0195136047.pdf

798 BASIC CONTROL SYSTEMS

wherevtis the terminal voltage applied to the motor,eais the back emf given by Equation (2),Ra

andLainclude the series resistance and inductance of the armature circuit and electrical source

put together, andτa=La/Rais theelectrical time constantof the armature circuit. Note that the

operatorpstands for (d/dt). The electromagnetic torque is given by Equation (1), and from the

dynamic equation for the mechanical system given by

Te=Jpωm+Bωm+TL (5)

the acceleration is then given by

(B+Jp)ωm=Te−TL (6)

or

B( 1 +τmp)ωm=Te−TL (7)

whereτm=J/Bis themechanical time constant. The load torqueTL, in general, is a function of

speed,Jis the combined polar moment of inertia of the load and the rotor of the motor, andBis

the equivalent viscous friction constant of the load and the motor.

Laplace transforms of Equations (4) and (7) lead to the following:

Ia(s)=

Vt(s)−Ea(s)

Ra( 1 +τas)

=

Vt(s)−Kmm(s)

Ra( 1 +τas)

(8)

m(s)=[Te(s)−TL(s)]

1

B

1

( 1 +τms)

(9)

The corresponding block diagram representing these operations is given in Figure E16.2.1(b) in

terms of the state variablesIa(s) andm(s), withVt(s) as input.

The application of the closed-loop transfer functionM(s), shown in Figure 16.2.3, to the

block diagram of Figure E16.2.1(b) yields the following transfer function relatingm(s)and

Vt(s), withTL=0:

m(s)

Vt(s)

=

Km/[Ra( 1 +τms)B( 1 +τms)]

1 +

[

Km^2 /Ra( 1 +τas)B( 1 +τms)

] (10)

With mechanical dampingBneglected, Equation (10) reduces to

m(s)

Vt(s)

=

1

Km[τis(τas+ 1 )+ 1 ]

(11)

whereτi=JRa/Km^2 is theinertial time constant, and the corresponding block diagram is shown

in Figure E16.2.1(c). The transfer function relating speed to load torque withVt=0 can be

obtained from Figure E16.2.1(c) by eliminating the feedback path as follows:

m(s)

TL(s)

=−

1 /J s

1 +( 1 /J s)

[

Km^2 /Ra( 1 +τas)

]=−

τas+ 1

Js( 1 +τas)+(Km^2 /Ra)

(12)

Expressing the torque equation for the mechanical system as

Te=Kmia=Jpωm+Bωm+TL (13)

then dividing byKm, and substitutingωm=ea/Km, we obtain

ia=

J

K^2 m

dea

dt

+

B

Km^2

ea+

TL

Km

(14)

Equation (14) can be identified to be the node equation for a parallelCeq–Geq–ZLcircuit with

Ceq=J/Km^2 ; Geq=B/Km^2 ; ZL=Kmea/TL (15)