Introduction to Control Theory Including Optimal Control
Nguyen Tan Tien - 2002.3
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3. State Space Formulation
3.1 Introduction (Review)
3.1.1 Eigenvalues and eigenvectors
Consider a matrix A of order n × n . If there exists a vector
x ≠ 0 and a scalar λ such that
Ax = λ x
⎡λ1 0
⎢
Λ = ⎢ 0 λ2
M M
⎢0 0
⎣
L
L
O
L
⎤
⎥
⎥ = diag{λ1 , λ2 ,L, λn }
λn ⎥⎦
0
0
M
is called eigenvalue matrix of A .
Hence
then x is called an eigenvector of A . λ is called an eigenvalue
of A .
The above equation can be written in the form
[ λ I − A]x = 0
Λ = X −1 A X
3.1.2 The Cayley-Hamilton theorem
Given a square matrix A (of order n × n ) and integers r and s ,
then
Ar A s = Ar + s = A s+r = A s Ar
where I is the unit matrix (of order n × n ). It is known that the
above homogeneous equation has non-trivial (that is, nonzero) solution only if the matrix [ λ I − A ] is singular, that is if
det(λ I − A) = 0
This is an equation in λ of great importance. We denoted it
by c(λ ) , so that
c(λ ) = det(λ I − A) = 0
It is called the characteristic equation of A . Written out in full,
this equation has the form
a12
⎡λ − a11
⎢ a 21 λ − a 22
c (λ ) = ⎢
M
⎢ M
an 2
⎣ a n1
L a1n ⎤
L a 2n ⎥ = 0
O
M ⎥⎥
L λ − a nn ⎦
On expansion of determinant, c(λ ) is seen to be a polynomial
of degree n in λ , having the form
c(λ ) = λn + b1λn−1 + K + bn−1λ + bn
= (λ − λ1 )(λ − λ2 ) L (λ − λn )
=0
λ1 , λ2 ,L, λn , the roots of c(λ ) = 0 , are the eigenvalues of A .
Assuming that A has distinct eigenvalue λ1 , λ2 ,L, λn , the
corresponding eigenvectors x1 , x 2 ,L, x n are linearly independent. The (partitioned) matrix
X = [x1 x 2 L x n ]
is called the modal matrix of A. Since
A xi = λ xi
where
(i = 1,2,L, n)
This property leads to the following definition.
Definition
Corresponding to a polynomial in a scalar variable λ
f (λ ) = λk + b1λk −1 + K + bk −1λ + bk
define the (square) matrix f ( A) , called a matrix polynomial,
by
f ( A) = A k + b1 A k −1 + K + bk −1 A + bk I
where I is the unit matrix of order n × n .
For example, corresponding to
f (λ ) = λ2 − 2λ + 3 and
A = ⎡ 1 1⎤ , we have
⎢⎣− 1 3⎥⎦
f ( A) = ⎡⎢ 1 1⎤⎥ ⎡⎢ 1 1⎤⎥ − 2 ⎡⎢ 1 1⎤⎥ + 3⎡⎢1 0⎤⎥ = ⎡⎢ 1 2⎤⎥
⎣− 1 3⎦ ⎣− 1 3⎦ ⎣− 1 3⎦ ⎣0 1⎦ ⎣− 2 5⎦
Of particularly interest to us are polynomials f having the
property that f ( A) = 0 . For every matrix A one such
polynomial can be found by the Cayley-Hamilton theorem
which states that: Every square matrix A satisfies its own
characteristic equations.
For the above example, the characteristic equation of A is
c(λ ) = (λ − λ1 )(λ − λ2 ) = λ2 − 4λ + 4
where λ1 , λ2 are eigenvalues of A . So that
c( A) = A 2 − 4 A + 4 I
= ⎡ 1 1⎤ ⎡ 1 1⎤ − 4 ⎡ 1 1⎤ + 4 ⎡1 0⎤
⎢⎣− 1 3⎥⎦ ⎢⎣− 1 3⎥⎦ ⎣⎢− 1 3⎥⎦ ⎣⎢0 1⎥⎦
= ⎡⎢0 0⎤⎥
⎣0 0 ⎦
it follows that
AX = ΛX
In fact the Cayley-Hamilton theorem guarantees the existence
of a polynomial c(λ ) of degree n such that c( A) = 0 .
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8
Chapter 3 State Space Formulation
Introduction to Control Theory Including Optimal Control
Nguyen Tan Tien - 2002.3
_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
3.2 State Space Forms
⎡ x1 ⎤
y = [1 0 0] ⎢ x 2 ⎥
⎢x ⎥
⎣ 3⎦
Consider the system equation in the form
y ( n ) + a1 y ( n −1) + K + a n −1 y& + a n y = u
(3.1)
It is assumed that y (0), y& (0), L , y ( n−1) (0) are known. If we
define x1 = y , x 2 = y& , L , x n = y ( n −1) , (3.1) is written as
x&1 = x 2
x& 2 = x 3
Let
x1 =
1 y + 1 &y&
5
5
1 ( −1 + 2i ) &y&
x 2 = 15 (2 + i ) y − 12 iy& + 10
1 (1 + 2i ) &y&
x 3 = 15 (2 − i ) y + 12 iy& − 10
(3.6)
where i 2 = −1 . Then
M
x& n −1 = x n
x& n = − a n x1 − a n−1 x 2 − K − − a1 x n + u
which can be written as a vector-matrix differential equation
⎡ x&1 ⎤ ⎡ 0
1
⎢ x& 2 ⎥ ⎢ 0
0
⎢ M ⎥=⎢ M
M
⎢ x& ⎥ ⎢ 0
0
−
1
n
⎢
⎥ ⎢
⎢⎣ x& n ⎥⎦ ⎣− a n − a n−1
(2) Diagonal form
L 0
0 ⎤ ⎡ x1 ⎤ ⎡0⎤
L 0
0 ⎥ ⎢ x 2 ⎥ ⎢0 ⎥
O M
M ⎥ ⎢ M ⎥ + ⎢M⎥ u
L 0
1 ⎥ ⎢ x n −1 ⎥ ⎢0⎥
L − a 2 − a1 ⎥⎦ ⎢⎢ x ⎥⎥ ⎢⎣1 ⎥⎦
⎣ n ⎦
(3.2)
(3.7)
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In general form, a MIMO system has state equations of the
form
that is, as x& = A x + B u , where x , A and B are defined in
equation (3.2). The output of the system
⎡ x1 ⎤
⎢ ⎥
y = [1 0 L 0] ⎢ x 2 ⎥
M
⎢x ⎥
⎣ n⎦
⎡ x&1 ⎤ ⎡2 0 0 ⎤ ⎡ x1 ⎤ 1 ⎡ 2 ⎤
⎢ x& 2 ⎥ = ⎢0 i 0 ⎥ ⎢ x 2 ⎥ + ⎢− 1 + 2i ⎥ u
⎢ x& ⎥ ⎢0 0 − i ⎥ ⎢ x ⎥ 10 ⎢ − 1 − 2i ⎥
⎦⎣ 3⎦
⎣
⎦
⎣ 3⎦ ⎣
and
⎡ x1 ⎤
y = [1 1 1] ⎢ x 2 ⎥
⎢x ⎥
⎣ 3⎦
(3.3)
⎧ x& = A x + B u
⎨
⎩ y = C x + Du
(3.14)
and a SISO system has state equations of the form as (3.4).
3.3 Using the transfer function to define state variables
that is, as, where C = [1 0 L 0] . The combination of
equations (3.2) and (3.3) in the form
It is sometimes possible to define suitable state variables by
considering the partial fraction expansion of the transfer
function. For example, given the system differential equation
⎧ x& = A x + B u
⎨
⎩y = C x
&y& + 3 y& + 2 y = u& + 3u
(3.4)
The corresponding transfer function is
are known as the state equations of the system considered. The
matrix A in (3.2) is said to be in companion form. The
components of x are called the state variables x1 , x 2 , L , x n .
The corresponding n-dimensional space called the state space.
Any state of the system is represented by a point in the state
space.
G(s) =
Hence
Y (s) = X 1 (s) + X 2 (s)
Example 3.1 _______________________________________
2U ( s )
s +1
U (s)
X 2 (s) = −
s+2
X 1 (s) =
Obtain two forms of state equations of the system defined by
&y&& − 2 &y& + y& − 2 y = u
where matrix A corresponding to one of the forms should be
diagonal.
On taking the inverse Laplace transforms, we obtain
x&1 = − x1 + 2u
x& 2 = −2 x 2 − u
(a) A in companion form
Let the state variable as x1 = y , x 2 = y& , x 3 = &y& . Then
⎡ x&1 ⎤ ⎡0 1 0⎤ ⎡ x1 ⎤ ⎡0⎤
⎢ x& 2 ⎥ = ⎢0 0 1 ⎥ ⎢ x 2 ⎥ + ⎢0⎥ u
⎢ x& ⎥ ⎢2 − 1 2⎥ ⎢ x ⎥ ⎢1 ⎥
⎦⎣ 3⎦ ⎣ ⎦
⎣ 3⎦ ⎣
and
Y ( s)
2
1
s+3
=
=
−
U ( s ) ( s + 1)( s + 2) s + 1 s + 2
or
(3.5)
⎡ x&1 ⎤ ⎡− 1 0 ⎤ ⎡ x1 ⎤ ⎡ 2 ⎤
⎢⎣ x& 2 ⎥⎦ = ⎢⎣ 0 − 2⎦⎥ ⎢⎣ x 2 ⎥⎦ + ⎢⎣− 1⎥⎦ u
⎡x ⎤
y = [1 1] ⎢ 1 ⎥
⎣x2 ⎦
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9
Chapter 3 State Space Formulation
Introduction to Control Theory Including Optimal Control
Nguyen Tan Tien - 2002.3
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We can of course make a different choice of state variables,
for example
x = 2e −3t + 4e −3t
=
U (s)
s +1
U (s)
X 2 (s) =
s+2
then
X 1 (s) =
3τ
0
= 2e −3t + 4e −3t
Y (s) = 2 X 1 (s) − X 2 (s)
t
∫ e τ dτ
( te − e
22 e −3t
9
3t
1
3
+ 19
1 3t
9
)
+ 43 t − 94
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To use this method to solve the vector matrix equation (3.15),
we must first define a matrix function which is the analogue of
the exponential, and we must also define the derivative and the
integral of a matrix.
x&1 = − x1 + u
&x 2 = −2 x 2 + u
∞
Definition Since e z =
∑ n! z
1
n
(all z ), we define
0
∞
Now the state equations are
eA =
∑ n! A
1
n
= A0 + A +
1 2
1
A + K = I + A + A2 + K
2!
2!
⎡ x&1 ⎤ ⎡− 1 0 ⎤ ⎡ x1 ⎤ ⎡1⎤
⎢⎣ x& 2 ⎥⎦ = ⎢⎣ 0 − 2⎦⎥ ⎢⎣ x 2 ⎥⎦ + ⎢⎣1⎥⎦ u
(where A0 ≡ I ) for every square matrix A .
⎡x ⎤
y = [2 − 1] ⎢ 1 ⎥
⎣ x2 ⎦
Example 3.4 _______________________________________
0
⎡ 1 0 0⎤
Evaluate e At given A = ⎢− 1 0 0⎥
⎢⎣ 1 1 1 ⎥⎦
3.4 Direct solution of the state equation
By a solution to the state equation
x& = A x + B u
(3.15)
We must calculate A 2 , A3 ,…
⎡ 1 0 0⎤
A 2 = ⎢ − 1 0 0 ⎥ = A = A3 = K
⎢⎣ 1 1 1⎥⎦
we mean finding x at any time t given u (t ) and the value of
x at initial time t 0 = 0 , that is, given x(t 0 ) = x 0 . It is
instructive to consider first the solution of the corresponding
scalar differential equation
It follows that
x& = a x + b u
e At = I + A t +
(3.16)
1 2 2 1 33
A t + A t +K
2!
3
1 2 1 3
⎛
⎞
= I + A ⎜ t + t + t + K⎟
2
!
3
⎝
⎠
given x = x0 at t = 0 . The solution of (3.15) is found by an
analogous method.
Multiplying the equation by the integrating factor e − at , we
obtain
= I + A(e t − 1)
t
0
0 ⎤
⎡1 0 0⎤ ⎡⎢ e − 1
⎥
0 ⎥
= ⎢0 1 0⎥ + ⎢ − e t + 1 0
t
t
t
⎣⎢0 0 1⎥⎦ ⎢⎣ e − 1 e − 1 e − 1⎦⎥
⎡ et
0
0⎤
⎥
⎢
= ⎢− e t + 1 1
0⎥
t
t
t
e
e
e
−
−
1
1
⎦⎥
⎣⎢
e − at ( x& − ax ) = e − at b u
or
d −at
[e x] = e −at b u
dt
Integrating the result between 0 and t gives
__________________________________________________________________________________________
e
−at
x − x0 =
t
∫e
− aτ
b u (τ ) dτ
0
that is
x=
e at x0
123
complementary function
+
t
∫ e b u(τ ) dτ
144424443
−a( t −τ )
(3.17)
0
particular int ergral
Example 3.3 _______________________________________
Solve the equation
x& + 3 x = 4t
given that x = 2 when t = 0 .
In fact, the evaluation of the matrix exponential is not quite as
difficult as it may appear at first sight. We can make use of the
Cayley-Hamilton theorem which assures us that every square
matrix satisfies its own characteristic equation.
One direct and useful method is the following. We know that
if A has distinct eigenvalues, say λ1 , λ2 ,L, λn , there exists a
non-singular matrix P such that
⎡λ1 0
⎢
P A P = ⎢ 0 λ2
M M
⎢0 0
⎣
−1
L
L
O
L
0⎤
0 ⎥ = diag{λ , λ ,L, λ } = Λ
1 2
n
M ⎥
λn ⎥⎦
We then have A = P Λ P −1 , so that
Since a = −3 , b = 4 and u = t , on substituting into equation
(3.17) we obtain
A 2 = ( PΛP −1 )( PΛP −1 ) = PΛ ( P −1P )ΛP −1 = PΛ2 P −1
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Chapter 3 State Space Formulation
Introduction to Control Theory Including Optimal Control
Nguyen Tan Tien - 2002.3
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e At = ⎡⎢ 2 2 ⎤⎥ e −t + ⎡⎢−1 −2⎤⎥ e −2t
⎣1 1⎦
⎣− 1 − 1⎦
and in general
A r = PΛr P −1
(r = 1,2,L)
If we consider a matrix polynomial, say f ( A) = A − 2 A + I ,
we can write it as
2
= P f (Λ ) P −1
Since (for the above example)
f ( λ ) = λ 2 − 2λ + I
⎤
⎡λ1 0
⎥
⎢
⎥ − 2 ⎢ 0 λ2
M M
⎥
⎢
λ2n ⎥⎦ ⎣ 0 0
0
0
M
⎡λ12 − 2
0
⎢
λ22 − 2λ2 + 1
=⎢ 0
⎢ M
M
⎢ 0
0
⎣
called
the
poles
and
Definition
Let A(t ) = [aij (t )], then
= P diag{ f (λ1 ), f (λ2 ),L, f (λn )}P −1
L
L
O
L
λ1 , λ2 ,L, λn are
e λ1t , e λ2t ,L, e λnt are called the modes of the system.
We now define the derivative and the integral of a
matrix A(t ) whose elements are functions of t .
= P ( Λ2 − 2Λ + I ) P −1
0
⎢
2
0
λ
2
=⎢
⎢M M
⎢0 0
⎣
eigenvalues
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f ( A) = PΛ2 P −1 − 2 PΛP −1 + P I P −1
⎡λ12
In the solution to the unforced system equation x = e At x 0 , the
L
L
O
L
0 ⎤ ⎡1 0
0 ⎥ + ⎢0 1
M ⎥ ⎢M M
λn ⎥⎦ ⎢⎣0 0
L
L
O
L
0⎤
0⎥
M⎥
1⎥⎦
⎤
L
0
⎥
L
0
⎥
⎥
O
M
L λ2n − 2λn + 1⎥⎦
(1)
d
d
A(t ) = A& (t ) = [ (aij )], and
dt
dt
(2)
∫ A(t )dt = [∫ a
that is, each element of the matrix is differentiated (or
integrated).
⎡ 6t sin 2t ⎤
For example, if A = ⎢ 2
, then
3 ⎥⎦
⎣t + 2
A& = ⎡⎢ 6 2 cos 2t ⎤⎥ , and
0 ⎦
⎣2t
∫
= diag{ f (λ1 ), f (λ2 ),L, f (λn )}
The results holds for the general case when f (x) is a
polynomial of degree n . In generalizing the above, taking
f ( A) = e At , we obtain
e At = P diag{e λ1t , e λ2t ,L, e λnt } P −1
ij (t ) dt ],
(3.18)
⎡ 3t 2
− 12 cos 2t ⎤
⎥ + C ( C is a constant matrix)
Adt = ⎢ 3
3t
⎥⎦
⎢⎣ 13 t + 2t
From the definition a number of rules follows: if α and β are
constants, and A and B are matrices,
d
(i)
(α A + β B ) = α A& + β B&
dt
(ii)
b
b
b
a
a
a
∫ (α A + β B)dt = α ∫ Adt + β ∫ Bdt
Example 3.5 _______________________________________
d
(iii)
( A B ) = A B& + A& B
dt
Find e At given A = ⎡ 0 2 ⎤
⎢⎣− 1 − 3⎥⎦
(iv) If A is a constant matrix, then
The eigenvalues of A are λ1 = −1 and λ2 = −2 . It can be
verified that P = ⎡ 2 1 ⎤ and that P −1 = ⎡ 1 1 ⎤ . ( P can
⎢⎣− 1 − 1⎥⎦
⎢⎣− 1 − 2⎥⎦
⊗ Note that although the above rules are analogous o the
rules for scalar functions, we must not be dulled into
accepting for matrix functions all the rules valid for scalar
functions. For example, although d ( x n ) = n x n−1 x& in general,
dt
be found from A = P Λ P −1 ). Using (3.18), we obtain
it is not true that
⎡ −t 0 ⎤ ⎡ 1 1 ⎤
e At = ⎡ 2 1 ⎤ ⎢e
⎢⎣− 1 − 1⎥⎦ 0 e −2t ⎥ ⎢⎣− 1 − 2⎥⎦
⎣
⎦
⎡ 2e −t − e −2t 2e −t − 2e −2t ⎤
= ⎢ −t
⎥
− 2t
− e −t + 2e −2t ⎦
⎣− e + e
then
e At in its spectral or modal form as
and
so that
(3.19)
where λ1 , λ2 ,L, λn are the eigenvalues of A and A1 , A2 ,L ,
An are matrices of the same order as the matrix A . In the
above example, we can write
d
dt
( A n ) = n A n−1 A& .For example,
⎡ 2
⎤
A = ⎢t 2t ⎥ ,
⎣0 3⎦
2
d
⎡ 3
⎤
( A 2 ) = ⎢4t 6t + 6⎥
0
0
dt
⎣
⎦
⎡4t 3 4t 2 ⎤
&
2 AA = ⎢
0 ⎥⎦
⎣ 0
when
From the above discussion it is clear that we can also write
e At = A1e λ1t + A2 e λ2t + K + An e λnt
d At
(e ) = Ae At
dt
d 2
A ≠ 2 AA& .
dt
We now return to our original problem, to solve equation
(3.15): x& = A x + B u . Rewrite the equation in the form
x& − A x = B u
⇒
e − At (x& − A x) = e − At ( B u )
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11
Chapter 3 State Space Formulation
Introduction to Control Theory Including Optimal Control
Nguyen Tan Tien - 2002.3
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or
With this notation equation (3.20) becomes
d − At
(e x) = e − At B u
dt
∫
e − At x(t ) − x(0) =
t
∫e
− Aτ
3.5 Solution of the state equation by Laplace transforms
B u (τ ) dτ
Since the state equation is in a vector form we must first
define the Laplace transform of a vector.
so that
t
∫e
A(t −τ )
B u (τ ) dτ
(3.20)
Definition
0
Example 3.6 _______________________________________
A system is characterized by the state equation
⎡ x&1 ⎤ ⎡ 0 2 ⎤ ⎡ x1 ⎤ ⎡0⎤
⎢⎣ x& 2 ⎥⎦ = ⎢⎣− 1 − 3⎥⎦ ⎢⎣ x2 ⎥⎦ + ⎢⎣1⎥⎦ u
If the forcing function u (t ) = 1 for t ≥ 0 and x(0) = [1 − 1]T .
Find the state x of the system at time t .
We have already evaluated e At in Example 3.5. It follows that
−2t
−t
−2t ⎤
⎡ −t
e At x(0) = ⎢ 2e −t − e −2t 2e −t − 2e −2t ⎥ ⎡ 1 ⎤
− e + 2e ⎦ ⎢⎣− 1⎥⎦
⎣− e + e
= e −2t ⎡⎢ 1 ⎤⎥
⎣− 1⎦
∫
t
∫
⎡
= ⎢2e
∫ ⎣ 2e
t
0
−(t −τ )
⎡ L [ x1 (t )] ⎤ ⎡ X 1 ( s ) ⎤
⎢ L [ x2 (t )]⎥ ⎢ X 2 ( s) ⎥
We define L [x(t )] = ⎢
⎥ = ⎢ M ⎥ = X( s )
M
⎢ L [ x (t )]⎥ ⎢ X ( s)⎥
n
⎣
⎦ ⎣ n ⎦
From this definition, we can find all the results we need. For
example,
⎡ L [ x&1 (t )] ⎤ ⎡ sX 1 ( s) − x1 (0) ⎤
⎥
⎢
⎥ ⎢
&
L [x& (t )] = ⎢ L [ x 2 (t )]⎥ = ⎢ sX 2 ( s) − x2 (0) ⎥
M
M
⎢ L [ x& (t )]⎥ ⎢ sX ( s) − x (0)⎥
n
n
⎦
⎣
⎦ ⎣ n
s X( s ) − x(0) = A X( s ) + B U ( s ) where U ( s ) = L [u (t )]
− 2e −2(t −τ ) ⎤ dτ
⎥
− e −(t −τ ) ⎦
−2(t −τ )
or
−t
− 2t ⎤
⎡
= ⎢1 − 2−et +−e2t ⎥
e
e
−
⎣
⎦
−t
−2t ⎤ ⎡
−t
−2t ⎤
⎡ −2t ⎤ ⎡
x(t ) = ⎢ e −2t ⎥ + ⎢1 − 2−et +−e2t ⎥ = ⎢1 − 2−et + 2−e2t ⎥
⎣− e ⎦ ⎣ e − e
⎦ ⎣ e − 2e
⎦
X( s) = ( sI − A) −1 x(0) + ( sI − A) −1 B U ( s)
__________________________________________________________________________________________
At
The matrix e in the solution equation (3.20) is of special
interest to control engineers; they call it the state-transition
matrix and denote it by Φ(t ) , that is,
Φ(t ) = e At
(3.21)
For the unforced system (such as u (t ) = 0 ) the solution (Eq.
(3.20)) becomes
x(t ) = Φ(t ) x(0)
so that Φ(t ) transforms the system from its state x(0) at some
initial time t = 0 to the state x(t ) at some subsequent time t hence the name given to the matrix.
Since e At e − At = I . It follows that [e At ]−1 = e − At .
Φ −1 (t ) = e − At = Φ(−t )
At − Aτ
Φ(t ) Φ(−τ ) = e e
=e
A(t −τ )
( sI − A) X( s ) = x(0) + B U ( s )
Unless s happens to be an eigenvalue of A , the matrix
( sI − A) is non-singular, so that the above equation can be
solved giving
Hence the state of the system at time t is
Also
⎡ x1 ⎤
⎢x ⎥
Let x(t ) = ⎢ 2 ⎥
M
⎢x ⎥
⎣ n⎦
Now we can solve the state equation (3.15). Taking the
transform of x& = A x + B u , we obtain
t
e A(t −τ ) B u (τ ) dτ = e A(t −τ ) ⎡⎢0⎤⎥ dτ
⎣1⎦
0
0
Hence
(3.22)
0
0
x(t ) = e At x(0) =
t
x(t ) = Φ(t ) x(0) = Φ(t − τ ) B u (τ ) dτ
On integration, this becomes
= Φ (t − τ ) .
(3.23)
and the solution x(t ) is found by taking the inverse Laplace
transform of equation (3.23).
Definition
( sI − A) −1 is called the resolvent matrix of the system.
On comparing equation (3.22) and (3.23) we find that
Φ (t ) = L
−1
{( sI − A) −1}
Not only the use of transforms a relatively simple method for
evaluating the transition matrix, but indeed it allows us to
calculate the state x(t ) without having to evaluate integrals.
Example 3.7 _______________________________________
Use Laplace transform to evaluate the state x(t ) of the system
describe in Example 3.6.
⎡ x&1 ⎤ ⎡ 0 2 ⎤ ⎡ x1 ⎤ ⎡0⎤
⎢⎣ x& 2 ⎥⎦ = ⎢⎣− 1 − 3⎥⎦ ⎢⎣ x2 ⎥⎦ + ⎢⎣1⎥⎦ u
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T z& = A T z + B u
y = CT z
For this system
( sI − A) = ⎡⎢ s −2 ⎤⎥ , so that
⎣1 s + 3⎦
( sI − A) −1 =
or as
1
⎡ s + 3 2⎤
s ( s + 3) + 2 ⎢⎣ − 1 s ⎥⎦
z& = A1 z + B1 u
(3.26)
y = C1 z
s+3
2
⎡
⎤
⎢ ( s + 1)( s + 2) ( s + 1)( s + 2) ⎥
⎥
=⎢
−1
s
⎢
⎥
⎢⎣ ( s + 1)( s + 2) ( s + 1)( s + 2) ⎥⎦
where A1 = T −1 A T , B1 = T −1 B, C1 = C T . The transformation
(3.25) is called state-transformation, and the matrices A and
A1 are similar. We are particular interested in the
1
2
2 ⎤
⎡ 2
−
⎢ s +1 − s + 2
+
+
s
1
s
2 ⎥
=⎢
1
1
1
2 ⎥
⎢−
⎥
+
−
+
s +1 s + 2⎦
⎣ s +1 s + 2
transformation when A1 is diagonal (usually denoted by Λ )
and A is in the companion form, such as (3.2)
So that
−2t
−t
−2t ⎤
⎡ −t
L −1{( SI − A) −1} = ⎢ 2e −t − e −2t 2e −t − 2e −2t ⎥ = Φ(t )
−
e
+
e
−
e
+
2
e
⎣
⎦
Hence the complementary function is as in Example 3.6. For
the particular integral, we note that since
⎡ x&1 ⎤ ⎡ 0
1
⎢ x& 2 ⎥ ⎢ 0
0
⎢ M ⎥=⎢ M
M
⎢ x& ⎥ ⎢ 0
0
1
n
−
⎥ ⎢
⎢
⎢⎣ x& n ⎥⎦ ⎣− a n − a n −1
It is assumed that the matrix A has distinct eigenvalues
λ1 , λ2 ,L, λn . Corresponding to the eigenvalue λi there is the
1
L {u (t )} = , then
s
eigenvector x i such that
1
1 ⎡ s + 3 2⎤ ⎡0⎤
( sI − A) −1 BU ( s ) =
( s + 1)( s + 2) s ⎢⎣ − 1 s ⎥⎦ ⎣⎢1 ⎥⎦
A x i = λi xi
2
⎡
⎤
⎢ s ( s + 1)( s + 2) ⎥
⎥
=⎢
1
⎢
⎥
⎣⎢ ( s + 1)( s + 2) ⎦⎥
V is called the modal matrix, it is non-singular and can be
used as the transformation matrix T above. We can write the n
equations defined by equation (3.27) as
AV = Λ V
−t
−2 t ⎤
⎡
{( sI − A) BU ( s )} = ⎢1 − 2−et +−e2t ⎥
−
e
e
⎣
⎦
−1
which is the particular integral part of the solution obtained in
Example 3.6.
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3.6 The transformation from the companion to the
diagonal state form
Note that the choice of the state vector is not unique. We now
assume that with one choice of the state vector the state
equations are
x& = A x + B u
y =Cx
(3.24)
where
⎡λ1 0
⎢
Λ = ⎢ 0 λ2
M M
⎢0 0
⎣
(3.28)
L
L
O
L
0⎤
0 ⎥ = diag{λ , λ ,L, λ }
1 2
n
M ⎥
λn ⎥⎦
From equation (3.28) we obtain
Λ = V −1 AV
(3.29)
The matrix A has the companion form
1
0
⎡ 0
⎢ 0
0
1
M
M
A=⎢ M
⎢ 0
0
0
⎢⎣− a0 − a1 − a 2
L
0 ⎤
L
0 ⎥
O
M ⎥
L
1 ⎥
L − a n−1 ⎥⎦
where A is the matrix of order n × n and B and C are matrices
of appropriate order.
so that its characteristic equation is
Consider any non-singular matrix T of order n × n . Let
det( sI − A) = λn + a n−1λn−1 + K + a1λ + a0 = 0 ]
x =T z
(3.27)
V = [x1 x 2 L x n ]
On taking the inverse Laplace transform, we obtain
L
(i = 1,2,L, n)
We define the matrix V whose columns are the eigenvectors
2
1 ⎤
⎡1
+
⎢ −
⎥
= ⎢ s s +1 s + 2 ⎥
1
1
⎢
⎥
−
⎣ s +1 s + 2 ⎦
−1
0 ⎤ ⎡ x1 ⎤ ⎡0⎤
L 0
0 ⎥ ⎢ x 2 ⎥ ⎢0 ⎥
L 0
O M
M ⎥ ⎢ M ⎥ + ⎢M⎥ u
1 ⎥ ⎢ x n −1 ⎥ ⎢0⎥
L 0
L − a 2 − a1 ⎥⎦ ⎢⎢ x ⎥⎥ ⎢⎣1 ⎥⎦
⎣ n ⎦
(3.25)
By solving this equation we obtain the eigenvalues
λ1 , λ2 ,L, λn . The corresponding eigenvectors have an
then z is also a state vector and equation (3.24) can be written
interesting form. Consider one of the eigenvalues λ and the
as
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corresponding eigenvector x = [α1 α 2 L α n ]T . Then the
equation A x = λ x , corresponds to the system of equations
⎧α 2
⎪
⎪α 3
⎨
⎪
⎪α n
⎩
so the transformation V −1x gives the new choice of the state
vector
⎡ z1 ⎤
0
4 ⎤ ⎡ y⎤
⎡ 4
⎢ z 2 ⎥ = 1 ⎢8 + 4i − 10i − 2 + 4i ⎥ ⎢ y& ⎥
⎢ z ⎥ 20 ⎢8 − 4i 10i − 2 − 4i ⎥ ⎢ &y&⎥
⎣
⎦⎣ ⎦
⎣ 3⎦
= λ α1
= λα2
M
= λ α n−1
[
Setting α1 = 1 , we obtain x = 1 λ λ2 L λ
the modal matrix in this case takes the form
1
⎡ 1
⎢ λ1
λ1
V =⎢ M
M
⎢ n−1 n−1
λ1
⎣λ1
]
n −1 T
L 1 ⎤
L λ1 ⎥
O M ⎥
⎥
L λ1n−1 ⎦
The state equations are now in the form of equation (3.26),
that is
. Hence
z& = A1z + B1u
y = C1z
where
A1 = V −1 AV = diag{2, i,−i}
(3.30)
1 ⎡ 2 ⎤
⎢− 1 + 2i ⎥
10 ⎢ − 1 − 2i ⎥
⎣
⎦
C1 = C V = [1 1 1]
B1 = V −1B =
In this form V is called Vandermonde matrix; it is nonsingular since it has benn assumed that λ1 , λ2 ,L, λn are
distinct. We now consider the problem of obtaining the
transformation from the companion form to diagonal form.
__________________________________________________________________________________________
Example 3.8 _______________________________________
Consider the system in state space form of Eq.(3.24)
Having chosen the state vector so that
x& = A x + B u
&y&& − 2 &y& _ y& − 2 y = u
y =Cx
is written as the equation (in companion form)
Taking the Laplace transforms we obtain
⎡ x&1 ⎤ ⎡0 1 0⎤ ⎡ x1 ⎤ ⎡0⎤
⎢ x& 2 ⎥ = ⎢0 0 1 ⎥ ⎢ x 2 ⎥ + ⎢0⎥ u
⎢ x& ⎥ ⎢2 − 1 2⎥ ⎢ x ⎥ ⎢1⎥
⎦⎣ 3⎦ ⎣ ⎦
⎣ 3⎦ ⎣
⎡ x1 ⎤
y = [1 0 0]⎢ x 2 ⎥
⎢x ⎥
⎣ 3⎦
s X (s) = A X (s) + B U (s)
Find the transformation which will transform this into a state
equation with A in diagonal form.
The characteristic equation of A is
det(λ I − A) = λ3 − 2λ2 + λ − 2 = (λ − 2)(λ2 + 1) = 0
that is λ1 = 2, λ2 = i and λ3 = −i .
From (3.30) the modal matrix is
⎡1 1 1 ⎤
V = ⎢2 i − i ⎥
⎢⎣4 − 1 − 1⎥⎦
and its inverse can be shown to be
V −1 =
0
4 ⎤
1 ⎡ 4
⎢8 + 4i − 10i − 2 + 4i ⎥
20 ⎢8 − 4i 10i − 2 − 4i ⎥
⎣
⎦
The transformation is defined by equation (3.25), that is
x = V z or z = V −1x . The original choice for x was
x = [ y y& &y&]
T
3.7 The transfer function from the state equation
(3.24)
Y (s) = C X (s)
⇒
X ( s ) = ( sI − A) −1 B U ( s )
and G ( s ) =
Y (s)
= C ( sI − A) −1 B
U (s)
(3.31)
Example 3.9 _______________________________________
Calculate the transfer function from the system whose state
equations are
x& = ⎡1 −2 ⎤ x + ⎡1 ⎤ u
⎢⎣3 − 4⎥⎦
⎢⎣2⎥⎦
y = [1 − 2]x
G ( s ) = [1 − 2]⎡⎢ s − 1 2 ⎤⎥
⎣ − 3 s + 4⎦
−1
⎡1 ⎤
⎢⎣2⎥⎦
−1
s+4
−2
⎤
⎡
⎢ ( s + 1)( s + 2) ( s + 1)( s + 2) ⎥ 1
⎥ ⎡ ⎤
= [1 − 2]⎢
s−2
3
⎥ ⎢⎣2⎥⎦
⎢
⎢⎣ ( s + 1)( s + 2) ( s + 1)( s + 2) ⎥⎦
=−
3s + 2
( s + 1)( s + 2)
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Example 3.10_______________________________________
Given that the system
x& = A x + B u
y =Cx
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Chapter 3 State Space Formulation
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has the transfer function G (s) , find the transfer function of the
system
z& = T −1 AT z + T −1 B u
y = CT z
If the transfer function is G1 ( s )
G1 ( s ) = C T ( sI − T −1 AT ) −1T −1 B
= C T {T −1 ( sI − A)T }−1T −1 B
= C ( sI − A) −1 B
= G(s)
so that G1 ( s ) = G ( s ) .
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Chapter 3 State Space Formulation