MIME 3380, Modeling and Control of Engineering Systems
Homework 5: Due Monday Feb 20, 2012
1. Problem 4.2 on page 126 of your text. Use Simulink to simulate this system. (Outputs required,if
this needs to be said!)
No real simulation problems expected!
2. Problem 4.15 on page 130 of your book.
J p  J p0 2 sin  cos   m p gl sin   0
Assuming small angles ( near 0), sin   
  0 2 
mp
Jp
gl
0

d   
m

p
gl  0 2
dt   
 J p
I  A  0


 m
  p gl   2
0
 J p
   0 2 
cos   1
mp
Jp
1
  
0   

1
mp

 2 
gl  0 2  0

Jp

gl
  0 2 
mp
Jp
gl  positive is unstable for all 0
The other equilibrium point is for the pendulum to be hanging:
Let       where   is a small angle
  0 2 sin      cos      
mp
Jp
gl sin     
sin       sin  cos    sin   cos    
cos       cos  cos   sin  sin    1


m
m
d
  0 2   p gl    0 2  p gl    

dt
Jp
J p 

The solution of y  cy  c , the form of the ODE above, is

y

1
e
2
ct
 e
ct
 

mp 
Thus the system is stable under the condition that  0 2 
gl   0, Otherwise it is unstable

J p 

Plotting


m
d
   0 2  p gl     =0 with 0 varied leads to the bifurcation diagram given


dt
J
p


3. Problem 5.4 on page 164 of your book.
 0
Exp 
 d
d   e t cos d t e t sin d t 
t
?
0   e t sin d t e  t cos d t 
0
0
0
0
 t
 t
 t
 t
d  e 0 cos d t e 0 sin d t   0 e 0 cos d  e 0 d sin d t


dt  e 0t sin d t e 0t cos d t   0 e 0t sin d t  e 0td cos d t
 0

 d
d   e t cos d t e t sin d t 


0   e  t sin d t e  t cos d t 
d At
e  Ae At  QED
dt
4. Problem 5.9 on page 166 of your book.
0e0t sin d t  e 0td cos d t 

0e 0t cos d  e 0td sin d t 
0
0
0
0
a
 a
Eigenvalues of 
:
 ab  b ab 
a   a
a 
 0   a
I  A  


    a    ab   ab  a 2b  0





 0    ab  b ab  b  ab   ab 
 2   a  ab    ab  a 2b  0

 a  ab   1
 a  ab 
2
 4  ab  a 2b  
 a  ab   1
2
For  <1:
2
 a  ab   1
a 2 1  6b  b 2   4ab  1   a  ab   a 2 1  6b  b 2   4ab  2
2
 a  ab 
2
2
a 2 1  6b  b 2   4ab
2
1 2

a 1  6b  b 2   4ab  1   a  ab   a 2 1  6b  b 2   4ab  2
2
2
2(a  ab)  4  a (1  b)  2
c=
Modeling the problem in Matlab to explore
a 10% government spending requirement:
a = 0.2500
b =0.5000
A=
0.2500 0.2500
-0.3750 0.1250
B=
0.2500
0.1250
C=
1 1
D=1
>> sys=ss(A,B,C,D,1) %Note: building a
discrete model
x1 x2
y1 1 1
d=
u1
y1 1
Sampling time (seconds): 1
Discrete-time state-space model.
>> step(0.1*sys)
>> title('Keysian Economic Model, a=0.25, b
= 0.5, G=10%')
>> grid
>> xlabel('Time in years (Ignore step output
"seconds")')
>> ylabel('Output in $')
a=
x1 x2
x1 0.25 0.25
x2 -0.375 0.125
b=
u1
x1 0.25
x2 0.125
>