Page 67 - Practical Control Engineering a Guide for Engineers, Managers, and Practitioners

42 Chapter Three

and make the following changes of variable

Y=gy
t=fT

fd(gy) +gy=gu
d(-rt)
dy +y=u
dT
We will not make much use of this variable change concept until
Chap. 5 when it becomes helpful in addressing round-off error.
3·2·2 Solution of the Continuous Time Domain Model
We will solve the model for the first-order process developed in
Sec. 3-2-1 for a variety of conditions. First, consider the case where the
process input, initially zero, is assigned the nonzero value of Uc at
time zero and held there for all time. The reader hopefully will recog-
nize that this situation is the same as giving the process a step change
in the process input at time zero. Hence, if the equation can be solved,
the step-change response will be obtained and we can compare the
results with the graph in Fig. 3-2.
For this case of a constant process input, Eq. (3-8) becomes

(3-10)

There are many ways of solving this equation and a couple of
them are reviewed in App. E. The solution for the case of a constant
input is

(3-11)

We will refer to this equation repeatedly in this book so the reader
should be comfortable with its development before proceeding.

Comments about the Solution
First, the readers should convince themselves that Eq. (3-11) behaves
just like the curve in Fig. 3-2 for the case of Y = 0, f= 10.0, g = 2.0, and
0
Uc = 1.0. To verify the time constant value, look where the process
output appears to reach 63% of the final value. It should take f seconds
to get there. Now, look at the final value. It should be two units greater
than the initial value of 0.0 because the process gain is 2.0.