A First Course in FUZZY and NEURAL CONTROL

58 CHAPTER 2. MATHEMATICAL MODELS IN CONTROL

However, the response continues to exhibit the small steady-state error of 10 −
9 .9502 = 0.049 8meters/second shown in Figure 2.22.

0

2

4

6

8

10

(^246810) t 12 14 16 18 20
Figure 2.22.y(t)=9. 9502 u(t)− 9. 9502 e−^1.^005 tu(t)
We see that there is no issue in this example relating to overshoot. The only
issue is that of steady-state error. It should be clear that just increasing the pro-
portional gain will not correct the steady-state error. To correct this, we need to
implement integral control. The reason for not needing derivative control should
be obvious from the analytical solution for the closed-loop solution; for example,
y(t)=9. 9502 u(t)− 9. 9502 e−^1.^005 tu(t)which is of the formC[1−e−at]u(t)in
which there is no term that can produce sinusoidal oscillation. The shape of the
response clearly shows only exponential behavior. Overshoot can occur only if
the system oscillates, that is, has second-order complex conjugate poles.
We now proceed to implement integral control by choosing values forKI> 0.
Choosing a value ofKP= 1000,KI=1,andwithKD=0,wegettheforward
transfer function as
G(s)=Gc(s)Gp(s)=
μ
KDs^2 +KPs+KI
s
∂μ
1
ms+b

∂

=

μ

1000 s+1

s

∂μ

1

1000 s+50

∂

=

s+0. 001

s(s+0.05)

The closed-loop transfer function therefore is given by

Y(s)

F(s)

=

G(s)

1+G(s)

=

s+0. 001

s(s+0.05)

1+ss(s+0+0.^001 .05)

=

s+0. 001

s^2 +1. 05 s+0. 001

A controller canonical form of state-variable realization can then be obtained
as follows. Let
Q(s)
Q(s)

Y(s)

F(s)

=

s+0. 001

s^2 +1. 05 s+0. 001

Upon choosing
Q(s)
F(s)

=

1

s^2 +1. 05 s+0. 001