Matlab And Simulink For Modeling And Control.ppt

Modeling and Control
with
Matlab and Simulink
Sch. of Elec. Eng., Wuhan Univ.
June 2003

Spring 2003
Modeling and Control with Matlab and Simulink 2
1. Introduction
With the help of an example, a DC motor,
the use of MATLAB and Simulink for modeling,
analysis and control design is demonstrated.
It is assumed that the reader already has
basic knowledge of MATLAB and Simulink.
The main focus is on the use of the Control
System Toolbox functions.

Spring 2003
2. Modeling a DC Motor
In this example we will learn how to
develop a linear model for a DC motor,
how to analyze the model under MATLAB
(poles and zeros, frequency response,
time-domain response, etc.), how to
design a controller, and how to simulate
the open-loop and closed-loop systems
under SIMULINK.

Spring 2003
2.1 Physical System
Consider a DC motor, whose electric circuit of the armature
and the free body diagram of the rotor are shown in Figure
1.

Spring 2003
Physical System
The rotor and the shaft are assumed to be rigid. Consider the
following values for the physical parameters:
moment of inertia of the rotor J = 0.01 kg  m2
damping (friction) of the mechanical
system
b = 0.1 Nms
(back-)electromotive force constant K = 0.01 Nm/A
electric resistance R = 1 
electric inductance L = 0.5 H

Spring 2003
Physical System
The input is the armature voltage V in Volts (driven by a
voltage source). Measured variables are the angular
velocity of the shaft  in radians per second, and the shaft
angle  in radians.

Spring 2003
System Equations
The motor torque, T, is related to the armature current, i,
by a constant factor K:
(1)
The back electromotive force (emf), Vb, is related to the
angular velocity by:
(2)

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2.3 Transfer Function
From Figure 1 we can write the following equations based on
the Newtons law combined with the Kirchhoffs law:
Using the Laplace transform, equations (3) and (4) can be
written as:

Spring 2003
Transfer Function
where s denotes the Laplace operator. From (6) we can
express I(s):
and substitute it in (5) to obtain:

Spring 2003
Transfer Function
This equation for the DC motor is shown in the block diagram
in Figure 2.

Spring 2003
Transfer Function
From equation (8), the transfer function from the input voltage,
V (s), to the output angle,  , directly follows:
From the block diagram in Figure 2, it is easy to see that the
transfer function from the input voltage, V (s), to
the angular velocity, , is:

Spring 2003
3 MATLAB Representation
The above transfer function can be entered into Matlab by
defining the numerator and denominator polynomials, using
the conventions of the MATLAB's Control Toolbox. The
coefficients of a polynomial in s are entered in a descending
order of the powers of s.
Example: The polynomial A = 3s3
+ 2s + 10 is in MATLAB
entered as: A = [3 0 2 10].
Furthermore, we will make use of the function conv(A,B),
which computes the product (convolution) of the polynomials A
and B. Open the M-file motor.m. It already contains the
definition of the motor constants:
J=0.01;
b=0.1;
K=0.01;
R=1;
L=0.5;

Spring 2003
MATLAB Representation
The transfer function (9) can be entered in MATLAB in a
number of different ways.
1. As Ga(s) can be expressed as Gv(s)  1/s, we can enter
these two transfer functions separately and combine them
in series:
aux = tf(K,conv([L, R],[J, b]));
Gv = feedback(aux,K);
Ga = tf(1,[1, 0])*Gv;
Here, we made use of the function feedback to create a
feedback connection of two transfer functions and the
multiplication operator *, which is overloaded by the LTI
class of the Control System Toolbox such that is computes
the product of two transfer functions.

Spring 2003
2. Instead of using convolution, the first of the above three
commands can be replaced by the product of
two transfer functions:
aux = tf(K,[L, R]) * tf(1,[J, b]);
3. Another possibility (perhaps the most convenient one) is to
define the transfer function in a symbolic way. First introduce a
system representing the Laplace operator s (differentiator) and
then enter the transfer function as an algebraic expression:
s = tf([1, 0],1);
Gv = K/((L*s + R)*(J*s + b) + K^2);
Ga = Gv/s;

Spring 2003
It is convenient to label the inputs and outputs by the names
of the physical variables they represent:
Gv.InputName = 'Voltage';
Gv.OutputName = 'Velocity';
Ga.InputName = 'Voltage';
Ga.OutputName = 'Angle';
Now by calling motor from the workspace, we have both the
velocity (Gv) and the position (Ga) transfer functions defined
in the workspace.

Spring 2003
3.1 Exercises
1. Convert Gv and Ga into their respective state-
space (function ss) and zero-pole-gain (function
zpk) representations.
2. What are the poles and zeros of the system? Is
the system stable? Why?

Spring 2003
4 Analysis
The Control System Toolbox offers a variety of functions that allow
us to examine the systems characteristics.
4.1 Time-Domain and Frequency Responses
As we may want plot the responses for the velocity and angle in
one figure, it convenient to group the two
transfer functions into a single system with one input, the voltage,
and two outputs, the velocity and the angle:
G = [Gv; Ga];

Spring 2003
Analysis
Another way is to first convert Ga into its state-space
representation and then add one extra output being equal to
the second state (the velocity):
G = ss(Ga);
set(G,'c',[0 1 0; 0 0 1], …
'd',[0;0], 'OutputName',{'Velocity'; 'Angle'});
Note that this extension of the state-space model with an
extra output has to be done in one set command in order to
keep the dimensions consistent.

Spring 2003
Analysis
Now, we can plot the step, impulse and frequency responses
of the motor model:
figure(1); step(G);
figure(2); impulse(G);
figure(3); bode(G);
You should get the plots given in Figure 3 and Figure 4.

Spring 2003
Analysis

Spring 2003
4.2 Exercise
1. Simulate and plot in MATLAB the time response of the
velocity and of the angle for an input signal cos(2t), where t
goes from 0 to 5 seconds.

Spring 2003
5 Control Design
Let us design a PID feedback controller to control the velocity
of the DC motor. Recall that the transfer function of a PID
controller is:

Spring 2003
5 Control Design
where u is the controller output (in our case the voltage V ),
e = uc - y is the controller input (the control error), and Kp, Kd, Ki
are the proportional, derivative and integral gains, respectively.
A block diagram of the closed-loop system is given in Figure 5.

Spring 2003
5.1 Proportional Control
First, try a simple proportional controller with some estimated
gain, say, 100. To compute the closed-loop
transfer function, use the feedback command. Add the
following lines to your m-file:
Kp = 100;
Gc = feedback(Gv*Kp,1);
Gc.InputName = 'Desired velocity';
Here Gc is the closed-loop transfer function. To see the
step response of the closed-loop system, enter:
figure(4); step(Gc,0:0.01:2);
You should get the plot given in Figure 6:

Spring 2003
To eliminate the steady-state error, an integral action must be
used. To reduce the overshoot, a derivative action can be
employed. In the following section, a complete PID controller
is designed.
5 Proportional Control
To eliminate the steady-state error, an integral action must be
used. To reduce the overshoot, a derivative action can be
employed. In the following section, a complete PID controller
is designed.

Spring 2003
5.2 PID Control
Let us try a PID controller. Edit your M-file so that it contains
the following commands:
Kp = 1;
Ki = 0.8;
Kd = 0.3;
C = tf([Kd Kp Ki],[1 0]);
rlocus(Ga*C);
Kp = rlocfind(Ga*C);
Gc = feedback(Ga*C*Kp,1);
figure(9); step(Gc,0:0.01:5)
The rlocus and rlocfind functions are used to select the overall
gain of the PID controller, such that the controller is stable and
has the desired location of the poles (within the defined ratio
among the Kp, Ki and Kd constants). If the design is not
satisfactory, this ratio can be changed, of course. We should
obtain a plot similar to the one in Figure 7:

Spring 2003
PID Control

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5.3 Exercise
1. Use the root locus and the Nyquist criterion to find out for
what value of the gain Kp the proportional controller for the
angle Ga(s) becomes unstable.

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6 SIMULINK Model
The block diagram from Figure 2 can be directly
implemented in SIMULINK, as shown in the figure Figure 8:

Spring 2003
SIMULINK Model
Set the simulation parameters and run the simulation to see the
step response. Compare with the response in Figure 3. Save
the file under a new name and remove the position integrator
along with the Graph and To Workspace blocks. Group the
block describing the DC motor into a single block and add a
PID controller according to Figure 5. The corresponding
SIMULINK diagram is given in Figure 9. Experiment with the
controller. Compare the responses with those obtained
previously in MATLAB.

Spring 2003
7 Obtaining MATLAB Representation from a SIMULINK Model
From a SIMULINK diagram, a MATLAB representation (state
space, transfer function, etc.) can be obtained.
The Inport and Outport blocks must be added to the
SIMULINK diagram, as shown in Figure 10.
Then we can use the linmod command to obtain a state-space
representation of the diagram:
[A,B,C,D] = linmod(filename);
where filename is the name of the SIMULINK file.

Spring 2003
Obtaining MATLAB Representation from a SIMULINK Model

Spring 2003
7.1 Exercise
1. Convert the four matrices A, B, C, and D into a
corresponding state space LTI object. Convert this one into a
transfer function and compare the result with the transfer
function entered previously in MATLAB.

Matlab And Simulink For Modeling And Control.ppt

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