Page 88 - Practical Control Engineering a Guide for Engineers, Managers, and Practitioners

Basic Co1cepts i1 Precess A1alysis 63

Assume that the set point S is given a step at time zero and that
Y(O) is zero. Since for t ~ 0 , S is a constant, the transform for S is then
(remember that for t < 0, S(t) = 0 ).

where Sc is the size of the set-point step.
Equation (3-39) becomes

Y= gks+gl sc (3-40)
2
1's +(1+ gk)s+ gl s

Question 3-8 What can the final value theorem tell us about whether this
controlled process will settle out with no offset?

Anlwlr Applying the final value theorem to Eq. (3-40) gives

5
Y(oo) = lim sY = lim s gks + gl c
2
•-tO .-tO -rs +(1+gk)s+gl s
=lim (gks+ gl)Sc 5 c
2
•-tO -rs +(1+ gk)s+ gl
So, the presence of integral control removes the offset.

Questloa 3-7 Using the result in App. F for the Laplace transform of the integral,
could you arrive at Eq. (3-40) starting with

dY
f-+Y=gU
dt
'
U(t) = ke(t) +I J due(u)
0
or

(3-41)

Anlwlr Applying the Laplace transform to Eq. (3-41) gives
- - e
-rsY+Y=gkt+gl-
s