EKT 119 ELECTRIC CIRCUIT II. Chapter 3: Frequency Response of AC Circuit Sem2 2015/2016 Dr. Mohd Rashidi Che Beson
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1 EKT 9 ELECTRIC CIRCUIT II Chapter 3: Frequency Response of AC Circuit Sem 05/06 Dr. Mohd Rashidi Che Beson
2 TRANSFER FUNCTION (TF Frequency response can be obtained by using transfer function.
3 DEFINITION: Transfer function, H( is a ratio between output and input. H ( Y X ( ( 3
4 TRANSFER FUNCTION H ( Y X ( ( Output signal Input signal H ( H ( 4
5 4 condition of TF: V H ( voltage gain V I H ( current gain I V ( H ( impedance I( I( H ( admi tan ce V ( o i o i ( ( ( ( Because there is no unit, they are called GAIN 5
6 6 KUTUB DAN SIFAR (POLES AND ZEROS Transfer function is written in fraction The numerator and denominator can be existed as a polynomial ( ( ( X Y H ( ( ( D N H
7 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 7
8 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. 8
9 Complex S-plane 9
10 0 POLES/ZEROS / ( / / ( / ( / / ( ( ( n n k k p z K H real zero real pole quadratic zero quadratic pole Poles/zeros at the origin
11 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 (-,-,,,0,etc Quadratic Zeros/poles:Zeros/poles that are not located at imaginary or real axis (-+, +5, 3-3, etc
12 EXAMPLE 4 ( H Simplified, ( 4 H
13 ZEROS Let numerator, N(=0 0 st zero: ω 0 nd zero: 0 3
14 POLE Let denominator, D(=0 4 0 pole:
15 FREQUENCY RESPONSE PLOT USING SEMILOG GRAPH 5
16 MAGNITUDE PLOT AND PHASE PLOT magnitude plot H ( vs frequency( phase angle plot ( vs frequency ( 6
17 HOW TO DO MAGNITUDE AND PHASE PLOT i. Transform the time domain circuit (t into freq. domain circuit (ω ii. Determine the TF, H(ω iii.plot the magnitude of that tf, H(ωagainst ω. iv.plot the phase of that tf, (º against ω. 7
18 BODE PLOTS Bode plots are semilog plots of magnitude (in decibels and phase (in degrees of a transfer function versus frequency 8
19 Logarithm DECIBEL SCALE log PP log P log P log P / P log P log P log P n n log P log 0 9
20 BODE PLOT CHARACTERISTIC FOR POLES AND ZEROS 0
21 Logarithm of tf: N ( H ( D( log H log N logd
22 GAIN Gain is measured in bels G number of bels log 0 P P
23 Decibel (db G db 0 log 0 P P G db 0 log 0 V V 3
24 TRANSFER FUNCTION H H He H 0 log db 0 H 4
25 Magnitude H 0 log 0 H (db /
26 6 GENERAL EQUATION OF TF Before draw, make sure the general equation of tf is obtained first: ] [ / ( ( ( ( n n n n p z K H
27 7 EX. ] [ ( ( ( ( H ] [ / ( ( ( ( n n n n p z K H COMPARE
28 H ] ( ( 0 ( ( [ Constant: K Zero at the origin: 0 Real zero: Quadratic zero: n 0 Real pole: Quadratic pole: n 0 8
29 BODE PLOT OF A CONSTANT,K 9
30 30 ( (GAIN constant ] [ / ( ( ( ( n n n n p z K H
31 CHARACTERISTICS Magnitude for constant is : H 0log K Phase angle for constant is: 0 3
32 BODE PLOT FOR CONSTANT magnitude plot phase plot 3
33 BODE PLOT FOR ZERO AT THE ORIGIN 33
34 34 ( ZERO AT THE ORIGIN (ω N ] [ / ( ( ( ( n n n n p z K H
35 CHARACTERISTIC OF Magnitude: (ω N Straight line with 0dB/dec of slope that has a value of 0 db at = H 0N (db/dec Phase: 90N 35
36 MAGNITUDE PLOT 36
37 PHASE PLOT 37
38 BODE PLOT OF POLE AT THE ORIGIN 38
39 (3 POLE AT THE ORIGIN /(ω (ω -N H( K( z n n ( ( / p [ ] n n 39
40 CHARACTERISTIC OF (ω -N Magnitude: Straight line with -0dB/dec of slope that has a value of 0 db at = H - 0N (db/dec Phase: 90N 40
41 MAGNITUDE PLOT 4
42 PHASE PLOT 4
43 BODE PLOT OF REAL ZERO 43
44 44 (4 REAL ZERO ] [ / ( ( ( ( n n n n p z K H
45 CHARACTERISTIC OF (+ω/z N Magnitude: H 0 0N(dB/dec z z Phase: 0 45 o 90 o 0 z 45
46 MAGNITUDE PLOT 46
47 PHASE PLOT 47
48 BODE PLOT OF REAL POLE 48
49 49 (5 REAL POLE ] [ / ( ( ( ( n n n n p z K H
50 CHARACTERISTIC OF (+ω/p -N Magnitude: H 0-0NdB/dec Phase: o o 0 p p p 50
51 MAGNITUDE PLOT 5
52 PHASE PLOT 5
53 BODE PLOT OF QUADRATIC ZERO 53
54 54 (6 QUADRATIC ZERO ] [ / ( ( ( ( n n n n p z K H
55 CHARACTERISTIC OF ( + n + n N Magnitude: H Phase: 0 40NdB/dec 0 90 o 80 o 0 n n n 55
56 MAGNITUDE PLOT 56
57 PHASE PLOT 57
58 BODE PLOT OF QUADRATIC POLE 58
59 59 (7 QUADRATIC POLE ] [ / ( ( ( ( n n n n p z K H
60 CHARATERISTIC OF ( + n + n -N Magnitude: H 0-40N db/dec k k Phase: o o 0 k 60
61 MAGNITUDE PLOT 6
62 PHASE PLOT 6
63 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. 63
64 EX. Draw the Bode plot for the given tf below: 00 H( ( ( 0 64
65 65 SOLUTION General equation: /0 ( / ( 0 /0 (0( / (( 00 0 ( ( 00 ( H
66 Magnitude of tf: H db 0 log log 0 0 log 0 0 log log 0 0dB: straight line 66
67 Phase of tf: 90 o tan tan 0 Zero at the origin Pole at Pole at 0 67
68 MAGNITUDE PLOT GUIDANCE z=0 ω=0. ω= ω=0 ω=00 0dB/dec 0dB/dec 0dB/dec 0dB/dec p= 0dB/dec -0dB/dec -0dB/dec -0dB/dec p=0 0dB/dec 0dB/dec -0dB/dec -0dB/dec Resultant =0dB/dec =0dB/dec =-0dB/dec =-0dB/dec 68
69 Magnitude plot z=0 0 Constant p= -0-0 p= - 69
70 PHASE PLOT GUIDANCE ω=0 ω=0. ω= ω=0 ω=00 z=0 90º 90º 90º 90º 90º p= 0º/dec -45º/dec -45º/dec -90 º -90 º p=0 0º/dec 0º/dec -45º/dec -45º/dec -90º Resultant 90º -45º/dec -90º/dec -45º/dec -90º Add all of the lines that having a slope only 70
71 Phase plot 90 o z= p=-0-90 o p= - 7
72 EX. Draw the Bode plot for the given tf below: H( 0 ( 7
73 SOLUTION General equation: ( 0 ( 5 /0 ( / 73
74 Magnitude of tf : H db 0log 0 5 0log 0 0 0log 0 0log 0 0 log 5 4dB : straight line 74
75 Phase of tf: 90 o tan tan 0 p =0 z = -0 p = - 75
76 MAGNITUDE PLOT GUIDANCE ω=0. ω= ω=0 ω=00 p=0-0db/dec -0dB/dec -0dB/dec -0dB/dec p= 0dB/dec -0dB/dec -0dB/dec -0dB/dec z=0 0dB/dec 0dB/dec 0dB/dec 0dB/dec Resultant -0dB/dec -40dB/dec -0dB/dec - 0dB/dec 76
77 H db Magnitude plot z=-0 constant p=- -40 p=0 77
78 Phase plot Guidance ω=0 ω=0. ω= ω=0 ω=00 p=0-90º -90º -90º -90º -90º p= 0º/dec -45º/dec -45º/dec -90 º -90 º z=0 0º/dec 0º/dec 45º/dec 45º/dec 90º Resultant -90º -45º/dec 0º/dec 45º/dec -90º Add all the lines that having a slope only 78
79 90 o Phase plot z= o -90 o p= - p=0-35 o 79
80 EXAMPLE 3 Draw the Bode plot for the given tf below: s Hs ( s 0s00 80
81 SOLUTION Standard equation: s Hs ( s 0s00 Replace s=ω and divide it with 00; H( 00 (
82 Magnitude of tf: H db 0log 0log 0 0 0log 0 /0 00 /00 0 log 00 40dB : straight line 8
83 Phase of tf: 0 90 tan - 00 z =0 ω n = 0 83
84 Magnitude plot H db z= ω ω n =0 constant 84
85 Phase plot z= ω ω n =0 85
86 EXAMPLE 4 Determine the tf: 86
87 ANSWER H( 0 4 ( (0 (00 87
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