# Project #1

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```					                                        EE-600
Project #4

Systems of Differential Equations

State Variable Equation Synthesis

A system may be described by a differential equation, a transfer function or by a
system of differential equations. Consider the differential equation

d 3 y (t )    d 2 y (t )    dy(t )            dx(t )
3
2      2
2         y (t )          2 x(t ).
dt            dt           dt                dt

The corresponding transfer function is

Y ( s)      s2
H ( s)            3               .
X ( s) s  2s 2  2s  1

The corresponding state-variable equations (in general form) are

w  Aw  cx.


y  bw.

We can determine the values of A,c and b using tf2ss (transfer function to
state-space):

>> [A,c,b] = tf2ss([1 2], [1 2 2 1]);

Verify that the matrices created by this MATLAB command are the same at those
created manually or analytically using the state-space creation technique
discussed in the lecture. MATLAB arranges the order of the rows of the state-
variable equations differently than in the lecture. The state-variable equations
can also be converted back to a transfer function using

>> [num, den] = ss2tf(A, c, b, 0);

1
(You need a zero in the non-existent ‘D’ argument.) The answer you get is
should correspond to the original transfer function. You can also manually

H ( s )  bsI  A  c.
1

You can also use MATLAB to do this formulation:

>> syms s;
>> b*inv(s*eye(3)-A)*c

Using Laplace transforms and the methods discussed in Project #2, find the
solution to the differential equation if x(t) = u(t).

Numerical Solutions to State Variable Equations

The MATLAB function ode45( ) can be used to numerically solve simultaneous
differential equations. In Project #1, this function was used to solve a first-order
differential equation. The general form of the function is

>> [t y] = ode45(‘de’, <time range>, <initial conditions>);

The de.m function specified the differential equation. Suppose that we had a
matrix differential equation,

 y1  0 0 1  y1  0

                
 y 2   0 2 0  y 2   0 u (t ).

                
 y 3  3 0 0  y3  1
               

The de.m file would look like this:

function ydot = de(t,y)

ydot = zeros(size(y));
ydot(1) = y(3);
ydot(2) = 2*y(2);
ydot(3) = 3*y(1)+1;

The <time range> is a two-element array for begin and end times for the
numerical simulation, e.g., [0 1] would correspond to the time from zero to one.

2
The <initial conditions> is an array corresponding to y1(0), y2(0), etc., e.g., [0 0
0] would be for a system initially at rest. Find the solution to the differential
equation in the previous section numerically if the input x(t) is a step function u(t).
Let the initial conditions be y(0) = 0, y’(0) = 0, y’’(0) = 0. By plotting, compare the
analytic solution (derived in the first section) to the numeric solution.

3

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