Page 85 - Practical Control Engineering a Guide for Engineers, Managers, and Practitioners

60 Chapter Three

where it is subtracted from the product of the process input U multi-
plied by the process gain g. This result is multiplied by 1/-r. The
resulting signal, which is sY(s) _or dY/dt, is then integrated via
1/s to form the process output Y(s) or Y, which is fed back, and
so on. This structure is similar to the analog computer patch-
board of the 1960s. (It is also similar to the block diagrams that
make up models in Matlab's Simulink.) This approach to block
diagrams will be used in a latter chapter (Chap. 6) when an
underdamped process is modified by feedback to present a better
face to the outside world.

3-3-2 Applying the New Tool to the Rrst-Order Model
Returning to Eq. (3-33), assume that the time domain function U(t) is
a step function having a constant value of Uc. Therefore, it will be
treated as a nonzero constant for t ~ 0 . As with all of our variables,
U(t) is assumed to be zero for t < 0. The Laplace transform for Uc (see
App. F and/ or the box given in Sec. 3-3) is

and Eq. (3-33) becomes
- g u
Y=---' (3-35)
'rS + 1 S

To invert this transform to get Y(t), Eq. (3-35) needs to be simpli-
fied to a point where we can recognize a familiar form and match it
up with a time domain function. Partial fractions can be used to split
Eq. (3-35) into two simpler terms. ~eferring to App. F the reader can
verify that the new expression for Y is

- g u
Y=---'
'rS + 1 S
_ gU, gU,
--s----1
s+-
'r
=gU,[~-~J

s+-
'r

We already know the time domain functions for the Laplace trans-
forms, namely,