ECE 5754 nonlinear systems Autumn 2014

Solutions set, homework # 1: Introduction

Problem 1. (5 points)

A mathematical model that describes a wide variety of physical nonlinear systems is thenth-order differential equation

y(n) = g(t, y, y, . . . , y(n1), u

)where u and y are scalar variables. With u as input and y as output, find a state spacemodel of the given system.

Solution

Select as state variables

x1 := y , xi := y(i1) , i = 2, . . . , n

to obtain the system in state-space form

x = f(x, u)

y = h(x)

where

f(x) =

x2x3...xn

g(t, x1, . . . , xn, u)

, h(x) = x1

Problem 2. (5 points)

The nonlinear dynamic equations for a single-link manipulator with flexible joint (ignoringdamping) is given by

Iq1 +MgL sin q1 + k(q1 q2) = 0Jq2 k(q1 q2) = u

where q1 R and q2 R are angular positions, I > 0 and J > 0 are moments of inertia,k > 0 is a spring constant, M > 0 is the total mass, L > 0 is a distance, and u R is atorque input. Choose state variables for this systems, and write the state equation. Findall the equilibrium points of the unforced system, i.e., when u = 0.

Solution

Define state variable as follows

x1 = q1 , x2 = q1 , x3 = q2 , x4 = q2

to obtain the state-space model

x = f(x, u)

where

f(x) =

x2

MgLI sinx1 kI (x1 x3)x4

kJ (x1 x3) + 1J u

When u = 0, the set of all equilibrium points is

E = {x R4 : x1 = npi , x2 = 0 , x3 = x1 , x4 = 0 , for all n = 0, 1, 2, . . .}

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Problem 3. (20 points)

Consider the nonlinear 2-dimensional system

x1 = x2 + x1(1 x21 x22

)x2 = x1 + x2

(1 x21 x22

)where > 0 is a small parameter (i.e., 0 < < 1).

1. Compute all equilibrium points of the system;

2. Using polar coordinates, prove that the system has a unique non-trivial (i.e., differentfrom an equilibrium solution) isolated closed orbit;

3. Prove that the closed orbit in question is a stable limit cycle, i.e., all trajectories thatoriginate sufficiently near the closed orbit approach the orbit asymptotically;

4. Argue that all trajectories of the system (not originating on the orbit itself) approachthe closed orbit asymptotically except one: which one?

Write the system compactly as follows:

x = Ax+ f(x)

where f(x) is regarded as a perturbation of the nominal linear system x = Ax. Com-pare the behavior of the system with the case in which = 0:

1. What kind of system is the nominal linear system?

2. What conclusions can you draw regarding the behavior of the nominal linear systemin presence of small nonlinear perturbations?

Solution

1. To obtain the solutions of the algebraic equation

0 = x2 + x1(1 x21 x22

)0 = x1 + x2

(1 x21 x22

)solve for the linear term in x2 in the first equation to obtain

x2 = x1(1 x21 x22

)Substitution into the second equations yields

0 = x1

(1 + 2

(1 x21 x22

)2)which has the unique solution x1 = 0. Consequently, the only equilibrium point is theorigin, (x1, x2) = (0, 0).

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2. Changing coordinates to polar coordinates :=x21 + x

22, = arctan(x2/x1), one

obtains the equivalent representation

= (1 2) (1) = 1 (2)

with state space X = R0 [0, 2pi) and initial condition 0 :=x21(0) + x

22(0), 0 =

arctan(x2(0)/x1(0)). The last equation of (1) yields (t) = 0 + t, hence the solutionstarting from any 0 > 0 and any 0 [0, 2pi) keeps on spinning around the originin the counter-clock wise direction, for all t 0. Note that for 0 = 0 the solutioncollapses into the equilibrium point at the origin of the Cartesian plane. Analysisof the first equation in (1) reveals that = 1 and = 1 are the only equilibriumpoints for this system (note that = 1 is not a valid equilibrium). As mentioned, = 0 corresponds to the origin, whereas = 1 corresponds to a closed (periodic)orbit of the system this is because the trajectory originating at 0 = 1 satisfies(t)2 = x21(t) +x

22(t) = 1 for all t 0 and travels on this set with constant angular

velocity (t) = 1.

3. To analyze the behavior of the trajectories of the first system in (1), notice that(t) > 0 for all (t) (0, 1) and (t) < 0 for all (t) (1,). As a result, alltrajectories except the ones originating at 0 = 0 and 0 = 1 approach the point = 1 asymptotically. Note that the convergence to the point = 1 can not happen infinite time, as otherwise uniqueness of the solutions at = 1 would be contradicted.As a result, almost all trajectories of the system in the Cartesian plane approachasymptotically the orbit = {x R2 : x21 + x22 = 1}, as shown in Figure 1 (the onlyone who does not apart from the orbit itself is the trajectory at the origin.)

x = y + epsilon x (1 x2 y2)y = x + epsilon y (1 x2 y2)

epsilon = 0.5

2 1.5 1 0.5 0 0.5 1 1.5 2

2

1.5

1

0.5

0

0.5

1

1.5

2

x

y

Figure 1: Phase plot of the system in Problem 3. x-axis: x1(t), y-axis: x2(t).

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It should be clear that the nominal system is a linear (harmonic) oscillator. The result ofthe problem establishes the well-known fact that the properties of the harmonic oscillator(existence of a continuum of closed orbits) do not persist under small perturbations.

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