1. Chapter 3: Frequency Response of AC Circuit Sem2 2015/2016
EKT 119 ELECTRIC CIRCUIT II
Chapter 3: Frequency Response of AC Circuit Sem2 2015/2016 1

2. Overview
This chapter will introduce the idea of the transfer function: a means of describing the relationship between the input and output of a circuit.
Bode plots and their utility in describing the frequency response of a circuit will also be introduced. 2 2

3. Frequency Response
Frequency response is the variation in a circuit’s behavior with change in signal frequency.
This is significant for applications involving filters.
Filters play critical roles in blocking or passing specific frequencies or ranges of frequencies.
Without them, it would be impossible to have multiple channels of data in radio communications. 3 3

4. Transfer Function
One useful way to analyze the frequency response of a circuit is the concept of the transfer function H(ω).
H(ω) is the frequency dependent ratio of a forced function Y(ω) to the forcing function X(ω). 4 4

5. TRANSFER FUNCTION (TF)
Frequency response can be obtained by using transfer function. 5

6. DEFINITION: Transfer function, H() is a ratio between output and input.
Being a complex quantity, H() has a magnitude |H()| and a phase φ. 6

7. TRANSFER FUNCTION
Output
signal
Input
signal 7

8. 4 condition of TF:
Because there is no
unit, they are called
GAIN 8

9. KUTUB DAN SIFAR (POLES AND ZEROS)
Transfer function can be written in fraction
The Numerator and Denominator can be existed as a polynomial 9

10. The roots of numerator also known as ZEROS. Zeros exist when N()=0
The roots of denominator also known as POLES.
Poles exist when D()=0 10

11. Poles and Zeros The symbol for pole is x The symbol for zero is o
Complex s-plane is used to plot poles and zeros. 11

12. Complex S-plane 12

13. EXAMPLE 1
For the circuit shown below, calculate the gain I0(ω)/Ii(ω) and its poles and zeros. 13

14. Solution
By current division,
The zeros are at
The poles are at 14

15. EXAMPLE 2
For the RC circuit shown below, obtain the transfer function Vo/Vs and its frequency response.
(a) Time domain RC circuit 15

16. Solution
(b) frequency domain RC circuit 16

17. 17

18. Transfer Function Standard Form
The transfer function can be written in terms of factors with real and imaginary parts. For example: 18 18

19. STANDARD FORM POLES/ZEROS
quadratic zero
Poles/zeros at the origin
real zero
real pole
quadratic pole 19

20. LOCATION OF POLES/ZEROS
Zeros/poles at the origin: Zeros/poles that are located at 0
Real Zeros/poles: Zeros/poles that are located at real axis
(-1,-2,1,2,10,etc)
Quadratic Zeros/poles:Zeros/poles that are not located at imaginary or real axis (- 1+j2, 2+j5, 3-j3, etc) 20

21. EXAMPLE 21

22. ZEROS
Let numerator, N()=0 22

23. POLE
Let denominator, D()=0 23

24. FREQUENCY RESPONSE PLOT
USING SEMILOG GRAPH 24

25. MAGNITUDE PLOT AND PHASE PLOT
phase angle plot 25

26. HOW TO DO MAGNITUDE AND PHASE PLOT
Transform the time domain circuit (t) into freq. domain circuit (ω)
Determine the TF, H(ω)
Plot the magnitude of that tf, H(ω)against ω.
Plot the phase of that tf, (º) against ω. 26

27. BODE PLOTS
Bode plots are semilog plots of magnitude (in decibels) and phase (in degrees) of a transfer function versus frequency.
Before we begin to construct Bode plots, we should take care of two important issues: the use of logarithms and decibels in expressing gain. 27

28. DECIBEL SCALE
Logarithm 28

29. BODE PLOT CHARACTERISTIC FOR POLES AND ZEROS29

30. Logarithm of tf: 30

31. GAIN
Gain is measured in bels 31

32. Decibel (dB) 32

33. TRANSFER FUNCTION 33

34. 34

35. GENERAL EQUATION OF TF
Before draw, make sure the general equation of TF is obtained first: 35

36. EX.
COMPARE 36

37. 37

38. BODE PLOT OF A CONSTANT,K38

39. (1) (GAIN)
constant 39

40. CHARACTERISTICS Phase angle for constant is:
Magnitude for constant is :
Phase angle for constant is: 40

41. BODE PLOT FOR CONSTANT
magnitude plot
phase plot f 41

42. BODE PLOT FOR ZERO AT THE ORIGIN42

43. (2) ZERO AT THE ORIGIN (jω)N43

44. CHARACTERISTIC OF (jω)N
Magnitude:
Straight line with 20dB/dec of slope that has a value of 0 dB at =1
Phase: 44

45. MAGNITUDE PLOT 45

46. PHASE PLOT 46

47. BODE PLOT OF POLE AT THE ORIGIN47

48. (3) POLE AT THE ORIGIN 1/(jω)N @ (jω)-N48

49. CHARACTERISTIC OF (jω)-N
Magnitude:
Straight line with -20dB/dec of slope that has a value of 0 dB at =1
Phase: 49

50. MAGNITUDE PLOT 50

51. PHASE PLOT 51

52. BODE PLOT OF REAL ZERO 52

53. (4) REAL ZERO 53

54. CHARACTERISTIC OF (1+jω/z1)N
Magnitude:
Phase: 54

55. MAGNITUDE PLOT 55

56. PHASE PLOT 56

57. BODE PLOT OF REAL POLE 57

58. (5) REAL POLE 58

59. CHARACTERISTIC OF (1+jω/p1)-N
Magnitude:
Phase: 59

60. MAGNITUDE PLOT 60

61. PHASE PLOT 61

62. BODE PLOT OF QUADRATIC ZERO62

63. CHARACTERISTIC OF (j2+2n+n2)N
Magnitude:
Phase: 63

64. MAGNITUDE PLOT 64

65. PHASE PLOT 65

66. BODE PLOT OF QUADRATIC POLE66

67. (7) QUADRATIC POLE 67

68. CHARATERISTIC OF (j2+2n+n2)-N
Magnitude:
Phase: 68

69. MAGNITUDE PLOT 69

70. PHASE PLOT 70

71. SUMMARY Table 14.3 page 623 71

72. SUMMARY Table 14.3 page 623 72

73. SUMMARY Table 14.3 page 623 73

74. HOW TO DRAW A BODE PLOT
While drawing the bode plot, every factor (i.e zeros/poles) were drawed separately on the semilog graph.
Finally, all of the factor are combined to form the answer. 74

75. EX.1
Draw the Bode plot for the given tf below: 75

76. SOLUTION
General equation: 76

77. Magnitude of tf: 77

78. Phase of tf:
Zero at the
origin
Pole at 2
Pole at 10 78

79. MAGNITUDE PLOT GUIDANCE
z=0
20dB/dec
20dB/dec
20dB/dec
20dB/dec
p=2
0dB/dec
-20dB/dec
-20dB/dec
-20dB/dec
p=10
0dB/dec
0dB/dec
-20dB/dec
-20dB/dec
Resultant
=20dB/dec
=0dB/dec
=-20dB/dec
=-20dB/dec 79

80. Magnitude plot z=0 Constant p= -10 p= -2 20 0.2 2 10 20 100 200 1 -2080

81. PHASE PLOT GUIDANCE
Add all of the lines that having a slope only 81

82. Phase plot
z=0
90o 20
100 2 10
0.2 1
200
p=-10
p= -2
-90o 82

83. EX.2
Draw the Bode plot for the given tf below: 83

84. SOLUTION
General equation: 84

85. Magnitude of tf : 85

86. Phase of tf:
p1=0
z1 = -10
p2= -2 86

87. MAGNITUDE PLOT GUIDANCE
-20dB/dec
-20dB/dec
-20dB/dec
-20dB/dec
p=2
0dB/dec
-20dB/dec
-20dB/dec
-20dB/dec
z=10
0dB/dec
0dB/dec
20dB/dec
20dB/dec
Resultant
-20dB/dec
-40dB/dec
-20dB/dec
- 20dB/dec 87

88. Magnitude plot HdB z=-10 constant p=-2 p=0 40 34 20 14 0.1 1 2 10 20
100 
-20
p=-2
-40
p=0 88

89. Phase plot Guidance
Add all the lines that having a slope only 89

90. Phase plot z=-10 p= -2 p=0 90o 0.2 1 2 10 20 100 200 -45o -90o -135o

91. EXAMPLE 3
Draw the Bode plot for the given tf below: 91

92. SOLUTION
Standard equation:
Replace s=jω and divide it with 100; 92

93. Magnitude of tf: 93

94. Phase of tf:
z1=0
ωn= 10 94

95. Magnitude plot ω z=0 ωn =10 constant HdB 40 20 0.1 1 10 100 -20 -40
-60 95

96. Phase plot
z=0 90 ω
0.1 1 10
100
-90
ωn =10
-180 96

97. EXAMPLE 4
Determine the TF(H): 97

98. ANSWER 98