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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 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
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 2.2 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
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)
Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 6 / 37
2.2 Solution of homogeneous state equations
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)
Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 7 / 37
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)
2.2 Solution of homogeneous state equations
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
2.2 Solution of homogeneous state equations
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)
Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 10 / 37
2.2 Solution of homogeneous state equations
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 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
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
Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 13 / 37
2.3 Matrix exponential function
How to calculate eAt?
Laplace transform method
Eigenvalue method
Cayley-Hamilton method
Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 14 / 37
2.3 Matrix exponential function
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
2.3 Matrix exponential function
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
Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 16 / 37
2.3 Matrix exponential function
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
Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 17 / 37
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 ...
2.3 Matrix exponential function
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
Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 19 / 37
2.3 Matrix exponential function
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.
Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 20 / 37
2.3 Matrix exponential function
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
Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 21 / 37
2.3 Matrix exponential function
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
Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 22 / 37
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
2.3 Matrix exponential function
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
Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 24 / 37
2.3 Matrix exponential function
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
�
Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 25 / 37
2.3 Matrix exponential function
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
Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 26 / 37
2.3 Matrix exponential function
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 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
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]
Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 29 / 37
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)
Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 30 / 37
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
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
Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 33 / 37
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
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
Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 35 / 37
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.
Prof J Wang (Tongji Uni) Chap 2. Solution of state equations Spring 2012 36 / 37
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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