Page 371 - Modern Control Systems

Section 5.9 Design Examples 345
The goal of the design is to choose K t and K so that (1) the percent overshoot
of the output to a step command, r{i), is less than or equal to 10%, (2) the steady-
state error to a ramp command is minimized, and (3) the effect of a step disturbance
is reduced. Since the system has an inner loop, block diagram reduction can be used
to obtain the simplified system of Figure 5.34(b).
The output due to the two inputs of the system of Figure 5.34(b) is given by

Y(s) = T(s)R(s) + [T(s)/K]T d(s), (5.67)

where
KG(s) L(s)
T(s) =
1 + KG(s) 1 + L(s)'
The error is

First, let us select K and K± to meet the percent overshoot requirement for a step
input, R(s) = A/s. Setting T d(s) = 0, we have

KG(s)

K A\ K A
(5.69)
s(s + K{) + K\s J s 2 + Kis + K\s

To set the overshoot less than 10%, we select £ = 0.6 by examining Figure 5.8 or
using Equation (5.16) to determine that the overshoot will be 9.5% for I = 0.6.
We next examine the steady-state error for a ramp, r{i) = Bt,t >: 0, using (Equa-
tion 5.28):

=
=
*• Jajfefe)} m- (570)

The steady-state error due to a unit step disturbance is equal to -1/K. (The
student should show this.) The transient response of the error due to the step dis-
turbance input can be reduced by increasing K (see Equation 5.68). In summary,
we seek a large K and a large value of KjK\ to obtain a low steady-state error for
the ramp input (see Equation 5.70). However, we also require £ = 0.6 to limit the
overshoot.
For our design, we need to select K. With £ = 0.6, the characteristic equation of
the system is
s 2 + 2£<o„s + oil = s 2 + 2(0.6)w„.v + K. (5.71)