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Automatic Control Systems
Lecture- DC Motor Modelling and Control
1
Emam FathyDepartment of Electrical and Control Engineering
email: [email protected]://www.aast.edu/cv.php?disp_unit=346&ser=68525
2
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.”
Armature Controlled D.C Motor
u
iaT
Ra La
J
B
eb
System constants
ea = armature voltageeb = back emfRa = armature winding resistanceLa = armature winding inductanceia = armature winding currentKb = back emf constantKt = motor torque constantJm = moment of inertia of the motorBm=viscous-friction coefficients of the motorJL = moment of inertia of the loadBL = viscous friction coefficient of the loadN1/N2 = gear ratio
oltageback-emf v e,edt
diLiRu bb
a
aaa
Mechanical Subsystem
BωωJTmotor
Input: voltage uOutput: Angular velocity
Electrical Subsystem
Armature Controlled D.C Motor
uia
T
Ra La
J
B
eb
Torque-Current:
Voltage-Speed:
atmotor iKT
The previous equations result in the following mathematical model:
Power Transformation:
ωKe bb
0at
baa
a
a
iKBωJ
uωKiRdt
diL
Kt: torque constant,
Kb: velocity constant For an ideal motor bt KK
Armature Controlled D.C Motor
uia
T
Ra La
J
B
eb
Taking Laplace transform of the system’s differentialequations with zero 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
Armature Controlled D.C Motor
Reduced Order Model
Assuming small inductance, La 0
abt
at
RKKBJs
RK
U(s)
Ω(s)
Armature Controlled D.C Motor
If output of the D.C motor is angular position θ then:
abt
at
RKKBJss
RK
U(s)
(s)
Which yields the following transfer function
Armature Controlled D.C Motor
)()( sssordt
d
uia
T
Ra La
J
θ
B
eb
Example
• Assume the following values for the physical parameters.
– moment of inertia of the rotor (J) = 0.01 kg.m^2/s^2
– damping ratio of the mechanical system (B) = 0.1 Nms
– electromotive force constant (K=Kb=Kt) = 0.01 Nm/Amp
– electric resistance (Ra) = 1 ohm
– electric inductance (La) = 0.5 H
– input (U): Source Voltage
– output (theta): position of shaft
– The rotor and shaft are assumed to be rigid 10
Design requirements
• The uncompensated motor can rotate at 0.1rad/sec with an input voltage of 1.
• For the unit step input , then the motor speedoutput should have:
– Settling time less than 2 seconds
–Overshoot less than 5%
– Steady-state error less than 1% 11
Uncompensated System
12
Uncompensated System
13
From the plot, when 1 volt is applied to the system; the
motor can achieve:
maximum speed of 0.1 rad/sec, ten times smaller
than our desired speed.
it takes the motor 3
seconds to reach its
steady-state speed; this
does not satisfy the 2
seconds settling time
criterion.
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
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
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 2
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 2
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