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Exercise~15)([143]) Consider the system

1 = z2 ( 2 2)

2 -Xl - X2 sat X 2 - X3= 3 = X3 sat(x~ -X)

where sat(middot) is the saturation function Show that the origin is the unique equiIibmiddot rium point and use V(z) =xT x to show that it is globally asymptotically stable

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Exercise ~Consider the system

1 = X2 - g(t)X1(X~ + X~) 2 = -Xl -9(t)X2(Z~ + x~)

where g(t) is ~ntinuously differentiable bounded and g(t) k gt 0 for all t gt o Is the origm UnIformly asymptotically stable Is it exponentially stable shy

o = X2 o = -ZI - Z2 sat(x~ - z~) =gt ZI =0

o = X3 sat(x~ - x~) =gt Z3 sat(-z~) =0 =gt Z3 =0

Hence the origin is the unique equilibrium point consider

Vex) = xTx = x~+x~+zi Vex) = 2[XIX2 - IX2 - x~sat(z~ - z) + zsat(z~ - z)]

= -2(z~ - z~)sat(x~ - z~) $ 0

11 is negative semidefinite v=0 =gt z~(t) == zi(t) =gt 3(t) == 0

Hence both X2(t) and Z3(t) must be constants This implies that 2(t) == O From the second state equation We conclude that Xl(t) == O Then the first state equation implies that Z2(t) == O Consequently Z3(t) == o IY LaSalles theorem (Corollary 32) the origin is globally asymptotically stable

~)Try Vex) =(x~ + x~) t

v = XIX2 - g(t)x~(x~ + 3) - XIZ2 - g(t)xi(xi + xi) = -g(t)(z~ + z~)2 $ -4kV2(x)

Hence Vet x) is negative definite and the origin is uniformly asymptotically stable From the inequality V lt -4kV2 we cannot conclude exponential stability Let us try linearization Assume get) is bounded-

_ 8f t r I _[-3g(t)xyen - g(t)x~ 1-2g(t)r1X2 ] I =[0 1]A(t) - 8x ( ) c=o - -1 - 2g(t)IX2 -3g(t)x~ - g(t)xyen c=o -1 0

A is a constant matrix that is not Hurwitz Hence by Theorem 313 we conclude that the origin is not

exponentially stable

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bt~ ~~ 91-1


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bt~ ~~ 91-1


P0-~OD1C ORB TS

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A-eso e--I~ 50eu~~ ~o+- S~ UU-- 1-4 ~~ -II(




~C(1) ~ 5-o1Yte 7r~ ~



1J-e ~6S c3b IV

~ + ampf2 ampL1 ~

C5 ~)t ~oleAl~~ ~ero 9-r)al ~~ l101-~




1 = z2 ( 2 2)

2 -Xl - X2 sat X 2 - X3= 3 = X3 sat(x~ -X)


~ -~

~ _ ~ +-b~ _~2~_x~

bL-D




1 = X2 - g(t)X1(X~ + X~) 2 = -Xl -9(t)X2(Z~ + x~)






= -2(z~ - z~)sat(x~ - z~) $ 0



~)Try Vex) =(x~ + x~) t







A-eso e--I~ 50eu~~ ~o+- S~ UU-- 1-4 ~~ -II(




~C(1) ~ 5-o1Yte 7r~ ~



1J-e ~6S c3b IV

~ + ampf2 ampL1 ~

C5 ~)t ~oleAl~~ ~ero 9-r)al ~~ l101-~




1 = z2 ( 2 2)

2 -Xl - X2 sat X 2 - X3= 3 = X3 sat(x~ -X)


~ -~

~ _ ~ +-b~ _~2~_x~

bL-D




1 = X2 - g(t)X1(X~ + X~) 2 = -Xl -9(t)X2(Z~ + x~)






= -2(z~ - z~)sat(x~ - z~) $ 0



~)Try Vex) =(x~ + x~) t







1 = z2 ( 2 2)

2 -Xl - X2 sat X 2 - X3= 3 = X3 sat(x~ -X)


~ -~

~ _ ~ +-b~ _~2~_x~

bL-D




1 = X2 - g(t)X1(X~ + X~) 2 = -Xl -9(t)X2(Z~ + x~)






= -2(z~ - z~)sat(x~ - z~) $ 0



~)Try Vex) =(x~ + x~) t






(3..2J Le- ~ :C~t)~CQ Pi---GnCLrL€ - Rutgers ECEgajic/pdffiles/Nonlinear... · 2015-06-19 · (1)...

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