Download dy dt + 2y = u(t)− u(t −1) dy dt = −2y + u(t)− u(t −1)

ECE 350 – Linear Systems I
MATLAB Tutorial #3
Using MATLAB to Solve Differential Equations
This tutorial describes the use of MATLAB to solve differential equations. Two methods
are described. The first uses one of the differential equation solvers that can be called
from the command line. The second uses Simulink to model and solve a differential
equation.
Solving First Order Differential Equations with ode45
The MATLAB commands ode 23 and ode 45 are functions for the numerical solution of
ordinary differential equations. They use the Runge-Kutta method for the solution of
differential equations. This example uses ode45. The function ode45 uses higher order
formulas and provides a more accurate solution than ode 23.
In this example we will solve the first order differential equation:
dy
+ 2y = u(t) − u(t − 1)
dt
over the range 0 ≤ t ≤ 5. We will assume that the initial value y(0) = 0. The first step is
to write the equation in the form:
dy
= −2y + u(t) − u(t − 1)
dt
Next we need to create a MATLAB m-file to calculate dy/dt. From the MATLAB
window select: File  New  M-File. Enter the following:
function dy = f(t,y)
dy = -2*y + (t>=0)-(t>=1);
This creates a function that calculates the value of dy (actually dy/dt). The function is
called “f” and the file should be named: f.m when it is saved since all MATLAB
functions should be saved in a file with the same name as the function. Convince
yourself that the right hand side of the equation for dy corresponds to the expression
above. Make sure that you save this function in your working directory or one that is in
your path.
Once the function for dy/dt has been created we need to specify the end points of the time
range (0 ≤ t ≤ 5) and the initial value of y (zero). This is done in the MATLAB command
window with the following commands:
>> t= [0 5];
>> inity=0;
The ode45 command can now be called to compute and return the solution for y along
with the corresponding values of t using:
>> [t,y]=ode45(@f, t, inity);
Note that the @f refers to the name of the function containing the function. The solution
can be viewing with:
>> plot(t,y)
as shown in figure 1.
Figure 1. Solution of First Order Differential Equation
Solving First Order Differential Equations with ode45
The ode45 command solves first order differential equations. In order to use this
command to solve a higher order differential equation we must convert the higher order
equation to a system of first order differential equations. For example, in order to solve
the second order equation:
d2 y
dy
+
3
+ 2y = cos 2t
dt 2
dt
with y(0) = 0 and
dy
=1
dt t =0
for 0 ≤ t ≤ 10 we must define a variable x such that:
x=
dy
dt
Thus we can rewrite the differential equation as:
d2 y
dy
= −3 − 2y + cos 2t
2
dt
dt
or in terms of x:
dx
= −3x − 2y + cos 2t
dt
We will use the ode45 command to solve the system of first order differential equations:
dx
= −3x − 2y + cos 2t
dt
dy
=x
dt
We place these two equations into a column vector, z, where:
⎡ x ⎤
z=⎢
⎥
⎢⎣ y ⎥⎦
and
⎡
⎢
dz ⎢
=
dt ⎢
⎢
⎣
dx
dt
dy
dt
⎤
⎥
⎥
⎥
⎥
⎦
A function is created in a MATLAB m-file to define dz/dt named f2 which contains:
function dz=f2(t,z)
dz=[-3*z(1)-2*z(2)+cos(2*t); z(1)];
Note that the first column of dz contains the equation for dx/dt and the second an
equation for dy/dt. The values of x and y are referred to as z(1) and z(2) since they are
the first and second elements of the vector z. The range of t values and the initial
conditions are input at the command line with:
>> t=[0 10];
>> initz=[1;
0];
and the equations are solved and plotted using:
>> [t,z]=ode45(@f2, t, initz);
>> plot(t, z(:,2))
Figure 2. Solution of Second Order Differential Equation
Modeling Linear Systems Using Simulink
Simulink is a companion program to MATLAB and is included with the student version.
It is an interactive system for simulating linear and nonlinear dynamic systems. It is a
graphical mouse-driven program that allows you to model a system by drawing a block
diagram on the screen and manipulating it dynamically. It can work with linear,
nonlinear, continuous time, discrete time, multivariable, and multirate systems. In this
section the basic operation of Simulink will be described. The next section will
demonstrate the use of Simulink to solve differential equations.
1. Open MATLAB and in the command window, type: simulink at the prompt.
Alternatively, there is a Simulink icon in the menu bar.
2. After a few seconds Simulink will open and the Simulink Library Browser will open
as shown in figure 3. It is important to note that the list of libraries may be different
on your computer. The libraries are a function of the toolboxes that you have
installed.
Figure 3. Simulink Library Browser
3. Click on the Create a New Model icon in the Library Browser window. An additional
window will open. This is where you will build your Simulink models.
4. Click on the “+” sign next to “Simulink” in the Library Browser. A list of sublibraries will appear including Continuous, Discrete, etc. These sub-libraries contain
the most common Simulink blocks.
5. Click once on the “Sources” sub-library. You should see a listing of blocks as shown
in the right column of figure 4.
Figure 4. Source Blocks in the Simulink Library
6. Scroll down this list until you see the Sine Wave icon. Click once on the icon to select
it and drag this icon into the blank model window.
7. Click once on the “Sinks” sub-library in the left part of the Library Browser. Click
and drag the “Scope” icon to the model window to the right of the Sine Wave block.
The model window should now appear as shown in figure 5. Make sure you have
used “Scope”, not “Floating Scope”.
Figure 5. Model Window with Sine Wave and Scope Blocks
8. Next we want to connect the Sine Wave to the Scope block. Move the mouse over
the output terminal of the Sine Wave block until it becomes a crosshair. Click and
drag the wire to the input terminal of the Scope block. The cursor will become a
double cursor when it is in the correct position. Release the mouse button and the
wire will snap into place. Your completed system should now appear as shown in
figure 6.
Figure 6. Completed System for Viewing Sine Wave
9. In the model window select Simulation Start. You will hear a beep when the
simulation is complete. Double click on the Scope icon and it will open and display
the output of the sine wave block. Try clicking on the binocular icon in the Scope
window. This should cause the axes to readjust as shown in figure 7.
Figure 7. Scope Display
10. Note that the period of the sine wave is just over six. What frequency does this
correspond to? Return to the model window and double click on the Sine Wave
block. The Sine Wave parameters window will open as shown in figure 8.
Figure 8. Sine Wave Block Parameters
11. The frequency was set to 1 rad/sec. Any of the parameters shown can be altered to
change the sine wave. Change the entry in the frequency parameter to: 2*pi*10.
This is a 10 Hz sine wave. Click OK and re-simulate. Use the binoculars icon to
expand the scope display. What do you observe? The sine wave should NOT be
displayed properly. This is because at this higher frequency the sampling rate has not
been set high enough.
12. Return to the model window and select: Simulation  Configuration Parameters.
The Stop Time for the simulation defaults to 10. Change it to 1, click OK, and re-run
the simulation. You should now be able to properly view the sine wave on the scope.
Use the binoculars to expand the display. Verify that the period of each cycle is one
0.1 second as you expect for a 10 Hz wave.
13. Another method for modifying the display is to specify the step size. Return to the
model window and select Simulation  Configuration Parameters. Note that the
Solver Type defaults to “variable-step”. Change the Solver Type to “fixed step” and
then enter the value 0.01 into the Fixed Step Size entry. Click OK and run the
simulation once again. Note that the display is now smooth.
14. Drag the Pulse Generator from the Source sub-library into the model window.
Double click the Pulse Generator block and modify the parameters as shown in figure
9 below. Click OK.
Figure 9. Pulse Generator Parameters
15. In the scope window, click on the Parameters icon (second from the left). Change the
number of axes to 2 and click OK. Return to the model window and note that the
scope now has two inputs. Connect a wire from the Pulse Generator to the second
input. Your model should now appear as shown in figure 10.
Figure 10. Model Window with Two Sources
16. Re-run the simulation and observe the two waveforms as shown in figure 11 below.
Figure 11. Two Input Scope Window
17. Open the Continuous sub-library. Drag the Integrator block into the model window
and reconfigure the diagram as shown in figure 12.
Figure 12. Simulink Diagram
18. Re-simulate the system and observe the output on the scope. Note that the second
trace is now the integral of the pulse train input.
Modeling a Differential Equation in Simulink
Now we will use Simulink to model the differential equations solved earlier. Beginning
with the first order differential equation:
dy
+ 2y = u(t) − u(t − 1)
dt
First we need to simulate the single one-second pulse. Since the simulation will run for
0≤t≤5, we can use the pulse generator to generate a five second pulse with a 1 second
pulse width. Since it will not repeat within the five seconds, this simulates a single pulse.
Remove the integrator from the model shown in figure 12 and connect the pulse
generator directly to the scope input. Change the pulse generator to have the settings as
shown in figure 13. The 20% pulse width corresponds to one second as needed.
Change the simulation time in the configuration parameters to five seconds and simulate
the system. Specify fixed-step samples of 0.01 seconds. Verify that the pulse generator
is outputting the desired pulse for the differential equation above.
Figure 13. Pulse Generator Settings for 1 Second Pulse
The differential equation above can be written as:
dy
= −2y + u(t) − u(t − 1) = −2y + p(t)
dt
where p(t) is the one second pulse. The right hand side of this equation can be modeled
in Simulink using the model shown in figure 14. The subtraction block and the gain
block are found in the Math Operations sub-library.
Figure 14. Simulink Model
The labels on the wires are inserted by double clicking on the wires and typing in the text.
By double clicking on the gain block you can set its constant to two. If the input to the
gain block is y, then the output of the subtractor is dy/dt. By passing this output through
an integrator, the input y is found. The model in figure 15 should now simulate the
original differential equation.
Figure 15. Simulink Model of First Order Differential Equation
Run the simulation for 5 seconds and confirm that the scope displays the same output as
we obtained in the first section of this tutorial.
We can also simulate the second order differential equation:
d2 y
dy
+ 3 + 2y = cos 2t
2
dt
dt
in Simulink. This equation can be re-written as:
d2 y
dy
= −3 − 2y + cos 2t
2
dt
dt
and modeled in Simulink as shown in figure 16. In order to get the three input subtractor,
use the two input subtractor selected above. Double click on the block and change the
“List of Signs” to: + - - . This will result in a block that adds the top input and subtracts
the second and third inputs.
Figure 16. Simulink Model
In order to get y and dy/dt we need to integrate the output of this summer twice. If we
add two integrators to the output of figure 16 we can feed the values of y and dy/dt back
to the inputs as shown in figure 17.
Figure 17. Simulink Model of 2nd Order Differential Equation
Simulate the circuit for 10 seconds. The output shown in figure 18 is obtained on the
scope.
Figure 18. MATLAB Scope Output
If we want to compare this output to that obtained in the earlier solution we can save the
output to the workspace. The simout block found in the Sinks sub-library is added as
shown in figure 19. Double click the simout block and change the format of the output to
an Array.
Figure 19. Simulink Model of 2nd Order Circuit
Simulate the model again. Note that an array named simout appears in the workspace.
Plot this output along with the output obtained for the command line solution of the
second order differential equation. The following commands are used:
>> z = [1; 0];
>> t = [0 10];
>> [t, z]=ode45(@f2, t, z);
>> plot(t,z(:,2))
>> hold on
>> plot(0:.01:10, simout, 'r')
The two solutions appear in figure 20. Note that the time values for the simout plot are
determined by the fixed spacing of 0.01 specified in the configuration parameters in
Simulink.
Figure 20. Two Solutions of the 2nd Order Differential Equation
The solutions are not the same. There are two reasons. First of all, the sinusoidal input is
a “sine” and not a “cos”. In order to convert a sine to a cos we need to apply a phase shift
of π/2. Double click on the Sine Wave block and enter: pi/2 for the phase. Re-simulate
and plot along with the command line solution as shown in figure 21.
Figure 21. Two Solutions of the 2nd Order Differential Equation
Now the steady state portions of the solutions are the same. However, the initial transient
response is not. This is because the two solutions were obtained with different initial
conditions. In the earlier command line solution the initial conditions: y(0) = 0 and
dy
= 1 were used. The integrator blocks in Simulink allow for the inclusion of
dt t =0
initial conditions. The second integrator outputs the value of y. Thus, the default initial
condition of zero is correct. However the first integrator outputs dy/dt. Double click on
the first integrator and change the initial condition to one. After simulating and replotting both solutions the curves in figure 22 are obtained. Note that the solutions now
match. This confirms that either method will provide a correct solution.
Figure 22. Two Solutions of the 2nd Order Differential Equation