1. Advanced Control Systems (ACS)
Lecture-5
Mathematical Modeling of Electrical & Electronic and Electromechanical Systems
Dr. Imtiaz Hussain
URL :

2. Outline of this Lecture
Part-I: Electrical Systems
Basic Elements of Electrical Systems
Equations for Basic Elements
Examples
Part-II: Electronic Systems
Operational Amplifiers
Inverting vs Non-inverting
Part-III: Electrmechanical Systems

3. Basic Elements of Electrical Systems
The time domain expression relating voltage and current for the resistor is given by Ohm’s law i-e
The Laplace transform of the above equation is

4. Basic Elements of Electrical Systems
The time domain expression relating voltage and current for the Capacitor is given as:
The Laplace transform of the above equation (assuming there is no charge stored in the capacitor) is

5. Basic Elements of Electrical Systems
The time domain expression relating voltage and current for the inductor is given as:
The Laplace transform of the above equation (assuming there is no energy stored in inductor) is

6. V-I and I-V relations Component Symbol V-I Relation I-V Relation
Resistor
Capacitor
Inductor

7. Example#1
The two-port network shown in the following figure has vi(t) as the input voltage and vo(t) as the output voltage. Find the transfer function Vo(s)/Vi(s) of the network. C
i(t)
vi( t)
vo(t)

8. Example#1
Taking Laplace transform of both equations, considering initial conditions to zero.
Re-arrange both equations as:

9. Example#1
Substitute I(s) in equation on left

10. Example#2
Design an Electrical system that would place a pole at -3 if added to another system.
System has one pole at
Therefore, C
i(t)
vi( t)
v2(t)

11. Example#3
Find the transfer function G(S) of the following two port network.
i(t)
vi(t)
vo(t) L C

12. Example#3
Simplify network by replacing multiple components with their equivalent transform impedance.
I(s)
Vi(s)
Vo(s) L C Z

13. Transform Impedance (Resistor)
iR(t)
vR(t) + -
IR(S)
VR(S)
ZR = R
Transformation

14. Transform Impedance (Inductor)
iL(t)
vL(t) + -
IL(S)
VL(S)
LiL(0)
ZL=LS

15. Transform Impedance (Capacitor)
ic(t)
vc(t) + -
Ic(S)
Vc(S)
ZC(S)=1/CS

16. Equivalent Transform Impedance (Series)
Consider following arrangement, find out equivalent transform impedance. L C R

17. Equivalent Transform Impedance (Parallel)

18. Equivalent Transform Impedance
Find out equivalent transform impedance of following arrangement. L2 R2 R1

19. Back to Example#3
I(s)
Vi(s)
Vo(s) L C Z

20. Example#3
I(s)
Vi(s)
Vo(s) L C Z

21. Operational Amplifiers

22. Example#4
Find out the transfer function of the following circuit.

23. Example#5
Find out the transfer function of the following circuit and draw the pole zero map.
10kΩ
100kΩ

24. 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

25. Example-6: 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.

26. Example-6: Loud Speaker

27. 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).

28. Example-2: Loud Speaker
Mechanical Free body Diagram Electrical Schematic

29. Example-6: Loud Speaker
Force on a current carrying conductor in a magnetic field is given by
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).
(1)

30. Example-6: Loud Speaker
Back EMF is given by
To find the transfer function X(S)/Ein(s) we have to eliminate current i.
(2)

31. Example-6: Loud Speaker
(2)
(1)
Taking Laplace transform of equations (1) and (2) considering initial conditions to zero.
Re-arranging equation (3) as
(3)
(4)

32. Example-6: Loud Speaker
Put I(s) in equation (4)
After simplification the transfer function is calculated as

33. Example-7: Capacitor Microphone (Home Work)
The system consists of a capacitor realized by two plates, one is fixed and the other is movable but attached to a spring.

34. 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.”

35. 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.

36. D.C Drives Motor Characteristics
Power is defined as the product of torque and angular velocity. 

37. Example-8: Armature Controlled D.C Motoria T Ra La J  B eb
Vf=constant
Input: voltage u
Output: Angular velocity 
Elecrical Subsystem (loop method):
Mechanical Subsystem

38. Example-8: Armature Controlled D.C Motoria T Ra La J  B eb
Vf=constant
Power Transformation:
Torque-Current:
Voltage-Speed:
where Kt: torque constant, Kb: velocity constant For an ideal motor
Combing previous equations results in the following mathematical model:

39. Example-8: Armature Controlled D.C Motor
Taking Laplace transform of the system’s differential equations with zero initial conditions gives:
Eliminating Ia yields the input-output transfer function

40. Example-8: Armature Controlled D.C Motor
Reduced Order Model
Assuming small inductance, La 0
which is equivalent to  B
The D.C. motor provides an input torque and an additional damping effect known as back-emf damping

41. Example-8: Armature Controlled D.C Motor
If output of the D.C motor is angular position θ then we know u ia T Ra La J θ B eb
Vf=constant
Which yields following transfer function

42. Example-9: Field Controlled D.C Motorif Tm Rf Lf J ω B Ra La ea ef
Applying KVL at field circuit
Mechanical Subsystem

43. Example-9: Field Controlled D.C Motor
Power Transformation:
Torque-Current:
where Kf: torque constant
Combing previous equations and taking Laplace transform (considering initial conditions to zero) results in the following mathematical model:

44. Example-9: Field Controlled D.C Motor
Eliminating If(S) yields
If angular position θ is output of the motor if Tm Rf Lf J θ B Ra La ea ef

45. Example-10
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.
Drive the overall transfer function of the system i.e. ΩL(s)/ Ea(s)
Determine the gear ratio such that the rotational speed of the load is reduced to half and torque is doubled.
15 v ia T Ra La Jm  Bm eb
Vf=constant JL N1 N2 BL L ea

46. 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

47. Example-10
Since armature inductance is negligible therefore reduced order transfer function of the motor is used.
15 v ia T Ra La Jm  Bm eb
Vf=constant JL N1 N2 BL ea L

48. Example-11
A field controlled D.C motor runs at 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.
Drive the overall transfer function of the system i.e. ΩL(s)/ Ef(s)
Determine the gear ratio such that the rotational speed of the load is reduced to 500 rpm. if Tm Rf Lf Jm ωm Bm Ra La ea ef JL N1 N2 BL L

49. Example-11 Position ServomechanismRa La N1 + kp -
JM BM + + + ia BL T JL e ea eb θ r c _ _ _ N2
if = Constant

50. 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

51. System Equations θ(S) Km = S(TmS+1) 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)

52. System Equations (contd…..)
Where
And
Also Km = K
RaBeq+KKb Tm
RaJeq
Jeq=Jm+(N1/N2)2JL
Beq=Bm+(N1/N2)2BL

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