Download Lecture-4: Modeling Real World Systems - Dr. Imtiaz Hussain

Modeling & Simulation of Dynamic
Systems
Lecture-4
Mathematical Modeling of Real World Systems
Dr. Imtiaz Hussain
email: [email protected]
URL :http://imtiazhussainkalwar.weebly.com/
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Modeling of Mechanical Systems
• Automatic cruise control
•
A throttle is the
mechanism by which
the flow a
of constant
a fluid is vehicle
The purpose of the cruise control system is to maintain
speed despite external disturbances, such as changes inmanaged
wind or road grade.
by obstruction.
• This is accomplished by measuring the vehicle speed, comparing it to the
desired speed, and automatically adjusting the throttle.
• The resistive forces, bv, due to rolling resistance and wind drag act in the
direction opposite to the vehicle's motion.
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Modeling of Mechanical Systems
u  mv  bv
• The transfer function of the systems would be
V (s)
1

U ( s ) ms  b
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Electromechanical Systems
• Electromechanics combines electrical and mechanical
processes.
• Devices which carry out electrical operations by using
moving parts are known as electromechanical.
–
–
–
–
–
Relays
Solenoids
Electric Motors
Electric Generators
Switches and e.t.c
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Example-1: Potentiometer
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Example-1: Potentiometer
• The resistance between the wiper
(slider) and "A" is labeled R1, the
resistance between the wiper and "B" is
labeled R2.
• The total resistance between "A" and "B"
is constant, R1+R2=Rtot.
• If the potentiometer is turned to the
extreme counterclockwise position such
that the wiper is touching "A" we will call
this θ=0; in this position R1=0 and
R2=Rtot.
• If the wiper is in the extreme clockwise
position such that it is touching "B" we
will call this θ=θmax; in this position
R1=Rtot and R2=0.
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Example-1: Potentiometer
• R1 and R2 vary linearly with θ between the
two extremes:
R1 

 max
Rtot
 max  
R2 
Rtot
 max
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Example-1: Potentiometer
• Potentiometer can be used to sense
angular position, consider the circuit
of figure-1.
Figure-1
• Using the voltage divider principle we
can write:
eout
R1
R1

ein 
ein
R1  R2
Rtot
eout 

 max
ein
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Example-2: Loud Speaker
• A voltage is typically applied across the terminals of the
loudspeaker and the "cone" moves in and out causing pressure
waves perceived as sound.
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Example-2: Loud Speaker
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Example-2: Loud Speaker
• The speaker consists of a fixed magnet
that produces a uniform magnetic field
of strength β.
• The speaker has a cone with mass (M),
that moves in the x direction.
• The cone is modelled with a spring (K)
to return it to its equilibrium position,
and a friction (B).
• Attached to the cone, and within the
magnetic field is a coil of wire or radius
"a." The coil consists of "n" turns and it
moves along with the cone.
• The wire has resistance (R) and
inductance (L).
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Example-2: Loud Speaker
• Mechanical Free body Diagram
f e  Mx  Bx  kx
Electrical Schematic
di
ein  iR  L  em
dt
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Example-2: Loud Speaker
f e  Mx  Bx  kx
di
ein  iR  L  em
dt
• Force on a current carrying conductor in a magnetic field is given by
f e  il
• Where ℓ is the total length of wire in the field.
• It is equal to the circumference of the coil (2·π·a) times the number of
turns (n).
• That is, ℓ=2·π·a·n).
i 2an  Mx  Bx  kx
(1)
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Example-2: Loud Speaker
di
ein  iR  L  em
dt
• Back EMF is given by
em  lx   2anv
di
 2anx  ein  iR  L
dt
(2)
• To find the transfer function X(S)/Ein(s) we have to eliminate current
i.
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Example-2: Loud Speaker
i 2an  Mx  Bx  kx
(1)
di
 2anx  ein  iR  L
dt
(2)
• Taking Laplace transform of equations (1) and (2) considering initial
conditions to zero.
 2anI( s )  Ms2 X ( s )  BsX ( s )  kX( s )
(3)
 2ansX ( s )  Ein ( s )  I ( s )R  LsI ( s )
(4)
• Re-arranging equation (3) as
Ms 2 X ( s )  BsX ( s )  kX ( s )
I (s) 
 2an
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Example-2: Loud Speaker
• Put I(s) in equation (4)
Ms 2 X ( s )  BsX ( s )  kX ( s )
Ms 2 X ( s )  BsX ( s )  kX ( s )
2ansX ( s )  Ein ( s )  R
 Ls
 2an
 2an
• After simplification the transfer function is calculated as
X (s)
2an
 3
Ein ( s ) ( s LM  s 2 ( RM  LB )  s( RB  LK  4 2 a 2 n 2 )  RK )
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Example-3: Capacitor Microphone
• The system consists of a capacitor realized by two plates, one is
fixed and the other is movable but attached to a spring.
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Example-3: Capacitor Microphone
• Equations of Electrical Subsystem
𝑅
𝑖
𝑢
𝑢𝑐
𝐶(𝑥)
𝑢 = 𝑅𝑖 + 𝑢𝑐
1
𝑢𝑐 =
𝐶
𝑖𝑑𝑡
1
𝑢 = 𝑅𝑖 +
𝐶
𝑖𝑑𝑡
where
Therefore,
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Example-3: Capacitor Microphone
• Free body diagram of moveable plate of Mass M
𝐹(𝑡)
M
𝐹𝑘 (𝑡)
𝐹𝑚 (𝑡)
𝐹 𝑡 = 𝐹𝑘 𝑡 + 𝐹𝑚 (𝑡)
𝐹 𝑡 = 𝑘𝑥 + 𝑀𝑥
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Example-3: Capacitor Microphone
• Capacitance equation
𝜀𝐴
𝐶=
𝑥
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Example-4: Electromagnetic Relay
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Example-5: Push Button
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Example-6: D.C Drives
• Speed control can be achieved using
DC drives in a number of ways.
• Variable Voltage can be applied to the
armature terminals of the DC motor .
• Another method is to vary the flux per
pole of the motor.
• The first method involve adjusting the
motor’s armature while the latter
method involves adjusting the motor
field. These methods are referred to as
“armature control” and “field control.”
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Example-6: D.C Drives
• Motor Characteristics
• For every motor, there is a specific Torque/Speed curve and Power curve.
• Torque is inversely proportional to the speed of the output shaft.
• Motor characteristics are frequently
given as two points on this graph:
• The stall torque,, represents the
point on the graph at which the
torque is maximum, but the shaft is
not rotating.
• The no load speed is the maximum
output speed of the motor.
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Example-6: D.C Drives
• Motor Characteristics
• Power is defined as the product of torque and angular velocity.
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Example-6.1: Armature Controlled D.C Motor
Ra
Input: voltage u
Output: Angular velocity 
La
B
u
ia
eb
T
J

Elecrical Subsystem (loop method):
dia
u  Ra ia  La
 eb ,
dt
Mechanical Subsystem
Tmotor  Jω  Bω
where eb  back-emf voltage
Example-6.1: Armature Controlled D.C Motor
Ra
La
Power Transformation:
Torque-Current:
Voltage-Speed:
B
u
Tmotor  K t ia
ia
eb
T
J

eb  K b ω
where Kt: torque constant, Kb: velocity constant For an ideal motor
Kt  Kb
Combing previous equations results in the following mathematical
model:
 dia
 Ra ia  K b ω  u
 La
dt

 Jω
   B-K t ia  0
Example-6.1: Armature Controlled D.C Motor
Taking Laplace transform of the system’s differential equations with
zero initial conditions gives:
La s  Ra I a(s)  K b Ω(s)  U(s)

Js  B Ω(s)-K t I a(s)  0
Eliminating Ia yields the input-output transfer function
Kt
Ω(s)

U(s) La Js 2  JRa  BLa s  BRa  K t K b
Example-6.1: Armature Controlled D.C Motor
Reduced Order Model
Assuming small inductance, La 0

K t Ra 
Ω(s)

U(s) Js  B  K t K b Ra 
which is equivalent to
K t K b Ra
B
K t u Ra

• The D.C. motor provides an input torque and an additional damping
effect known as back-emf damping
Example-6.1: Armature Controlled D.C Motor
If output of the D.C motor is angular position θ then we know
Ra
d

dt
or
La
B
( s )  s ( s )
ia
u
eb
J
T
θ
Which yields following transfer function

K t Ra 

U(s) sJs  B  K t K b
(s)
Ra 
Example-6.2: Field Controlled D.C Motor
Ra
Rf
if
ef
Lf
Tm
J
B ω
Applying KVL at field circuit
ef  if Rf  Lf
Mechanical Subsystem
Tm  Jω  Bω
di f
dt
La
ea
Example-6.2: Field Controlled D.C Motor
Power Transformation:
Torque-Current: Tm  K f i f
where Kf: torque constant
Combing previous equations and taking Laplace transform (considering
initial conditions to zero) results in the following mathematical model:

 E f ( s )  R f I f ( s )  sL f I f ( s )


 Js( s )  B( s )  K f I f ( s )
Example-6.2: Field Controlled D.C Motor
Eliminating If(S) yields
Kf
Ω(s)

E f (s) Js  B ( L f s  R f )
If angular position θ is output of the motor
Ra
Rf
if
ef
Lf
Tm
La
J
B θ
(s)
E f (s)

Kf
sJs  B ( L f s  R f )
ea
Example-6.3
An armature controlled D.C motor runs at 5000 rpm when 15v applied at the
armature circuit. Armature resistance of the motor is 0.2 Ω, armature
inductance is negligible, back emf constant is 5.5x10-2 v sec/rad, motor torque
constant is 6x10-5, moment of inertia of motor 10-5, viscous friction coeffcient
is negligible, moment of inertia of load is 4.4x10-3, viscous friction coeffcient of
load is 4x10-2.
La
Ra
ea 15 v
ia
N1
Bm
eb
T
Jm
BL

JL
N2
L
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 is
reduced to half and torque is doubled.
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
Example-6.3
Since armature inductance is negligible therefore reduced order transfer
function of the motor is used.
Kt
ΩL(s)

U(s)
J eq Ra  Beq La s  Beq Ra  K t K b

La
Ra
ea 15 v

ia
N1
Bm
eb
T
Jm
BL

JL
N2
2
N 
J eq  J m   1  J L
 N2 
L
2
N 
Beq  Bm   1  BL
 N2 
Example-6.4
A field controlled D.C motor runs at 10000 rpm when 15v applied at the field
circuit. 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, viscous
friction coefficient is 0.003, moment of inertia of load is 4.4x10-3, viscous
friction coefficient of load is 4x10-2.
Ra
Rf
ef
if
Lf
La
ea
Tm
Bm ωm
N1
Jm
BL
JL
N2
L
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 is
reduced to 500 rpm.
Position Servomechanism
+
+
e
kp
r
_
La
Ra
+
ea
_
N1
+
ia
JM
BM
T
eb
_
BL
θ
JL
N2
-
if = Constant
c
Numerical Values for System constants
r = angular displacement of the reference input shaft
c = angular displacement of the output shaft
θ = angular displacement of the motor shaft
K1 = gain of the potentiometer shaft = 24/π
Kp = amplifier gain = 10
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
K = 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
n= gear ratio = N1/N2 = 1/10
System Equations
e(t)=K1[ r(t) - c(t) ]
or
E(S)=K1 [ R(S) - C(S) ]
(1)
Ea(s)=Kp E(S)
(2)
Transfer function of the armature controlled D.C motor Is given by
θ(S)
Ea(S)
=
Km
S(TmS+1)
System Equations (contd…..)
Where
K
Km =
RaBeq+KKb
And
Tm
RaJeq
=
RaBeq+KKb
Also
Jeq=Jm+(N1/N2)2JL
Beq=Bm+(N1/N2)2BL
Output gear
ia
o
b
Sun Ray
L
1/n
A
Solar axis

o
Vehicle axis
B
m
Rf
ib
R
R
m
_
Servo
+
+
eo
_
+
es
_
Amplifier
Ka
+
D.C Motor
ea M
_
_
Tachometer et
+
Input & Output variables
Solar axis

r
o
Center of
output gear
Vehicle axis
Error Discriminator
ia(t) - ib(t)
2I
-C/L – W/2L -C/L + W/2L
- W/2L

W/2L
C/L-W/2L
-2I
ia= W/2 + L tan
ib = W/2 - L tan
(ia-ib)/ = 2L
(A)
(B)
C/L+W/2L
Operational amplifier & Servo Amplifier
The out of op-amp is
eo =-RF (ia-ib)
Transfer function is given by:
eo/(ia-ib)= -RF
Similarly output of servo amplifier is
es = -K ea
Transfer function is given by:
es / ea = -K
D.C Motor & Output Gear
Transfer function of the D.C motor is given by:
θm / ea = Ki /(S2 Ra J + S Ra B + Ki Kb)
Output Gear
θo = 1/n θm
θo/θm = 1/n
Tachometer
et = S kt θm
et/θm = S kt
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END OF LECTURES-4
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