linearisation & lyapunov equation - university of … tutorial on eg4321/eg7040 nonlinear...

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

2nd Tutorial on EG4321/EG7040 Nonlinear ControlLinearisation & Lyapunov Equation

Dr. Angeliki Lekka1

1Control Systems Research GroupDepartment of Engineering, University of Leicester

March 2, 2017

Dr. Angeliki Lekka ([email protected]) Linearisation & Lyapunov March 2, 2017 1 / 22

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Important things to remember

Linearisation of Nonlinear Systems

if all eigenvalues of A are strictly in the LHP, then linearised system isstrictly stable, hence equilibrium point is stable

if at least one of the eigenvalues of A is in the RHP, then linearisedsystem is unstable, hence equilibrium point is unstable

if the eigenvalues of A are in the LHP but one or more are on the jωaxis, then conclusions w.r.t. stability cannot be derived from thelinearised system


.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Important things to remember

From Lecture 4

If matrix A has negative eigenvalues, there exists a matrix P > 0 satisfyingthe Lyapunov equation A′P + PA = −Q

Steps to solve the Lyapunov equation

choose a positive definite matrix Q

Usually, Q = I is chosen (there is a reason for this)

solve for P from the Lyapunov equation

P should be square and symmetric (if it is not, we can turn it to one)

check whether P is positive definite

Definiteness defined only for square matrices


.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Exercise Sheet 3, Question 1 (1)

Example

If A has the form

A =

[0 1−1 −1

]determine whether

1 A symmetric positive definite solution P exists to the Lyapunovequation A′P + PA = −I

2 If a solution does exist, find the numerical value of P


.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.


To satisfy the 1st part, the eigenvalues of A must have negative real parts;we must solve the characteristic equation det(λI − A) = 0

det(λI − A) =

∣∣∣∣[ λ 00 λ

]−

[0 1−1 −1

]∣∣∣∣=

∣∣∣∣[ (λ− 0) (0− 1)(0− (−1)) (λ− (−1))

]∣∣∣∣=

∣∣∣∣[ λ −11 λ+ 1

]∣∣∣∣= λ(λ+ 1)− (1(−1)) = 0

= λ2 + λ+ 1 = 0 (1)


.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.


We can solve Eq.1 using the quadratic formula λ1,2 =−β±

√β2−4αγ2α

λ1,2 =−1±

√(−1)2 − 4

2

=−1±

√(−3)

2

=−1± j

√3

2(2)

Since the eigenvalues have negative real parts, a positive definite solutionto the Lyapunov equation does exist


.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.


To answer the 2nd part, we need to solve the Lyapunov EquationA′P + PA = −I numerically:

[0 1−1 −1

]′ [P11 P12

P21 P22

]+

[P11 P12

P21 P22

] [0 1−1 −1

]=

[−1 00 −1

]⇒

[0 −11 −1

] [P11 P12

P21 P22

]+

[P11 P12

P21 P22

] [0 1−1 −1

]=

[−1 00 −1

]⇒

[0 · P11 + (−1 · P21) 0 · P12 + (−1 · P22)1 · P11 + (−1 · P21) 1 · P12 + (−1 · P22)

]+

[0 · P11 + (−1 · P12) 1 · P11 + (−1 · P12)0 · P21 + (−1 · P22) 1 · P21 + (−1 · P22)

]=

[−1 00 −1

]


.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.


⇒[

−P21 −P22

P11 − P21 P12 − P22)

]+

[−P12 P11 − P12

−P22 P21 − P22

]=

[−1 00 −1

]⇒

[−P21 − P12 P11 − P12 − P22

P11 − P21 − P22 −2P22 + P12 + P21

]=

[−1 00 −1

]Since P is chosen symmetric, i.e. P12 = P21, we get:

⇒[

−2P12 P11 − P12 − P22

P11 − P12 − P22 −2P22 + 2P12

]=

[−1 00 −1

]


.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.


So, now we must solve

−2P12 = −1 (3)

P11 − P12 − P22 = 0 (4)

−2P22 + 2P12 = −1 (5)

From Eq.3, P12 = 1/2

Substituting P12 in Eq.5, yields −2P22 + 212 = −1 ⇒ P22 = 1

Substituting P12 and P22 in Eq.4, we getP11 − 1/2 − 1 = 0 ⇒ P11 = 3/2

and hence,

P =

[3/2 1/21/2 1

]Dr. Angeliki Lekka ([email protected]) Linearisation & Lyapunov March 2, 2017 9 / 22

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Exams 2015, Question 1c (1)

Example

A mass-spring-damper system with nonlinear damping is described by:

x1 = x2

x2 = −K

Mx1 −

B

Mx2(1 + x22 )

where K is the spring stiffness, B is the damping coefficient and M is themass.

1 Linearise this system about the point x1 = x2 = 0 and giveexpressions for the “A” matrix of the system

2 Determine whether a symmetric positive definite solution exists to theLyapunov equation A′P + PA = −I , where “A” is your answer to part1. If it is solvable find P. If not explain your answer.


.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.


To linearise the nonlinear mass-spring-damper system we must calculatethe Jacobian matrix A:

A =∂f

∂x

∣∣∣∣x1=x2=0

=

[∂f1∂x1

∂f1∂x2

∂f2∂x1

∂f2∂x2

]∣∣∣∣∣x1=x2=0

=

[0 1

− KM − B

M − 3BM x22

]∣∣∣∣x1=x2=0

Evaluating the Jacobian matrix at x1 = x2 = 0, yields:

A =

[0 1

− KM − B

M

]


.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.


To satisfy the 1st part, the eigenvalues of A must have negative real parts;we must solve the characteristic equation det(λI − A) = 0

det(λI − A) =

∣∣∣∣[ λ 00 λ

]−[

0 1

− KM − B

M

]∣∣∣∣=

∣∣∣∣[ (λ− 0) (0− 1)

(0− (− KM )) (λ− (− B

M ))

]∣∣∣∣=

∣∣∣∣[ λ −1KM λ+ B

M

]∣∣∣∣= λ

(λ+

B

M

)+

K

M= 0 (6)


.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.


To solve Eq.6 we use the quadratic formula like before, so we get:

λ1,2 =−β ±

√β2 − 4αγ

2α

λ1,2 =− B

M ±√

BM

2 − 4 KM

2

Since the eigenvalues have negative real parts, a positive definite solutionto the Lyapunov equation does exist


.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.


To answer the 2nd part, we need to solve the Lyapunov EquationA′P + PA = −I numerically:[

0 1

− KM − B

M

]′ [P11 P12

P21 P22

]+

[P11 P12

P21 P22

] [0 1

− KM − B

M

]=

[−1 00 −1

]⇒

[0 − K

M

1 − BM

] [P11 P12

P21 P22

]+

[P11 P12

P21 P22

] [0 1

− KM − B

M

]=

[−1 00 −1

]⇒

[0 · P11 + (− K

MP21) 0 · P12 + (− KMP22)

1 · P11 + (− BMP21) 1 · P12 + (− B

MP22)

]+

[0 · P11 + (− K

MP12) 1 · P11 + (− BMP12)

0 · P21 + (− KMP22) 1 · P21 + (− B

MP22)

]=

[−1 00 −1

]


.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.


⇒[

− KMP21 − K

MP12 − KMP22 + P11 − B

MP12

P11 − BMP21 − K

MP22 P12 − BMP22 + P21 − B

MP22

]=

[−1 00 −1

]Since P is chosen symmetric, i.e. P12 = P21, we get:

⇒[

−2KM P12 P11 − B

MP12 − KMP22

P11 − BMP12 − K

MP22 2P12 − 2BM P22

]=

[−1 00 −1

]


.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.


So, now we must solve

−2K

MP12 = −1 (7)

P11 −B

MP12 −

K

MP22 = 0 (8)

2P12 −2B

MP22 = −1 (9)

From Eq.7, P12 =12MK

Substituting P12 in Eq.9, yields

21

2

M

K− 2B

MP22 = −1 ⇒ −2B

MP22 = −1− M

K

⇒ P22 =

(1 +

M

K

)M

2B⇒ P22 =

1

2

M

B

(K +M

K

)Dr. Angeliki Lekka ([email protected]) Linearisation & Lyapunov March 2, 2017 16 / 22

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.


Now, substituting P12 and P22 in Eq.8, we get

P11 −B

MP12 −

K

MP22 = 0

⇒ P11 −1

2

M

K

B

M− 1

2

M

B

K

M

(K +M

K

)= 0

⇒ P11 −1

2

B

K− 1

2

K

B

(K +M

K

)= 0

⇒ P11 −1

2

(B

K+

K

B

K +M

K

)= 0

⇒ P11 =1

2

(B

K+

K +M

B

)


.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.


Hence,

P =

[12

(BK + K+M

B

)12MK

12MK

12MB

K+MK

]


.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Suggestions for personal study/revision

Be familiar with stability concepts w.r.t the eigenvalues of a matrix

Revise on derivatives and partial derivatives, so you are able tocompute Jacobian matrices (perform linearisation)

Be familiar with the concepts of positiveness & negativeness ofmatrices (ability to solve the Lyapunov equation)

Be familiar with the chain rule - particularly useful throughout thecourse


.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Appendix

Derivative Chain Rule (The Outside-Inside Rule)

The derivative of a composite function can be calculated using the chainrule, i.e. the derivative of a composite function is equal to the derivativeof the outside function evaluated at the inside function times thederivative of the inside function, e.g. V (x) = f (g(x))

V (x) = f ′(g(x)) · g(x)′


.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Appendix

Example

V (x) = (3x2 + 1)5

V (x) = 5(3x2 + 1)4(

d

dx(3x2 + 1)

)= 30x(3x2 + 1)4

Example

V (x) = (sin(x))3

V (x) = 3(sin(x))2(

d

dxsin(x)

)= 3cos(x)sin2(x)


linearisation & lyapunov equation - university of … tutorial on eg4321/eg7040 nonlinear...

Documents