SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS
Example:
Mathematical model of a mechanical system is defined as a system of differential equations
as follows:
x 1  20 x 1  15 x 2  1.5 f  0
x 2  12 x 1  5 x 2  2 f
where f is input, x1 are x2 outputs.
At t=0 x1=2 and x2=-1.
a) Find the eigenvalues of the system.
b) If f is a step input having magnitude of 3, find x1(t).
c) If f is a step input having magnitude of 3, find x2(t).
d) Find the response of x1 due to the initial conditions.
e) Find the response of x2 due to the initial conditions.
f) How do you obtain [sI-A]-1 with MATLAB?
SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS
Let us obtain the State Variables Form so as to 1st order derivative terms are left-hand
side and non-derivative terms are on the right-hand side.
x 1  20 x 1  15 x 2  1.5 f
x 2  12 x 1  5 x 2  2 f
x 1   20 15 x 1  1.5
   
 x    2  f
x
12
5
 2 
  2  
A
State Variables Form
B
 s 0  20 15 s  20  15





0
s
12
5

12
s

5

 
 

sI  A  
det sI  A   (s  20) * (s  5)  (12) * (15)  s2  15 s  280
D(s)
s  5 15 
1
[sI  A]  2
s  15 s  280  12 s  20
1
SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS
a) Eigenvalues are roots of the polynomial D(s) or eigenvalues of the matrix A.
or
D(s)  s2  15 s  280  0
 15  152  4 * 1 * (280)
s 1 ,2 
2 *1
s1  25.8371
s2  10.8371
Initial Conditions
b) x1(t) due to the forcing
X 1 (s)
1 x 1 (0)
1 1.5

  sI  A  
  sI  A   F(s)
X 2 (s)G
x 2 (0)
2 
General Solution
Solution due to the initial conditions Solution due to the input
Particular Solution
Homogeneous Solution
15  1.5 3
 X1( s ) 
s  5
1

  2
 12 s  20  2  s
X
(
s
)
s

15
s

280

 
 2 P
X1P ( s ) 
1
s 2  15 s  280
[( s  5 )* 1.5  15* 2 ]
3
4.5 s  67.5

s s 3  15 s 2  280 s
clc;clear;
num=[4.5 67.5];
den=[1 15 -280 0];
[r,p,k]=residue(num,den)
SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS
X 1P ( s ) 
 0.0515
0.2925
0.2411


s  25.8371 s  10.8371 s  0
8
8
x 10
7
6
5
x1(t)
4
3
x1P ( t )  0.0515 e
 25 .8371 t
10 .8371 t
 0.2925 e
 0.2411
2
1
0
System is instable because of the positive root.
-1
0
0.5
1
t(s)
c) x2(t) due to input
6 s  174
1
3
X 2Ö (s)  2
[15 * 1.5  (s  20) * 2]  3
s  15 s  280
s s  15 s2  280 s
2
Laplace transform of x2p
0.02
0.6014
0.6214
X 2P( s ) 


s  25.8371 s  10.8371 s  0
x2 P ( t )  0.02 e  25 .8371 t  0.6014 e10 .8371 t  0.6214
16
8
x 10
14
12
10
x2(t)
clc;clear;
num=[6 174];
den=[1 15 -280 0];
[r,p,k]=residue(num,den)
1.5
8
6
4
2
0
0
0.5
1
t(s)
1.5
2
SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS
d) x1 due to the initial conditions.
15   2 
X1 (s)
s  5
1

  2
 12 s  20  1
X
(
s
)
s

15
s

280
 2 h

 
X1h (s) 
2 s  25
(s  5) * 2  15 * (1)

s2  15 s  280
s2  15 s  280
clc;clear;
num=[2 -25];
den=[1 15 -280];
[r,p,k]=residue(num,den)
x 1h (t)  2.0907e 25.8371t  0.0907e10.8371t
e) x2 due to the initial conditions
12 * 2  (s  20) * (1)
s4
X 2h (s) 

s2  15 s  280
s2  15 s  280
clc;clear;
num=[-1 4];
den=[1 15 -280];
[r,p,k]= residue(num,den)
x 2h (t)  0.8136e 25.8371t  0.1864e10.8371t
f) [sI-A]-1 with Matlab.
clc;clear;
syms s;
i1=eye(2)
A=[-20 15;12 5];
a1=inv(s*i1-A)
pretty(a1)
SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS
Example:
Mathematical model of a system is given below. Where V(t) is input, q1(t) and q2(t) are
outputs.
3(q1  q2 )  V  2q 1
a) Write the equations in the form of state variables.
0.8q2  3(q1  q2 )  0
b) Write Matlab code to obtain eigenvalues of the system.
c) Write Matlab code to obtain matrix [sI-A]-1.
d) Results of (b) and (c) which are obtained by computer are as follows:
4 s2  15
6s
6 
1


2
[sI  A]1 
15
4
s

6
s
4
s

6
2
 s1,2  0.75  1.7854i, s3
s (4 s  6 s  15) 
2
 15 s

15
s
4
s
 6 s

q1 (0)  5
At t=0
q2 (0)  3 and V(t) is a step input having magnitude of 2.
q 2 (0)  0.4
Find the Laplace transform of
q 2 due to the initial conditions.
e) Find the Laplace transform of q1 due to the input.
2
0
V2(t)
t (s)
SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS
3(q1  q2 )  V  2q 1
0.8q2  3(q1  q2 )  0
a) State variables are q1, q2 and q 2  u.
System of differential equations is arranged so as to 1st order derivative terms are lefthand side and non-derivative terms are on the right-hand side.
q 1  1.5 q1  1.5 q2  0.5 V
q 2  u
u  q2  3.75 q1  3.75 q2
State variables
0 q1  0.5
q 1   1.5 1.5
  
   
0
1 q2    0  V
q 2    0
 u   3.75  3.75 0  u   0 
  
   
A
b) Matlab code which gives the eigenvalues of the system.
A=[-1.5 1.5 0;0 0 1;3.75 -3.75 0]; eig(A)
c) Matlab code which produces [sI-A]-1
clc;clear
A=[-1.5 1.5 0;0 0 1;3.75 -3.75 0];
syms s;
i1=eye(3);
sia=inv(s*i1-A);
pretty(sia)
B
SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS
d)
Q 1 (s)
q1 (0)



1 
Q 2 (s)  [sI  A] q2 (0)
 U(s) 
 u(0) 

h


Q 2h (s)  Uh (s) 
4 s2  15
6s
6  5 
Q 1 (s)
1



 
2
15
4
s

6
s
4
s

6
Q 2 (s) 
2

  3
s
(
4
s

6
s

15
)
2
 U(s) 
 15 s

15
s
4
s
 6 s 0.4 

h



1
2
5
*
15
s

(

3
*

15
s
)

0
.
4
*
(
4
s
 6s)
2
s (4 s  6s  15)
2
2
1
.
6
s

75
s

45
s

2
.
4
s
1
.
6
s
 122.4s 1.6s  122.4
Q 2h (s)  Uh (s) 

 2
2
2
s (4 s  6s  15)
s(4s  6s  15) 4s  6s  15
e) Q 1 (s)
0.5


2
1 
Q
(
s
)

[
sI

A
]
0
 2 
 
 U(s) 
 0 s

Ö
 
4 s2  15
6s
6  0.5
Q 1 (s)
1



 2
2
Q
(
s
)

15
4
s

6
s
4
s

6
 2 
2

 0 s
s
(
4
s

6
s

15
)
2
 U(s) 
 15 s

15
s
4
s
 6 s   0 

Ö

4 s2  15
1
2

2
Q 1Ö (s) 
0.5 * (4s  15) *   4
s (4 s2  6s  15) 
s  4 s  6 s3  15s2
SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS
Example: Write the equation of motion of the mechanical system given below in the State
Variables Form. Force applied on the system is F(t)=100 u(t) (a step input having magnitude
100 Newtons) and at t=0 x0=0.05 m and dx/dt=0. Find x(t) and v(t).
d2x
dx
m 2  c  k x  F(t)
dt
dt
State variables are x and v=dx/dt .
x  v
v  x  
m=20 kg
c
k
1
v  x  F(t)
m
m
m
1  x   0 
x   0
   
 
 F(t)

v   k / m  c / m v  1 / m
Matlab program to obtain eigenvalues:
>>a=[0 1;-250 -2];eig(a)
c=40 Ns/m
k=5000 N/m
1  x   0 
x   0
   
 
 F(t)

v   250  2 v  0.05
SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS
1  x   0 
x   0
   
 
 F(t)

v   250  2 v  0.05
Applying Laplace transform and arranging,
1  X(s)  0 
s X(s)  x 0   0


 V(s)  0.05 F(s)
sV
(
s
)

v

250

2
0


  

1  X(s) x 0   0 
X(s)  0
s   
  
 F(s)

V(s)  250  2 V(s) v 0  0.05
1  X(s) x 0   0 
1 0 X(s)  0
s
 
  
 F(s)


0 1 V(s)  250  2 V(s) v 0  0.05
X(s) x 0   0 
sI  A         F(s)
V(s) v 0  0.05
X(s)
1 x 0 
1  0 
   sI  A    sI  A 
 F(s)
V
(
s
)
v
0
.
05
 
 0


Solution due to the
initial conditions
Solution due to the input
1 
 s
sI  A   

250
s

2


sI  A 
1
 s  2 1
1

Det(sI  A)  250 s 
Det(sI  A)  s2  2 s  250
SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS
sI  A
1
 s  2 1
1
 2
s  2 s  250  250 s 
X(s)
1

  2
V(s) s  2 s  250
F(s) 
100
s
 s  2 1 0.05
1

 250 s   0  s2  2 s  250



0.05 s2  0.1 s  5
X(s) 
s (s2  2s  250)
clc;clear;
syms s;
A=[0 1;-250 -2];
i1=eye(2); %unit matix with dimension 2x2
siA=s*i1-A;
x0=[0.05;0]; %Initial conditions
B=[0;0.05];
Fs=100/s;
X=inv(siA)*x0+inv(siA)*B*Fs;
pretty(X)
V(s)  
 s  2 1  0  100
 250 s  0.05 s



7.5
s2  2s  250
For x(t) ;
clc;clear;
num=[0.05 0.1 5];
den=[1 2 250 0];
[r,p,k]=residue(num,den)
X(s) 
0.015  0.001i
0.015  0.001i
0.02


s  (1  15.7797i) s  (1  15.7797i)
s
SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS
Steady-state value (Final value)
0.015  0.001i
0.015  0.001i
0.02
X(s) 


s  (1  15.7797i) s  (1  15.7797i)
s
t
x(t)  Ae cos(15.7797t  )  0.02
A  2 * abs(0.015  0.001i)  0.0301
  angle(0.015  0.001i)  0.0666 rad
x(t)  0.0301e t cos(15.7797t  0.0666)  0.02
0.06
Initial value, x0
0.05
0.04
x(t)
0.03
0.02
0.01
0
-0.01
0
1
2
3
t (s)
4
5
6
SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS
V(s)  
For v(t)
clc;clear;
num=[-7.5];
den=[1 2 250];
[r,p,k]=residue(num,den)
0.2376i
 0.2376i

s  (1  15.7797i) s  (1  15.7797i)
A  2 * abs(0.2376i)  0.4752
v(t)  Ae  t cos(15.7797t  )
v(t)  0.4752e t cos(15.7797t  1.57)
  angle(0.2376i)   / 2 rad
0.4
0.3
0.2
0.1
0
v(t)
V(s) 
7.5
s2  2s  250
-0.1
-0.2
-0.3
-0.4
-0.5
0
1
2
3
t (s)
4
5
6
SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS
Example: Mathematical model of a mechanical system having two degrees of freedom is
given below. If F(t) is a step input having magnitude 50 Newtons, find the Laplace
transforms of x and θ.
mx  2k x  2k R   F(t)
mR 2   c R2   2k R x  3k R2   0
10x  4000 x  800   F(t)
R=0.2 m
m=10 kg
k=2000 N/m
0.4   0.8   800 x  240   0
c=20 Ns/m
x
State variables
x  v
  
F(t)
10
    2   2000 x  600 

v  x  400 x  80  

x  v
  
0
 x   0
    0
0
  
 
80
 v   400


   2000  600
0  x   0 
0 1      0 
     F(t)
0 0   v  0.1

0  2   0 
1
SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS
 X(s) 
x 
0
 (s) 
 
0





1
1 

  [sI  A]    [sI  A]  F(s)
 V(s) 
v 
0.1
(s)

 0 
0
clc;clear
A=[0 0 1 0;0 0 0 1;-400 80 0 0;2000 -600 0 -2];
syms s;
eig(A)
i1=eye(4);
sia=inv(s*i1-A);
pretty(sia)
D(s)  s 4  2s 3  1000 s2  800 s  80000
If the initial conditions are zero, only the solution due to the input exists. For F(s) 
Eigenvalues:
[sI  A]1 
System is stable since real parts of all
eigenvalues are negative.
50
s
SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS
 X(s) 
 (s) 




 V(s) 
(s)
0
0
  50
 
0.1 s
 0 
0.1 * (s2  2 s  600)
5s2  10s  3000
50
X(s)  4

s  2 s3  1000 s2  800 s  80000 s s 5  2 s 4  1000 s3  800 s2  80000 s
(s) 
0.1 * 2000
50
10000

s 4  2 s 3  1000 s2  800 s  80000 s s 5  2 s 4  1000 s 3  800 s2  80000 s