Page 86 - Practical Control Engineering a Guide for Engineers, Managers, and Practitioners

Basic Concepts in Process Analysis 61

1 1
and
s 1
s+-
'r
The first transform is for a step (or a constant) and the second is
for an exponential. So, by inspection, we can write the time domain
form as

Now, if the reader remembers Eq. (3-11), she will see that a second way
has been obtained to solve the differential equation [Eq. (3-10)].
3·3·3 The Laplace Transform of Derivatives
According to the recipe, the derivative in Eq. (3-31) was replaced by the
operator s. App. F shows that the basis for this comes directly from the
definition of the Laplace transform, which, for a quantity Y(t), is

(3-36)

Note that e-''Y(t) is integrated from t = 0 to t = oo. It may seem
like a technicality but the integration starts at zero so the value of
the quantity Y(t) fort < 0 is of no interest and is assumed to be zero.
If the quantity has a nonzero initial value, say Y , then strictly speak-
0
ing we have to look at it as

Y = limt-H Y(t) = Y(O+)
0
That is, Y is the initial value of Y(t) when t = 0 is approached from
0
the right or from positive values of t. So, effectively, a nonzero initial
value corresponds to a step change at t = 0 from the Laplace trans-
form point of view. This subtlety comes into play when one evaluates
the Laplace transform of the derivative, as in

L{dY} = r-dt -st dY
dt Jo e dt

The evaluation of this equation presents a bit of a challenge so I
put the gory details in App. F for the reader to check if she wishes.
However, after all the dust settles the result is

dY}
-
L dt =sY-Y(O+) (3-37)
{