1. Feedback Control Systems (FCS) Dr. Imtiaz Hussain email: imtiaz.hussain@faculty.muet.edu.pkimtiaz.hussain@faculty.muet.edu.pk URL :http://imtiazhussainkalwar.weebly.com/ Lecture-10-11 Mathematical Modelling of Real World Systems 1

2. Modelling of Mechanical Systems 2 Automatic cruise control The purpose of the cruise control system is to maintain a constant vehicle speed despite external disturbances, such as changes in wind or road grade. 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. A throttle is the mechanism by which the flow of a fluid is managed by obstruction.

3. Modelling of Mechanical Systems 3 The transfer function of the systems would be

4. Modelling of Mechanical Systems 4 Consider a simple pendulum shown below.

5. 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 5

6. Introduction Micro-electromechanical (MEMS) systems is the technology of very small devices. It merges at the nano-scale into nanoelectromechanical systems(NEMS) and nanotechnology. 6

7. Example-1: Potentiometer 7

8. 8 The resistance between the wiper (slider) and "A" is labeled R 1, the resistance between the wiper and "B" is labeled R 2. The total resistance between "A" and "B" is constant, R 1 +R 2 =R tot. 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 R 1 =0 and R 2 =R tot. If the wiper is in the extreme clockwise position such that it is touching "B" we will call this θ=θ max ; in this position R 1 =R tot and R 2 =0.

9. Example-1: Potentiometer 9 R 1 and R 2 vary linearly with θ between the two extremes:

10. Example-1: Potentiometer 10 Potentiometer can be used to sense angular position, consider the circuit below. Using the voltage divider principle we can write:

11. Example-2: Loud Speaker 11 A voltage is typically applied across the terminals of the loudspeaker and the "cone" moves in and out causing pressure waves perceived as sound.

12. Example-2: Loud Speaker 12 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).

13. Example-2: Loud Speaker 13 Mechanical Free body Diagram Electrical Schematic

14. Example-2: Loud Speaker 14 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)

15. Example-2: Loud Speaker 15 Back EMF is given by To find the transfer function X(S)/E in (s) we have to eliminate current i. (2)

16. Example-2: Loud Speaker 16 Taking Laplace transform of equations (1) and (2) considering initial conditions to zero. Re-arranging equation (3) as (2) (1) (3) (4)

17. Example-2: Loud Speaker 17 Put I(s) in equation (4) After simplification the transfer function is calculated as

18. Example-3: Capacitor Microphone 18 The system consists of a capacitor realized by two plates, one is fixed and the other is movable but attached to a spring.

19. Example-4: Electromagnetic Relay 19

20. Example-5: Push Button 20

21. 21 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.”

22. 22 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.

23. 23 Example-6: D.C Drives Motor Characteristics Power is defined as the product of torque and angular velocity.

24. Mechanical Subsystem Input: voltage u Output: Angular velocity  Elecrical Subsystem (loop method): Example-6.1: Armature Controlled D.C Motor u iaia T RaRa LaLa J  B ebeb V f =constant

25. Torque-Current: Voltage-Speed: Combing previous equations results in the following mathematical model: Power Transformation: where K t : torque constant, K b : velocity constant For an ideal motor Example-6.1: Armature Controlled D.C Motor u iaia T RaRa LaLa J  B ebeb V f =constant

26. Taking Laplace transform of the system’s differential equations with zero initial conditions gives: Eliminating I a yields the input-output transfer function Example-6.1: Armature Controlled D.C Motor

27. Reduced Order Model Assuming small inductance, L a  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-6.1: Armature Controlled D.C Motor

28. If output of the D.C motor is angular position θ then we know Which yields following transfer function Example-6.1: Armature Controlled D.C Motor u iaia T RaRa LaLa J θ B ebeb V f =constant

29. Applying KVL at field circuit Example-6.2: Field Controlled D.C Motor ifif TmTm RfRf LfLf J ω B RaRa LaLa eaea e f Mechanical Subsystem

30. Torque-Current : Combing previous equations and taking Laplace transform (considering initial conditions to zero) results in the following mathematical model: Power Transformation: where K f : torque constant Example-6.2: Field Controlled D.C Motor

31. If angular position θ is output of the motor Eliminating I f (S) yields Example-6.2: Field Controlled D.C Motor ifif TmTm RfRf LfLf J θ B RaRa LaLa eaea e f

32. 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. 1.Drive the overall transfer function of the system i.e. Ω L (s)/ E a (s) 2.Determine the gear ratio such that the rotational speed of the load is reduced to half and torque is doubled. Example-6.3 15 v iaia T RaRa LaLa JmJm  BmBm ebeb V f =constant JLJL NN1NN1 NN2NN2 BLBLBLBL LL eaea

33. System constants e a = armature voltage e b = back emf R a = armature winding resistance = 0.2 Ω L a = armature winding inductance = negligible i a = armature winding current K b = back emf constant = 5.5x10 -2 volt-sec/rad K t = motor torque constant = 6x10 -5 N-m/ampere J m = moment of inertia of the motor = 1x10 -5 kg-m 2 B m =viscous-friction coefficients of the motor = negligible J L = moment of inertia of the load = 4.4x10 -3 kgm 2 B L = viscous friction coefficient of the load = 4x10 -2 N-m/rad/sec gear ratio = N 1 /N 2

34. Since armature inductance is negligible therefore reduced order transfer function of the motor is used. Example-6.3 15 v iaia T RaRa LaLa JmJm  BmBm ebeb V f =constant JLJL NN1NN1 NN2NN2 BLBLBLBL LL eaea

35. 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. 1.Drive the overall transfer function of the system i.e. Ω L (s)/ E f (s) 2.Determine the gear ratio such that the rotational speed of the load is reduced to 500 rpm. Example-6.4 ifif TmTm RfRf LfLf JmJm ωmωm BmBm RaRa LaLa eaea e f JLJL NN1NN1 NN2NN2 BLBLBLBL LL

36. + k p - JLJL _ iaiaiaia eebeeb RaRaRaRa LaLa + T rc eaeaeaea _ + e _ + NN1NN1 NN2NN2 BLBLBLBL θ i f = Constant Position Servomechanism JMBMJMBM

37. 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 K 1 = gain of the potentiometer shaft = 24/ π K p = amplifier gain = 10 e a = armature voltage e b = back emf R a = armature winding resistance = 0.2 Ω L a = armature winding inductance = negligible i a = armature winding current K b = back emf constant = 5.5x10 -2 volt-sec/rad K = motor torque constant = 6x10 -5 N-m/ampere J m = moment of inertia of the motor = 1x10 -5 kg-m 2 B m =viscous-friction coefficients of the motor = negligible J L = moment of inertia of the load = 4.4x10 -3 kgm 2 B L = viscous friction coefficient of the load = 4x10 -2 N-m/rad/sec n= gear ratio = N 1 /N 2 = 1/10

38. e(t)=K 1 [ r(t) - c(t) ] or E(S)=K 1 [ R(S) - C(S) ] E a (s)=K p E(S) Transfer function of the armature controlled D.C motor Is given by (1) (2) θ (S) E a (S) = KmKm S(T m S+1) System Equations

39. System Equations (contd…..) Where And Also KmKm = K R a B eq +KK b TmTm = R a J eq R a B eq +KK b J eq =J m +(N 1 /N 2 ) 2 J L B eq =B m +(N 1 /N 2 ) 2 B L

40. AB _ + Servo Amplifier K a M Output gear Solar axis Vehicle axis D.C Motor Tachometer Sun Ray iaiaiaia iibiib RfRf R R eoeoeoeo eseseses + + _ _ eaea + _ + _ eeteet  1/n oo mm mm L o b

41. Input & Output variables Center of output gear Vehicle axis Solar axis  oooo rrrr

42. Error Discriminator  2I -2I i a (t) - i b (t) -C/L – W/2L -C/L + W/2L - W/2L W/2L W/2L C/L-W/2LC/L+W/2L i a = W/2 + L tan  i b = W/2 - L tan  (i a -i b )/  = 2L (B) (A)

43. Operational amplifier & Servo Amplifier The out of op-amp is e o =-R F (i a -i b ) Transfer function is given by: e o /(i a -i b )= -R F Similarly output of servo amplifier is e s = -K e a Transfer function is given by: e s / e a = -K

44. D.C Motor & Output Gear Transfer function of the D.C motor is given by: θ m / e a = K i /(S 2 R a J + S R a B + K i K b ) Output Gear θ o = 1/n θ m θ o /θ m = 1/n

45. Tachometer e t = S k t θ m e t /θ m = S k t

46. END OF LECTURES-10 To download this lecture visit http://imtiazhussainkalwar.weebly.com/ 46