Control Example using Matlab
Cruise Control
Modeling a Cruise Control System
•
The
inertia
of the wheels is
neglected
•
Aerodynamic Drag
is
neglected
–
is proportional to the square
of the car’s speed
Physical setup and system equations
•
The problem is reduced to the simple mass and damper system
•
It is assumed that
friction
is opposing the motion of the car
–
is proportional to the car's speed
Modeling a Cruise Control System
•
Using Newton's law, the dynamic equation
for this system is:
u
b
m
x
x
where u is the force from the engine.
•
There are several different ways to describe a system of
linear differential equations.
•
To calculate the Transfer Function, we shall use the Stare
Space representation then transform it to TF using
ss2tf
State

space equations
•
The
state

space representation
is given by the
equations:
where
is an n by 1 vector representing the
state
(commonly position and velocity variables in mechanical systems),
u is a scalar representing the
input
(commonly a force or torque in mechanical systems), and
y is a scalar representing the
output
.
Du
C
y
Bu
A
dt
d
x
x
x
x
State Equation for our CC model
The state vector is [ x, ] and y is
x
x
x
The equation
u
m
1
m
b
x
x
is
u
b
m
x
x
or
u
0
x
1
0
y
u
m
1
0
x
m
b
0
1
0
x
x
x
x
Design requirements
•
For this example, let's assume that
m =
1000
kg
b =
50
Nsec/m
u =
500
N
•
Design requirements
–
When the engine gives a
500
Newton force, the car will reach a
maximum velocity of
10
m/s
(
22
mph).
–
An automobile should be able to accelerate up to that speed in less
than
5
seconds
. Since this is only a cruise control system, a
10
%
overshoot
on the velocity will not do much damage. A
2
% steady

state error
is also acceptable for the same reason.
–
Rise time <
5
sec
Overshoot <
10
%
Steady state error <
2
%
Matlab representation and open

loop response
m =
1000
;
b =
50
;
u =
500
;
A = [
0 1
;
0

b/m]
B = [
0
;
1
/m]
C = [
0 1
]
D =
0
step(A,u*B,C,D)
this step response does not meet the design criteria placed on the
problem. The system is
overdamped
, so the overshoot response is
fine, but the
rise time is too slow
.
P controller
•
The first thing to do in this problem is to
transform
the state

space equations into
transfer function
form.
m =
1000
; b =
50
; u =
500
; A = [
0 1
;
0

b/m] B = [
0
;
1
/m] C = [
0 1
] D =
0
[num,den]=
ss
2
tf
(A,B,C,D)
•
Next, close the loop and add some
proportional control
to see
if the response can be improved.
k =
100
;
[numc,denc] = cloop(k*num,den,

1
);
%Closed

loop transfer
function
t =
0
:
0.1
:
20
;
step
(
10
*numc,denc,t) %for
10
m/sec the signal used by
step
axis
([
0 20 0 10
])
Get the Transfer Function
•
For
m =
1000
; b =
50
; u =
500
; A = [
0 1
;
0

b/m] B = [
0
;
1
/m] C = [
0 1
] D =
0
[num,den]=
ss
2
tf
(A,B,C,D)
•
Matlab should return the following to the command window:
num =
0 0.0010 0
den =
1.0000 0.0500 0
•
This is the way Matlab presents the TF
0.001
s +
0

s^
2
+
0.05
s +
0
Step function
•
The step function is one of most useful functions in
Matlab for control design.
•
Given a system that can be described by either a transfer
function or a set of state

space equations,
the response
to a step input
can immediately be plotted.
•
A step input can be described as a
change in the input
from
zero to a finite value at time t =
0
.
•
By default, the step command performs a unit step (i.e.
the input goes from zero to one at time t =
0
).
•
The basic command to use the step function is one of the
following
–
(depending if you have a set of state

space equations or a
transfer function form):
step(A,B,C,D)
step(num,den)
P controller
•
The steady
state error is more than 10%,
and
the rise time is still too slow
.
•
Adjust the proportional gain to make the
response better.
but you will not be able to make
the
steady state value go to 10m/sec
without getting rise times that are too
fast.
•
You must always keep in mind that you are designing a real system,
and for a cruise control system to respond
0

10
m/sec
in less than half
of a second is unrealistic.
•
To illustrate this idea, adjust the k value to equal
10
,
000
, and you
should get the following velocity response:
P controller
•
The solution to this problem is to add some
integral control
to eliminate the steady state
error.
•
Adjust k
until you get a reasonable rise time.
–
For example, with k =
600
, the response should now
look like:
PI controller
k =
600
;
ki =
1
; % small to get feeling for integral response
num
1
= [k ki]; % PI has two gains
den
1
= [
1 0
];
num
2
= conv(num,num
1
); % convolution used to multiply
the
% polynomials representing the
gains
den
2
= conv(den,den
1
);
[numc,denc] = cloop(num
2
,den
2
,

1
);
t =
0
:
0.1
:
20
;
step(
10
*numc,denc,t)
axis([
0 20 0 10
])
Remember,
integral control makes the transient response worse
, so adding too
much initially can make the response unrecognizable.
PI controller
•
Now you can adjust
the ki value to reduce
the steady state error.
•
With
ki = 40
and
k
adjusted up a little
more to
800
, the
response looks like
the following:
As you can see, this step
response
meets all of the
design criteria,
and
therefore no more iteration
is needed.
It is also noteworthy that
no derivative control was
needed
in this example
Modeling a Cruise Control System
•
ךוכיחה חוכ תוחפ עינמה חוכל הווש היצרניאה חוכ
.
Physical setup and system equations
Let’s now add Drag and assume the friction to be non velocity
dependent (the static friction, proportional to normal force).
)
(
)
(
)
(
2
t
Bv
t
Cu
dt
t
dv
M
)
(
)
(
)
(
2
t
u
t
dt
t
d
רמולכ
:
M

תינוכמה תסמ
B

ךוכיח םדקמ
C

תינוכמה לש עונמה חכ
M
CB
B
C
v
max
max
v
v
1
)
(
0
t
u
רשאכו
:
ב תכרעמ תינב
Simulink
•
רובע
C =
6250
[N]
B =
2.5
[N/(m/s)
2
]
M =
1250
[Kg]
הרעה
:
ןיב הנתשמ הרקבה תוא
0
ל

1
,
ב שמתשהל יאדכ ןכל

Saturation
Very Short Simulink Tutorial
•
In the Matlab command window write
simulink
.
•
The window that has opened is the
Simulink Library
Browser
.
–
It is used to choose various Simulink modules to use in your simulation.
•
From this window, choose the
File
menu
, and then
New
(
Model
).
–
Now we have a blank window, in which we will build our model.
–
This blank window and the library browser window, will be the windows
we’ll work with.
•
We choose components from the library browser, and then drag
them to our work window.
–
We’ll use only the Simulink library (also called toolbox) for now.
Simulink Tutorial
As we can see, the Simulink library is divided into several
categories:
1.
Continuous
–
Provides functions for continuous
time, such as integration, derivative, etc.
2.
Discrete
–
Provides functions for discrete time.
3.
Funcitons & Tables
–
Just what the name says.
4.
Math
–
Simple math functions.
5.
Nonlinear
–
Several non

linear functions, such as
switches, limiters, etc.
6.
Signals & Systems
–
Components that work
with signals. Pay attention to the
mux/demux
.
7.
Sinks
–
Components that handle the outputs of the
system (e.g.
display it on the screen
).
8.
Sources
–
Components that generate source signals
for the system.
Simulink Tutorial: First simulation
•
Drag the
Constant
component
from the Simulink library
(
Sources)
to the work window,
and then drag the
Scope
module
(
Sinks)
in the same manner.
•
Now,
click the little triangle
(output port) to the
right of the
Constant
module
, and
while
holding the mouse button down,
drag the mouse to the
left side of
the scope
(input port) and then
release it.
You should see a
pointed arrow being drawn.
•
Double click the
Constant
module to open its dialog window.
•
Now you can change this module’s parameters.
•
Change the constant value to
5
.
Simulink Tutorial: First simulation
•
Double

click the scope to view its window.
•
You can choose
Simulation Parameters
…
from the
Simulation
to change the time
limits for your simulation.
•
Choose
Start
from the
Simulation
menu
(or press
Ctrl+T
, or click the play button
on the toolbar) to start the simulation.
Simulink Tutorial: First simulation
•
Choose
Start
from the
Simulation
menu
(or press
Ctrl+T
, or click the play button
on the toolbar) to start the simulation.
•
Now this is a rather silly simulation. All
it does is output the constant value
5
to the
graph (the x

axis represents the simulation
time).
•
Right

click on the scope’s graph window
and choose
Autoscale
to
get the following
result:
Simulink Tutorial
•
Now let’s try something a little
bit more complicated.
•
First, build the following
system (you can find the
Clock
module in the
Sinks
category
.
•
The
Trigonometric Function
and
Sum
modules reside in the
Math
category
):
Simulink Tutorial
•
Now, let’s see if the
derivative is really a cosine.
•
Build the following system
(the
Derivative
module is
located in the
Continuous
category):
We can see that this is indeed a
cosine, but something is wrong
at the beginning.
Simulink Tutorial
•
This is because at time
0
,
the derivative has no
prior information for
calculation
–
there’s
no initial value
for
the derivative,
–
so
at the first time step
,
the derivative assumes
that its input has a
constant value (and so the
derivative is
0
).
An other example
•
Let’s say that we have a differential equation that we want
to model. The equation is:
2
5
.
0
A
A
•
How can we solve this numerically using Simulink?
5
.
0
0
A
We’ll notice
2
simple facts:
1.
If we have A, then we have A
'
(multiplication).
2.
If we have A
'
, then we have A (integration).
We can get out of this loop by using the initial condition. We know that
A
0
=
0.5
, so now we can calculate A
'
and then recalculate the new
A, and so on.
Simulink Tutorial
•
double

click the
Integrator
and choose
Initial Condition Source:
External
.
•
Note that pressing
ctrl
when clicking the mouse
button on a line,
allows
you to split it into
2
lines
:
•
Set the simulation
stop time to
3.5
seconds
–
the solution goes to infinity
•
and see the results in the scope:
ב תכרעמ תינב
Simulink
Automatic Cruise Control
•
רובע
C =
6250
[N]
B =
2.5
[N/(m/s)
2
]
M =
1250
[Kg]
הרעה
:
ןיב הנתשמ הרקבה תוא
0
ל

1
,
ב שמתשהל יאדכ ןכל

Saturation
Automatic Cruise Control
•
בכרה לש היצלומיסה תא הנבנ
(plant)
Automatic Cruise Control
Automatic Cruise Control
•
הלחתהה יאנת רובע
היוצר תוריהמו
,
תכרעמה תוגהנתה תא ןחב
–
רקב םע
P
,
רקב םע
I
רקב םעו
PI
(
דרפנב הרקמ לכ
.)
–
הרעה
:
תויהל הרומא היוצר הבוגת
:

הריהמ
.

לש ילמיסקמ ךרע
(
t
)
הלעי אל
"
ידמ
"
בצמב שרדנה ךרעה לעמ
דימתמ
.

איה דימתה בצמה תאיגש
"
ספא
( "
ןתינש לככ הנטק
.)
5
.
0
)
0
(
8
.
0
r
Automatic Cruise Control
P controller
–
No wind
Automatic Cruise Control
P controller
–
with wind
Automatic Cruise Control
I controller
–
No wind
Automatic Cruise Control
PI controller
–
No wind
Automatic Cruise Control
PI controller
–
with wind
Automatic Cruise Control
Automatic Cruise Control
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