Page 87 - Practical Control Engineering a Guide for Engineers, Managers, and Practitioners

62 Chapter Three

Thus, the Laplace transform of a derivative of a quantity is equal
to s times the Laplace transform of that quantity Y, minus that quan-
tity's initial value Y(Q+). In our example and in our recipe box, ~e
stipulated that the initial value was to be zero, so replacing Y by s Y is
the correct way to take the transform of the derivative, just as we
proposed in the previous section. In most of this book, the initial
value of transformed variables will be assumed to be zero.
The Laplace transform of the second-order derivative:

2
L{~:n= sL{Y}-sY(O)- ~; lo
(3-38)
2
= s Y -sY(o+)- Y(O+)
That is, the Laplace transform of the second derivati~e of a quan-
2
tity is s times the Laplace transform of that quantity, Y, minus the
initial value of that quantity times s, minus the initial value of that
quantity's first derivative.
Thus, when the initial conditions are all zero, the various deriva-
tives can be transformed by replacing the derivative by Laplace trans-
form of the quantity times the appropriate power of s.

3-3-4 Applying the Laplace Transform to the Case
with Proportional plus Integral Control
Equation (3-21) can now easily be transformed. Start with the time
domain equation derived earlier

Apply th~ Laplace transform rules and get an algebraic equation
solvable for Y

2
('rs + (1 + gk)s + gl) Y = (gks + gl)S

where Y and S have been factored out.
Solving for Y gives

Y= gks+gl S=GS
2
-rs + (1 + gk)s + gl
G= gks+gl (3-39)
2
-rs + (1 + gk)s + gl
where G represents the transfer function from S to Y.