Feedback Control Systems (FCS)

Dr. Imtiaz Hussainemail: imtiaz.hussain@faculty.muet.edu.pk

URL :http://imtiazhussainkalwar.weebly.com/

Lecture-7Mathematical Modeling of Real World Systems

1

Modelling of Mechanical Systems

2

• Automatic cruise control

• The purpose of the cruise control system is to maintain a constant vehiclespeed despite external disturbances, such as changes in wind or road grade.

• This is accomplished by measuring the vehicle speed, comparing it to thedesired speed, and automatically adjusting the throttle.

• The resistive forces, bv, due to rolling resistance and wind drag act in thedirection opposite to the vehicle's motion.

Modelling of Mechanical Systems

3

bvvmu

• The transfer function of the systems would be

bmssU

sV

1

)(

)(

Electromechanical Systems

• Electromechanics combines electrical and mechanicalprocesses.

• Devices which carry out electrical operations by usingmoving parts are known as electromechanical.

– Relays

– Solenoids

– Electric Motors

– Switches and e.t.c

4

Potentiometer

5

Potentiometer

6

• R1 and R2 vary linearly with θ between thetwo extremes:

totRRmax

1

totRRmax

max

2

Potentiometer

7

• Potentiometer can be used to senseangular position, consider the circuitof figure-1.

• Using the voltage divider principle wecan write:

intot

inout eR

Re

RR

Re 1

21

1

inout eemax

Figure-1

totRRmax

1

8

D.C Drives

• Speed control can be achieved usingDC drives in a number of ways.

• Variable Voltage can be applied to thearmature terminals of the DC motor .

• Another method is to vary the flux perpole of the motor.

• The first method involve adjusting themotor’s armature while the lattermethod involves adjusting the motorfield. These methods are referred to as“armature control” and “field control.”

oltageback-emf vwhere e,edt

diLiRu bb

aaaa

Mechanical Subsystem

BωωJTmotor

Input: voltage uOutput: Angular velocity

Electrical Subsystem (loop method):

Example-2: Armature Controlled D.C Motor

uia

T

Ra La

J

B

eb

Torque-Current:

Voltage-Speed:

atmotor iKT

Combing previous equations results in the following mathematical model:

Power Transformation:

ωKe bb

0at

baaa

a

i-KBωJ

uωKiRdt

diL

where Kt: torque constant, Kb: velocity constant For an ideal motor

bt KK

Example-2: Armature Controlled D.C Motor

uia

T

Ra La

J

B

eb

Taking Laplace transform of the system’s differential equations withzero initial conditions gives:

Eliminating Ia yields the input-output transfer function

btaaaa

t

KKBRsBLJRJsL

K

U(s)

Ω(s)

2

0(s)IΩ(s)-KBJs

U(s)Ω(s)K(s)IRsL

at

baaa

Example-2: Armature Controlled D.C Motor

Reduced Order Model

Assuming small inductance, La 0

abt

at

RKKBJs

RK

U(s)

Ω(s)

Example-2: Armature Controlled D.C Motor

If output of the D.C motor is angular position θ then we know

abt

at

RKKBJss

RK

U(s)

(s)

Which yields following transfer function

Example-3: Armature Controlled D.C Motor

)()( sssordt

d

uia

T

Ra La

J

θ

B

eb

Applying KVL at field circuit

Example-3: Field Controlled D.C Motor

ifTm

Rf

Lf JωB

Ra La

eaef

dt

diLRie

f

ffff

Mechanical Subsystem

BωωJTm

Torque-Current: ffm iKT

Combing previous equations and taking Laplace transform (considering initial conditions to zero) results in the following mathematical model:

Power Transformation:

)()()(

)()()(

sIKsBsJs

sIsLsIRsE

ff

fffff

where Kf: torque constant

Example-3: Field Controlled D.C Motor

If angular position θ is output of the motor

Eliminating If(S) yields

Example-3: Field Controlled D.C Motor

)( ff

f

f RsLBJs

K

(s)E

Ω(s)

ifTm

Rf

Lf J

θB

Ra La

eaef

)( ff

f

f RsLBJss

K

(s)E

(s)

An armature controlled D.C motor runs at 5000 rpm when 15v applied at thearmature circuit. Armature resistance of the motor is 0.2 Ω, armatureinductance is negligible, back emf constant is 5.5x10-2 v sec/rad, motor torqueconstant is 6x10-5, moment of inertia of motor 10-5, viscous friction coefficientis negligible, moment of inertia of load is 4.4x10-3, viscous friction coefficientof load is 4x10-2.

1. Drive the overall transfer function of the system i.e. ΩL(s)/ Ea(s)

2. Determine the gear ratio such that the rotational speed of the load isreduced to half and torque is doubled.

Example-4

15 via

T

RaLa

Jm

Bm

eb

JL

N1

N2

BL

L

ea

System constants

ea = armature voltage

eb = back emf

Ra = armature winding resistance = 0.2 Ω

La = armature winding inductance = negligible

ia = armature winding current

Kb = back emf constant = 5.5x10-2 volt-sec/rad

Kt = motor torque constant = 6x10-5 N-m/ampere

Jm = moment of inertia of the motor = 1x10-5 kg-m2

Bm=viscous-friction coefficients of the motor = negligible

JL = moment of inertia of the load = 4.4x10-3 kgm2

BL = viscous friction coefficient of the load = 4x10-2 N-m/rad/sec

gear ratio = N1/N2

Since armature inductance is negligible therefore reduced order transferfunction of the motor is used.

Example-4

15 via

T

RaLa

Jm

Bm

eb

JL

N1

N2

BL

L

ea

btaeqaeqaeq

tL

KKRBsLBRJ

K

U(s)

(s)Ω

Lmeq JN

NJJ

2

2

1

Lmeq B

N

NBB

2

2

1

A field controlled D.C motor runs at 10000 rpm when 15v applied at the fieldcircuit. Filed resistance of the motor is 0.25 Ω, Filed inductance is 0.1 H,motor torque constant is 1x10-4, moment of inertia of motor 10-5, viscousfriction coefficient is 0.003, moment of inertia of load is 4.4x10-3, viscousfriction coefficient of load is 4x10-2.

1. Drive the overall transfer function of the system i.e. ΩL(s)/ Ef(s)

2. Determine the gear ratio such that the rotational speed of the load isreduced to 500 rpm.

Example-5

ifTm

Rf

Lf

JmωmBm

Ra La

eaef

JL

N1

N2

BL

L

+

kp

-

JL

_

ia

eb

RaLa

+

Tr c

ea

_

+

e

_

+

N1

N2

BL

θ

if = Constant

JM

BM

Example-5

Numerical Values for System constants

r = angular displacement of the reference input shaftc = angular displacement of the output shaftθ = angular displacement of the motor shaftK1 = gain of the potentiometer shaft = 24/πKp = amplifier gain = 10ea = armature voltageeb = back emfRa = armature winding resistance = 0.2 ΩLa = armature winding inductance = negligibleia = armature winding currentKb = back emf constant = 5.5x10-2 volt-sec/radK = motor torque constant = 6x10-5 N-m/ampereJm = moment of inertia of the motor = 1x10-5 kg-m2

Bm=viscous-friction coefficients of the motor = negligibleJL = moment of inertia of the load = 4.4x10-3 kgm2

BL = viscous friction coefficient of the load = 4x10-2 N-m/rad/secn= gear ratio = N1/N2 = 1/10

e(t)=K1[ r(t) - c(t) ]or

E(S)=K1 [ R(S) - C(S) ]

Ea(s)=Kp E(S)

Transfer function of the armature controlled D.C motor Is given by

(1)

(2)

θ(S)

Ea(S)=

Km

S(TmS+1)

System Equations

System Equations (contd…..)

Where

And

Also

Km =K

RaBeq+KKb

Tm =RaJeq

RaBeq+KKb

Jeq=Jm+(N1/N2)2JL

Beq=Bm+(N1/N2)2BL

END OF LECTURES-7

To download this lecture visit

http://imtiazhussainkalwar.weebly.com/

25