Lecture: Continuous-time linear systems
Automatic Control 1
Continuous-time linear systems
Academic year 2010-2011
Prof. Alberto Bemporad (University of Trento) Automatic Control 1 Academic year 2010-2011 1 /42
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Automatic Control 1 : Continuous-time linear systems, Lecture notes of Psychology
Dynamical models, Ordinary differential equations with Examples, Continuous-time linear systems.
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Lecture: Continuous-time linear systems
Automatic Control 1
Continuous-time linear systems
Academic year 2010-
Lecture: Continuous-time linear systems Dynamical systems
Dynamical models
A dynamical system is an object (or a set of objects) that evolves over time,
possibly under external excitations.
Examples: a car, a robotic arm, a population of animals, an electrical circuit,
a portfolio of investments, etc.
The way the system evolves is called the dynamics of the system.
A dynamical model of a system is a set of mathematical laws explaining in a
compact form and in quantitative way how the system evolves over time,
usually under the effect of external excitations.
Main questions about a dynamical system:
1 Understanding the system (“How X and Y influence each other ?”)
2 Simulation (“What happens if I apply action Z on the system ?”)
3 Design (“How to make the system behave the way I want ?”)
Lecture: Continuous-time linear systems Dynamical systems
Dynamical models
Working on a model has almost zero cost compared to real experiments (just
mathematical thinking, paper writing, computer coding)
However, a simulation (or any other inference obtained from the model) is as
better as the dynamical model is closer to the real system
Conflicting objectives:
1 Descriptive enough to capture the main behavior of the system
2 Simple enough for analyzing the system
“Make everything as simple as possible, but not simpler.”
- Albert Einstein
Albert Einstein
(1879-1955)
Making a good model is an art! (that you are learning ...)
Lecture: Continuous-time linear systems Differential equations
Ordinary differential equations (ODEs)
First order differential equation (=the simplest dynamical model):
˙ x ( t ) = ax ( t ) a ∈ R, ˙ x ¨
dx
dt
x ( 0 ) = x 0 x 0
∈ R
Its unique solution is x ( t ) = e
at x 0
0 0.2 0.4 0.6 0.8 1
0
1
2
3
4
x(t)
t
a>
a=
a<
Lecture: Continuous-time linear systems Differential equations
First order differential equations with inputs
Introduce the forcing signal u ( t )
˙ x ( t ) = ax ( t ) + bu ( t ) a , b ∈ R, u ( t ) ∈ R
x ( 0 ) = x 0
x 0
∈ R
The unique solution x ( t ) is
x ( t ) = e
at x 0 ︸︷︷︸
natural response
t
0
e
a ( t − τ ) bu ( τ ) dτ
forced response
x ` ( t ) = e
at x 0 effect of the initial condition
x f ( t ) =
t
0
e
a ( t − τ ) bu ( τ ) dτ effect of the input signal
Lecture: Continuous-time linear systems Differential equations
Examples
x ( t ) = voltage x ( t ) = velocity
x(t)
C
R
u(t)
x(t)
M
!
u(t)
u ( t ) − RC ˙ x ( t ) − x ( t ) = 0 − βx ( t ) + u ( t ) = M ˙ x ( t )
˙ x ( t ) = −
1
RC
x ( t ) −
1
RC
u ( t ) ˙ x ( t ) = −
β
M
x ( t ) +
1
M
Lecture: Continuous-time linear systems Linear systems
Example: RLC circuit
x!(t)
C
u(t)
R
x"(t)
L
u ( t ) − Rx 1
( t ) − L
dx 1 ( t )
dt
− x 2
( t ) = 0 Kirchhoff’s voltage law
x 1 ( t ) = C
dx 2 ( t )
dt
Kirchhoff’s current law
Rewrite as the 2
nd order linear system
dx 1 ( t )
dt
R
L
x 1 ( t ) −
1
L
x 2 ( t ) +
1
L
u ( t )
dx 2 ( t )
dt
1
C
x 1
( t )
or in matrix form
˙ x ( t ) =
R
L
1
L
1
C
A
x ( t ) +
1
L
B
u ( t )
Lecture: Continuous-time linear systems Linear systems
Example: Mass-spring-damper system
x!(t), x"(t)
M
u(t)
K
˙ x 1
( t ) = x 2
( t ) velocity = derivative of traveled distance
M ˙ x 2
( t ) = u − βx 2
( t ) − Kx 1
( t ) Newton’s law
Rewrite as the 2
nd order linear system
dx 1 ( t )
dt
= x 2 ( t )
dx 2 ( t )
dt
β
M
x 2 ( t ) −
K
M
x 1 ( t ) +
1
M
u ( t )
or in matrix form
˙ x ( t ) =
K
M
β
M
A
x ( t ) +
1
M
B
u ( t )
Lecture: Continuous-time linear systems Linear algebra recalls
Linear algebra recall
The eigenvalues of A ∈ R
n × n are the roots λ 1 ,... , λ n of its characteristic
polynomial
det( λ i I − A ) = 0, i = 1, 2,... , n
An eigenvector of A is any vector v i
∈ R
n such that
Av i = λ i v i
for some i = 1, 2,... , n.
Diagonalization of A :
λ 1 0 ... 0
0 λ 2 ... 0
0 0 ... λ n
=^ T
− 1 AT , T =
v 1 | v 2 |... | v n
(not all matrices A are diagonalizable, see Jordan normal form)
Lecture: Continuous-time linear systems Linear algebra recalls
Linear algebra recall
Example:
A =
, det( λI − A ) =
λ − 1 − 3
5 λ − 2
= λ
2 − 3 λ + 17
Eigenvalues: λ 1
3
2
- j
p 59
2
, λ 2
3
2
− j
p 59
2
Complex numbers recall:
Imaginary unit : j ¨
p
− 1
Cartesian form : c = a + jb , c ∈ C, a , b ∈ R
Real part of c : ℜ c = a
Imaginary part of c : Im c = b
Conjugate of c : ¯ c = a − jb
Polar form : c = ρe
jθ , ρ ≥ 0, θ ∈ R
Modulus or magnitude : | c | =
p
a
2
- b
2 = ρ
Angle or phase : ∠ c = θ
Complex exponential : e
c = e
a + jb = e
a e
jb = e
a (cos b + j sin b )
Lecture: Continuous-time linear systems Linear algebra recalls
Eigenvalues and modes
Let u ( t ) ≡ 0 and assume A diagonalizable
The state trajectory is the natural response
x ( t ) = e
At x ( 0 ) = Te
Λ t T
− 1 x 0 ︸ ︷︷ ︸
α
= [ v 1
... v n
]
e
λ 1 t ... 0
0 ... e
λn t
α
î
v 1 e
λ 1 t
... v n e
λn t
ó
α 1
. . . α n
n ∑
i = 1
α i e
λi t v i
where v i =eigenvector of A , λ i =eigenvalue of A , α = T
− 1 x ( 0 ) ∈ R
n
The evolution of the system depends on the eigenvalues λ i of A , called modes
of the system (sometimes we also refer to e
λ i t as the i -th mode)
A mode λ i is called excited if α i
Lecture: Continuous-time linear systems Linear ordinary differential equations
Differential equations of order n
dy
( n ) ( t )
dt
n
- a n − 1
dy
( n − 1 ) ( t )
dt
n − 1
- · · · + a 1
˙ y ( t ) + a 0
y ( t ) = 0
By setting x 1
( t ) ¨ y ( t ), x 2
( t ) ¨ ˙ y ( t ),... , x n
( t ) ¨ y
n − 1 ( t ), this is equivalent to the
system of n first-order equations
˙ x 1 ( t ) = x 2 ( t )
˙ x 2 ( t ) = x 3 ( t )
˙ x n ( t ) = − a 0 x 1 ( t ) +... − a n − 1 x n ( t )
x ( 0 ) = [ y ( 0 ) ˙ y ( 0 )... y
n − 1 ( 0 )]
′
Example:
¨ y ( t ) + 2˙ y ( t ) + 5 y ( t ) = 0
x 1 ( t ) = y ( t )
x 2 ( t ) = ˙ y ( t )
⇒
˙ x 1 ( t ) = x 2 ( t )
˙ x 2 ( t ) = − 5 x 1 ( t ) − 2 x 2 ( t )
x ( 0 ) = [ y ( 0 ) ˙ y ( 0 )]
′
Lecture: Continuous-time linear systems Linear ordinary differential equations
Examples of state-space realizations
Example 1
¨ y ( t ) − 2˙ y ( t ) + y ( t ) = u ( t ) + 2˙ u ( t )
⇒
d
dt
x ( t ) =
0 1
− 1 2
x ( t ) +
0
1
u ( t )
y ( t ) =
î
1 2
ó
x ( t )
Double check:
˙ y =
î 1 2
ó
˙ x =
î 1 2
ó
0 1
− 1 2
x ( t ) +
0
1
u ( t )
=
î − 2 5
ó
x ( t ) + 2 u ( t )
¨ y =
î
− 2 5
ó
˙ x + 2˙ u =
î
− 5 8
ó
x ( t ) + 5 u ( t ) + 2˙ u ( t )
¨ y ( t ) − 2˙ y ( t ) + y ( t ) =
î
− 5 8
ó
x ( t ) + 5 u ( t ) + 2˙ u ( t ) − 2
Äî
− 2 5
ó
x ( t ) + 2 u ( t )
ä
î
1 2
ó
x ( t )
=
î
− 5 + 4 + 1 8 − 10 + 2
ó
x ( t ) + ( 5 − 4 ) u ( t ) + 2˙ u ( t )
= u ( t ) + 2˙ u ( t ) ok!
Lecture: Continuous-time linear systems Linear ordinary differential equations
Alternative state-space realization method
In the following special case (=no input derivatives)
dy
( n ) ( t )
dt
n
- a n − 1
dy
( n − 1 ) ( t )
dt
n − 1
- · · · + a 1 ˙ y ( t ) + a 0 y ( t ) = b 0 u ( t )
we can define the following states
x 1 = y → ˙ x 1 = x 2
x 2 = ˙ y → ˙ x 2 = x 3
. .
. =
. . .
x n =
d n − 1 y
dt n − 1 → ˙ x n =
d n y
dt n = − a n − 1
dy ( n − 1 ) ( t )
dt n − 1 − · · · − a 1 ˙ y ( t ) − a 0 y ( t ) + b 0 u ( t )
and therefore set
A =
0 1 0 ... 0 0 0 1 ... 0
. . .
. . .
. . .
. . . 0 0 0 ... 1 − a 0 − a 1 − a 2 ... − an − 1
, B =
0 0
. . . 0 b 0
C = [ 1 0 0 ... 0 ] , D = 0