0195136047.pdf

796 BASIC CONTROL SYSTEMS

Potentiometer Servoamplifier Servomotor

−

Js^2 + Fs

REK^1 C

p Ka +Ksi Km

Ea Td

Figure 16.2.14Block diagram of servomechanism with integral-error control.

M(s)=

C(s)

R(s)

=

Qi+sK

Js^3 +Fs^2 +Ks+Qi

(16.2.34)

whereQi=KiKpKmandK=KpKaKm, as defined earlier. Since the integral term stands alone

without combining with the viscous-frictionFterm (as was the case with error-rate and output-rate

controls), its influence on system performance differs basically from the previous compensation

schemes. Also note that the order of the system is changed from second to third, because of the

integral-error compensation. The inclusion of the integral term implies that a third independent

energy-storing element is present. The position lag error, which exists with error-rate and output-

rate control, disappears with integral-error control. This is a characteristic of integral control

which greatly improves steady-state performance and system accuracy. However, it may make

the dynamic behavior more difficult to cope with successfully.

Let us now present some illustrative examples.

EXAMPLE 16.2.1

Since dc motors of various types are used extensively in control systems, it is essential for

analytical purposes that we establish a mathematical model for the dc motor. Let us consider the

case of a separately excited dc motor with constant field excitation. The schematic representation

of the model of a dc motor is shown in Figure E16.2.1(a). We will investigate how the speed

of the motor responds to changes in the voltage applied to the armature terminals. The linear

analysis involves electrical transients in the armature circuit and the dynamics of the mechanical

load driven by the motor. At a constant motor field currentIf, the electromagnetic torque and the

generated emf are given by

Te=Kmia (1)

ea=Kmωm (2)

whereKm=kIfis a constant, which is also the ratioea/ωm. In terms of the magnetization curve,

eais the generated emf corresponding to the field currentIfat the speedωm. Let us now try to find

the transfer function that relatesm(s)toVt(s).

Solution

The differential equation for the motor armature currentiais given by

vt=ea+La

dia

dt

+Raia (3)

or

Ra( 1 +τap)ia=vt−ea (4)