a boundary layer
Describing-function compensationReshape \(G(j\omega)\) so it no longer crosses the \(-1/N\) locusOnly as reliable as the describing function itself
Anti-windupStop the integrator accumulating while the actuator is saturatedNone worth mentioning — it should be standard practice

Sliding mode deserves a closer look, because Example 2 has already drawn it. Define a switching function \(s(\mathbf{x})=\dot x+\lambda x\) and choose \(u=-M\,\text{sgn}(s)\). Wherever \(s\dot s<0\) the control drives the state toward the surface \(s=0\), and once there the switching is infinitely fast and the state is pinned to the surface. On the surface \(\dot x=-\lambda x\), so the motion decays as a first-order system with time constant \(1/\lambda\) — governed entirely by the designer's choice of \(\lambda\) and completely independent of the plant. That insensitivity to plant parameters is the point of the method, and it makes sliding mode the most robust controller in this book. Its defect is equally sharp: real switching is not infinitely fast, so the state chatters about the surface at whatever frequency the hardware permits, exciting unmodelled dynamics. Replacing the sign function by a saturation of small width — a boundary layer — trades a little of the robustness for a great deal of smoothness.

Where this leads. Every method in this chapter answers the question "will it behave?" rather than "is this the best I can do?" Chapter 30 changes the question. Optimal control asks for the input that minimises a cost, and its solution for a linear plant with a quadratic cost — the LQR — turns out to be exactly the state feedback of Chapter 26 with the gain chosen by a matrix equation of Lyapunov type. Robust control asks for a controller that works for every plant in a set, which is the sector idea of this section carried into the linear world. Adaptive control asks for a controller that learns its own gains, which makes the closed loop nonlinear by construction — and so its stability proofs are Lyapunov proofs, of exactly the kind Section 29-7 set out.
Section 29-9

Worked Examples

1 Two equilibria of a damped pendulum

Problem. A pendulum obeys \(\ddot\theta+b\dot\theta+\dfrac{g}{L}\sin\theta=0\) with \(b=1\) and \(g/L=10\). Find its equilibria and classify each.

Solution. With \(x_1=\theta\), \(x_2=\dot\theta\), the state equations are \(\dot x_1=x_2\) and \(\dot x_2=-x_2-10\sin x_1\). Equilibria need \(x_2=0\) and \(\sin x_1=0\), so \(x_1=n\pi\) — infinitely many, of two distinct kinds. The Jacobian is

Working
\[ A=\begin{bmatrix}0&1\\-10\cos x_1&-1\end{bmatrix} \]

At the hanging position \(x_1=0\): \(\cos x_1=1\), so \(A=\begin{bmatrix}0&1\\-10&-1\end{bmatrix}\) and \(\lambda^{2}+\lambda+10=0\), giving \(\lambda=-0.5\pm j3.12\). Complex with negative real part — a stable focus, and comparing with Chapter 8's standard form, \(\omega_n=\sqrt{10}=3.16\) rad/s and \(\zeta=0.5/3.16=0.158\). At the inverted position \(x_1=\pi\): \(\cos x_1=-1\), so the characteristic equation becomes \(\lambda^{2}+\lambda-10=0\) and \(\lambda=2.70,\,-3.70\).

Real roots of opposite sign: a saddle point, unstable for every value of \(b\). The two incoming directions of that saddle form the separatrix dividing initial conditions that swing back from those that go over the top. Notice what the linearisation cannot tell us — the eigenvalues at \(x_1=0\) hold only for small \(\theta\), and say nothing about how large a push the pendulum survives. That is a question for Section 29-7.

2 On–off control and the birth of sliding mode

Problem. A satellite of unit inertia is controlled by thrusters: \(\ddot x=u\) with \(u=\pm1\). Take first \(u=-\text{sgn}(x)\) and then \(u=-\text{sgn}(x+\dot x)\), starting from \(x(0)=1\), \(\dot x(0)=0\). Describe the motion in the phase plane.

Solution. Eliminate time with the slope relation of Section 29-4. With \(x_1=x\), \(x_2=\dot x\), \(\;x_2\,dx_2=u\,dx_1\), so every arc is a parabola. For \(u=-\text{sgn}(x_1)\):

Position feedback alone
\[ x_1\gt0:\;\; \tfrac12x_2^{2}=-x_1+C \qquad\qquad x_1\lt0:\;\; \tfrac12x_2^{2}=x_1+C' \]

From \((1,0)\) the constant is \(C=1\), so the arc reaches the \(x_2\) axis at \(x_2=-\sqrt2=-1.414\). The left-hand arc then carries it to \((-1,0)\), and by symmetry back again. The trajectory is closed — a sustained oscillation of amplitude 1 that never decays, exactly as the figure in Section 29-4 shows. On–off control alone cannot bring the satellite to rest.

Now add rate feedback, switching on \(s=x_1+x_2\). From \((1,0)\) we have \(s>0\), so \(u=-1\) and the same parabola \(\tfrac12x_2^2=1-x_1\) is followed until it meets \(s=0\), that is until \(x_2=-x_1\):

Reaching the switching line
\[ \tfrac12x_2^{2}=1+x_2 \;\Longrightarrow\; x_2^{2}-2x_2-2=0 \;\Longrightarrow\; x_2=1-\sqrt3=-0.732,\quad x_1=+0.732 \]

On the line, \(\dot s=x_2+u=x_2-\text{sgn}(s)\). For \(|x_2|<1\) the sign term dominates, so \(s\dot s<0\) on both sides: the control switches infinitely fast and the state is trapped on the line. It then slides into the origin obeying \(\dot x_1=-x_1\), a first-order decay whose time constant is set by the designer, not by the plant. Two lines of phase-plane geometry have produced sliding-mode control.

3 A limit cycle caused by saturation

Problem. A unity-feedback loop contains a saturation of unit slope limiting at \(\pm1\), followed by \(G(s)=\dfrac{12}{s(s+1)(s+2)}\). Predict the limit cycle.

Solution. First locate the phase crossover of \(G\). Writing the denominator as \(j\omega(2-\omega^{2})-3\omega^{2}\), the imaginary part vanishes when \(\omega(2-\omega^2)=0\), so \(\omega=\sqrt2\) rad/s and there

Working — the critical point on G
\[ G(j\sqrt2)=\frac{12}{-3(2)}=-2 \]

The saturation's describing function is real, so \(-1/N\) lies on the negative real axis and the intersection must be at \(G=-2\). Setting \(-1/N(A)=-2\) gives \(N(A)=0.5\), and with \(K=1\), \(S=1\), \(u=1/A\):

Solving for the amplitude
\[ \frac{2}{\pi}\left[\sin^{-1}u+u\sqrt{1-u^{2}}\right]=0.5 \;\Longrightarrow\; \sin^{-1}u+u\sqrt{1-u^{2}}=0.7854 \]

Solving numerically, \(u=0.4039\), so \(A=1/u=2.48\). The loop hunts at \(\omega=1.414\) rad/s with an amplitude of about \(2.5\) at the saturation input. Sanity-check the result against Section 29-6: without the saturation the linear loop has \(|G|=2\) at phase crossover, so it would be unstable — the oscillation grows until the saturation has cut the effective gain from 1 to 0.5, and there it stops. The limit cycle is stable, because increasing \(A\) further reduces \(N\), moving \(-1/N\) leftward out of the encircled region.

4 A relay with hysteresis

Problem. Replace the saturation of Example 3 by a relay of output \(\pm1\) with hysteresis \(h=0.2\), and take \(G(s)=\dfrac{6}{s(s+1)(s+2)}\). Find the limit cycle, and compare with the ideal relay.

Solution. Start with the shape of the critical locus. Writing \(\phi=\sin^{-1}(h/A)\), the describing function is \(N=\dfrac{4M}{\pi A}e^{-j\phi}\), so

The hysteresis locus is a horizontal line
\[ -\frac{1}{N}=-\frac{\pi A}{4M}e^{\,j\phi}=-\frac{\pi A}{4M}\cos\phi-j\,\frac{\pi A}{4M}\sin\phi=-\frac{\pi}{4M}\sqrt{A^{2}-h^{2}}\;-\;j\,\frac{\pi h}{4M} \]

The imaginary part does not depend on \(A\) at all: the locus is the horizontal line \(\operatorname{Im}=-\pi h/(4M)=-0.157\). Intersect it with the Nyquist plot, whose parts are \(\operatorname{Re}G=-18/[9\omega^{2}+(2-\omega^{2})^{2}]\) and \(\operatorname{Im}G=-6(2-\omega^{2})/\{\omega[9\omega^{2}+(2-\omega^{2})^{2}]\}\). Setting \(\operatorname{Im}G=-0.157\) and solving numerically gives \(\omega=1.242\) rad/s, and at that frequency \(\operatorname{Re}G=-1.278\). Equate real parts:

Solving for the amplitude
\[ \frac{\pi}{4}\sqrt{A^{2}-0.04}=1.278 \;\Longrightarrow\; \sqrt{A^{2}-0.04}=1.627 \;\Longrightarrow\; A=1.64 \]

Compare with the ideal relay, \(h=0\): the locus becomes the negative real axis, the intersection is at \(\omega=\sqrt2=1.414\) rad/s where \(G=-1\), and \(\pi A/4=1\) gives \(A=4/\pi=1.27\). Hysteresis has lowered the hunting frequency from \(1.41\) to \(1.24\) rad/s and raised the amplitude from \(1.27\) to \(1.64\) — the phase lag it introduces makes the oscillation slower and larger, which is the general price of adding memory to a switching element.

5 Where linearisation fails and Lyapunov succeeds

Problem. Determine the stability of the origin for \(\dot x_1=x_2-x_1(x_1^{2}+x_2^{2})\), \(\dot x_2=-x_1-x_2(x_1^{2}+x_2^{2})\).

Solution. Try linearisation first. The cubic terms have zero Jacobian at the origin, so \(A=\begin{bmatrix}0&1\\-1&0\end{bmatrix}\), whose eigenvalues are \(\pm j\). Pure imaginary: by the rule of Section 29-3 the test is inconclusive — the linearised system is a centre, and the discarded cubic terms decide the outcome. Take the obvious energy-like candidate \(V=x_1^{2}+x_2^{2}\), which is positive definite and radially unbounded, and differentiate along the trajectories.

Working
\[ \dot V=2x_1\dot x_1+2x_2\dot x_2=2x_1x_2-2x_1^{2}\!\left(x_1^{2}+x_2^{2}\right)-2x_1x_2-2x_2^{2}\!\left(x_1^{2}+x_2^{2}\right) \]
\[ \dot V=-2\left(x_1^{2}+x_2^{2}\right)^{2}=-2V^{2} \]

The cross terms \(2x_1x_2\) cancel exactly — they are the rotation the linearisation saw — and what remains is negative definite everywhere except at the origin. Since \(V\) is positive definite and radially unbounded, the origin is globally asymptotically stable. The cubic terms that linearisation threw away were the entire mechanism: they are a nonlinear damping that pulls every trajectory inward while the linear part spins it around.

6 Solving the Lyapunov equation

Problem. For \(A=\begin{bmatrix}0&1\\-2&-3\end{bmatrix}\), solve \(A^{T}P+PA=-I\) and use \(P\) to settle stability.

Solution. Write \(P=\begin{bmatrix}p_{11}&p_{12}\\p_{12}&p_{22}\end{bmatrix}\) and form the product. With \(A^{T}=\begin{bmatrix}0&-2\\1&-3\end{bmatrix}\),

Working
\[ A^{T}P+PA=\begin{bmatrix}-4p_{12} & p_{11}-3p_{12}-2p_{22}\\ p_{11}-3p_{12}-2p_{22} & 2p_{12}-6p_{22}\end{bmatrix}=\begin{bmatrix}-1&0\\0&-1\end{bmatrix} \]

Three equations in three unknowns. The first gives \(p_{12}=0.25\); the third gives \(0.5-6p_{22}=-1\), so \(p_{22}=0.25\); the second then gives \(p_{11}=3(0.25)+2(0.25)=1.25\).

Result and test
\[ P=\begin{bmatrix}1.25&0.25\\0.25&0.25\end{bmatrix},\qquad p_{11}=1.25\gt0,\qquad \det P=0.3125-0.0625=0.25\gt0 \]

Both leading minors are positive, so by Sylvester's criterion \(P\) is positive definite and the system is asymptotically stable. The Lyapunov function is \(V=1.25x_1^{2}+0.5x_1x_2+0.25x_2^{2}\), and its level curves are ellipses that every trajectory crosses inward. Confirm independently: the characteristic equation is \(\lambda^{2}+3\lambda+2=0\) with roots \(-1\) and \(-2\). The value of the Lyapunov route is that it never formed that polynomial, which is why it generalises to systems that have none.

Review

Chapter Summary

No superposition

Response depends on amplitude; equilibria may be many; limit cycles and jump resonance become possible.

The catalogue

Saturation, dead zone, backlash, relay. Single-valued gives real \(N\); multi-valued gives phase lag.

Linearisation

Jacobian eigenvalues settle local stability — except on the imaginary axis, and except for hard nonlinearities.

Phase plane

Exact for second order. Six singular points: node, focus, saddle, each stable or not; centre undecided.

Describing function

\(N(A)=Y_1/A\,\angle\phi_1\): relay \(4M/\pi A\); saturation falls from \(K\) to \(0\). Needs a low-pass \(G\).

Limit cycles

\(G(j\omega)=-1/N(A)\). Each crossing gives \(A\) and \(\omega\); the direction of increasing \(A\) decides stability.

Lyapunov

\(V\) positive definite with \(\dot V\) negative definite proves stability. For linear systems, \(A^{T}P+PA=-Q\).

Sector criteria

Circle and Popov guarantee stability for a whole sector of nonlinearities; conservative but rigorous.

Practice

Problems

For every describing-function problem, check that the linear part is low-pass at the predicted frequency before trusting the answer. Difficulty rises down the list.

  1. Classify the singular point at the origin of \(\ddot x+3\dot x+2x=0\), and then of \(\ddot x+3\dot x-2x=0\).
  2. Find and classify all equilibria of \(\dot x_1=x_2\), \(\dot x_2=-x_1+x_1^{3}-x_2\).
  3. An ideal relay has output \(\pm4\). Write its describing function and state the value of \(-1/N\) when \(A=2\).
  4. A saturation has slope 2 and limits at \(\pm3\). Compute \(N(A)\) for \(A=6\) and express it as a fraction of the small-signal gain.
  5. A relay of output \(\pm1\) drives \(G(s)=\dfrac{10}{s(s+2)(s+5)}\). Predict the amplitude and frequency of the limit cycle.
  6. For the loop of Problem 5, decide whether the predicted limit cycle is stable using the rule of Section 29-6.
  7. Repeat Problem 5 with a relay having a dead zone of \(\pm0.5\). Show that a limit cycle exists only if the loop gain exceeds a threshold, and find it.
  8. Solve \(A^{T}P+PA=-I\) for \(A=\begin{bmatrix}0&1\\-1&-2\end{bmatrix}\) and confirm stability by Sylvester's criterion.
  9. Show that \(V=x_1^{2}+x_2^{2}\) proves the origin of \(\dot x_1=-x_1+x_1x_2\), \(\dot x_2=-x_2-x_1^{2}\) asymptotically stable, and state what region your argument covers.
  10. A double integrator \(1/s^{2}\) is driven by \(u=-\text{sgn}(x+\lambda\dot x)\). Sketch the phase portrait and find the settling behaviour once sliding begins.
  11. Explain why a dead zone in the forward path leaves a residual steady-state error even when the linear part is type 1, and why saturation does not.
  12. A servo shows a persistent \(3\) Hz buzz of constant amplitude regardless of the command. List the three measurements you would make to decide whether the cause is backlash, relay hysteresis, or a marginally stable linear loop.
Tip: before any calculation, ask which of three questions the problem is really posing. "Will small disturbances die away?" is answered by linearisation and takes two minutes. "Will the machine hunt, and how badly?" is answered by the describing function, and the whole of the work is finding where \(G(j\omega)\) crosses the \(-1/N\) locus — so find the phase crossover first, because for every single-valued nonlinearity that is where the intersection must be. "Is it stable for every disturbance?" is answered only by Lyapunov, and the candidate \(V\) is very often the physical energy. Choosing the wrong tool costs far more time than any arithmetic in this chapter.