Modern Control Engineering

610 Chapter 8 / PID Controllers and Modified PID Controllers

where

This controller involves one pole at the origin and two zeros. If we assume that the two zeros are

located at the same place (a double zero), then can be written as

Then the characteristic equation for the system becomes

or

s(s+1)(s+5)+5K(s+a)^2 =0

which can be rewritten as

s^3 +(6+5K)s^2 +(5+10Ka)s+5Ka^2 =0 (8–8)

If we place the double zero between s=–3ands=–6,then the root-locus plot of

may look like the one shown in Figure 8–41. The speed of response should be fast, but not faster

than necessary, because faster response generally implies larger or more expensive components.

Therefore, we may choose the dominant closed-loop poles at

s=–3_j2

(Note that this choice is not unique. There are infinitely many possible closed-loop poles that we

may choose from.)

Since the system is of third order, there are three closed-loop poles. The third one is located

on the negative real axis to the left of point s=–5.

Let us substitute s=–3+j2into Equation (8–8).

(–3+j2)^3 +(6+5K)(–3+j2)^2 +(5+10Ka)(–3+j2)+5Ka^2 =0

Gc1(s)Gp(s)

1 +Gc1(s)Gp(s)= 1 +

K(s+a)^2

s

5

(s+1)(s+5)

= 0

Gc1(s)=K

(s+a)^2

s

Gc1(s)

Gc1(s)=Kpa 1 +

1

Ti s

+Td sb

Root-Locus Plots of (s + a)^2 /(s^3 + 6s^2 + 5s)

witha = 3, a = 4, a = 4.5, and a = 6

Real Axis

Imag Axis

2

− 4

− 6

− 8

− 14 − 12 − 10 − 8 − 6 − 4 −20 2

4

− 2

0

6

8

a = 6 a = 4.5

a = 4

a = 3

Figure 8–41

Root-locus plots of

5K(s+a)^2 /Cs(s+1)

(s+5)Dwhena=3,

a=4, a=4.5,and

a=6.

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