1. Chapter 2 Modeling in the frequency domain
Youngjoon, Han

2. 2.1 Introduction

3. Definition and examples
Laplace Transform
Definition and examples
Unit Step Function u(t)

4. Laplace transform thorems-Properties

5. Laplace transform Pairs

6. Test
During analyzing a circuit, input as a sinusoidal function must be Laplace transformed. It is to say that f(t) = e-atcoswtu(t). What is F(s)?
F(s)=2/s2 , what is f(t)?

7. Partial Fraction

8. Partial Fraction

9. Partial fraction; repeated factor

10. Partial fraction; repeated factor

11. Roots of the Denominator of F(s) are Complex or Imaginary

12. The Transfer Function
C(s)=G(s)R(s)
Output=Transfer Function * Input

13. Example 2.4
dc(t)/dt+2c(t)=r(t)
What is transfer function or gain?
sC(s)+2C(s)=R(s) for zero initial condition
(s+2)C(s)=R(s)
G(s)=C(s)/R(s)=1/(s+2)
Example 2.5 If r(t)=1, R(s)=1/s
C(s)=1/[s(s+2)]=A/s+B/(s+2)

14. Solution to the example 2.5
1/[s(s+2)]=A/s+B/(s+2)
A=1/2, B=-1/2
C(t)=½-½e-2t

15. Electric Network Transfer Function

16. The Analogy of Physical variable
Type
Rate
Effort
Electrical
I (current)
Voltage V
Mech. (Trans)
V (velocity)
Force f
Mech. (Rotational)
w(ang. Vel.)
Torque T
Fluid
q (flow rate)
Pressure p
Thermal
Q (heat flow rate)
Temp T

17. Analogy Quantity= (rate) dt Type Rate Quantity Electrical I (current)
Charge Q
Mech. (Trans)
V (velocity)
Displ x
Mech. (Rotational)
w(ang. Vel.)
Angular disp q
Fluid
q (flow rate)
Mass m
Thermal
Q (heat flow rate)
Heat energy Q
Quantity= (rate) dt

18. Transfer Function for RLC Circuit (1)

19. Transfer Function for RLC Circuit(1)

20. Transfer Function for RLC Circuit (1)

21. Transfer Function for RLC Circuit (2)
(Ls+R+1/Cs)I(s)=V(s)
Vc(s)=(1/Cs)I(s)=V(s)/[LCs2+RCS+1]
G(s)=Vc(S)/V(s)=(1/LC)/[s2+(R/L)S+1/LC]

22. Complex Circuits Using Mesh Analysis
Replace passive element values with their impedances
Laplace transform
Assume current in each mesh
Kirchhoff’s voltage law around each mesh
Solve the simultaneous equations
Transfer function

23. Mesh Analysis method Mesh 1 Mesh 2 Impedance Laplace form
Assume currents

24. Mesh Analysis method Write equations around the meshes
Sum of impedance around mesh 1
Sum of applied voltages around the mesh
Sum of impedance common to two meshes
Sum of impedance around mesh 2

25. Mesh Analysis method
Determinant

26. Mesh Analysis method

27. Complex Circuits Nodal Analysis
Number of nodes equal to the simultaneous equations
Use Kirchhoff current law
Sum currents flowing from each node
Conductance=1/resistance
Admittance Y(s)
Y(s)=1/impedance=1/Z(s)
Y(s)=I(s)/V(s)

28. Nodal Analysis Method i1 + i2 +i3=0 i3 + i4 =0 i3 i1 i2 i4
Kirchhoff current law at these two nodes i2 i4
i1 + i2 +i3=0
i3 + i4 =0

29. Nodal Analysis Method
Kirchhoff current law
conductance

30. Nodal Analysis Method

31. Nodal Analysis Method
Real part of admittance is conductance, imaginary part is susceptance; 1/R =G has only real part

32. Operational Amplifier
High input impedancedoesn’t load, Zi= ∞ (ideal)
The Currents into both terminals are zero
The voltage across the input terminals is zero
Low output impedance, Z0=0 (ideal)
High constant gain amplification A= ∞

33. Operational Amplifier and Inverting OP-Amp

34. Non-inverting OP-Amp
Noninverting Op-amp

35. Mechanical System Variables for translation movement

36. Mass, spring, and damper system

37. Free-body diagram of mass, spring, and damper system
impedances

38. Transfer Function of mass, spring, and damper system

39. Impedance for mechanical components
 [ Sum of impedance] X(s) = [ Sum of applied forces]

40. Two-degrees-of-freedom translational mechanical system

41. Two-degrees-of-freedom translational mechanical system

42. Two-degrees-of-freedom translational mechanical system

43. Two-degrees-of-freedom translational mechanical system

44. Three-degrees-of-freedom translational mechanical system

45. Three-degrees-of-freedom translational mechanical system

46. Skill-Assessment Exercise 2.8

47. Mechanical System variables for rotational movement

48. Transfer function – two equation of motion

49. Transfer function – two equation of motion

50. Transfer function – two equation of motion

51. Transfer function – two equation of motion

52. Three-degrees-of-freedom rotational system

54. Skill-Assessment Exercise 2.9

55. Transfer Function For Gear System
F, V must be compatible
Energy=T1 Ө1= T2 Ө2
T2/T1=Ө1/ Ө 2= r2/r1= N2/N1

56. Transfer functions for angular displacement and torque

57. Rotational system driven by gears
Mechanical Impedance of destination axis = ([#of destination teeth]/ [# of source teeth])2 * Mechanical Impedance of source axis
(Js2+Ds+K) (N1/N2)q1=(N2/N1)T1
[J (N1/N2)2 s2+D (N1/N2)2 s+K (N1/N2)2 ] q1=T1
T2=(N2/N1)T1
 q2=(N1/N2)q1

58. Example 2.21

59. Gear train

60. Transfer function –Gear with loss
Target gear
Source gear

61. Armature controlled DC servomotor
Amature
F=BlI

62. Armature controlled DC servomotor
Stationary permanent magnets
Voltage e=Blv; v is velocity
Armature, rotor

63. Electromechanical System
Voltage is proportional to velocity
Electromotive force (emf) vb(t)
vb(t)=Kbdqm(t)/dt
Kb; back emf constant
dqm(t)/dt; wm(t) angular velocity
Vb(s)=Kbsqm(s)
Around the armature loop
RaIa(s)+ LasIa(s)+ Vb(s)= Ea(s)

64. Typical equivalent mechanical loading on a motor
Tm(s)= (Jms2+ Dms) qm(s)

65. Electromechanical System
The torque is proportional to the armature current
Tm(s)=KtIa(s)  Ia(s)=Tm(s)/Kt
Applying Ia(s)=Tm(s)/Kt to RaIa(s)+ LasIa(s)+ Kbsqm(s)= Ea(s)
(Ra+Las) Tm(s)/Kt+ Kbsqm(s)= Ea(s)
Applying Tm(s)= (Jms2+ Dms) qm(s)
(Ra+Las) (Jms2+ Dms) qm(s) /Kt+ Kbsqm(s) =Ea(s)

66. Electromechanical System
For small La, (La<<R)
[(Ra /Kt)(Jms+Dm) +Kb]sqm(s) = Ea(s)
G(s)= qm(s)/Ea(s)=K/[s(s+a)]
= {Kt /(Ra Jm)}/[s{s+(Dm+ KtKb /Ra )/ Jm}]

67. DC motor driving a rotational mechanical load
Jm=Ja+ JL(N1/N2)2
Dm=Da+ DL(N1/N2)2

68. Electromechanical System
From the previous equation, put La=0
(Ra+Las) Tm(s)/Kt+ Kbsqm(s)= Ea(s)
RaTm(t)/Kt+ Kbwm(t)= ea(s)
Tm(t)=-(Kt Kb /Ra) wm+ (Kt/Ra)ea(t)
Tm(t)
Torque-speed curve wm

69. Torque-speed curves with an armature voltage, ea, as a parameter
(Kt/Ra)ea(s)
Kt/Ra=Tstall/ea
Kb=ea(s)/wno-load
ea(s)/ Kb
Tm(t)=-(Kt Kb /Ra) wm+ (Kt/Ra)ea(s)

70. DC motor and load

71. DC motor and load
From the graph
Gear ratio

72. Nonlinearities A linear System possesses two properties:
superposition and homogeneity

73. Nonlinearities

74. Linearization about a point A
f(x) – f(x0) ≈ ma(x-x0)
 δf(x) ≈ maδx
Equilibrium point A

75. Example 2.26 Linearization for f(x)=5cosx at x= p/2
Using Taylor series expansion
Linearization for f(x)=5cosx at x= p/2
f(x) – f(p/2 ) = df(x)/dx|x=p/2 (x - p/2 )
 f(x)=-5(x-p/2)

76. Example 2.27
X”+2X+cos X=0 at X=p/4
Solution; X= dx+ p/4
cos(dx+ p/4)=cos(p/4)-sin(p/4) dx+ …
dx” +2 dx+[cos(p/4)-sin(p/4) dx]=0
dx” +2 dx’-20.5/2 dx= -20.5/2

77. Example 2.28(Nonlinear electrical network)
Ldi/dt+10 ln(ir/2)-20=v(t)
ir=2e0.1vr
vr=10 ln(ir/2)

78. Example 2.28(Nonlinear electrical network)
Taylor series
i0=14.78, L=1

79. Antenna Azimuth Position Control System- Schematic

80. Case Studies – Antenna Control: Transfer Function
Subsystems of the antenna azimuth position control System

81. Case Studies – Antenna Control: Transfer Function
Input Potentiometer; Output Potentiometer
Preamplifier
Power Amplifier

82. Case Studies – Antenna Control: Transfer Function
Motor and Load

83. Case Studies – Antenna Control: Transfer Function