Chapter 2...chapter 2 two-di mensional systems 9 3hduvrq(gxfdwlrq/wg $oouljkwvuhvhuyhg...

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Chapter 2

Two-Dimensional Systems

9

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10 CHAPTER 2.



http://TestBankSolutionManual.eu/Solution-for-Nonlinear-Control-by-Hassan-K-Khalil-International-version-

11




12 CHAPTER 2.




13




14 CHAPTER 2.




15




16 CHAPTER 2.




17




18 CHAPTER 2.




19




20 CHAPTER 2.




21




22 CHAPTER 2.




23




24 CHAPTER 2.

2.3 Equilibrium points:

0 = x1[ν(x2)− u], 0 = −(α− 1)x1ν(x2) + (α− x2)u

From the first equation, x1 = 0 or ν(x2) = u. For x1 = 0, x2 = α. By differentiationof ν(x2), it can be seen that max{ν(x2)} = 1/[γ + 2

√

β(1 − β − γ)] = 1.22. Thus,for u > 1.22, there is a unique equilibrium point at x = (0, α) = (0, 23). Foru < 1.22, the equation ν(x2) = u has two solutions, which are the roots of

(1− β − γ)x22 + (γ − 1/u)x2 + β = 0

For each roots x2, x1 is given by x1 = (α− x2)/(α− 1).

(a) For u = 0.5, there are two equilibrium points at (0, 23) and (1.033, 0.275). Asecond root of the foregoing equation at x2 = 35.475 was rejected because itresults in a negative x1.

∂f

∂x=

[(ν(x2)− u) x1ν

′(x2)−(α− 1)ν(x2) (−(α− 1)x1ν

′(x2)− u)

]

∂f

∂x

∣∣∣∣(0,23)

=

[0.16359 0

∗ −0.5

]

, λ = 0.16359,−0.5 =⇒ Saddle

∂f

∂x

∣∣∣∣(1.033,0.275)

=

[1.05× 10−4 1.323

−11 −29.596

]

, λ = −0.5,−29.105 =⇒ Stable node

The phase portrait is shown in Figure 2.12. The stable trajectories of thesaddle form a separatrix; all trajectories to its right converge to the stablenode.




25

x ’ = x (y/(0.39 + 0.57 y + 0.04 y2) − 0.5)y ’ = − (23 − 1) x y/(0.39 + 0.57 y + 0.04 y2) + (23 − y) 0.5

0 0.5 1 1.5 2 2.5 3 3.5 4

0

5

10

15

20

25

x

y

Figure 2.12: Exercise 2.3 (a).




26 CHAPTER 2.

(b) For u = 1, there are three equilibrium points at (0, 23), (1, 1), and (0.6023, 9.75).

∂f

∂x

∣∣∣∣(0,23)

=

[−0.3364 0

∗ −1

]

, λ = −0.3364,−1 =⇒ Stable node

∂f

∂x

∣∣∣∣(1,1)

=

[0 0.35

−22 −8.7

]

, λ = −1,−7.7 =⇒ Stable node

∂f

∂x

∣∣∣∣(0.6023,9.75)

=

[0 −0.0216

−22 −0.5243

]

, λ = 0.4754,−0.9997 =⇒ Saddle

The phase portrait is shown in Figure 2.13. The stable trajectories of thesaddle form a separatrix; all trajectories to its right converge to (1, 1), whilethose on the left converge to (0, 23).

x ’ = x (y/(0.39 + 0.57 y + 0.04 y2) − 1)y ’ = − (23 − 1) x y/(0.39 + 0.57 y + 0.04 y2) + (23 − y) 1

0 0.5 1 1.5 2 2.5 3 3.5 4

0

5

10

15

20

25

x

y

Figure 2.13: Exercise 2.3 (b).




27

(c) For u = 1.5, there is a unique equilibrium point at (0, 23).

∂f

∂x

∣∣∣∣(0,23)

=

[−0.83641 0

∗ −1.5

]

, λ = −0.83641,−1.5 =⇒ Stable node

The phase portrait is shown in Figure 2.14. All trajectories converge to thestable node.

x ’ = x (y/(0.39 + 0.57 y + 0.04 y2) − 1.5)y ’ = − (23 − 1) x y/(0.39 + 0.57 y + 0.04 y2) + (23 − y) 1.5

0 0.5 1 1.5 2 2.5 3 3.5 4

0

5

10

15

20

25

x

y

Figure 2.14: Exercise 2.3 (c).




28 CHAPTER 2.

2.4x1 = x2, x2 = −x1 − εh′(x1)x2

There is a unique equilibrium point at (0, 0).

∂f

∂x=

[0 1

−1− εh′′(x1)x2 −εh′(x1)

]

,∂f

∂x

∣∣∣∣(0,0)

=

[0 1−1 −εh′(0)

]

(a) h′(v) = −1 + 3v2 − v4 + 114v

6.

∂f

∂x

∣∣∣∣(0,0)

=

[0 1−1 1

]

, λ =1

2(1± j

√3) =⇒ Unstable focus

The phase portrait is shown in Figure 2.15. There are three limit cycles. theinner one is stable, the intermediate one unstable, and the outer one stable.All trajectories inside the unstable limit cycle, except the equilibrium point,approach the inner stable limit cycle. All trajectories outside the unstablelimit cycle approach the outer stable limit cycle.

x ’ = yy ’ = − x − y ( − 1 + 3 x2 − x4 + x6/14)

−6 −4 −2 0 2 4 6

−6

−4

−2

0

2

4

6

x

y

Figure 2.15: Exercise 2.4 (a).




29

(b) h′(v) = −1 + 3v2 − v4 + 112v

6.

∂f

∂x

∣∣∣∣(0,0)

=

[0 1−1 1

]

, λ =1

2(1± j

√3) =⇒ Unstable focus

The phase portrait is shown in Figure 2.16. There is a stable limit cycle. Alltrajectories, except the equilibrium point, approach the limit cycle.

x ’ = yy ’ = − x − y ( − 1 + 3 x2 − x4 + x6/12)

−6 −4 −2 0 2 4 6

−6

−4

−2

0

2

4

6

x

y

Figure 2.16: Exercise 2.4 (b).




30 CHAPTER 2.

2.5 (a)

C1v1 =v2 − v1 − g(v2)

R1, C2v2 =

v1 + g(v2)− v2R1

− v2R2

x1 =1

C1 + C2(C1v1+C2v2) and x2 = v2 =⇒ v1 =

(C1 + C2)x1 − C2x2C1

x1 =C1v1 + C2v2C1 + C2

= − x2(C1 + C2)R2

x2 = v2 =1

C2

[v1 + g(v2)− v2

R1− v2R2

]

=1

C1C2R1[(C1 + C2)x1 − C2x2] +

1

C2

[g(x2)− x2

R1− x2R2

]

=C1 + C2

C1C2R1(x1 − x2) +

1

C2R1g(x2)−

1

C2R2x2

(b)

x1 = − 1

2CRx2, x2 =

2

CR(x1 − x2) +

1

CRg(x2)−

1

CRx2

In the τ = t/(CR) time scale,

x1 = −1

2x2, x2 = 2(x1 − x2) + g(x2)− x2 = 2x1 − 3x2 + g(x2)

where xi denotes dxi/dτ . Equilibrium points:

0 = x2, 0 = 2x1 − 3x2 + g(x2)

There is a unique equilibrium point at (0, 0).

∂f

∂x=

[0 − 1

22 −3 + g′(x2)

]

∂f

∂x

∣∣∣∣(0,0)

=

[0 −0.52 0.234

]

, λ = 0.117± j0.9931 =⇒ Unstable focus

The phase portrait is shown in Figure 2.17. There are two limit cycles; theinner one is stable and the outer one is unstable. All trajectories inside theunstable limit cycle, except the equilibrium point, approach the stable limitcycle.




31

x ’ = − 0.5 yy ’ = 2 x − 3 y + 3.234 y − 2.195 y3 + 0.666 y5

−4 −3 −2 −1 0 1 2 3 4

−4

−3

−2

−1

0

1

2

3

4

x

y

Figure 2.17: Exercise 2.5.




32 CHAPTER 2.

2.6 (a) Set

g(x1) = −x1 +4U2

27(1− x1)2

The equilibrium points are the solutions of g(x1) = 0, which can be writtenas

x1(1− x1)2 =

4U2

27

By differentiation, it can be shown that, within the interval [0, 1], the left-hand side of the foregoing equation has a maximum 4/27 at x1 = 1

3 . Thusfor U < 1, the equation g(x1) = 0 has two solutions xa <

13 and xb >

13 . By

sketching g(x), it can be seen that g′(xa) < 0 and g′(xb) > 0.

∂f

∂x=

[0 1

g′(x1) −2ζ

]

The characteristic equation is

λ2 + 2ζλ− g′(x1) = 0

Since g′(xa) < 0, the equilibrium point (xa, 0) is a stable node or stable focus.Since g′(xb) > 0, the equilibrium point (xb, 0) is a saddle.

(b)

x1 = x2, x2 = −x1 − x2 +1

12(1− x1)2

The equation

x1(1− x21) =1

12

has three solutions x1 = 1.2574, 0.6388, and 0.1037. The solution x1 = 1.2574is outside the set {0 ≤ x1 ≤ 1 − δ}. Hence, the two equilibrium points are(0.1037, 0) and (0.6388, 0).

∂f

∂x

∣∣∣∣(0.1037,0)

=

[0 1

−0.7685 −1

]

, λ = −0.5± j0.72 =⇒ Stable focus

∂f

∂x

∣∣∣∣(0.6388,0)

=

[0 1

2.5375 −1

]

, λ = 1.1696,−2.1696 =⇒ Saddle

The phase portrait is shown in Figure 2.18. Because the solution is restrictedto x1 ≥ 0, there is a region whose boundary is formed of three segments:

(1) the stable trajectories of the saddle;

(2) the trajectory that converges to the stable focus and touches the verticalline x1 = 0;

(3) the segment of the line x1 = 0 that connects the trajectories in (1) and(2).

All trajectories inside this region converge to the stable focus. Trajectoriesoutside the region diverge to ∞ or leave the region of validity {0 ≤ x1 ≤ 1−δ}.




33

x ’ = yy ’ = − x − y + 1/(12 (1 − x)2)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

x

y





34 CHAPTER 2.

2.7 (a) With u = 0, the system is

x1 = x2, x2 = sinx1

Equilibrium points:

0 = x2, 0 = sinx1 =⇒ x = (nπ, 0), n = 0,±1,±2, . . .

For even n,∂f

∂x

∣∣∣∣(nπ,0)

=

[0 11 0

]

, λ = 1,−1 =⇒ Saddle

For odd n,∂f

∂x

∣∣∣∣(nπ,0)

=

[0 1−1 0

]

, λ = ±j

Linearization fails, but it can be seen from the phase portrait (Figure 2.19)that the equilibrium point is a center.

x ’ = y y ’ = sin(x)

−8 −6 −4 −2 0 2 4 6 8

−4

−3

−2

−1

0

1

2

3

4

x

y


(b)

E = 1 + cosx1 +1

2x22

E = −x2 sinx1 + x2(sinx1 + u cosx1) = ux2 cosx1 = k(2− E)(x2 cosx1)2

E < 2 and x2 cosx1 6= 0 =⇒ E > 0




35

(c) The closed-loop system is

x1 = x2, x2 = sinx1 + 0.1(1− cosx1 −1

2x22)x2(cosx1)

2

The equilibrium points remain as (nπ, 0). For even n,

∂f

∂x

∣∣∣∣(nπ,0)

=

[0 11 0

]

, λ = 1,−1 =⇒ Saddle

For odd n,

∂f

∂x

∣∣∣∣(nπ,0)

=

[0 1−1 0.2

]

, λ = 0.1± j0.99 =⇒ Unstable focus

The phase portrait is shown in Figure 2.20. The stable and unstable trajecto-ries of the saddles at (π, 0) and (−π, 0) define a region that includes the origin(0, 0) on its boundary. A trajectory starting inside this region oscillates, ap-proaching the boundary of the region gradually. It comes closer and closer tothe origin. When it is close enough we can switch to another controller thatmakes origin a stable node or stable focus. Such controller can be designedby linearization since it needs to work only in the neighborhood of the origin.

x ’ = yy ’ = sin(x) + 0.1 (2 − E) y (cos(x))2

E = 1 + cos(x) + 0.5 y2

−8 −6 −4 −2 0 2 4 6 8

−4

−3

−2

−1

0

1

2

3

4

x

y





36 CHAPTER 2.




37




38 CHAPTER 2.




39




40 CHAPTER 2.




41




42 CHAPTER 2.




43

2.10 Equilibrium points:

0 = u− 1

3u3 − w + I, 0 = b0 + b1u− w

=⇒ b0 − I + b1u = u− 1

3u3

The two sides of the preceding equation are sketched in Figure 2.27 for b1 > 1 andb1 < 1. For b1 > 1 there is a unique solution. For b1 < 1, there are three cases,depending on the value of b0 − I. In case A there is a unique solution on the leftbranch of the curve u − 1

3u3. In case B, there are three solutions on the three

branches of the curve u− 13u

3, where the middle branch has positive slope while thetwo outer branches have negative slopes. In case C, there is a unique solution onthe right branch of the curve u− 1

3u3 that has negative slope. Since the derivative

of u − 13u

3 is 1 − u2, for a solution on the left or right branches, 1 − u2 < 0, whilea solution in the middle branch satisfies 1 − u2 > 0. The Jacobian matrix at the

u

u − u3/3

b0 − I + b

1 u

b1 > 1

u

A

B

C

b1 < 1


equilibrium point (u, w) is given by

∂f

∂x

∣∣∣∣(u,w)

=

[(1 − u2) −1εb1 −ε

]

and its characteristic equation is

λ2 + (ε+ u2 − 1)λ+ ε(b1 + u2 − 1) = 0

(a) For b1 > 1 there is a unique equilibrium point (u, w). If it is on one of thetwo outer branches of the curve u− 1

3u3, then 1− u2 < 0. Hence, ε+ u2 − 1

and b1 + u2 − 1 are positive. The equilibrium point is a stable node or stablefocus. If the equilibrium point is in the middle branch, then 1 − u2 > 0. Inthis case the signs of the coefficients of the characteristic equation depend onthe values of ε and b1.




44 CHAPTER 2.

(b) For b1 < 1, there are three cases as discussed earlier. In cases A and C, there isa unique equilibrium point on a branch of the curve u− 1

3u3 that has negative

slope. Therefore, as in part (a), each of these equilibrium points is eitherstable node or stable focus. In case B, there are three equilibrium points.The two equilibrium points on the outer branches will be stable nodes orstable foci. For the equilibrium point on the middle branch, 1− u2 > 0. Thesigns of the coefficients of the characteristic equation depend on the values ofε and b1.

(c) For b1 = 1.5 and I = 0, the equilibrium point is (−1.5444,−0.31656) and theeigenvalues of the Jacobian matrix are −1.2552 and −0.22984. Hence, it isa stable node. The phase portrait is shown in Figure 2.28. All trajectoriesconverge to the stable node.

x ’ = x − x3/3 − y + I y ’ = 0.1 (2 + 1.5 x − y)

I = 0

−4 −3 −2 −1 0 1 2 3 4

−4

−3

−2

−1

0

1

2

3

4

x

y

Figure 2.28: Exercise 2.10 (c): b1 = 1.5; I = 0.




45

For b1 = 1.5 and I = 2, the equilibrium point is (0, 2) and the eigenvalues ofthe Jacobian matrix are 0.8405 and 0.0595. Hence, it is unstable node. Thephase portrait is shown in Figure 2.29. The equilibrium point is surrounded bya stable limit cycle. All trajectories, except the equilibrium point, approachthe limit cycle.

x ’ = x − x3/3 − y + I y ’ = 0.1 (2 + 1.5 x − y)

I = 2

−3 −2 −1 0 1 2 3

−2

−1

0

1

2

3

4

x

y

Figure 2.29: Exercise 2.10 (c): b1 = 1.5; I = 2.




46 CHAPTER 2.

(d) For b1 = 0.5 and I = 0, the equilibrium point is (−2.0905, 0.95476) and theeigenvalues of the Jacobian matrix are −3.3547 and −0.1154. Hence, it isa stable node. The phase portrait is shown in Figure 2.30. All trajectoriesconverge to the stable node.

x ’ = x − x3/3 − y + I y ’ = 0.1 (2 + 0.5 x − y)

I = 0

−3 −2 −1 0 1 2 3

−2

−1

0

1

2

3

4

x

y

Figure 2.30: Exercise 2.10 (d): b1 = 0.5; I = 0.




47

For b1 = 0.5 and I = 2, There are three equilibrium points at (0, 2), (1.2247, 2.6124),and (−1.2247, 1.3876). For the equilibrium point (0, 2) the eigenvalues of theJacobian matrix are 0.9525 and −0.05249. It is a saddle. For the equilibriumpoint (1.2247, 2.6124) the eigenvalues of the Jacobian matrix are −0.3± j0.1.It is a stable focus. For the equilibrium point (−1.2247, 1.3876) the eigenval-ues of the Jacobian matrix are −0.3 ± j0.1. It is a stable focus. The phaseportrait is shown in Figure 2.31. The stable trajectories of the saddle forma separatrix that divides the plane into two haves. All trajectories in theright half converge to (1.2247, 2.6124) and those on the left half converge to(−1.2247, 1.3876). Although the equilibrium point is a stable focus it appearsin the portrait as if it was a stable node because the absolute value of theimaginary part of the eigenvalues is smaller that the absolute value of the realpart.

x ’ = x − x3/3 − y + I y ’ = 0.1 (2 + 0.5 x − y)

I = 2

−3 −2 −1 0 1 2 3

0

0.5

1

1.5

2

2.5

3

3.5

4

x

y

Figure 2.31: Exercise 2.10 (d): b1 = 0.5; I = 2.




48 CHAPTER 2.




49




50 CHAPTER 2.

2.12 (a)

z1 =√γx1, z2 =

x2√γ, τ = γt

dz1dτ

=1

γ

dz1dt

=1√γ

dx1dt

=x2√γ= z2

dz2dτ

=1

γ

dz2dt

=1

γ√γ

dx2dt

=1

γ√γ[0.1x32 − γ2x1 − γx2] = 0.1z32 − z1 − z2

(b) There is a unique equilibrium point at the origin. The eigenvalues of theJacobian matrix at the origin are −0.5 ± j0.866. The equilibrium point is astable focus. The phase portrait is shown in Figure 2.33. There is unstablelimit cycle that encloses the origin. All trajectories inside the limit cycleconverge to the origin.

x ’ = yy ’ = − x − y + 0.1 y3

−4 −3 −2 −1 0 1 2 3 4

−4

−3

−2

−1

0

1

2

3

4

x

y


(c) In the z1-z2 plane, the limit cycle intersects the z1-axis at a = (4, 0) andb = (−4, 0). It intersects the z2-axis at c = (0, 3.5) and d = (0,−3.5). Whenmapped into the x1-x2 plane, the points a, b, c, d map into

a′ =

(4√γ, 0

)

, b′ =

(−4√γ, 0

)

, c′ = (0, 3.5√γ) , d′ = (0,−3.5

√γ)

As γ increases, the limit cycle shrinks in the x1 direction and expands in thex2 direction.




Chapter 2...chapter 2 two-di mensional systems 9 3hduvrq(gxfdwlrq/wg $oouljkwvuhvhuyhg...

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