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Problem 2.09PP
Modify the equation of motion for the cruise control in Example, Eq., so that it has a control law; 
that is, let
u = K (vr-v ), (2.108) 
where
vr= reference speed. (2.109)
/C= constant. (2.110)
This is a “proportionarcontrol law in which the difference between \^rand the actual speed is
used as a signal to speed the engine up or slow it down. Revise the equations of motion with vr 
as the input and v as the output and find the transfer function. Assume that/n = 1500 kg and b = 
70 N -sec/m, and find the response for a unit step In vr using Matlab. Using trial and error, find a 
value of K that you think would result in a control system In which the actual speed converges as 
quickly as possible to the reference speed with no objectionable behavior.
70 N -sec/m, and find the response for a unit step In vr using Matlab. Using trial and error, find a 
value of K that you think would result in a control system In which the actual speed converges as 
quickly as possible to the reference speed with no objectionable behavior.
Eq.
b u
V H— V = —.
m m
EXAMPLE 2.1 A Simple System; Cruise Control Model
1. Write die equedon of motion for the ipeed m l forward motion of 
lhecaribownmFig.2.1 aasnmiiig that the en^ne imparta a force ■ ae 
abown. Take the Laplace transform of the reauiting differential eqoadon 
and find the transfer function between the input« and the output v.
2. Use Matlab to 6nd the response of the velocity of the car for the case 
in which the inpntjumpe from being « = Oat time f = 0 to a constant
Hgui«2.1
Cruise control model
« B 500 N thereafter. Assume that the car a 
visoous drag coeflkient, b B 50 N*secAn.
I « is 1000 kg and
Solution
1. Equations of motion: For sunpBdty we asanme that the rotmkNinl 
inertia o f the wheeb b negligible and that there b ftiction retarding 
the motion of the car that b proportional to the car’s speed with a 
proportionality constant, The car can then be approximated for 
modeling poipoeea using the ftee-body diagram men m Fig. 2.2, which 
defines coordinmes, shows all forces acting on the body (heavy lineaX 
and imficates the accdetaCion (dashed line). The coordinate of the car’a 
positioo X u the distaoce from the refotenoe line sbo«m and b chosen ao 
that positive isto the r i ^ Note that in this case the meitial acceleration 
u amply the second derivative of x (that is, a * x) because the car 
position b measured wkh respect to « inertial reference frame. The 
equation of motion b found mung Eq. (2.1). The friction force acts 
opposite to the direction motiMi; therefore k b drawn rqiposhe to the 
direction o f positive motion m l eiketcd as a negative fioroe in Eq. (2.1). 
Theresukb
(2.2)
^ b . M
X + — X = — . 03)
For the case of the automotive cruise control where the variable of 
interest b the speed, v ( ^ ) , the equbion of motion becomes
* “ 0 .4 )V + —V B —.
H k sobtion of wch an equation win be cowered in detail in Q upter 3;
however, the essence b that you aaaume a aolotionoftfae form v s
pvea an input o f the form ■ b Then, tince v b the
differential equation chi be written aa^
03)
Rgurt2,2
r>co body diagram for t::
0 6 )
a7)
Step response wtth Hatlab
The c* term C H iceb out, and we find that
V -!■ yg _ m
« . . + 1 '
For reasons that will become efetf in Quqker 3, this b often written 
using capital letters to signify that k b the‘Itunsforro'’of the solution, or
v w _ i
* + h '
This expresskm of die dUferential equation (2.4) b called the tran ti’er 
function and wiD be used extensively m later chapters. Note that, in 
essence, we have aubstkuled i fof dfdt in Eq. (Z4X Thb transfer Auh 
tion serves as a mwh model that relates the ch ’s velocity to the forces 
propelling the car, that b , inputs from the accelerator pedal. Transfer 
functions of a system will be used in later chapters to design feedback 
corXrollers such as a cruise control device found in many modern cars. 
X Time reaponae: The dynamics ofn system can be prescribed to Matlab 
in terms of its transfer function as can be seen in the Matlab statemenb 
below that implemeiits EN magnitude. The Mep responae b shown m Fig. 2.3.
Rgure2.3 
Response of the car 
velodtytoastepinti
Step-by-step solution
step 1 of 7
Consider the equation of motion.
V + — V ® — IT
Where, 
u is the input 
V is the output 
m is the mass 
b is the friction force 
Substitute JC (v^-v) foru.
m m
. b K K
V + — V * — V ,------- V
m m m
. b K K
V + — V + — V = — V-
m m m
A block diagram of the scheme is shown in Figure 1.
Step 2 of 7
1
u m
. b . K
Step 3 of 7
Figure 1
Step 4 of 7
Consider the transfer function of the closed loop system.
y (s )
K (x ) b K
S + — + —
m m
Write the inputs for MATLAB. 
K
m
den
I m mj
Step 5 of 7
Consider K=100,200,1000,5000 
Write the MATLAB program, 
m = 1500; 
b = 70;
k = [100 200 1000 5000]; 
hold on 
t=0:0.2:50; 
for i=1:length(k)
K=k(i);
num =K/m;
den = [1 b/m+K/m];
sys=tf(num,den);
y = step(sys,t);
plot(t,y)
grid
end
hold off
Step 6 of 7
The output for the MATLAB code is given in Figure 2.
Step 7 of 7
Thus, the larger the K is the better the performance with no objectionable behavior for any of the 
cases. Increasing K also results in the need for higher acceleration is less obvious. In Figure 2, 
for K - 1,000 , there is a response in 5 seconds and the steady state error is 5%.