Chap2 Student

Chapter 2. Solution of State Equations

Modern Control Theory (Course Code: 10213403)

Professor Jun WANG

(�� Ç)

Department of Control Science & Engineering

School of Electronic & Information Engineering

Tongji University

Spring semester, 2012

Outline

1 2.1 Introduction

2 2.2 Solution of homogeneous state equations

3 2.3 Matrix exponential function

4 2.4 State transition matrix

5 2.5 Solution of nonhomogeneous state equations

6 2.6 Simulations with MATLAB

Outline

1 2.1 Introduction






2.1 Introduction

Introduction

In Chapter 1, a linear time-invariant system is described by

State equation: x(t) = Ax(t) + Bu(t), x(0) = x0, t > 0

Output equation: y(t) = Cx(t) + Du(t)

In order to analyze the characteristics of the system, we need to

solve the equations.

The key point is to obtain the solution of the state x(t) from the

state equation.

We decompose the solution of x(t) into two parts (Why?):

Self-excited dynamics

x(t) = Ax(t), x0 , 0

Forced dynamics

x(t) = Ax(t) + Bu(t), x0 = 0

Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 4 / 37

Outline

1 2.1 Introduction






2.2 Solution of homogeneous state equations

Solution of homogeneous state equations

For a general state-space system Σ(A, B, C, D). If u(t) = 0, we can

obtain the homogenous state equation

x(t) = Ax(t)

x(0) = x0, t > 0

The above equation expresses the inherent characteristics of the

system.

The corresponding system is called autonomous system.

The homogenous state equation can be solved by

◮ Laplace transform approach

◮ Linear algebraic approach (to be more specific, Matrix exponential

approach)



Laplace transform approach

Let’s begin with the following scalar differential equation

x = ax

Taking the Laplace transform of the equation, we obtain

sX(s) − x(0) = aX(s), where X(s) = L[

x(t)]

∴ X(s) =x(0)

s − a

The inverse Laplace transform of the equation gives the solution

x(t) = eatx(0)


Can we extend the solution to the vector differential equations

x(t) = Ax(t)

Taking the Laplace transform of both sides of the equation, we obtain

sX(s) − x(0) = AX(s), where X(s) = L[

x(t)]

∴ (sI − A)X(s) = x(0)

Premultiplying both sides by (sI − A)−1, we obtain

X(s) = (sI − A)−1 x(0)

L−1

−−→ x(t) = L−1[

(sI − A)−1]

x(0)

∵ (sI − A)−1=

I

s+

A

s2+

A2

s3+ · · ·

∴ L−1[

(sI − A)−1]

= I + At +A2t2

2!+

A3t3

3!+ · · · , eAt

Hence, x(t) = eAtx(0)


Linear algebraic approach

Let’s begin with the following scalar differential equation

x = ax

We may assume a solution x(t) of the form

x(t) = b0 + b1t + b2t2 + · · · + bktk + · · ·

By substituting it to the differential equation, we have

b1 + 2b2t + 3b3t2 + · · · + kbktk−1 + · · · =

a(

b0 + b1t + b2t2 + · · · + bktk + · · ·

)

If the assumed solution is to be true, the above equation must hold for

any t.Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 9 / 37


By equating the coefficients of the equal powers of t, we have

b1 = ab0

b2 =1

2ab1 =

1

2a2b0

b3 =1

3ab2 =

1

3 × 2a3b0

...

bk =1

k!akb0

Since x(t) = b0 + b1t + b2t2 + · · · + bktk + · · · , we have x(0) = b0.

Hence the solution x(t) can be written as

x(t) =

(

1 + at +1

2!a2t2 + · · · +

1

k!aktk + · · ·

)

x(0)

= eatx(0)



We shall now solve the vector-matrix differential equation

x(t) = Ax(t)

By analogy with the scalar case, we assume that the solution is in the

form of a vector power series in t, i.e.

x(t) = b0 + b1t + b2t2 + · · · + bktk + · · · , with x(0) = b0

By substituting the solution into the differential equation, we have

b1 + 2b2t + 3b3t2 + · · · + kbktk−1 + · · · =

A(

b0 + b1t + b2t2 + · · · + bktk + · · ·

)

Therefore, we finally get the solution

x(t) =

(

I + At +1

2!A2t2 + · · · +

1

k!Aktk + · · ·

)

x(0)

= eAtx(0)Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 11 / 37

Outline

1 2.1 Introduction






2.3 Matrix exponential function

Characteristics of matrix exponential function

eAt, I + At +

A2t2

2!+ · · · +

Aktk

k!+ · · ·

Properties

limt→0

eAt = I

deAt

dt= AeAt = eAtA

eA(t+τ) = eAt· eAτ

(

eAt)−1

= e−At

(

eAt)m

= eA(mt)

e(A+F)t = eAt· eFt = eFt

· eAt iff AF = FA where A, F ∈ Rn×n



How to calculate eAt?

Laplace transform method

Eigenvalue method

Cayley-Hamilton method



Laplace transform method

eAt = L−1[

(sI − A)−1] Example

Compute eAt where A =[

0 10 −2

]

.

Solutions

Since sI − A =[

s 00 s

]

−[

0 10 −2

]

=[

s −10 s+2

]

.

we obtain

(sI − A)−1 =

[

1s

1s(s+2)

0 1s+2

]

∴ eAt = L−1[

(sI − A)−1]

=[

1 12 (1−e−2t)

0 e−2t

]

�Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 15 / 37


Eigenvalue method

Eigenvalue method

Suppose J is the Jordan canonical form of A, i.e. P−1AP = J

eAt = PeJtP−1

How to prove? Let’s begin from the definition!!!

eAt = ePJP−1t

= I + PJP−1t +(PJP−1)2t2

2!+ · · · +

(PJP−1)ktk

k!+ · · ·

= P

(

I + Jt +J2t2

2!+ · · · +

Jktk

k!+ · · ·

)

P−1

= PeJtP−1



How can we use eAt = PeJtP−1 to simplify the calculation of eAt?

If A has distinct eigenvalues λ1, λ2, · · · , λn, i.e.

J = Λ =

λ1 0

λ2

. . .

0 λn

we can easily calculate

eJt =

eλ1t 0

eλ2t

. . .

0 eλnt

=⇒ eAt = P

eλ1t 0

eλ2t

. . .

0 eλnt

P−1


Cayley-Hamilton method

Arthur Cayley

(British, 1821-1895)

Field Mathematian

Inst University of Cambridge

Known for Projective geometry, Group

theory, CayleyHamilton theorem

Notable awards Copley Medal (1882)

William R. Hamilton

(Irish, 1805-1865)

Field Physicist, astronomer, and

mathematician

Inst Trinity College, Dublin

Known for Hamilton’s principle,

Hamiltonian mechanics,

Hamilton-Jacobi equation,

CayleyHamilton theorem ...


Cayley-Hamilton theorem

Theorem (Cayley-Hamilton Theorem)

Every matrix satisfies its own characteristic equation, that is, if the

characteristic equation of a given matrix A is

∆(λ) = |λI − A| = λn + an−1λ

n−1 + · · · + a1λ+ a0 = 0

then

∆(A) = An + an−1An−1 + · · · + a1A + a0I = 0



How to use Cayley-Hamilton Theorem?

Consider a general polynomial of degree p

F(s) = f0 + f1s + f2s2 + · · · + fpsp

Assume that ∆∗(s) is another polynomial of degree n and n < p

∆∗(s) = a0 + a1s + · · · + ansn

Dividing F(s) by ∆∗(s) results in

F(s) = Q(s)∆∗(s) + R(s)

where Q(s) is the quotient polynomial and R(s) is the remainder

polynomial of degree (n − 1) or less.



Similarly, the matrix polynomial F(A) can be written as

F(A) = Q(A)∆∗(A) + R(A)

where A is an n × n matrix with the characteristic polynomial ∆∗(·).

According to the Cayley-Hamilton Theorem, we have ∆∗(A) = 0, i.e.

F(A) = R(A) , a0I + a1A + · · · + an−1An−1

This equality is applicable for convergent infinite series (i.e. p → ∞).

The definition of matrix exponential yields

eAt , I + At +A2t2

2!+ · · · +

Aktk

k!+ · · ·

By applying for the Cayley-Hamilton Theorem, we have

eAt = a0(t)I + a1(t)A + · · · + an−1(t)An−1



Compute eAt by the Cayley-Hamilton Method

eAt = a0(t)I + a1(t)A + · · · + an−1(t)An−1

If A has n distinct eigenvalues λ1, λ2, · · · , λn, we have n equations

eλit = a0(t) + a1(t)λi + · · · + an−1(t)λn−1i , i = 1, 2, · · · , n


Compute eAt by the Cayley-Hamilton Method

eAt = a0(t)I + a1(t)A + · · · + an−1(t)An−1

When λi is a repeated eigenvalue with algebraic multiplicity mi,

we can form the following mi linearly independent equations:

eλit = a0(t) + a1(t)λi + · · · + an−1(t)λn−1i

deλt

dλ

∣

∣

∣

∣

λ=λi

=d

dλ

[

a0(t) + a1(t)λ + · · · + an−1(t)λn−1]

∣

∣

∣

∣

λ=λi

...

dmi−1eλt

dλmi−1

∣

∣

∣

∣

λ=λi

=dmi−1

dλmi−1

[

a0(t) + a1(t)λ + · · · + an−1(t)λn−1]

∣

∣

∣

∣

λ=λi


Examples of the Cayley-Hamilton method

Example

Calculate eAt by the Cayley-Hamilton method, where A =

−1 0

0 −2

.

Solutions

It is obvious that A has two eigenvalues λ1 = −1 and λ2 = −2.

From the Cayley-Hamilton theorem, we have

eAt = a0(t)I + a1(t)A

eλ1t = a0(t) + a1(t)λ1

eλ2t = a0(t) + a1(t)λ2



By takeing the values of λ1 and λ2 to the above equations, we have

e−t = a0(t) − a1(t)

e−2t = a0(t) − 2a1(t)−→

a0(t) = 2e−t − e−2t

a1(t) = e−t − e−2t

Hence

eAt = a0(t)I + a1(t)A

=(

2e−t − e−2t)

·

1 0

0 1

+(

e−t − e−2t)

·

−1 0

0 −2

=

e−t 0

0 e−2t

�



Example

Calculate eAt where A =

−1 0 0

0 −2 1

0 0 −2

.

Solutions

It is obvious that A has three eigenvalues λ1 = −1 and λ2,3 = −2.

From the Cayley-Hamilton theorem, we have

eAt = a0(t)I + a1(t)A + a2(t)A2

eλ1t = a0(t) + a1(t)λ1 + a2(t)λ21

eλ2t = a0(t) + a1(t)λ2 + a2(t)λ22

teλ2t = a1(t) + 2a2(t)λ2



By takeing the values of λ1 and λ2 to the above equations, we have

e−t = a0 − a1 + a2

e−2t = a0 − 2a1 + 4a2

te−2t = a1 − 4a2

−→

a0 = 4e−t − 3e−2t − 2te−2t

a1 = 4e−t − 4e−2t − 3te−2t

a2 = e−t − e−2t − te−2t

For simplicity, let denote x = e−t and y = e−2t.

Hence

eAt = a0I + a1A + a2A2

= a0 ·

[

1 0 00 1 00 0 1

]

+ a1 ·

[

−1 0 00 −2 10 0 −2

]

+ a2

[

1 0 00 4 −40 0 4

]

=

e−t 0 0

0 e−2t te−2t

0 0 e−2t

�Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 27 / 37

Outline

1 2.1 Introduction






2.4 State transition matrix

State-transition matrix

We can write the solution of the homogenous state equation x = Ax as

x(t) =Φ(t)x(0)

whereΦ(t) is an n × n matrix and is the unique solution of

Φ(t) = AΦ(t), Φ(0) = I

To verify this, note that

x(0) =Φ(0)x(0) = x(0)

x(t) = Φ(t)x(0) = AΦ(t)x(0) = Ax(t)

For a linear time-invariant system, we have

Φ(t) = eAt = L−1[

(sI − A)−1]


2.4 State transition matrix

Properties of state-transition matrices

For the time-invariant system x = AX, the state-transition matrix

Φ(t) = eAt

has the following properties

Φ(0) = I

Φ(t) = AΦ(t) =Φ(t)A

Φ−1(t) =Φ(−t)

Φ(t1 + t2) =Φ(t1)Φ(t2) =Φ(t2)Φ(t1)[

Φ(t)]m

=Φ(mt)

Φ(t2 − t1)Φ(t1 − t0) =Φ(t2 − t0)


Outline

1 2.1 Introduction






Solution of nonhomogeneous state equations

Let’s consider the nonhomogeneous state equation described by

x = Ax + Bu, or x − Ax = Bu

Premultiply both sides of this equation by e−At

e−At (x − Ax) =d

dt

[

e−Atx(t)]

= e−AtBu

Integrating the equation between 0 and t gives

e−Atx(t) − x(0) =

∫ t

0e−AτBu(τ)dτ

i.e. x(t) = eAtx(0) +

∫ t

0eA(t−τ)Bu(τ)dτ

The above equation can also be written as

x(t) =Φ(t)x(0) +

∫ t

0Φ(t − τ)Bu(τ)dτ

2.5 Solution of nonhomogeneous state equations

Laplace transform approach

Let’s consider the nonhomogeneous state equation described by

x = Ax + Bu

The Laplace transform of the equation yields

sX(s) − x(0) = AX(s) + BU(s) =⇒ (sI − A)X(s) = x(0) + BU(s)

∴ X(s) = (sI − A)−1x(0) + (sI − A)−1BU(s)

= L[

eAt]

x(0) + L[

eAt]

BU(s)

The inverse Laplace transform of the equation can be obtained by use

of the convolution integral as follows

L−1

−−→ x(t) = eAtx(0) +

∫ t

0eAt−τBu(τ)dτ


Outline

1 2.1 Introduction






2.6 Simulations with MATLAB

Matlab commands used in this chapter

Command Description

inv Matrix inverse

det Determinant of a square matrix

eig Eigenvalues and eigenvectors

exp Exponential function

ode45 Solve differential equations, medium order method


2.6 Simulations with MATLAB

A simple SS model with Simulink

Let’s build up a simple state-space model with Simulink. The input is

a step signal and the output is displayed on a “scope”.

Step

x' = Ax+Bu y = Cx+Du

State-Space Scope

The State-Space block implements a

system whose behavior is defined by

Σ(A, B, C, D). The parameter of the model

can be entered by double clicking the

block.


Chapter 2. Solution of State Equations

Modern Control Theory (Course Code: 10213403)

Professor Jun WANG

(�� Ç)

Department of Control Science & Engineering

School of Electronic & Information Engineering

Tongji University

Spring semester, 2012

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