State-Space Model
In general, the dynamic equations of a lumped-parameter
continuous system may be represented by
x& = f ( x (t ), u(t ), t )
y = g( x( t ), u( t ), t )
state equation
output equation
where f and g are nonlinear vector-valued functions. Using a
linearized technique, the dynamic equations for linear time
invariant system can be represented by
x& = Ax + Bu
y = Cx + Du
where A ∈ R n ×n , B ∈ R n ×m and C ∈ R n × p are system state
matrix, input matrix and output matrix, respectively.
x& = An ×n x + Bn ×mu
y = C p ×n x + D p ×m u
For single-input single-output systems, m = p = 1. This form
is call the state space form of the dynamic system. In a brief
form, the system dynamic behaviour is described by
(A, B ,C ,D ) or ⎡ A B ⎤ . The block diagram is shown as
⎢C D ⎥
⎣
⎦
D
B
u(t)
+
x(t)
+
C
+
A
+
y(t)
Canonical Forms of the State-space
Description
Controllable canonical form
1
⎡ 0
⎢ 0
0
⎢
x& (t ) = ⎢ .
.
⎢
0
⎢ 0
⎢⎣ − p0 − p1
y (t ) = [γ 0 γ 1 ...
⎤
⎡ 0⎤
⎢ 0⎥
...
0
0 ⎥
⎥
⎢ ⎥
...
.
. ⎥ x (t ) + ⎢ . ⎥u (t )
⎥
⎢ ⎥
...
0
1 ⎥
⎢ 0⎥
⎢⎣1⎥⎦
... − pn − 2 − pn −1 ⎥⎦
γ n − 2 γ n −1 ]x (t )
...
0
0
Controllable Canonical Form
γn -1
γ
x2
xn
u
1
1
s
-p n -1
1
s
1
s
-p
1
-p 0
1
x1
1
s
y
γ
0
Observable Canonical Form
⎡0
⎢1
⎢
x& (t ) = ⎢ .
⎢
⎢0
⎢⎣0
y (t ) = [0
0 ... 0
0 ... 0
.
...
.
0 ... 0
0 ... 1
0 ... 0
⎡ β0 ⎤
⎤
⎢ β ⎥
⎥
⎢ 1 ⎥
⎥
⎥ x (t ) + ⎢ . ⎥u (t )
⎥
⎢
⎥
β
− pn − 2 ⎥
⎢ n −2 ⎥
⎢⎣ β n −1 ⎥⎦
− pn −1 ⎥⎦
1]x (t )
− p0
− p1
.
βn -1
β
1
x1
u
β
0
1
s
x n -1
1
s
1
s
xn
1
s
-p n -1
-p1
-p 0
y
1
State Transformation
The choice of state variables for a particular system is not
unique.
x& (t ) = Ax (t ) + Bu(t )
y (t ) = Cx (t ) + Du(t )
⇔
z(t ) = Px (t )
−1
x ( t ) = P z( t )
$ (t ) + Bu
$ (t )
z&(t ) = Az
$ (t ) + Du
$ (t )
y (t ) = Cz
(nonsingular linear transformation)
z&(t ) = PAP −1 z(t ) + PBu(t )
y (t ) = CP −1 z(t ) + Du(t )
In general, P can be time dependent, as P(t), and
z&(t ) = [ P& (t ) P(t ) + P(t ) AP −1 (t )]z (t ) + P (t ) Bu(t )
y (t ) = CP −1 (t ) z (t ) + Du (t )
Discrete time state-space equations
x ( k + 1) = Ax ( k ) + Bu ( k )
y ( k ) = Cx ( k ) + Du ( k )
State-space Description and Transfer
Function Description
Models of dynamic components or systems can be found in
the form of a set of ordinary differential equations. The
representation of a system can be formulated into a
transfer function form (with zero initial conditions) or a
state-space form (becomes a set of first order differential
equations).
Transfer function form: shows the relationship between the
system inputs and outputs. It is called the input-output
description, or the external description, of a system.
State-space form: shows the structural information of the
system using system states. It is, therefore, called the
internal description of the system.
ss2tf
x& (t ) = Ax (t ) + Bu (t )
y (t ) = Cx (t ) + Du (t )
( sI − A) X ( s ) = BU ( s )
Y ( s ) = CX ( s ) + DU ( s )
Y ( s)
= C ( sI − A) −1 B + D
U ( s)
= CΦ ( s ) B + D
= G( s)
Where Φ ( s ) = ( sI − A) −1 = adj ( sI − A) = Q ( s )
det( sI − A)
(the resolvent matrix)
P( s )
P(s) is the characteristic polynomial of A, and
P(s) = s n + pn −1s n −1 + . . . . . .
+ p1s + p0
= ( s − λ 1 )( s − λ 2 ) . . . . . . ( s − λ n )
Q( s) = Qn −1s n −1 +
......
+ Q1s + Q0
where
Qn −1 = I
1
tr ( AQi )
n −1
= AQi + Pi I
Pi = −
Qi −1
(C.T.Chen 1984)
A and Qi commute and AQ0 + P0 I = 0
and λi denotes the eigenvalue of the matrix A, which corresponding
to the natural behaviour (modes) of the system.
Solution of the state space equation
x& = Ax + Bu
(*)
Taking the Laplace transform of (*),
sX ( s ) − x (0) = AX ( s ) + BU ( s )
therefore
X ( s ) = [ sI − A]−1 x (0) + [ sI − A]−1 BU ( s )
(**)
The inverse Laplace transform of (**) results in the solution
t
x (t ) = e x (0) + ∫ e A( t −τ ) Bu(τ )dτ
At
0
where e At
A2 t 2
Ak t k
= exp( At ) = I + At +
+ ...... +
+ ...
2!
k!
is called the state transition matrix, denoted as Φ (t ) , and
t
x (t ) = Φ (t ) x (0) + ∫ Φ (t − τ ) Bu(τ )dτ
0
tf2ss
1
Y ( s) Y ( s) ξ ( s)
nm −1s m −1 + ...... + n1s + n0
=
= N ( s)
= m
G( s) =
U ( s) ξ ( s) U ( s)
D ( s ) s + d m −1s m −1 + ...... + d1s + d 0
Y ( s) = nm−1 s m−1ξ ( s) +......+ n1 sξ ( s) + n0ξ ( s)
U ( s) = s mξ ( s) + d m−1 s m−1ξ ( s) +......+ d1 sξ ( s) + d 0ξ ( s)
therefore,
y ( t ) = nm −1ξ ( m −1) ( t )+. . . . . .+ n1ξ& ( t ) + n0ξ ( t )
u( t ) = ξ ( t )( m ) + d m −1ξ ( t )( m −1) +. . . . . .+ d1ξ& ( t ) + d 0ξ ( t )
Let
x1 (t ) = ξ (t )
x2 (t ) = x&1 (t ) = ξ& (t )
......
x m −1 (t ) = x& m − 2 (t ) = ξ (t )( m − 2 )
x m (t ) = x& m −1 (t ) = ξ (t )( m −1)
then, the state-space form of this system can be represented as
1
⎡ 0
⎢ 0
0
⎢
x& (t ) = ⎢ .
.
⎢
0
⎢ 0
⎢⎣ − d 0 − d1
y (t ) = [n0 n1 ...
0 ⎤
⎡ 0⎤
⎢ 0⎥
...
0
0 ⎥
⎥
⎢ ⎥
...
.
. ⎥ x (t ) + ⎢ . ⎥u (t )
⎥
⎢ ⎥
...
0
1 ⎥
⎢ 0⎥
⎢⎣1⎥⎦
... − d m −2 − d m−1 ⎥⎦
nm −2 nm −1 ]x (t )
...
0
Remark
Poles of the transfer function are the eigenvalues of the
system matrix A, provided that there is no pole-zero
cancellation in C(sI-A) -1B+D. Zero is the value of s such that
if the input u(t)= e stu0 is applied to the system, the output y(t)
= 0.
⎛ sI − A − B ⎞⎛ X ( s ) ⎞ ⎛ 0 ⎞ ⎛ 0 ⎞
⎜⎜
⎟⎟⎜⎜
⎟⎟ = ⎜⎜
⎟⎟ = ⎜⎜ ⎟⎟
D ⎠⎝ U ( s ) ⎠ ⎝ Y ( s ) ⎠ ⎝ 0 ⎠
⎝ C
therefore, zero is the value of s, which makes
⎛ sI − A − B ⎞
⎜⎜
⎟⎟ rank deficient.
D ⎠
⎝ C
Example:
k1
k2
( s + 3)
G ( s) =
=
+
( s + 1)( s + 2) s + 1 s + 2
s-plane
system zeros: -3
system poles: -2, -1
-3
-2
-1
residues: k1 = 2, k2 = -1
( s + 1)( s + 3)
k1 =
( s + 1)( s + 2)
( s + 2)( s + 3)
k2 =
( s + 2)( s + 1)
=2
s =−1
= −1
s =−2
Pole-zero diagram
Time response of the system
Consider a linear time invariant lumped system described as
d 2 y (t )
dy (t )
du (t )
+3
+ 2 y (t ) = 3
− u(t )
2
dt
dt
dt
The Laplace transform of this equation is obtained as the
following.
s 2Y ( s ) − sy (0− ) − y& (0− ) + 3[ sY ( s ) − y (0− )] + 2Y ( s )
= 3[ sU ( s ) − u(0− )] − U ( s )
⇒
( s + 3) y (0− ) + y& (0− ) − 3u(0− )
3s − 1
Y ( s) =
+
U ( s)
2
2
s4
+2
3s4
+4
2 4443
s 4
+4
3s2
+4
2 43
14444
1
zero- input response
due to the initial state
zero- state response
due to the input
Zero-input Response
-
homogeneous solution
( s + 3) y ( 0 − ) + y& ( 0 − ) − 3u( 0 − )
Y ( s) =
s 2 + 3s + 2
k1
k2
=
+
s+1 s+ 2
y ( t ) = k1e − t + k2 e −2 t
where
k1 = 2 y ( 0 − ) + y& ( 0 − )
k2 = −[ y ( 0 − ) + y& ( 0 − )]
The roots -1, and -2 of the denominator is called the modes of
the system. Thus, the modes essentially govern the zero-input
response of the system. The denominator polynomial is called
the characteristic polynomial of a system.
Zero-state Response
- Transfer function
Y ( s) =
3s − 1
U ( s)
2
s + 3s + 2
The transfer function determines completely the zero-state
responses of the LTIL system.
1
Let U ( s ) =
,the y(t) can be obtained as
s
4
− 3 .5 − 0 .5
Y ( s) =
+
+
s +1 s + 2
s
−t
− 2t
y (t ) = 41
e4
−
3
.
5
e
4244
3−
due to the
poles of G(s)
0{
.5
due to the
pole of U(s)
In this case, all system poles are excited by the input U(s).
If let U ( s ) = s + 1 , then the system pole -1 will not be not excited
s−2
by the input.
7
5
Y ( s) =
+
4( s + 2 ) 4( s − 2 )
7 − 2t 5 2t
y (t ) = e + e
4
4
It can be concluded that whether a pole will be excited or not
depends on whether U(s) has a zero to cancel it. The Laplace
ω0
1
transforms of the unit-step function and sine function 2
s + ω02
s
have no zero. Therefore, either input will excite all poles of any
linear time invariant lumped system. They can be applied on
system identification as testing signals.
strictly proper transfer function: deg{D(s)} > deg{N(s)}.
The system suppresses high-frequency noise. The output
is continuous even input is discontinuous.
biproper transfer function: deg{D(s)} = deg{N(s)}. The
ratio of the information-bearing signal and noise will not
be altered. A discontinuous input signal will results in a
discontinuous output.
improper transfer function: deg{D(s)} < deg{N(s)}. The
system can amplify high-frequency noise. A continuous
input signal may results in a discontinuous output.
Imag
s-plane
LHP
RHP
Real
Dominant poles: the pole which has longest time constant,
(near to the imaginary axis).
End of this Chapter
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