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Problem 4.26PP
We wish to design an automatic speed control for an automobile. Assume that (1) the car has a 
mass /n of 1000 kg, (2) the accelerator is the control U and supplies a force on the automobile of 
10 N per degree of accelerator motion, and (3) air drag provides a friction force proportional to 
velocity of 10 N • sec/m .
(a) Obtain the transfer function from control input U to the velocity of the automobile.
(b) Assume the velocity changes are given by
s+0,02 «+ao2
where V is given in meters per second. U is in degrees, and W is the percent grade of the road. 
Design a proportional control law U = -kPV that will maintain a velocity error of less than 1 m/sec 
in the presence of a constant 2% grade.
where V is given in meters per second. U is in degrees, and W is the percent grade of the road. 
Design a proportional control law U = -kPV that will maintain a velocity error of less than 1 m/sec 
in the presence of a constant 2% grade.
(c) Discuss what advantage (if any) integral control would have for this problem.
(d) Assuming that pure integral control (that is, no proportional tenn) is advantageous, select the 
feedback gain so that the roots have critical damping (C= 1 )•
Step-by-step solution
step 1 of 6
(a)
Consider the mass of the car.
ffl = l,000kg... (1)
Consider the accelerator is control U and supplies a force on the automobile.
=10N perdegreeof acelerator....(2)
Consider the air drag provides a friction force proportional to velocity. 
D=10N.sec/m...... (3)
Consider the relationship between the mass, acceleration and force.
m r = X ; ^ ’
n a = K ^ - D x ...... (4)
Where.
u is displacement,
X is velocity, 
x is acceleration.
Modify equation (4).
m v - K ju - D v ...... (5)
Where.
V is velocity, 
vis acceleration.
Apply Laplace transform in equation (5). 
m sy(s) = K ,U {s)-D V {s) 
m sy(s)+ D y(s) = /C.U(s)
(ms + D )y (s ) = /C,U(s)
(6)
1/ (s) (»« + D)
Calculate the transfer function of the automobile system. 
Substitute equations (1), (2) and (3) in equation (6).
10
U (s) (1,000s-MO) 
l ' ( s ) 1,000(0.01)
I / ( s ) ° l,000 (s+0.01)
y (s ) 0.01
u ( s ) ” ( i+ 0 .0 1 )
(7)
Hence, the transfer function of the automobile system is
y (s )_ 0.01
t / ( s ) ( s+ 0 .0 1 )
Step 2 of 6
(b)
Draw the general block diagram of automobile speed control system.
fy
Figure 1
Step 3 of 6
Write the output equation of error detector from Figure 1.
£ (s ) = » ; ( s ) - F ( s )
f M J
£ ( s ) = i ; ( s ) -
1+—^ 1+—^
V. s+ a s+a
(8)
Consider the velocity changes of the system.
W .......(9)F ( i ) . — !— LF(s)+ H
' ’ j+ 0 .0 2 ' ’ j+ 0 .0 2
Step 4 of 6
Compare equation (8) from equation (9) and find the parameters a, A and B. Substitute a, A and 
6 in equation (8).
0.0S
£ (s ) = i ; ( s ) - S+0-02 s+0-02 G ( , )
* -
1 +
», J+0.02
1+ -
= > ; ( * ) -
s+0.02
0.05
t + 0 .0 2 + t, > s + 0 .0 2 + t, ^ 7
1. s+0.02 s+0.02 ;
( s + 0 .0 2 + i t , )F , ,( s ) - 0 . 0 8 .......(12)
Hence, the proportional constant value is * , > 0.0̂
Step 5 of 6
(c)
The clear advantage of integral control is [zero steady state eiTor| for step input.
Step 6 of 6
(d)
Calculate integral constant value.
Consider the value of proportional constant.
* , S - ■(13)
Substitute equation (13) in equation (10).
( » + 0 .0 2 ) l( ,( 5 ) -0 .0 5 g ( ^ )
i+ 0 . 0 2 + ^
, ■ j ( j+ 0 .0 2 ) F , , ( i ) - 0 .0 S jG ( t )
j ’ + 0.02s+ t,
.(14)
Write the general characteristics equation of the second order system. 
* ’ +2t«>,4+,’ - 0 ...... (15)
Compare the equation (15) with denominator of equation (14).
* ’ + 25o>,j+