Download Homework 1 Scores: Differential equation 30% Derivation of the

47453 S Digital Control Theory, 2004/2005 Autumn
The solution must contain the
Homework 1
The Figure 1 below shows a schematic diagram of an automobile and its two suspension systems. As the car
moves along the road, the vertical displacements at the tires act as the motion excitation to the automobile
suspension system. Replacing the two suspension systems by a simple model (in Figure 2), the suspension
system involves the mass of the wheel (m1), the spring constant of the wheel (k1), the mass of the automobile
(m2), the spring constant (k2) and the shock absorber constant (b). It can be further simplified as shown in
Figure 3. The parameters are the following:
In Figure 2:
In Figure 3:
47453 S Digital Control Theory, 2004/2005 Autumn
Differential equation
Derivation of the transfer function
Derivation of the state-space model
The Simulink model window (as shown below)
Running scope window (as shown below)
b = 7 Ns/m, k1 = 0.6 N/m, k2 = 0.5 N/m, m1 = 5kg, m2 = 200 kg.
b = 7 Ns/m, k = 0.5 N/m, m = 205 kg.
Complete the following tasks:
1) Choose the system in Figure 2 or Figure 3.
2) Derive the mathematical model of the chosen system:
a. Differential equation.
b. Transfer function between the displacement xi and xo in Figure 3 or between the
displacement u and y in Figure 2.
c. Derive the state-space model.
1) Use Matlab/Simulink to demonstrate the behaviour of the system in response to a step input.
Reference: Ogata (1990): Modern control engineering. Prentice-Hall.
Figure 1 Automobile with front and rare suspension systems.
Scores:
Differential equation
Derivation of the transfer function
Derivation of the state-space model
The Simulink model window (as shown below)
Running scope window (as shown below)
Figure 2 Simplified suspension system.
Figure 3 Further simplified suspension system.
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47453 S Digital Control Theory, 2004/2005 Autumn
47453 S Digital Control Theory, 2004/2005 Autumn
The state-space model based on Appendix:
Solution
x1 = x 0 − βo x i
Modelling translation systems, Newton’s second law can be applied as
m ⋅ a = ∑ F , where m is the mass, a is the acceleration and F is the force.
The force of a damper is proportional to the speed of compression as
F = b ⋅ x& ,
where b is the damper constant and x& is the speed of compression.
The force of a spring is proportional to the compression as
F = k ⋅ ∆x ,
where b is the damper constant and ∆x is the compression.
The equation of motion for the system:
or
The state-space model using Matlab command:
Taking Laplace transform of this last equation, assuming zero
initial values, we obtain:
» [a,b,c,d]=tf2ss([7 0.5],[205 7 0.5])
(ms
a=
-0.0341 -0.0024
1.0000
0
b=
1
0
c=
0.0341 0.0024
d=
0
2
)
+ bs + k X 0 (s) = (bs + k )X i (s)
Hence, the transfer function is given by:
X 0 (s)
bs + k
=
X i (s) ms 2 + bs + k
Applying b = 7, k = 0.5, m = 205:
G (s) =
b


1 x  

m
b  1  + 
x
−  x 2 
k b b i

 −
m
m m m
 x1 
x 0 = [1 0] 
x 2 
 x& 1   0
 x&  = − k
 2   m
0
1   x 1   0.0341
 x& 1  
 x&  = − 0.0024 − 0.0341  x  + 0.0013 x i

 2  
 2 
 x1 
x 0 = [1 0] 
x 2 
Simple model
m&x& 0 + b(x& 0 − x& i ) + k (x 0 − x i ) = 0
m&x& 0 + bx& 0 + kx 0 = bx& i + kx i
x 2 = x& 0 − β o x& i − β1 x i
X o (s)
7s + 0.5
=
X i (s) 205s 2 + 7s + 0.5 .
βo = 0
b b
b
β1 = − 0 =
m m
m
k b b k
k b b
β2 = −
− 0= −
m mm m
m mm
 x& 1  − 0.0341 − 0.0024  x 1  1
 x&  =  1
  x  + 0  x i
0
 2   
 2 
 x1 
x 0 = [0.0341 0.0024] 
x 2 
One should keep in mind that there is no unique solution for state-space model, but
several different state-space models can be obtained for the same system; see
Appendix.
47453 S Digital Control Theory, 2004/2005 Autumn
47453 S Digital Control Theory, 2004/2005 Autumn
The state-space model using Matlab command:
Complex model:
The equations of motion for the displacements x and y are:
» [a,b,c,d]=tf2ss([4.2 0.3],[1000 1435 222.5 4.2 0.3])
m1&x& + b(x& − y& ) + k 1 (x − u ) + k 2 ( x − y) = 0
m 2 &y& + b(y& − x& ) + k 2 (y − x ) = 0
a = -1.4350
1
0
0
and
Taking Laplace transform of this last equation, assuming zero initial
values, we obtain:
(m s
(m s
+ bs + k 1 + k 2 )X(s) = (bs + k 2 )Y(s) + k 1 U(s)
2
+ bs + k 2 )Y(s) = (bs + k 2 )X (s)
2
1
2
Eliminating X(s) from the last two equations, we have:
k 1 (bs + k 2 )
Y (s)
.
=
U (s) m1 m 2 s 4 + b(m1 + m 2 )s 3 + [k 1 m 2 + k 2 (m1 + m 2 )]s 2 + k 1 bs + k 1 k 2
Applying b = 7, k1 = 0.6, k2 = 0.5, m1 = 5, m2 = 200:
G (s) =
Y(s)
0.6(7s + 0.5)
=
U(s) 1000s 4 + 1435s 3 + 222.5s 2 + 4.2s + 0.3 .
The state-space model:
0

 x&   k 1 + k 2
&x&  −
m1
 =
0
 y&  
   k2
&y&   m

2
1
b
m1
0
b
m2
−
x 
 x& 
y = [0 0 1 0]  + 0u
 y
 
 y& 
0
k2
m1
0
k
− 2
m2
0 
0
b  x  k 
   1
m1   x&   m 

+
u
1   y   01 
b    
−
y&
 
m 2     0 
-0.2225 -0.0042 -0.0003
0
0
0
1
0
0
0
1
0
b = [ 1 0 0 0]’
c= 0
0 0.0042
d= 0
0.0003
 x& 1  − 1.435 − 0.2225 − 0.0042 − 0.0003  x 1  1
 x&   1
  x  0 
0
0
0
 2 = 
 2  +  x i
 x& 3   0
  x 3  0 
1
0
0
  
   
0
1
0
  x 4  0 
 x& 4   0
 x1 
x 
x 0 = [0 0 0.0042 0.0003] 2 
x3 
 
x 4 
47453 S Digital Control Theory, 2004/2005 Autumn
Simulink models:
Remark: When using Simulink, it is better to use the variable names such as “num”
and “den” or “a”, “b”, “c” and “d” in the transfer function and the state-space blocks.
The reason is: the variables are 10 digits exact while the variables written by hand (for
example 0.0013) are only 4-6 digits long. Using hand written variables may yields
incorrect simulation results.
47453 S Digital Control Theory, 2004/2005 Autumn
Appendix: Deriving state-space model
47453 S Digital Control Theory, 2004/2005 Autumn
47453 S Digital Control Theory, 2004/2005 Autumn
47453 S Digital Control Theory, 2004/2005 Autumn