Advanced Control Systems (ACS) Lecture-5 Mathematical Modeling of Electrical & Electronic and Electromechanical Systems Dr. Imtiaz Hussain email: imtiaz.hussain@faculty.muet.edu.pk URL :http://imtiazhussainkalwar.weebly.com/

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

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

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

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

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

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)

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

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

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)

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

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

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

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

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

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

Equivalent Transform Impedance (Parallel)

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

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

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

Operational Amplifiers

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

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

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

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.

Example-6: Loud Speaker

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).

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

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)

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)

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)

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

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.

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.”

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.

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

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

Example-8: Armature Controlled D.C Motor ia 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:

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

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

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

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

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:

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

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

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-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

Example-11 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. 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

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

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 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)

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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