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EEE 586 HW # 1 SOLUTIONS

Problem 1.4Pick as states x = [q; _q]. Then

dx

dt =

x2M1(x1)[uC(x1; x2)x2 Dx2 g(x1)]

Problem 1.5Pick as states x = [q1; _q1; q2; _q2]. Then

dx

dt =

0BB@

x2MgL

I sin(x1) kI(x1 x3)

x4kJ

(x1 x3) 1Ju

1CCA

Problem 1.6

Pick as states x = [q1; _q1; q2; _q2]. Then

dx

dt

=

0

[email protected](x1)[h(x1; x2)x2+ K(x1 x3)]x4J1K(x1 x3) + J1u

1

CCA

Problem 1.7A state space realizaton for G is

_x=Ax + Bu ; y = C x

Substitutingu = r z = r (t; y)

_x=Ax B(t;Cx) + Br ; y= Cx

Problem 1.10

a.

Denee =i

o; then _e=

_o= y. Collecting the equations forz and e we get the claimed state-spacerepresentation.

b. At he equilibrium points _z = 0; _e= 0. Therefore, the following equations should hold simultaneously:

Aze+ B sin(ee) = 0

Cze = 0

SinceA is invertible, ze =A1B sin(ee) and, hence,CA1B sin(ee) = 0. But G(0) =CA1B6= 0, so itmust be that sin(ee) = 0, or ee = k; k2 Z. From this it now follows that ze = 0 so the equilibria of thesystem are

ze = 0; ee = k

c. For this transfer function, _e= z, so

e = _z= 1

z 1

sin(e)

= 1

_e 1

sin(e)

which is similar to the pendulum equation.

Problem 1.11 a. Dene e = i o; then _e = _o =y. Collecting the equations forz and e we get theclaimed state-space representation.

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b. At he equilibrium points _z = 0; _e= 0. Therefore, the following equations should hold simultaneously:

Aze+ B sin(ee) = 0

Cze = 0

SinceA is invertible, ze =A1B sin(ee) and, hence,CA1B sin(ee) = 0. But G(0) =CA1B6= 0, so itmust be that sin(ee) = 0, or ee = k; k

2 Z. From this it now follows that ze = 0 so the equilibria of the

system areze = 0; ee = k

c. For this transfer function, _e= z, so

e = _z= 1

z 1

sin(e)

= 1

_e 1

sin(e)

which is similar to the pendulum equation.

Problem 1.13

1. At the equilibrium,x2= 0 and

x1+ x

31=6 = 0, implying thatx1= 0 orx1=

p

6.

The state matrix of the linearization is 0 11 1

;

0 12 1

the rst one corresponding to the zero equilibrium and the second to the other two. The eigenvalues of therst are0:5 j0:866 so the equilibrium type is stable focus. The eigenvalues of the second are 1;2 so theequilibrium type is a saddle.

2. At the equilibrium, x1 = x2 and 0:1x1 2x1 x21 0:1x31 = 0 so either x1 = x2 = 0 or x1 = x2 = rootsof the quadratic 0:1x21+ x1+ 1:9.

The state matrix of the linearization is

1 1

0:1 2x1 0:3x2

1 2 Substituting the values ofx1 at the equilibrium we compute the eigenvalues of this matrix to determine theequilibrium type:

x1= x2= 0, eigenvalues0:908;2:09, stable node. x1= x2= 7:45, eigenvalues1:5j1:18, stable focus. x1= x2= 2:55, eigenvalues 0:37;3:37, saddle.3. The obvious equilibrium is the origin (0; 0). When x26= 0 then, x2 = 2 + 2x1, from which x1 = 0 or

x1= 3. The corresponding values forx2 are 2 and -4, respectively. On the other hand, when x2 = 0,x1 canbe either 1 or 0. So the equilibrium points are

(0; 0); (1; 0); (0; 2); (3;4)The corresponding linearization state matrices are

1 00 2

;

1 10 2

;

3 04 2

;

9 34 2

Computing their eigenvalues we have the following characterization of equilibria:

At (0,0), eigenvalues 1; 2, unstable node.

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At (1,0), eigenvalues1; 2, saddle. At (0,2), eigenvalues3;2, stable node. At (-3,-4), eigenvalues3; 10, saddle.Notice that there is one more case to be examined, when x1= 1. This case leads to a singular dierential

equation where one of the equations becomes algebraic. For this example, the algebraic equation can be writtenas x2 = (1 +x1) + a(x1)(1 + x1)2 wherea(:) is a suitable function (found by substituting the expression backinto the equations).

4. The only equilibrium is the origin for which the linearization matrix has eigenvalues 0:5j0:866 and theequilibrium is an unstable focus.

Problem 1.15

Dierentiating theL cos andL sin terms we get

my+ mL( cos _2 sin ) = HmL( sin + _2 cos ) = V mg

Solving for H; Vand substituting, we obtain

I = mgL sin mL2mLy cos My = F my mL cos + mL _2 sin k _y

Collecting the second derivatives in the left-hand side,

I+ mL2 mL cos mL cos M+ m

y

=

mgL sin

F+ mL _2 sin k _y

Now, the inverse of the left-hand side matrix is

M+ m mL cos mL cos I+ mL2

1

()

where () is the determinant of the matrix, i.e.,

() = (I+ mL2)(m + M)m2L2 cos2 = I m + IM+ m2L2 sin2 + MmL2

which is always positive and satises the stated inequality (>(I+ mL2)M+ mI).

Problem 1.19

a.

d

dt

Z h0

A()d=!i kp

gh

A(h) _h=u kp

gh

Letx = h

_x= 1A(x) [u kpgx]; y= xb. x=p pa = gh

_x= g

A(x=g)[u kpx]; y = x=(g)

c. At equilibrium, 0 =uss kpgxss, yss = xss = r. Hence, uss= kpgr.

Problem 1.22

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a.

_x1 = dx1f dx1+ r1_x2 = dx2f dx2+ r2

The assumptions r1= x1,r2= r1=Y, and x1f= 0 yield

_x1 = ( d)x1_x2 = d(x2f x2) x1=Y

b. When = mx2=(km+ x2), the equilibrium equations are

0 =

mx2km+ x2

d

x1

0 = d(x2f x2) mx2x1Y(km+ x2)

From the rst equation,

x1 = 0 or mx2km+ x2

=d ) x2= kmdm

d

Substituting x1 in the second equilibrium yields x2 = x2f. Substituting x2 = kmd=(m d) in the secondequilibrium equation yields

x1= Y

x2f kmd

m d

Hence, there are two equilibrium points

Y

x2f kmd

m d

; kmd

m d

and [0; x2f]

c. When =mx2=(km+ x2+ k1x2

2), the equilibrium equations are

0 = mx2

km+ x2+ k1x2

2 d x1

0 = d(x2f x2) mx2x1Y(km+ x2+ k1x22)

From the rst equation, x1 = 0 or x2 is the root ofd= (x2). Sinced < maxx20f(x2)g, this equation hastwo roots, say x2a; x2b.

Substituting x1 = 0 in the second equilibrium yields x2 = x2f. Substituting x2 = x2a in the secondequilibrium yields

(.x2f x2a) (x2a)x1=Y = 0 ) x1= Y(x2f x2a)since(x2a) = d. Similarly, substituting x2= x2b in the second equilibrium equation yields x1= Y(x2f x2b).

Hence, there are three equilibrium points

[Y(x2f

x2a) ;x2a] and [Y(x2f

x2b) ; x2b] and [0; x2f]

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EEE 586 HW # 2 SOLUTIONS

Problem 2.1

1. Clearly, at an equilibrium x1 = x2. Substituting in the rst equation we get(x1 x31) = 0, so the

equilibria are (0; 0), (1; 1), (1; 1). To determine the type, we look at the linearization:

A

=

@f

@x = 1 + 6x21 1

1 1 whose eigenvalues at the three equilibria are: f1 jg, f4:828; 0:828g, f4:828; 0:828g, respectively. Thus,the origin is a stable focus and the other two are saddle points.

3. Here we distinguish the following cases:

1. x1= 0,x2= 0, yielding the equilibrium (0; 0).

2. x1= 0, 2 h(x) = 0, yielding the equilibrium (0; 2).

3. 1 x1 2h(x) = 0,x2= 0, yielding the equilibrium (1; 0).

4. 1 x1 2h(x) = 0, 2 h(x) = 0, yielding the equilibrium (3; 4).

To determine the type, we look at the linearization:

A=

@f

@x=

1 x1 2

x21+x1

+x1(1 + 2 x2(1+x1)2

) 2 x21+x1x2

x2(1+x1)2

2 2 x21+x1

!

whose eigenvalues at the four equilibria are: f1; 2g (unstable node), f2; 3g (stable node), f12g (saddle),f7:772; 0:772g(saddle), respectively.

5. At an equilibrium _x= 0. The possible solutions are given by

0 = (x1 x2)(1 x21 x

22); 0 = (x1+x2)(1 x

21 x

22)

The locus of 1 =x21+ x22 is a circle and these are not isolated equilibria. The other solution is (0; 0) which is an

isolated equilibrium [email protected]

@x

= 1 1

1 1 and whose eigenvalues are 1 j. Hence, (0; 0) is an unstable focus.

6. We have0 = x31+x2; 0 = x1 x

32

x2= x31) x1(1 x

81) = 0 ) x1= 0 or x

81= 1

The latter has two real roots at x1 =1. Hence, there are three equilibrium points at (0; 0), (1; 1), (1; 1).The jacobian is

@f

@x =

3x21 1

1 3x22

Evaluating at the three equilibria, (0; 0) has a Jacobian with eigenvalues 1; 1 and it is a saddle. (1; 1) and(1; 1) have a Jacobian with eigenvalues 2; 4 and it is a stable node.

Problem 2.2

1. At an equilibrium _x= 0. The possible solutions are (0; 0), (2; 0), (2; 0). To determine the type, we lookat the linearization:

A=

@f

@x=

0 1

1 + 5x4

1

16 1

!

whose eigenvalues at the three equilibria are: f0:5 j0:866g, f1:56; 2:56g, f1:56; 2:56g, respectively. Thus,the origin is a stable focus and the other two equilibria are saddle points.

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2. At an equilibrium _x= 0. The possible solutions are (0; 0), (1; 2), (1; 2). To determine the type, we lookat the linearization:

A=

@f

@x=

2 x2 x1

4x1 1

whose eigenvalues at the three equilibria are: f2; 1g, f0:5 j1:94g, f0:5 j1:94g, respectively. Thus, theorigin is a saddle point and the other two equilibria are stable foci.

3. At an equilibrium _x = 0 so x2 = 0 implying that (x1) = 0. The only solution is (0; 0). To determinethe type, we look at the linearization:

A=

@f

@x=

0 1

@@x1 1 + @@x2

=

0 1

0:5 1 + 0:5

whose eigenvalues are f0:25 j0:66g. Thus, the origin is a stable focus.

Problem 2.7

y(t) k1ea(tt0) +

Z tt0

ea(t)[k2y() +k3]d

y(t) +k3=k2 k3=k2+k1ea(tt0) + Z tt0

ea(t)k2[y() +k3=k2]d

Denez(t) = eat[y(t) +k3=k2]. Then

z(t) k3=k2eat +k1e

at0| {z }(t)

+

Z tt0

k2z()d

using BG lemma (t) +

Z tt0

()k2ek2(t)d

convert back to y

y(t) k1ea(tt0) + Z

t

t0

k3e(ak2)(t)d+ Z

t

t0

k1k2ea(tt0)ek2(t)d

From this, the desired result is obtained after performing the integrations.

Problem 2.12

It is enough to show that the operator T : IRn 7!IRn, T(x) =D1(D A)x+b, x 2 IRn is a contraction(due to the strict diagonal dominant condition, D1 exists), that is, kT(x) T(y)k = kx yk, 8x, y 2 IRn,0

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Problem 2.15

The system of equations can be written as:" 1 + x1

1+x21

1 + x21+x2

2

# x1x2

=

12

notice that the matrix on the right is strict diagonal dominant since the diagonal elements are bounded by34

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The quadratic in the brackets can be written as x>P x which is upper-bounded by max(P)kxk2. Hence, for

suciently large c, on the surface kxk = c, max(jx1j; jx2j) can be made arbitrarily large and rV f will benegative. (The same can be shown by completing the squares.) Hence, trajectories starting in M= fxjV(x) cgstay inM. Furthermore,Mcontains a single equilibrium point which is an unstable focus. It follows from P-Bcriterion that there is a periodic orbit in M.

Problem 2.20

= 1 +a 6= 0

By Bendixsion's criterion, there are no periodic orbits.2. The equilibrium points are the roots of

0 = x1(1 +x21+x

22); 0 =x2(1 +x

21+x

22)

The system has an isolated equilibrium at the origin and a continuum of equilibria on the unit circle x21+x22= 1.

It can be easily veried that the origin is a stable node. Transform the system into polar coordinates x1= r cos ,x2= r sin . Then it follows that

_r= r(1 r2)

For r < 1 every trajectory starting inside the unit circle approaches the origin as t ! 1. For r > 1, everytrajectory starting outside the unit circle diverges to 1. Hence, there are no limit cycles. (An argument usingno theorem but basic analysis on the trajectories.)

3. x1 must be zero from _x2= 0. But then _x1 = 1 6= 0. So the system has no equilibrium and by Lemma 3.2the index of any closed curve is zero and, therefore, it cannot be a closed orbit.

4. The equilibria are the line x2= 0. By Lemma 3.2, a periodic orbit must enclose at least one equilibrium.But then it will intersect the equilibrium line, and once the state becomes equal to the equilibrium, it willremain there. This contradicts the assumption of a periodic orbit, hence no such orbit exists.

5. The equilibria are x2 = 0 and x1 = k; k 2 Z. All are saddle points (the linearization matrix is[0 1; 1 0]). Hence, by Lemma 3.2, there can be no periodic orbits.

Problem 2.21

a. Let V(x) =x2 x1+bx1+a

. The functionV(x) is negative in D and the curve V(x) = 0 is the boundary of

D (@D). Evaluating,rV fjV=0 = cx1(x1+a)

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Problem 2.28 a. To apply P-B we note that the origin is the only equilibrium (see Reference [195] for arather complicated analysis, or resort to numerical evaluation). For the stability of the equilibrium, we formthe linearization matrix

A=1

1 +

1 +

Its eigenvalues are the roots ofs2 + 2s(1 ) + 2(22 + 1=2). The last term has negative discriminant,

so it is always positive. By the Routh criterion, the roots of the polynomial will be in the left half plane i1 >0. There will be two roots in the right half plane when >1, and the equilibrium will be competelyunstable (focus or node, to be determined later).

Next, consider the coordinate transformationx xand dene the functionV(x) = x21 + x22. Its derivative

[email protected]

@xf=

2

[x21+x

22 (x1tanh(x1) +x2tanh(x2)) (x1tanh(x2) +x2tanh(x1))]

Hence, _V 2[V4jxj], so it is negative forV >1622. The latter establishes the set M= fxjV 1622g

that is positively invariant and, by the P-B criterion, contains a periodic orbit.b.,c. The phase portrait of the system can be easily produced with MATLAB with a standard simulation

model (see attached Figures). Dierent initial conditions provide dierent trajectories and their number andlocations should be chosen to provide as complete of a picture as necessary. An overlay of the vectoreld usingthe quiver command is also helpful.

>> [X1 X2]=meshgrid(-3:.25:3,-3:.25:3);

>> F2=-X2+tanh(2*X1)+tanh(2*X2);

>> F1=-X1+tanh(2*X1)-tanh(2*X2);

>> quiver(X1,X2,F1,F2)

For > 1 the trajectories show a clear convergence to a limit cycle while for < 1 (part c), they showconvergence to the only equilibrium.

d. Asvaries the equilibrium changes from a stable focus 0 < 1.

Problem 2.29 a. To apply P-B we note that, from the second equation, the equilibrium should satisfy eitherx1= 0 or

x21+x2

1

= 1. But for the former, _x1= a so it cannot be an equilibrium. For the latter, the rst equation

leads to a 5x1= 0. Hence, the system has one equilibrium at

xe= a=5

1 + (a=5)2

Next, to analyze the stability of the equilibrium, we form the linearization matrix

A=

24 1 4x21+x21 + 8x21x2(1+x21)2 4x11+x21

b

1 x21+x2

1

+

2bx21x2

(1+x21)2

bx11+x2

1

35

Evaluating atx =xe we get

A=

" 5 8(a=5)

2

1+(a=5)2 4(a=5)1+(a=5)2

2b(a=5)2

1+(a=5)2 b(a=5)1+(a=5)2

#=

a=5

1 + (a=5)2

3(a=5) 5(a=5) 4

2b(a=5) b

Lettingc = 3(a=5) 5(a=5) , the eigenvalues ofA are a=51+(a=5)2 (a positive constant) times the roots ofs

2 +

s(b c) bc + 8b(a=5) = s2 + s(b c) + a + 25=a. Applying the y the Routh criterion, the roots of the polynomialwill be in the left half plane ib c >0, or b >3(a=5) 25=a. There will be two roots in the right half planewhenb 0, then _x1 > 0. Similarly, when x2 = 0 and x1 >0, then _x2 >0. Hence, trajectories cannotcross the boundary of the rst quadrant. Further, we notice that x1 is bounded since for large x1 (x1> a and

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x2 > 0), then _x1 < 0. Similarly, when x2 is large (x2 > 1 + a2 >1 + x21 and x1 > 0), then _x2 < 0. Hence, the

box a > x1 > 0, 1 +a2 > x2 > 0 is positively invariant, containing one unstable focus/node equilibrium for

b < 3(a=5) 25=a; by the P-B criterion, the system has a periodic orbit.b.,c. The phase portrait of the system can be easily produced with MATLAB with a standard simulation

model, as in Problem 2.28. Dierent initial conditions provide dierent trajectories and their number andlocations should be chosen to provide as complete of a picture as necessary. An overlay of the vectoreld usingthe quiver command is also helpful. Whena = 10, b = 2, the trajectories converge to a limit cycle, while whenb= 4 (above the critical value) they converge to the stable focus equilibrium.

d. Constructing a plot of the eigenvalues of the linearization matrix for a = 10 b = [0 : 100], we nd thatthe equilibrium is an unstable node for smallb (b < 0:2), bifurcating to an unstable focus until the critical valueof periodic solutions (b= 3:5), where it bifurcates to a stable focus, and then to a stable node for large valuesofb (b > 55:8).

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-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

-3

-2

-1

0

1

2

V and V`-level sets for Pr. 2.17.3 (detail)

V' > 0

|x1+2x

2|2= 3/2

x-trajectory

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-10 -5 0 5 10 15

-10

-5

0

5

10

V and V`-level sets for Pr. 2.17.3

V-level set

V(x) = |z|2/4=16

V-decreasing

V-decreasing

V-increasing(numerical eval.)

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EEE 586 HW # 3 SOLUTIONS

Problem 3.11. f(x) =x2 + jxj is not continuously dierentiable at the origin (jxj has a discontinuous derivative). It is

locally Lipschitz since its derivative (left and right limits) is bounded on any compact set. It is continuous sinceboth terms are. It is not globally Lipschitz since the derivative ofx2 is not bounded.

3. f(x) = sin xsgnx= sinx

x

(xsgnx) = sinx

x jxj. Is not continuously dierentiable at the origin. It is glob.L.

since its derivative (left and right limits) is bounded. Hence it is loc.L. and continuous.6. f(x) = tan x is not continuous at =2 but it is at 0. It is only locally continuously dierentiable. It is

locally Lipschitz around 0.8. f(x) = [x1+ajx2j; (a+b)x1+bx21 x1x2] is not continuously dierentiable at 0 (unless a=0). It is

continuous everywhere and locally Lipschitz. It is not globally Lipschitz because of the x21 andx1x2 terms.

Problem 3.3Letf1; f2 be locally Lipschitz:jfi(x) fi(y)j kijx yj, for allx; y2 D.Then,jf1(x) + f2(x) f1(y) f2(y)j jf1(x) f1(y)j + jf2(x) f2(y)j (k1+ k2)jx yj, in D. Hence

f1+f2 is Lipschitz.Also, jf1(x)f2(x)f1(y)f2(y)j jf1(x)f2(x)f1(y)f2(y)f1(x)f2(y)j jf1(x)jjf2(x)f2(y)j+jf2(y)jjf1(x)

f1(y)j Kjx yj, inD , as long as f1; f2 are bounded in D. That is the case, for example, iffi are continuousandD is compact. In general, f1f2 is locally LipschitzFinally,jf2 f1(x) f2 f1(y)j =jf2[f1(x)] f2[f1(y)]j k2jf1(x) f1(y)j k2k1jx yj. So f1 f2 isLipschitz in the same domain.

Problem 3.4kKxk is Lipschitz in x and, using 3.3, g(x)kKxk and g(x)K(x) are Lipschitz on any compactset. (Note: continuity on compact sets implies boundedness.)

Wheng(x)kKxk > , then 1=kKxk < g(x)=so it is bounded on compact sets. That means thatKx=kKxkis Lipschitz.

It remains to show continuity at the boundary. Considerg(x)kKxk =,g(y)kKyk =+, for some small >0. Then g(x)Kx= Ky =kKyk =g(x)kKxkKx=kKxk Ky =kKyk = (1 =)Kx=kKxk Ky=kKyk.Since Kx=kKxk is Lipschitz, the norm of the last expression is less than =+Lkx yk. Since this can bemade arbitrarily small (for suciently smallx y and ), we can establish the continuity at the boundary.

Sofis locally Lipschitz on any compact set.

Problem 3.7 ([1, 2])First, the function f(x) = 11+g>(x)g(x)g(x) is bounded for all x.

kf(x)k = 11 +g>(x)g(x) g(x)

= kg(x)k1 + kg(x)k2 1;it is continuous dierentiable inD= ft0 t t0+a;kx x0k bg. Then there exists a unique solution x(t)on the interval t0 t t0+, for some >0 (Thm. 3.1).

Next, we extend the interval of existence to innity: We know that, ifx(t) is a solution of the dierentialequation on a t-interval, then there exists a maximal t-interval of existence. We prove by contradiction thatin our case the maximal t-interval is t

2[t0;

1). Assume the contrary, that is, the maximal t-interval is nite

t 2 [t0; ], this implies that the solution to the dierential equation escapes to innity in nite time limt! x(t) = 1so,

x(t) = x0; +

Z tt0

f(x)dt; t 2 [t0; ]

limt!

x(t) = x0+ limt!

Z tt0

f(x)dt

limt!

x(t) x0+ 1( t0)

1

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but the last inequality does not hold since the right side is nite. Then we can conclude that the maximalt-interval is [t0; 1).

Finally, this implies that the solution on an interval [t0; t0 +T] belongs to a compact set of the formD =fx :kx x0k rg where r =T. In that set f is loc.L. and, by Thm. 3.3, the solution is unique. Thisholds for arbitraryT.

Problem 3.8

1. LetD= x 2 IR2 j2x1+x2j 0 and starting in the rst quadrant, pick c large enough such that2x1+x2> 1 (c 18). Then

_V(x) = _x1x2+x1_x2

= x22+x21 x1sat(2x1+x2)

= x22+x21 x1 =

c2 +x41 x31x21

The functionc2 + x41

x31

has a minimum at x1= 3

4and as long as c > 3

p3

16 then _V(x)> 0 which implies

that the trajectories cannot reach the origin.

3. From the previous section we have seen that the region of attraction for the equilibrium is not IR2, hencethe equilibrium cannot be global.

Problem 3.11The Van der Pol equation:

_x=f(x; ); x(t0) = x0; f(x) =

x2

x1+

1 x21

x2

(1)

From Section 3.3 in the textbook and using =:

A (t; 0) = @@x

f(x; )x=x(t;0);=0

=

0 11 2x2x10

1 x21

0

B(t; 0) = @

@f(x; )

x=x(t;0);=0

=

0

1 x21

x2

The nal equations are:

_x = f(x; 0); x(t0) = x0

_x = A (t; 0) x+B(t; 0) ; x(t0) = 0

Problem 3.14 a. As in Pr. 3.11.

b.

_r = 1

2p

x>x2x>f(x) =

1

r

1

x>x+x1tanh(x1) x1tanh(x2) +x2tanh(x1) +x2tanh(x2)

= 1

r+x1+x2

r tanh(x1) +

x1+x2r

tanh(x2)

1

r+jx1+x2j

r j tanh(x1)j +j x1+x2j

r j tanh(x2)j

1

r+ 2p

2

2

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sincej tanh(:)j 1 and

jx1+x2jr

p

(x1+x2)2

r

rx21+x

22+ 2x1x2r2

r

1 +2x1x2

r2

p2

For the last inequality, notice thatx21+ x22 2x1x2 = (x1 x2)2 0 ) x21+ x22= r2 >2x1x2. Similarly for the

term

j x1+x2

j=r.

c. Immediate, using the comparison lemma and the solution of the ODE for r in part (b).

Problem 3.20Letfbe Lipschitz onD with constantk, and consider >0. Take= =k. Then for jxyj < ,

jf(x) f(y)j kjx yj < k=k=

Sinceis independent ofx; y, it follows that f is u.c.

Problem 3.23Setg() = f(x); 0 1. SinceD is convex, x 2 D for 0 1. Then,

g0() = f0(x)dx

d =f0(x)x

f(x) = f(x) f(0) =g(1) g(0) =Z 10

g0() d=Z 10

f0(x) d x

References

[1] Braun, M.,Dierential equations and their applications, Springer-Verlag 1983.

[2] Hale, J. K., Ordinary dierential equations, Krieger Publishing Company 1980.

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EEE 586 HW # 4 SOLUTIONS

Problem 4.3Obtain a Lyapunov function from the corresponding linearization.(1)A = [1 0;0 1], solveA>P+ P A= I, soP= 12 I andV(x) = 12 x>x. Then _V = x21+ x21x2 x22

12 x>x x21( 12 jx2j). So, locally infjx2j Q(r)jxj, whereQ(r) =

1 r=2

r=2 1

implying that _V is n.d. for r P+P A =I, so P = 12 I (A isI+ a skew symmetric matrix) and

V(x) = 12 x>x. Then _V = x1x2 x21(1x>x) + x2x1 x22(1x>x) = x>x(1x>x). So, locally in fjxj x > 1, implying that the system has a periodic solution. This is a stable limit cycle in reverse time,corresponding to an unstable limit cycle in forward time. This implies that the region of attraction cannot beglobal.

|||||-(3)A = [0 1; 11], solveA>P+P A= I, so P= 12 [3 1; 1 2] andV(x) = 12 x>P x. Then _V = 3x1x2 +x22

3x31x2 x22x21 x21 x1x2 x41 +x31x2 2x2x1 2x22 +2x2x31 +2x22x21. Collecting terms, _V = x21 x22 x41 +x21x22.So, locally infjx1x2j r < 2g, _V cV, for some c(r) > 0 implying local ES. An estimate of the region ofattraction is the largest Lyapunov level set contained within the hyperbolas jx1x2j P +P A =I, so P = 12 [3 1;1 2] and V(x) = 12 x>P x. Then _V =

3x2

1 x

1x

23x

1x

2x2

2+ 2x2

1x

1x3

2+ 4x

1x

22x4

2. Collecting terms, _V =

x2

1 x2

22x4

2 x

1x3

2. So,

locally infjx2j2 r < 2g, _V cV, for some c(r) > 0 implying local ES. This Lyapunov function oers noconclusion for global stability. We notice that the cross term x1x2 is responsible for the x1x

32 term that makes

the derivative indenite for largex. If we eliminate it and adjust the coecients to cancel the cross termsx1x2from the derivative, we get V = 2x21+x

22, _V =4x21 4x1x2+ 4x2x1 2x42 =4x21 2x42, which is negative

denite and shows global asymptotic stability. Notice, however, that in order to get this result, this Lyapunovfunction sacrices the local exponential stability. Roughly, the reason is that the local result is due to the 2x1coupling term in the second ODE, but the global attractiveness comes from thex32 term.

Problem 4.2

LetV(x) = 12 x2. Then _V =axp+1 +xg(x).

1. Whenp is odd and a 0 infjxj < a=2kg. Hence, 0 is unstable.

Whenp is even and a 6= 0, let V(x) = ax, for whichfx: V >0g has 0 on its boundary and contains pointsarbitrarily close to 0. Then _V = a2xp +xg(x) a22xp + a

2

2xp

1 2kjxja2

> 0 infjxj < a2=2kg. Hence, 0 isunstable.

Problem 4.6

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1. The system is:

_x1 = x22_x2 = x2+x21+x1x2

The linearization around the origin produces

x=0

= 0 0

0 1

which has an eigenvalue equal to zero and another equal to1.Using the center manifold equation we get

@@x1

h(x1)h2(x1) +h(x1) x21 x1h(x1) = 0

Leth(x1) = ax21 and substitute in the center manifold equation

x21(a 1) x31(a+ 2a3x21) = 0 ) a= 1; h(x1) = x21+O(jx1j3)

Then substitute x2

h(x1) in the reduced system equation

_x1= x41+O(jx1j5)

The origin of the reduced system is unstable, hence the origin of the original system is an unstable equilibrium.

Problem 4.7Let V = 12 x>P x. Then, _V = x>P Q= x>with the obvious choice P=Q1. Hence _V 0 for small y only. y2(y) =y2 +y4 >0 for all y. Hence, 0 islocally AS.

Problem 4.8

V = x2

1

u +x22, u= 1 +x

21 is locally pdf and not RU. Taking the derivative, _V =

12x21

u4

4x2

2

u2 >0 on the hyperbola. Whilethe substitution is straightforward, the expressions are quite long. Here, a Matlab evaluation is helpful to showthat the inner product is positive meaning that the vector eld always points to the right of the hyperbola.

The last observation implies that initial conditions starting to the right of the hyperbola remain there. Hence0 cannot be GAS.

Problem 4.9

a. Ifx1= 0,V(x) = x2

2

1+x22

+x22! 1 asjx2j ! 1. Ifx2= 0,V(x) = x2

1

1+x21

+x21! 1asjx1j ! 1.b. On the line x1= x2,V(x) =

x21

1+x21

6! 1 asjx1j = jx2j ! 1. Hence,V is not RU.

Problem 4.10Using the suggested expression,

x>P f+ f>P x=x>Z 1

0

"P

@f(x)

@x +

@ f(x)

@x

>P

#d

!x x>x

Next,V =f>P f 0, for anyf. It is PD iV = 0 , x= 0. This is equivalent tof= 0 , x= 0, i.e., 0 is aunique equilibrium of _x= f(x). We show this by contradiction. Suppose that there is q6= 0 such thatf(q) = 0.Then, from Part 1, q>P f(q) +f(q)>P q q>q. But the left hand side is 0 so q = 0 which is a contradiction.

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Next, we need to show that V is RU. That is, asjxj ! 1, so must f. Again, suppose that it is not soandjf(x)j < c as x grows. From Part 1, x>Pf

x>x 12 . But if f were bounded, then the fraction satises

x>Pfx>x

jPjcjxj ! 0 which is a contradiction.Thus, we have that V is PD and RU. Furthermore, _V f>([email protected]@x + @[email protected]

>P)f f>If 0 in U =fx :jx1j >jx2jg, that has zero on its boundary (we write 0 2 @U). Hence, 0 is an unstable equilibrium.2. (corrected) This solution is based on more elementary observations:

Consider the setx31 + x2 >= 0. On its boundary in the 1st quadrant, _x2 0. Also, consider the setx61 x32 >= 0. On its boundary in the 1st quadrant, _x1 0. The intersection of these two sets is therefore apositively invariant set. Since both elements of _x are positive, any trajectory starting inside it moves towardshigher values (the equilibrium at [1,1]). Since the same is true for trajectories starting arbitrarily close to 0, weconclude that 0 is unstable.

Problem 4.15

At an equilibrium, _x= 0, hence,8 0:6. So we choose= 1=0:6 = 1:67 and conclude that

the system is absolutely stable for the sector [0,1.67].Case 3: The Nyquist plot should be inside the diskD(; ). One can dene several possible disks. For

example, starting with the center at 0.2, the radius to achieve inclusion is found as 0.9. Thus, 1== 1:1!= 0:91 and 1== 0:7 ) = 1:43. Hence, the system is Absolutely stable for the sector [0:91; 1:43].

Other meaningful choices, e.g., to maximize the radius of the disk, would involve an iterative solution.7. The Nyquist plot is to the right of the vertical line at -0.341. Then the system is absolutely stable forthe sector [0, 2.93]. Alternatively, the Nyquist plot lies inside the disk D(-1/1.1, 1/0.5), so the system is A.S. inthe sector [-0.91, 2]

Problem 7.10

4. Fora A; (a sin ) = 0; ) (a) = 0. For a > A,

(a sin ) =

0 for 0 or B for < <

where= sin1(A=a). Thus,

(a) = 4aZ =2

B sin d= 4Ba cos = 4Bar

1A2

a2

Problem 7.11

4.

G(jw) = jw

w2 jw + 1=w2 +jw(1 w2)

(1w2)2 + w2

Im[G(jw)] = 0 ) w = 1. Then, Re[G(jw)] = 1 and the equation 1 + (a)Re[G] = 0 has the uniquesolutiona = (8=5)1=4. Thus, we expect the system to exhibit a periodic solution with amplitude approximatelya= (8=5)1=4 and frequency approximately 1rad=s.

5.

G(jw) = jw

w2 jw + 1=w2 +jw(1 w2)

(1w2)2 + w2

Im[G(jw)] = 0 ) w= 1. Then,Re[G(jw)] = 1 and the equation 1 + (a)Re[G] = 0 has the unique solutiona= (8=5)1=4 (See Problem 5.10.1). Thus, we expect the system to exhibit a periodic solution with amplitudeapproximately a = (8=5)1=4 and frequency approximately 1rad=s.

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EEE 586 HW # 8+ SOLUTIONS

Problem 8.15a. Using V(x) = 5x21+ 2x1x2+ 2x

22, we have

_V = (10x1+ 2x2)x2+ (2x1+ 4x2)[x1 x2 (2x2+ x1)(1 x22)]=

2x21+ 4x1x2

2x22

2(x1+ 2x2)

2(1

x22)

Forjx2j 1 we have _V 2(x1 x2)2 0. Thus, _V = 0 ) x1 x2 0 ) _x1 _x2 0 ) 3x2(2 x2)2 0 )x2 0. Hence, by LaSalle's theorem, we conclude that the origin is asymptotically stable.

b. Observe rst that _V 0 for allx 2 S. To show that Sis a region of attraction, we should show that it ispositively invariant. For this, it is enough to show that trajectories at the boundary ofSdo not leave S. For theboundaries that belong on the Lyapunov level set V(x) = 5 this is automatic since _V 0. For the boundariesthat belong to the absolute value conditionjx2j 1 we evaluate the derivative ofjx2j2 on the constraint:

d

dtx22= 2x2_x2

jx2j=1

= 2x2(x1+ x2)

It follows that the right-hand side is nonpositive on the boundaries so trajectories that start in S stay in S.Since the set is compact and _V 0 inS, by LaSalle's theorem, all trajectories starting inside it, must approachthe largest invariant set infx 2 Sj _V = 0g. From part (a), the largest invariant set is the origin, hence Sis anestimate of the region of attraction.

Problem 8.16Consider the Lyapunov function candidate

V(x) =1

2x22+

Z x10

(y y3) dy= 12

x22+1

2x21

1

4x41

which is positive denite in the regionjx1j P x, we get

_V = x>"

P

A 1

2 BB>P

+

A 1

2 BB>P>

P#

x + 2x>P Dg(t; y)

= 1

x>P DD>P x 1

x>C>Cx x>Qx + 2x>P Dg(t; y)

1

x>P DD>P x 1

kyk2 x>Qx + 2kx>P Dkkkyk

1kzk2 1

kyk2 x>Qx + 2kzkkyk; after settingz = D>P x

x>Qx; for

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Hence, the origin is GAS.

Problem 10.6 (1{4)

1. H(s) = ss2s+1 . Since H(s) is not Hurwitz we have the case that the nonlinearity sector [ ; ] should

have 0< < . Using the loop transformation with >1, GT(s) =G(s)= (1 + G(s)), we need to ndK such that ZT(s) =I+ KGT(s) is strictly positive real. In this case,GT is PR and therefore K GT is

PR for any K > 0. Hence,ZT is SPR for any K >0. So the nonlinearity sector would be [; ] for anyxed >1 and arbitrarily large.

Alternatively, using the small-gain version of the circle criterion (Nyquist plot contained in the circle) letthe feedback gain be c. The transformed linear system isHc(s) =

ss2+(c1)s+1 . For large c, the L2 gain

ofHc (peak magnitude) is slightly larger than 1=(c 1), say 1=(c 1 ), where >0 signies a smallconstant. Hence the sector of the transformed nonlinearity is [c+ 1 +; c 1 ] yielding an originalnonlinearity sector [1 + ; 2c 1 ], which is identical to the previous result.

2. H(s) = 1(s+1)(s+2) . In this case H(s) is Hurwitz and we may use nonlinearities of the type 0 = < . We

compute min! 2:5. Notice that the lower limit is zero when c = 8:74 forwhich the sector is [0; 17:48], same as before.

3. H(s) = 1(s+1)2

. Using the same procedure as in part 2, the nonlinearity sectors are any closed subsector

of (1; 2c + 1), for1< c 1, or (c 2pc; c + 2pc) for c >1.Notice that in these simple problems, it is possible to derive analytical expressions for the L2 gains and,hence, the sectors in terms of the loop-transformation constant. In general, we only expect a numericalevaluation of such expressions.

4. H(s) = s2

0:5

(s+1)(s2+1) . After a loop transformation the transformed system would be stable for 0 < c < 2(Routh). Using the following MATLAB script, we can evaluate the sector bounds for dierent values ofc.

h=tf([1 0 -.5],conv([1 1],[1 0 1]))

c=[0:.1:2];lb=0*c;hb=0*c;

for i=1:length(c)

mag=bode(feedback(h,c(i)));

m=1/max(mag);

lb(i)=c(i)-m;hb(i)=c(i)+m;

end

plot(c,lb,c,hb)

The largest radius sector occurs at c 1:2 and is (0:6880; 1:7120) and the system will be absolutely stablefor nonlinearities in any closed subsector. On the other hand, to nd the sector with the largest upperlimit, we observe that when the peak magnitude is at DC, the sector takes the form ( 2 + 2c; 2). Thesmallestc for this is approximately 1.7, yielding the sector (1:4; 2). (Keep in mind that these values areproduced numerically and small discrpancies may occur, especially near the stability boundary.)

Problem 10.11

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From equation (10.10),

P A + A>P = L>L P )P A + A>P+

2P+

2P = L>L

SinceD is nonsingular this implies that D + D> is nonsingular, hence W>Wis nonsingular (from (10.12)), and

W is nonsingular. Then

P

A +

2I +

A +

2I

>P = L>W W1 W>1 W>L

P

A +

2I +

A +

2I

>P = L>WW>W1 W>L

P

A +

2I +

A +

2I

>P = L>WD+ D>1 W>L

Using (10.11)L>W=C> P B we obtain

P

A +

2I +

A +

2I

>P+

C> P B D+ D>1 C B>P

Problem 10.12

(a). The original system is given by

_x = Ax + BLsat

L1F x

(1)

Adding and subtracting BF x we get

_x = Ax + BLsat

L1F x

+ BF x BF x_x = (A + BF)x + B

Lsat

L1F x

F xNext, deningu1= Lsat

L1y

F x,y = F x, the system becomes_x = (A + BF)x + Buy = F xu = r u1; r 0

which has the desired form.

(b). Starting with (1), we can write the closed-loop system as an interconnection of a transfer matrix givenbyG(s) = [A;BL; L1F; 0] and the nonlinearity (y) = sat(y) with a negative feedback. The sector forthe saturation nonlinearity is [0; 1] but, sinceG is unstable, it is clear that 0 is not admissible and we canonly expect a local stability result (absolute stability with nite domain). To show the feasibility of sucha result we can make the following argument.

A condition for asymptotic stability for any nonlinearity in the sector [ kmin; kmax] would be \ZT(s) isSPR," where

ZT(s) = I+ (kmax kmin) G(s) [I+ kminG(s)]1Since F is a stabilizing feedback gain, G(s) [I+ kminG(s)]

1represents a stable system with nite gain

forkminsuciently close to 1. Moreover, choosing kmax kmin suciently small, i.e.,kmaxalso close to 1,the whole second term can be made arbitrarily small and therefore ZTshould be SPR.

The issue is now to choose kmax; kmin to maximize the domain of absolute stability. For this kmax shouldbe larger than 1 and kmin should be as small as possible.

Alternatively, using a loop transformation, we can write GT(s) = G(s) [I+ G(s)]1

, where GT(s) isstable. GT(s) is absolutely stable for nonlinearities in a sector [ 1 ; 1 ], where is the gain of GT(s)

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and depends on . Then the original system G(s) is is absolutely stable for nonlinearities in a sector[ 1

; + 1

]. To nd a suitable we could solve the minimization problem

min

1

s.t. 1

It should be kept in mind that the solutions of these optimization problems may not be unique and/or theminimizers may not be admissible. Still, a feasible solution can be selected within a prescribed tolerancefrom the minimum (inmum). Then, using Theorem 10.2 and Problem 10.11, we can compute the matrixP that denes the associated Lyapunov function and describe the region of attraction as the maximallevel set contained in the region were the Lyapunov derivative is negative denite.

(c). Using the matrices of Problem 5.2.d and the procedure described above, we nd that = 0:8037 and= 5:0929, corresponding to the sector [0:6074; 1]. It now follows that the Lyapunov derivative is negativein the region

xj jF xj1= 10:6074 = 2:7105

. The matrix P, dening the Lyapunov function, is computedby solving the Riccati equation of Problem 10.11 with

A = 1 0:6074

0:6074 2:054

; B =

1:414 00 1:414

; C =

0 0:2776

0:2776 0:5275

; D =I

This solution is

P=

1:0183E2 1:1187E2

1:1187E2 3:2676E2

The maximal level set that is contained in the constraint set

xj jF xj1= 10:6074 = 2:7105

(a polytope)can now be computed along the lines of Problem 5.2 (HW 5). Here we adopt a computational approach,employing the contour function of Matlab to draw the constraint set and level sets for various constants.An example of such a procedure is:

p=[1.0183 1.1187;1.1187 3.2676]*1e-2;f=[0 1;1 -1.9];

[X,Y] = meshgrid(-4:.1:4, -4:.1:4);

V=X.^2*p(1,1)+X.*Y*2*p(1,2)+Y.^2*p(2,2);

y1=f(1,1)*X+f(1,2)*Y;y1=abs(y1);

y2=f(2,1)*X+f(2,2)*Y;y2=abs(y2);

clf, hold off

contour(X,Y,y1,[2.71 2.71]),hold on

contour(X,Y,y2,[2.71,2.71])

contour(X,Y,V,[1 1]*0.01)

Repeating the last command with dierent multipliers we nd the estimate for the region of attraction asx 2 IR2 jV(x) 0:013g

Problem 10.16LetN= f2 IRpjk k kg,k = 1;:::;pbe a convex parallelepiped. Any element 2 Ncan be written

as a convex combination of the 2p vertices such that

=2pXk=1

k(k);

2pXk=1

k= 1; 0 k 1; 8k= 1;:::;p

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with (k) being the k-th vertex.1 Then we can write every element aij() as

aij() =2pXk=1

kaij

(k)

With this, we can write the matrixA() as

A() =2pXk=1

kAk

(k)

whereAk

(k)

is the value ofaij

(k)

for a given k .Next, letv(x) = x>P x. Then

_V(x) = x>

A>()P+ P A()

x

= x>

2pXk=1

kA>k

(k)

P+ P2pXk=1

kAk

(k)!

x

Using the assumption

A>k (k)P+PAk

(k) I, we multiply the k-th inequality by k and perform asummation to obtain2pXk=1

kA>k

(k)

P+ P2pXk=1

kAk

(k)

I2pXk=1

k

It now follows that

_V(x) x>

I2pXk=1

k

!x= x>x

which is negative denite. ThereforeA() is Hurwitz for any 2 Nand, more important, all systems admitthe same Lyapunov function.

Problem 10.33 (1,3) For odd, time-invariant, memoryless nonlinearities

(a) = 2

a

Z 0

(a sin )sin d

1. (y) = y5

(a) = 2

a

Z 0

a5 sin6 d=2a4

1

6sin5 cos

0

+5

6

1

4sin3 cos

0

+3

4

1

2 1

4sin 2

0

= 2a4

5

6

3

4

2

= 5a4

8

1It is straightforward to show that any summation of the above form results in an element of N. On the other hand, any elementofNcan be written as a linear combination of the vertices with nonnegative coecients whose sum is 1. To show this, consider anelement2N and a vertix (1). Let 1 be the intersection of the boundary ofNand the line connecting and (1). Then is aconvex combination of (1) and 1, i.e., = 1(1) + 111, with 1+ 11 = 1 and 1; 11 0. Next, repeat this expansion for1, starting with the next vertix (2), etc., until all the vertices are used. While the properties of the expansion coecients followeasily from the fact that each step is a convex combination, a complete proof can be obtained by an induction argument. Noticethat such expansions are not unique.

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3. From the graph of the function we can write

(y) =

A + ky; y 0A + ky; y 1. Case a 1.

(a) = 2a

Z 0

a sin2 d= 2a

ha2 a

4sin 2

i

0= 1

Case a >1.

(a) = 2

a

Z 0

(a sin )sin d= 4

a

Z 2

0

(a sin )sin d

= 4

a

"Z 0

a sin2 d+

Z 2

sin d

#; = arcsin

1

a

= 4

a

a2 a

4sin 2

0 (cos )j2

= 4

aa

2 +1

2cos

= 2

"arcsin

1

a

+

1

a

r1 1

a2

#

where we used = arcsin1a

= arccos

q1 1

a2

to obtain the last expression. Notice that 0

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Problem 12.21. Linearization at the origin yields

A=

1 11 0

; B =

01

; C= (0 1) ;

The state feedback u = Kx = [7 4]xassigns the closed-loop eigenvalues at1; 2.Using the observer-based controller

_x= (A BK HC)x + Hy; u= Kx

whereH= [43; 12], which assigns the observer eigenvalues at5; 6, stabilizes the origin, at least locally.(ROA computation can be performed as usual, by nding the largest level set V(x) = cinside _V(x) 0. The

latter is guaranteed to hold in a nbhd of the origin since the linearization of the feedback system is exponentiallystable. One can now try various initial conditions to obtain insight on the consrvatism of this estimate. Also,by the nite gain property of the exponentially stable linearization, the system will have bounded response tosmall commands.)

2. Linearization at the origin yields

A=0@ 1 1 01 0 1

0 0 0

1A ; B= 0@ 001

1A ; C= (0 1 0) ;The state feedback u = Kx = [16 17 7]x assigns the closed-loop eigenvalues at1; 2; 3.

Using the observer-based controller

_x= (A BK HC)x + Hy; u= Kx

whereH= [335; 19;210], which assigns the observer eigenvalues at 5; 6; 7, stabilizes the origin, at leastlocally.

3. Linearization at the origin yields

A=0@ 1 1 01 1 0

1 0 21A ; B= 0@ 01

0

1A ; C= (1 0 0) ;The state feedback u = Kx = [1 2 0]x assigns the closed-loop eigenvalues at1; 2; 3.

Using the observer-based controller

_x= (A BK HC)x + Hy; u= Kx

where H= [9; 21; 1], which assigns the observer eigenvalues at5; 6; 2 and, hence, stabilizes the origin, atleast locally.

(ROA computation can be performed as usual, by nding the largest level set V(x) = cinside _V(x) 0. Thelatter is guaranteed to hold in a nbhd of the origin since the linearization of the feedback system is exponentiallystable. One can now try various initial conditions to obtain insight on the consrvatism of this estimate. Also,

by the nite gain property of the exponentially stable linearization, the system will have bounded response tosmall commands.)