Page 75 - Practical Control Engineering a Guide for Engineers, Managers, and Practitioners

50 Chapter Three

If you can accept this, then Eq. (3-22) immediately yields

lim -+oo Y(t) = Yss
1
However, for future reference, we need to slightly expand the
above manipulations and try to find a partial solution to Eq. (3-22).
We follow the same path (in App. E) that led to the solution of the
first-order differential equation in Eq. (3-10). Assume that the solu-
tion consists of a transient part Y, and a steady-state part Yss, as in

The transient portion of the solution Y, satisfies

2
d Y. dY.
-r---f+(l+ gk)-d I+ glY, = 0 (3-23)
dt t
and the steady-state part satisfies

(which we already suggested). If we let t -+ oo then the derivatives in
Eq. (3-23) will go to zero, as should Y, that is,

t-+ 00
dY,
--+0
dt
Y, -+0

If the transient part of the solution goes to zero then all that is left
is Y .. which is the same as saying

This means that as t -+ oo, Eq. (3-22) simplifies to

so,

So, when t -+ oo, Y, goes away and the steady-state part remains and
it is equal to the set point.