Table Of ContentUniversity of California at Berkeley
Class Notes for ME134:
Automatic Control Systems
by
Anouck Girard
Masayoshi Tomizuka
Roberto Horowitz
ME134 Instructor:
Professor Karl Hedrick
5104 Etcheverry Hall
khedrick@me.berkeley.edu
Department of Mechanical Engineering
University of California at Berkeley
Fall 2008
© 2004-2008 A. Girard, M. Tomizuka and R. Horowitz
1
Introduction: Automatic Control Systems
Recent trends in the development of modern civilization have been in the direction of
greater control.
With the advent of the steam engine, and the material improvements brought about by the
industrial revolution, man has had available greater quantities of power for his use. To
use this power effectively, he has been required to learn how to control and how to
regulate it.
As part of the control process, certain standards have to be established. The performance
of the equipment is compared to these standards, and, according to the difference,
appropriate action is taken to bring about a closer correspondence between the desired
objectives and the actual performance.
The need for good control is present in many phases of our existence. We will limit
ourselves in this class to a study of problems pertaining to the field of engineering
applications of automatic control.
The problem is to determine the desired objectives, and the best ways of producing those
objectives.
open-loop Figure 1 is a block diagram showing an open-loop control system. The input is
system sometimes called the reference, while the output is sometimes called the controlled
variable. Disturbances can be present in the system.
2
Mathematical Modeling of Systems
Concept
Given a physical system, find a model (mathematical representation) that accurately
predicts the behavior (the output for a given input) of the system.
Key Points
To analyze and design control systems, we need quantitative mathematical models of
systems. The dynamic behavior of systems can be described using differential equations.
If the equations can be linearized, then the Laplace transform can be used to simplify the
solution.
The input/output relationship for linear components and sub-systems can be described in
the form of transfer functions.
System Representation and Modeling
In order to analyze (and subsequently control) most dynamic systems, it is essential to
attain a reasonable understanding of how a system functions. To achieve this objective,
we formulate mathematical models that help us describe the behavior of systems.
Mathematical models generally serve two purposes:
a. They are used in conjunction with analytical techniques to develop control
schemes for the systems that they represent.
1
b. They are used as a design tool in computer simulation studies. In this context, the
model is used as the control object to test and on which to evaluate possible
control schemes. This procedure is generally more efficient, cheaper and less time
consuming than to test the control schemes on the actual system.
energetic Here we will lay the groundwork for the formulation of mathematical models for
systems
energetic (dynamic) systems. Of paramount importance in our methodology are the
state of the
concept of state of the system and the selection of a set of state variables that describe the
system
state state of the system and adequately serve our modeling objectives.
variables
State Variables and State Equations
Consider a single input, single output (SISO) dynamic system, as shown schematically in
dynamic
system figure 1. Since the system is dynamic, the output y(t) at time t‡ 0 is a function of the past
as well as the present inputs: u . Notice that u represents the entire history of the
[0,t] [0,t]
input variable u(t ) from t = 0 to t = t. It is obviously a problem to keep track of the entire
system history in order to be able to predict the future system output. Thus, it is important
to determine what the minimum amount of information needed, at time t, to predict the
output of the system immediately after time t.
u(t) y(t)
Dynamic System
input output
Figure 1. SISO Dynamic System.
The state variables are the minimum set of variables such that knowledge of these
state
variables at time k, together with the current (present) state of the system, is sufficient to
variables:
determine the future state and output of a system.
minimum
amount of
information The state variables represent the minimum amount of information that needs to be
needed to
retained at any time t in order to determine the future behavior of a system. Although the
predict the
number of state variables is unique (that is, it has to be the minimum and necessary
output
number of variables), for a given system, the choice of state variables is not.
Choosing state variables and deriving models is in general a non-trivial problem. In many
cases, there is no universal or general procedure. For engineering energetic systems, the
choice of state variables still depends on the problem. However, there exists a fairly
systematic procedure for selecting state variables and deriving models.
The order of the system designates the number n of state variables needed (minimum and
system order
necessary) to describe the system.
2
The set of n differential or difference first-order equations that govern the relationship
model = state
equations + between the input to a system and the n state variables are the state equations. Together
output with the output equation, they constitute the model of the system.
equation
Denoting the n state variables of a system as x, x, …, x and defining x& = dx dt, the
1 2 n
state equations can be written as:
x& (t)= f (x (t),x (t),...,x (t),u(t))
1 1 1 2 n
x& (t)= f (x (t),x (t),...,x (t),u(t))
2 2 1 2 n
...
x& (t)= f (x (t),x (t),...,x (t),u(t))
n n 1 2 n
Note that the right-hand side of the state equations is only a function of the state variables
and the input at time t. Define the nth order vectors:
Ø x ø Ø f ø
Œ 1œ Œ 1œ
Œ x œ Œ f œ
x = 2 and f = 2
Œ œ Œ œ
... ...
Œ œ Œ œ
º x ß º f ß
n n
We can write the system of n state equations in vector form as:
x&(t) = f(x(t),u(t))
When the dynamic system is linear, we can write the above equation as follows:
linear system
x&(t) = A(t)x(t)+b(t)u(t)
where A(t) is a n by n matrix and b(t) is an n by one vector.
Ø b (t)ø
Ø a (t) ... a (t)ø Œ 1 œ
Œ 11 1n œ Œ b (t)œ
A(t)= Œ ... ... œ and b(t) = 2 .
Œ œ
...
Œº a (t) ... a (t)œß Œ œ
n1 nn º b (t)ß
n
If A and b are constants, then the system is an nth order Linear Time Invariant (LTI)
linear time
invariant system.
(LTI) system
x&(t) = Ax(t)+bu(t)
The output equation is given by the algebraic equation:
3
y(t) = h(x(t),u(t))
When the system is LTI, the output equation can be written as:
y(t)= cx(t)+du(t)
[ ]
where c = c c ... c is a one by n vector and d is a scalar.
1 2 n
Mass Spring Damper Example
Consider the mass-spring-damper example shown in figure 2, where x is the position of
the mass, v = x& is the velocity of the mass, u is the input force applied to the mass, k is
the coefficient of elasticity of the spring, and b is the viscous damping coefficient of the
damper.
x
k
u
m
b
Figure 2. Mass-Spring-Damper System.
Assuming that the system is linear and time invariant, a model of the system can be
derived as follows:
a. Define the state of the system to be the position and the velocity of the mass,
r [ ]
x = x v T.
b. Use Newton’s laws to derive the equations of motion for the system:
mv&= f + f +u(t), where f and f are the forces exerted by the spring and
k v k v
damper respectively.
c. Use the constitutive relations for the spring and damper, f =- kx and f = - bv.
k v
d. Combine these equations and write them in matrix form:
4
d Ø x(t)ø Ø 0 1 ø Ø x(t)ø Ø 0 ø
Œ œ = Œ œ Œ œ +Œ œ u(t)
dt º v(t)ß º - k/m - b/mß º v(t)ß º 1/mß
[ ]Ø x(t)ø
y(t) = 1 0 Œ œ
º v(t)ß
Note that this is not the only state equation model that could have been used to describe
the system.
Difference Equations Example
y(k) = u(k-2)
The associated state equations would be:
x (k) = u(k-1)
1
x (k) = x (k-1)
2 1
The output equation would be:
y(k) = x (k-1) = u(k-2)
1
Re-writing:
x (k+1) = u(k)
1
x (k+1) = x (k)
2 1
y(k) = x (k)
2
In matrix form:
Ø x (k +1)ø Ø 0 0ø Ø x (k)ø Ø 1ø
Œ 1 œ = Œ œ Œ 1 œ +Œ œ u(k)
º x (k +1)ß º 1 0ß º x (k)ß º 0ß
2 2
[ ]Ø x (k)ø
y(k)= 0 1Œ 1 œ
º x (k)ß
2
(cid:222) x(k+1) = A.x(k) + b.u(k)
y(k) = C.x(k)
5
Modeling of Dynamic Systems: General Approach
A dynamic (or energetic) system is a collection of energy storage elements, power
dissipative elements, power sources, transformers and transducers. For successful
modeling of energetic systems, it is important to know the characteristics of each of these
elements.
In this class, we will model energetic systems as lumped parameter systems. That is, we
will model the system such that each point in the system embodies the properties of the
region immediately surrounding it. Lumped parameter systems are described by a finite
number of state variables.
Example of a Lumped System: Mass-Spring System
x
k
m
Figure 3. Mass-Spring System.
mx&&+kx= 0
The properties of the surrounding region are lumped (concentrated) at each point. The
state equations describing lumped parameter systems can be written as finite order
ordinary differential equations.
Many energetic systems cannot be modeled as lumped parameter systems. These systems
must be modeled as distributed parameter systems.
Example of a Distributed System: The Cantilever Beam
u(x,t): deflection
x
Figure 4. Cantilever beam.
6
¶ 2u(x,t) 1 ¶ 2u(x,t)
- = 0
¶ x2 a2 ¶ t2
Each little “chunk” of mass acts as an elastic segment. The properties of mass and
elasticity cannot be separated from each other. The state equations that describe
distributed parameter systems are partial differential equations.
In the next pages, we will present a unified approach for modeling mechanical, electrical,
fluid and/or thermal lumped parameter systems. The first step in modeling energetic
systems is to break the system up into elements, then find the basic relations that describe
the individual elements that form the system. Physical laws are generally used to obtain
these relations.
Mechanical systems: Newton’s Laws
Isaac Newton, a 17th-18th century English physicist and mathematician (also the guy
who had an apple hit his head supposedly), stated three laws that basically describe the
statics and dynamics of objects. Statics is the study of forces on an object at rest.
Dynamics is the study of how forces affect the motion of a body.
Newton's First Law of Motion (Law of Inertia): Every body continues in its state
of rest or of uniform speed in a straight line unless it is compelled to change that
state by forces acting on it.
Newton's Second Law of Motion: The acceleration of an object is directly
proportional to the net force acting on it and is inversely proportional to its mass.
The direction of the acceleration is in the direction of the applied net force.
Newton's Third Law of Motion (Law of Action-Reaction): Whenever one object
exerts a force on a second object, the second exerts an equal and opposite force
on the first.
The Law of the Conservation of Momentum: The total momentum of an isolated
system of bodies remains constant.
Electric Systems: Kirchhoff’s Rules
There are only two relatively simple rules known as Kirchhoff's rules (developed by
Gustav Robert Kirchhoff (1824-1887)).
• Point Rule: The algebraic sum of the currents toward any branch point is zero.
7
(cid:229)
I = 0
• Loop Rule: The algebraic sum of the potential differences in any loop, including
those associated with EMFs and those of resistive elements, must equal zero.
(cid:229) (cid:229)
x - IR= 0
Here is a good set of guidelines when you are working with these laws:
1. Choose any closed loop in the network and designate a direction to traverse the
loop when you apply the loop rule (If you have a diagram, draw it on the
diagram).
2. Go around the loop in that direction, adding EMF's and potential differences. An
EMF is positive if you go from (-) to (+) (which is in the direction of the field E in
the source) and negative when you go from (+) to (-). A -IR term is counted
negative if the resistor is traversed in the same direction as the assumed current,
positive if backwards.
3. Equate the sum of found in the previous step to zero.
4. If you need to, choose another loop to obtain a different relationship. The number
of loop equations you have will be one less than the number of loops you have.
You will be setting up a system of equations in order to find the current in each
loop.
5. The last equation should be a branch equation, where I + I = I (substitute the
1 2 3
currents for whichever branch you choose).
6. Solve the systems of equations. If one of the currents you solve is negative, it
simply means that the current goes in the direction opposite the way you assumed.
Thermal Systems: The Laws of Thermodynamics
Thermodynamics is the study of the inter-relation between heat, work and internal energy
of a system.
The British scientist and author C.P. Snow had an excellent way of remembering the
three laws:
1. You cannot win (that is, you cannot get something for nothing, because matter
and energy are conserved).
2. You cannot break even (you cannot return to the same energy state, because there
is always an increase in disorder; entropy always increases).
3. You cannot get out of the game (because absolute zero is unattainable).
8
