Lecture 5 - Physical Modeling 1.

Introduction

K. J. Åström • Dynamics is a major issue in control, since both the pro-

cesses to be controlled and the controllers are dynamical
1. Introduction systems.
2. The Concept of State • Dynamics has been described by differential equations
3. Linear State Models and Linearization relating inputs to outputs.

4. Physical Modeling • How much of the past do we need to consider?

5. Homework • Is there some way to minimize the information about the

past that we need?
Theme: Physical modeling, state models, linearization • Newton’s idea!

Static and Dynamic Models The Concept of State

Static models • The Great Modelers: Tycho Brahe, Kepler and Newton
y = f (u) • To predict the future motion of the planets it is enough to
know their current positions and the velocities.
The output signal y(t) at time t depends only on the value of • The state is the least information about a system that is
the input at time t. required for the prediction of its future development.
The output of a dynamic system at time t depends on past • Philosophical consequences, causality, predestination,
values of the input, for example (chaos).

t • A detailed description of the physics of a system, balances
y ( t) = h( t − τ )u(τ ) dτ of mass, momentum and energy.
0
• The state is the least number of variables required to
describe storage of mass, momentum and energy.
• Synonyms: state models, internal descriptions, white
boxes.

c K. J. Åström September, 2000 1
 Cruise Control A Simple Water Tank
A simple model of a car on a sloping road tells how its position How do level h and outflow q out depend on the inflow q in?
y depends on the throttle. Let the mass be m and assume that
Assume: Constant density
the propelling force is proportional to the throttle. We find

d2 y d y  d y 
m − c   = F − mg sin θ
dt2 dt dt dV
= qin − qout Massbalance
dt
Two states y and d y/dt.
h
V= A( h) dh Geometry
0

qout = a 2g h Energybalance

Many ways to choose the state.

Analysis and Simplification An Inverted Pendulum

Choosing h as a state variable we find y

dh 1  Momentum balance
= ( qin − a 2g h)
dt A( h) (Newton’s Equation)

qout = a 2g h
d2θ
J − mg l sin θ + mul cos θ = 0
dt2 θ
One function A(h) and one parameter a. L

Steady state relation h

Two states θ and dθ /dt.
qout = qin x

q2in 
h= Normalize with ω 0 = mg / J , introduce τ = ω 0 t and ū = u/g
2g a 2
q then
Not influenced by A! in
d2θ
Run ICtools or SysQuake − sin θ + u cos θ = 0
dτ 2

c K. J. Åström September, 2000 2
 Pendulum on a Cart Standard Model
Equations of motion Momen- x A system with finite number of states can be described by
tum balances
dx
J pθ¨ + ml x cos θ − mg l sin θ = 0 O = f ( x, u)
dt
mlθ¨ cos θ − mlθ˙ 2 sin θ + M ẍ = F y = g( x, u)

Momentum balance character- • x state

ized by four variables: • u input, control variable
θ , θ˙ , x and ẋ
• y output, measured variable
Four states are enough!
The model (a nonlinear ordinary differential equation (ODE)),
tells that the rate of change of the state at time t is uniquely
given by the state at time t, and the input at time t. If the
state is known at time t, old values of x do not give any extra
information.

Standard Model - Equilibrium Solutions Difficulties with Nonlinear Equations

Given the system • Solutions may not exist for all t. Example
dx dx 1
= f ( x, u) = x2 , x ( t) =
dt dt 1− t
y = g( x, u)
• There may be many solutions. Example
find constant values x 0 and u0 that satisfy the equation. Putting 2
dx/dt = 0 gives 
 t if t ≥ 0,
dx √ 
= x, x ( t) = 0 if t ≤ a
f ( x0 , u0 ) = 0 dt 


( t − a)2 if t > a

Compare with the water tank!

Bad modeling!
Not easy do discover!
• Numerical solutions require care

c K. J. Åström September, 2000 3
 Inverted Pendulum Inverted Pendulum
The model: The model:

J pθ¨ − mg sin θ = mu cos θ J pθ¨ − mg sin θ = mu cos θ

Introduce state variables x 1 = θ and x2 = θ˙ , then Introduce state variables x 1 = θ and x2 = θ˙ , then
dx1 dx1
= x2 = x2
dt dt
dx2 mg m dx2 mg m
= sin x1 + u cos x1 = sin x1 + u cos x1
dt Jp Jp dt Jp Jp

Find the stationary solution! Stationary solutions for u = 0 gives sin x 1 = 0 and x2 = 0.
The Audience is Thinking ... Two cases:
θ = x1 = 0 and θ˙ = x2 = 0 (pendulum up)
θ = x1 = π and θ˙ = x2 = 0 (pendulum down)

Expansion of Standard Model Linearization

We have used a very compact notation: dx/dt = f ( x, u), • Nonlinear systems are difficult.
y = g( x, u). It is important to know what this means. Writing • Purpose of control is to keep variables close to desired
all components of the vectors we get values.
dx1
= f 1 ( x1 , x2 , . . . , xn , u1 , u2 , . . . , u p)
• Approximate by considering small deviations from equilib-
dt rium
dx2
= f 2 ( x1 , x2 , . . . , xn , u1 , u2 , . . . , u p) • This is called linearization
dt
..
. • Major simplification
dxn • Approximation improves with quality of control system
= f n ( x1 , x2 , . . . , xn , u1 , u2 , . . . , u p)
dt
• Procedure:
y1 = g1 ( x1 , x2 , . . . , xn , u1 , u2 , . . . , u p)
y2 = g2 ( x1 , x2 , . . . , xn , u1 , u2 , . . . , u p) First determine the equilibria
.. Approximate the equations around the equilibria
.
yr = g1 ( x1 , x2 , . . . , xn , u1 , u2 , . . . , u p)

c K. J. Åström September, 2000 4
 Linearization of Static System Linearization of Dynamic Systems
Consider the system A curve is approximated by Start with
its tangent
dx
y = g(u) = f ( x, u)
10

dt
y = g( x, u)
A Taylor series expansion 5

around u = u 0 gives
Find the equilibria u = u 0 , x = x0 y = y0 by solving

y

y = g(u0 ) + g (u0 )(u − u0 ) + . . . 0

f ( x0 , u0 ) = 0

The linearized model

−5
0
u
0.5 1 1.5 2 2.5 3 3.5 4 Notice that there may be several solutions!
When is the approximation Decide what operating condition you want!
y − y0 = g  (u0 )(u − u0 ) good?

Linearization of Dynamic Systems ... Linearization of Dynamic Systems ...

dx dδ x dx
= f ( x, u) = = f ( x0 + δ x, u0 + δ u)
dt dt dt
y = g( x, u) f f
 f ( x0 , u0 ) + ( x0 , u0 )δ x + ( x0 , u0 )δ u
x u
Approximate around the equilibrium! y0 + δ y = g( x0 + δ x, u0 + δ u)
g g
x = x0 + δ x, u = u0 + δ u, y = y0 + δ y  y0 + ( x0 , u0 )δ x + ( x0 , u0 )δ u
x u

Hence
dδ x f f
dx f f = ( x0 , u0 )δ x + ( x0 , u0 )δ u = Aδ x + B δ u
= f ( x0 + δ x, u0 + δ u)  f ( x0 , u0 ) + ( x0 , u0 )δ x + ( x0 , u0 )δ u dt x u
dt x u g g
g g δy= ( x0 , u0 )δ x + ( x0 , u0 )δ u = Cδ x + Dδ u
y = g( x0 + δ x, u0 + δ u)  y0 + ( x0 , u0 )δ x + ( x0 , u0 )δ u x u
x u


c K. J. Åström September, 2000 5
 Linearization of Dynamic Systems A Remark on Notations
For small deviations around an equilibrium the system f
A= ( x0 , u0 )
dx x
= f ( x, u)
dt
Component-wise
y = g( x, u)
 f 1 ( x1 , x2 , . . . , xn , u1 , u2 , . . . , u p)
can be approximated by a11 =
 x1
 f 1 ( x1 , x2 , . . . , xn , u1 , u2 , . . . , u p)
dδ x a12 =
= Aδ x + B δ u  x2
dt
..
δ y = Cδ x + Dδ u .
 f n( x1 , x2 , . . . , xn , u1 , u2 , . . . , u p)
an1 =
 x1
f f
A= ( x0 , u0 ) B= ( x0 , u0 ) ..
x u .
g g  f n( x1 , x2 , . . . , xn , u1 , u2 , . . . , u p)
C= ( x0 , u0 ) D= ( x0 , u0 ) ann =
x u  xn

The Water Tank The Inverted Pendulum

  States x1 = θ = y and x2 = θ˙
dh 1
= ( qin − a 2g h) qout = qin = q0 = a 2g h0
dt A( h) dx1
 = x2
q20 dt
qout = a 2g h h0 =
2g a 2 dx2 mg m
= sin x1 + u cos x1
dt J J
Assume constant cross section A, introduce h = h 0 + δ h
 √
write in standard form
dδ h a 2g 1 a 2g h 0 1
=− δ h + δ qin = − δ h + δ qin dx x2
dt 2A h0 A 2Ah0 A = f ( x, u) =
 √ mg m
2g a 2g h 0 q0 dt sin x1 + u cos x1
δ qout = a δh = δh = δh J J
h0 h0 h0
Two stationary solutions
Time constant
θ = x1 = 0 and θ˙ = x2 = 0 (pendulum up)
2Ah0 Total water volume [m ] 3
T= = 2 θ = x1 = π and θ˙ = x2 = 0 (pendulum down)
q0 Flow rate [m3 /s]

Physical interpretation! 
c K. J. Åström September, 2000 6
 The Inverted Pendulum ... Linear Dynamical Systems - The State Model

x2 dx
f ( x, u) = mg = Ax + Bu
J sin x1 + m
J u cos x1 dt
y = Cx + Du

f 0 1 f 0
• Variables denote deviations from equilibrium
= mg , =
x cos x1 − m
u sin x1 0 u m
cos x1
• Think scalar and interpret as vectors
J J J

Evaluate for u = 0, x 1 = 0 och x2 = 0 (pendulum up)

Solution
f 0 1 f 0
t
A= = B= =
x mg
0 u m x( t) = e At x(0) + e A( t− s) Bu( s) ds
J J 0

t
Evaluate for u = 0, x 1 = π och x2 = 0 (pendulum down) y( t) = Ce At x(0) + C e A( t− s) Bu( s) ds + Du( t)
0
f 0 1 f 0
A= = B= = All information in the matrices A, B , C and D .
x − mJg 0 u − mJ

The Matrix Exponential Calculating with the Matrix Exponential

What is the meaning of e At? The matrix exponential is defined as
1 1 1 1 1
e At = I + At + ( At)2 + . . . + An tn + . . . e At = I + At + ( At)2 + ( At)3 . . . + ( At)n + . . .
2 n! 2 3! n!

If A can be diagonalized A = T Λ T −1 , then Differentiate!

d At 1 1
e At = T ( I + Λ t +
1 2 2
Λ t + ... +
1 n n
Λ t + . . . ) T −1 e = A + At + ( At)2 . . . + ( At)n−1 + . . . = Ae At
2 n! dt 2 ( n − 1)!

eλ 1 t 0 ... 0 t
 0
 λ
e 2t ... 0 
 −1 Differentiation of x(t) = e At x(0) + 0 eA(t−τ ) Bu(τ )dτ
= Te T Λt −1
=T
 ..
T

 .  gives
0 0 ... eλ n t
dx
= Ax + Bu
where λ i are the eigenvalues of the matrix A, i.e. the solutions dt
to the equation det(λ I − A) = 0

c K. J. Åström September, 2000 7
 Vector and Matrix Notations Relation between Input and Output
• Very compact and practical notation
dx
• Numerical calculations supported by nice software = Ax + Bu
dt
• Learn to formulate and interpret y = Cx + Du

• Essentially the same as for scalar equations Input-output relation

• BUT remember that AB = B A! for matrices
t
y( t) = Ce x(0) + C
At
e A( t− s) Bu( s) ds + Du( t)
0

Compare with first order systems! Take Laplace transforms

sX ( s) − x(0) = AX ( s) + B U ( s) X ( s) = ( sI − A)−1 x(0) + ( sI − A)−1 B U ( s)

Y ( s) = C X ( s) + DU ( s) Y ( s) = C X ( s) + DU ( s)

The transfer function is G (s) = D + C(sI − A) −1 B

Summary
• Obtaining dynamics from physics
• The concept of state
• The standard model for nonliner finite dimensional systems
dx
= f ( x, u), y = g( x, u)
dt

• Linearization and linear time invariant (LTI) systems

• The standard model for linear time invariant systems
dx
= Ax + Bu, y = Cx + Du
dt

• Vector and matrix notations, the matrix exponential e At

• Compact notation with computational tools

c K. J. Åström September, 2000 8