A First Course in FUZZY and NEURAL CONTROL

54 CHAPTER 2. MATHEMATICAL MODELS IN CONTROL

2.18. Using Newtonís law of motion, the model equations for this system are

m

dv

dt

+bv = f

y = v

wherefis the force produced by the engine andvis the measured velocity of the
automobile. For this example, Let us assume thatm= 1000kg,b=50Newton
seconds/meter, andf= 500Newtons, a constant force produced by the engine.
When the engine produces a 500 Newton force, the car will reach a maximum
velocity of 10 meters/second. An automobile should be able to accelerate up to
that speed in less than 5 seconds. For this simple cruise control system, we can
assume an allowable overshoot of 10% on the velocity and a 2% steady-state
error.
The next step is tofind the transfer function of the system above. For this,
we need to take the Laplace transform of the model equations. Whenfinding the
transfer function, zero initial conditions must be assumed. Performing Laplace
transforms of the two equations gives

msV(s)+bV(s)=F(s)

Y(s)=V(s)

Since our output is the velocity, Let us expressV(s)in terms ofY(s)and obtain

msY(s)+bY(s)=F(s)

The plant transfer function therefore becomes

Y(s)

F(s)

=

1

ms+b

Substitutingm= 1000 kg,b=50Newton seconds/meter assumed for the auto-
mobile, the transfer function is

Y(s)

F(s)

=

1

1000 s+50

=

(1/1000)

s+0. 05

For a constant force of 500 Newtons produced from the start, with the automo-
bile initially at rest, this represents a step input where

F(s)=

500

s

We can now compute the open-loop response of the system to examine how well
the system behaves without any controller action.
The open-loop response of the plant is that produced by the engine force
acting on the mass of the automobile. In this analysis, there is no controller
action as there is no feedback. Therefore, the output response is

Y(s)=

(1/1000)

s+0. 05

F(s)=

∑

(1/1000)

s+0. 05

∏∑

500

s

∏

=

0. 5

s(s+0.05)