A First Course in FUZZY and NEURAL CONTROL

2.7. PROPORTIONAL-INTEGRAL-DERIVATIVE CONTROL 57

Let us selectKP = 100as a start and see what happens. Substituting values
forKP,m,andbwe get the closed-loop transfer function as

Y(s)

F(s)

=

100

1000 s+ 50 + 100

=

100

1000 s+150

=

0. 1

s+0. 15

Since we are interested in obtaining a desired velocity of 10 meters/second,
we need to provide a step input of amplitude 10 meters/second. Therefore,
usingF(s)=10/s, we obtain the closed-loop responseY(s)as,

Y(s)=

0. 1

s+0. 15

F(s)=

μ

0. 1

s+0. 15

∂μ

10

s

∂

=

1

s(s+0.15)

=

1

0. 15

s

−

1

0. 15

s+0. 15

=

6. 6667

s

−

6. 6667

s+0. 15

The time-domain solution for this isy(t)=6. 6667 u(t)− 6. 6667 e−^0.^15 tu(t),
shown in Figure 2.21.

0

2

4

6

8

10

20 40 t 60 80 100

Figure 2.21. Time-domain solution

Once again, we see that choosingKP = 100still does not satisfy our rise
time criterion of less than 5 seconds. However, we do see an improvement over
the open-loop response shown earlier. ChoosingKP= 1000, and carrying out
an analysis similar to the above shows that the closed-loop response satisfies the
performance criteria for rise time.

Y(s)

F(s)

=

1000

1000 s+ [50 + 1000]

=

1000

1000 s+ 1050

=

1

s+1. 05

Y(s)=

1

s+1. 05

F(s)=

∑

1

s+1. 05

∏∑

10

s

∏

=

10

s(s+1.05)

=

9. 9502

s

−

9. 9502

s+1. 005

y(t)=9. 9502 u(t)− 9. 9502 e−^1.^005 tu(t)