Page 354 - Modern Control Systems

328 Chapter 5 The Performance of Feedback Control Systems

since Y(s) = T(s)R(s). Note that

KiG c(s)G(s) (TS + l)KiG c(s)G(s)
r(s) =
1 + H(s)G c(s)G(s) TS + 1 + K!G c(s)G(sy
and therefore,

_ 1 + rs(l - KjGMGjs))
* {S) TS + 1 + KtfMGis) {S) '

Then the steady-state error for a unit step input is

1
= lim s E(s) = ———^ , ,^,, ,. (5.36)
e ss ss w v J
, - o 1 + K x lim G e(s)G(s)
s—»0
We assume here that

lim sG c(s)G(s) = 0.
s—0

EXAMPLE 5.4 Steady-state error
Let us determine the appropriate value of K\ and calculate the steady-state error
for a unit step input for the system shown in Figure 5.21 when

G c(s) = 40 and G(s) = y ^

and

We can rewrite H(s) as

2
H{s) =
0.1s + 1

= Kj = 2, we can use Equation (5.36) to determine
Selecting K x
1 1 = 1
€ss
~ 1 + Id lim G c(s)G(s) " 1 + 2(40)(1/5) " 17'
A—»0
or 5.9% of the magnitude of the step input. •

EXAMPLE 5.5 Feedback system
Let us consider the system of Figure 5.24, where we assume we cannot insert a gain
Kx following R(s) as we did for the system of Figure 5.21. Then the actual error is
given by Equation (5.35), which is