Applied Electromagnetics - ECE 351
Author: Benjamin D. Braaten
North Dakota State University
Department of Electrical and Computer Engineering
Fargo, ND, USA.
Last updated: 7/22/2015
1
TABLE OF CONTENTS
CHAPTER 1.
1.1.
TRANSMISSION LINES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
Transmission Line Propagation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
1.1.1.
1.2.
1.3.
1.4.
1.5.
1.6.
1.7.
Transmission Line Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
Lossless Propagation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.2.1.
Wave velocity and characteristic impedance . . . . . . . . . . . . . . . . . . . . . . 12
1.2.2.
Phase constant, phase velocity and wavelength . . . . . . . . . . . . . . . . . . . . 13
1.2.3.
Voltage and current along the transmission line . . . . . . . . . . . . . . . . . . . 14
Examples: Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
1.3.1.
Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
1.3.2.
Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
Wave Reflections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
1.4.1.
Reflection coefficient and transmission coefficient . . . . . . . . . . . . . . . . . . 18
1.4.2.
Reflected power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Examples: Reflection Along the Transmission Line . . . . . . . . . . . . . . . . . . . . . . . . 20
1.5.1.
Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
1.5.2.
Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
1.5.3.
Example 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
The Coaxial Transmission Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
1.6.1.
High frequency analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
1.6.2.
Low frequency analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
Two-wire Transmission Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2
1.7.1.
High frequency analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
1.7.2.
Low frequency analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
1.8.
Voltage Standing Wave Ratio (VSWR) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
1.9.
Examples: Standing Wave Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
1.9.1.
Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
1.9.2.
Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
1.10.
Transmission Line of Finite Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
1.11.
Examples: Finite Length Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
1.12.
1.13.
1.11.1.
Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
1.11.2.
Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
Review of Decibel (dB) and Phasors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
1.12.1.
Decibels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
1.12.2.
Phasors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
1.12.3.
Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
1.12.4.
Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
Scattering Matrices or S-parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
1.13.1.
1.14.
Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
The Smith Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
1.14.1.
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
1.14.2.
Reflection coefficient example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
1.14.3.
Input impedance example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
1.14.4.
Voltage minimum and maximum example . . . . . . . . . . . . . . . . . . . . . . . 42
1.14.5.
Stub matching example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3
1.15.
1.14.6.
Load impedance example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
1.14.7.
Load impedance example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
Transient Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
1.15.1.
Analytical solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
1.15.2.
Voltage reflection diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
CHAPTER 2.
2.1.
ELECTROSTATIC FIELDS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
Vector Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
2.1.1.
Vector notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
2.1.2.
Dot product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
2.1.3.
Cross product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
2.1.4.
Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
2.1.5.
Coordinate conversion example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
2.1.6.
Coordinate conversion example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
2.1.7.
Coordinate conversion example 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
2.2.
Coulomb’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
2.3.
Electric Field Intensity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
2.3.1.
Field from a single point charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
2.3.2.
Field from a charge distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
2.3.3.
Total charge example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
2.3.4.
Field from a line charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
2.3.5.
Field from a sheet of charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
2.3.6.
Streamlines and sketches of fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
4
2.4.
Electric Flux Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
2.5.
Gauss’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
2.6.
2.5.1.
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
2.5.2.
Example of Gauss’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
2.5.3.
Application of Gauss’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
2.5.4.
Point charge example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
2.5.5.
Line charge example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
2.6.1.
Example of computing the divergence in a region . . . . . . . . . . . . . . . . . . 68
2.6.2.
Solving for volume charge in a region . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
2.7.
The Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
2.8.
Energy and Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
2.8.1.
Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
2.8.2.
Differential work example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
2.8.3.
Line integral example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
2.8.4.
Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
2.8.5.
Point charge example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
2.8.6.
Potential field of a line charge example . . . . . . . . . . . . . . . . . . . . . . . . . . 71
2.8.7.
Potential fields due to surface and volume charges . . . . . . . . . . . . . . . . . 71
2.8.8.
Potential due to a ring of charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
2.8.9.
Potential gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
2.8.10.
Gradient example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
2.8.11.
The dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
5
CHAPTER 3.
3.1.
PROPERTIES OF DIELECTRICS AND CONDUCTORS . . . . . . . . . . . . 77
Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
3.1.1.
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
3.1.2.
Convection current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
3.1.3.
Continuity of current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
3.1.4.
Metallic conductors (resistance) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
3.1.5.
Resistance example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
3.2.
Boundary Conditions for Perfect Conductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
3.3.
Boundary Conditions for Perfect Dielectrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
3.4.
Method of Images . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
3.5.
Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
3.5.1.
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
3.5.2.
Capacitance between finite parallel plates . . . . . . . . . . . . . . . . . . . . . . . . 85
3.5.3.
Capacitance between two parallel plates with two dielectrics example
86
3.6.
Poisson’s and Laplace’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
3.7.
Uniqueness Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
3.8.
Parallel Plate Capacitance Example Using Laplace’s Equation . . . . . . . . . . . . . . 88
3.9.
Non-Parallel Plate Capacitance Example Using Laplace’s Equation . . . . . . . . . . 89
CHAPTER 4.
4.1.
MAGNETOSTATIC FIELDS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
Biot-Savart Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
4.1.1.
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
4.1.2.
Biot-Savart law example of an infinite line of current. . . . . . . . . . . . . . . 92
6
4.2.
Magnetic Flux Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
4.3.
Ampere’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
4.3.1.
Magnetic field from an infinitely long current . . . . . . . . . . . . . . . . . . . . . 93
4.3.2.
Magnetic field in a coax . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
4.4.
Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
4.5.
Stokes Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
4.6.
Magnetic Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
4.7.
Solid Conductor Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
4.8.
Scalar and Vector Magnetic Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
4.9.
Coaxial Scalar Potential Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
4.10.
Vector Magnetic Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
4.11.
Magnetic Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
4.12.
Force on a Charge Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
4.13.
Force on a Differential Current Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
4.14.
Force on a Square Conducting Loop Example . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
4.15.
Force Between Differential Current Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
4.16.
Magnetic Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
4.17.
Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
4.18.
Inductance of a Coax Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
CHAPTER 5.
TIME-VARYING FIELDS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
5.1.
Faraday’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
5.2.
Induced Voltage on a Coil Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
5.3.
Displacement Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
7
5.4.
Maxwell’s Equations for Time-Varying Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
5.5.
Wave Propagation in Free Space (Lossless) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
5.6.
Wave Propagation in Lossy Dielectrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
CHAPTER 6.
6.1.
6.2.
6.3.
TOPICS ON ELECTROMAGNETIC COMPATIBILITY . . . . . . . . . . . . . . 120
Coupling Between Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
6.1.1.
General Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
6.1.2.
Capacitive coupling (low frequencies) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
6.1.3.
Inductive coupling (low frequencies) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
6.1.4.
Equations for both inductive and capacitive coupling . . . . . . . . . . . . . . 124
Shielding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
6.2.1.
Reducing capacitive coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
6.2.2.
Reducing inductive coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
6.2.3.
Summary of transmission line coupling equations . . . . . . . . . . . . . . . . . . 131
Twisted pair . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
6.3.1.
Inductive coupling - unbalanced . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
6.3.2.
Capacitive coupling - unbalanced . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
6.3.3.
Balanced case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
8
CHAPTER 1. TRANSMISSION LINES
Transmission lines (TL) are used to deliver power from a source to a load. This can be done
by using coaxial lines or conductors made out of wire. When distances are large enough between
the source and the load, we use TL theory to find the voltage (V) and current (I) along the TL-line.
These TLs are usually categorized as electrically small or electrically large structures. This then
results in the following two methods to analyze the propagation of a wave along a TL:
Lumped Elements - if the time delay between the source and load is negligible (L << λ/10).
e
Distributed Elements - if the time delay between the source and load is not negligible (L>λ/10).
S1
{ {
Lossless Transmission Line
S2
I+
V + = V0
V0
I
R
_
Represents wave front
(propagating along the TL)
a)
Equivalent ciruit of the lossless TL
S1
I+
V0
L1
C1
I
L3
L2
C2
S2
C3
R
_
b)
Figure 1. a) Propagation of a wave along a TL; the equivalent circuit of the lossless TL.
1.1. Transmission Line Propagation
First, consider the lossless TL shown in Fig. 1 a). If switch S1 is closed, a wave travels down
the TL. If switch S2 is closed right before the wave arrives at the load R, a wave will be reflected
back towards the source. The amount reflected back depends on R (or if S2 is open or closed). The
equivalent circuit for the TL in Fig. 1 a) is shown in Fig. 1 b). When S1 is closed, the current in
L1 begins to build. This then charges C1 . C1 then begins to supply current to L2 as it approaches
max voltage...etc. This then leads to a wave propagating down the TL towards the load R.
9
1.1.1. Transmission Line Equations
Next, consider the per-unit equivalent (lumped element model) circuit of a short TL shown
in Fig. 2. R represents the conductor loss, L represents the line inductance, C represents the line
capacitance and G represents the loss in the dielectric between the conductors. Now we want to
derive expressions for V (z) and I(z) on the TL shown in Fig. 2 in terms of R, L, C and G.
1/2R∆z
1/2L∆z KCL 1/2L∆z
+ I
V
-
1/2R∆z
IG
G∆z KVL
Ic
C∆z
I+∆I +
V+∆V
-
∆z
Figure 2. Per-unit equivalent circuit of a TL.
First, using KVL on the circuit shown in Fig. 2:
⇒
1
1
∂I 1
∂
1
V = R∆zI + L∆z
+ L∆z (I + ∆I) + R∆z(I + ∆I) + V + ∆V.
2
2
∂t 2
∂t
2
(1.1)
(
)
∂I ∂∆I
V
1
1 ∂I 1
1
V
∆V
= RI + L
+ L
+
+ R(I + ∆I) +
+
.
∆z
2
2 ∂t 2
∂t
∂t
2
∆z
∆z
(1.2)
∆V
∂I
1
1 ∂∆I
= −RI − L
− R∆I − L
.
∆z
∂t
2
2 ∂t
(1.3)
⇒
Next, evaluating the limit ∆z → 0 (i.e., lim∆z→0 ) we have I + ∆I → I, ∆I → 0 and
∆V
∆z
→
∂V
∂z
.
This then reduces 1.3 to the following:
(
)
∂V
∂I
= − RI + L
.
∂z
∂t
10
(1.4)
Next, using KCL on the circuit shown in Fig. 2:
I = IG + IC + I + ∆I
(
)
(
)
∆V
∂
∆V
= G∆z V +
+ C∆z
V +
+ I + ∆I.
2
∂t
2
⇒
⇒
(1.5)
(
)
(
)
I
∆V
∂
∆V
I
∆I
=G V +
+C
V +
+
+
.
∆z
2
∂t
2
∆z ∆z
(1.6)
)
(
)
(
∂
∆V
∆I
∆V
= −G V +
−C
V +
.
∆z
2
∂t
2
(1.7)
Next, evaluating the limit ∆z → 0 (i.e., lim∆z→0 ) we have V + ∆V → V , ∆V → 0 and
∆I
∆z
→
∂I
.
∂z
This then reduces 1.7 to the following:
(
)
∂I
∂V
= − GV + C
.
∂z
∂t
(1.8)
Equations 1.4 and 1.8 are referred to as the telegraphist’s equations. Their solutions lead to the
wave equations on the TL. Next, differentiating (1.4) w.r.t. z and (1.8) w.r.t. t we get:
∂ 2V
∂I
∂2I
=
−R
−
L
∂z 2
∂z
∂t∂z
(1.9)
∂ 2I
∂V
∂ 2V
= −G
−C 2 .
∂z∂t
∂t
∂t
(1.10)
and
Next, substituting (1.8) and (1.10) into (1.9) gives
∂ 2V
∂V
∂ 2V
=
LC
+ (LG + RC)
+ RGV.
2
2
∂z
∂t
∂t
11
(1.11)
Similar substitutions result in the following expression:
∂ 2I
∂ 2I
∂I
=
LC
+ (LG + RC)
+ RGI.
2
2
∂z
∂t
∂t
(1.12)
Equations (1.11) and (1.12) represent the general wave equations for the TL in Fig. 2.
1.2. Lossless Propagation
1.2.1. Wave velocity and characteristic impedance
For lossless propagation we have R = G = 0 in Fig. 2. This then simplifies (1.11) and (1.12)
to the following:
∂2V
∂ 2V
=
LC
∂z 2
∂t2
(1.13)
∂ 2I
∂ 2I
=
LC
.
∂z 2
∂t2
(1.14)
and
Solving the second order partial differential equations in (1.13) and (1.14) results in the following
assumed solutions:
V (z, t) = f1 (t − z/ν) + f2 (t + z/ν) = V + + V −
(1.15)
where ν represents the wave velocity, V + represents the forward traveling wave and V − represents
the backward traveling wave or reflected wave. The t − z/ν represents the forward traveling wave.
As t increases, z must also increase to sustain f (0) (the wave front). To solve for the wave velocity
we consider the solution of V (z, t) to
∂2V
∂z 2
2
= LC ∂∂tV2 . Without loss of generality we consider only f1 .
⇒
∂V (z, t)
∂f1
∂f1
∂(t − z/ν)
−1 ′
=
=
=
f
∂z
∂z
∂(t − z/ν)
∂z
v 1
(1.16)
′
where f1 is the partial derivative w.r.t. the argument. Similarly we have
∂f1
∂f1
∂V (z, t)
∂(t − z/ν)
′
=
=
= f1 ,
∂t
∂t
∂(t − z/ν)
∂t
(1.17)
1 ′′
∂ 2 f1
= 2 f1
2
∂z
ν
(1.18)
12
and
∂ 2 f1
′′
= f1 .
2
∂t
(1.19)
Notice that (1.18) and (1.19) are simply derivatives of f1 w.r.t. z and t, respectively. Substituting
(1.18) and (1.19) into (1.13) results in the following expression:
∂2V
1 ′′
∂ 2V
′′
=
f
=
LCf
=
LC
1
∂z 2
ν2 1
∂t2
(1.20)
1 ′′
′′
f1 = LCf1 .
2
ν
(1.21)
or
′′
Canceling f1 results in the following expression for the wave velocity ν:
1
ν=√
.
LC
(1.22)
Similarly for (1.14) we have the following solution:
[
]
1
1
I(z, t) =
f1 (t − z/ν) − f2 (t + z/ν) = I + + I − =
V (z, t)
Lν
Z0
where
√
Z0 =
L
.
C
(1.23)
(1.24)
Z0 is referred to as the characteristic impedance of the TL. Notice the expressions for the wave
velocity and characteristic impedance are written entirely in terms of L and C. This indicates
that ν and Z0 are dependent only on the dimensions of the physical structure and not time and
frequency.
1.2.2. Phase constant, phase velocity and wavelength
In this section we derive the expressions for the voltage and current along the TL when a
steady-state sinusoidal source is used to drive the TL. Start by defining f1 = f2 = V0 cos(ωt + ϕ).
From the previous section we have t = t ± z/νp where νp is referred to as the phase velocity.
13
⇒
V (z, t) = |V0 | cos[ω(t + z/νp ) + ϕ]
= |V0 | cos[ωt ± βz + ϕ].
(1.25)
⇒
V + = Vf (z, t) = |V0 | cos[ωt + βz + ϕ]
(1.26)
V − = Vb (z, t) = |V0 | cos[ωt − βz + ϕ]
(1.27)
and
where
β=
ω
.
νp
(1.28)
β is called the phase constant of the TL. If t = 0 (i.e., fix time and look at the spatial variation)
then we have Vf (z, 0) = |V0 | cos[βz]. Vf (z, 0) represents a periodic function that repeats w.r.t. a
value of z. Denote this value as λ and call it the wavelength of the wave. This then gives βλ = 2π.
⇒
λ=
νp
2π
= .
β
f
(1.29)
Now, if z = 0 (fix position and look at time variation) then we have V (0, t) = |V0 | cos[ωt]. This is
illustrated in Fig. 3. Notice that the sinusoid repeats every 2π.
⇒
T =
1
f
(1.30)
where T is the period of the sinusoid.
1.2.3. Voltage and current along the transmission line
Here we want to represent the voltage along the TL as complex functions. Using Euler’s
identity we have e±jx = cos(x) ± j sin(x).
14
V(0,t)
+
V0
2π
ωt
Figure 3. Sinusoidal voltage along the TL.
⇒
[
]
[
]
1 jx
−jx
±jx
cos(x) = Re e
= e +e
2
and
[
sin(x) = ±Im e
]
±jx
(1.31)
[
]
1 jx
−jx
=±
e −e
2j
(1.32)
⇒
V (z, t) = |V0 | cos[ωt ± βz + ϕ]
[
]
1
j(ωt±βz+ϕ)
−j(ωt±βz+ϕ)
=
|V0 | e
+e
2
[
][
]
1
jϕ
−jϕ
j(ωt±βz)
−j(ωt±βz)
|V0 | e + e
e
+e
=
2
[
]
j(ωt±βz)
−j(ωt±βz)
= V0 e
+e
.
(1.33)
[
Note that V0 was used to represent the complex voltage magnitude
1
|V |
2 0
]
jϕ
e
+e
−jϕ
in (1.33).
Next, define the following
Vc (z, t) = V0 e±jβz ejωt
(1.34)
Vs (z) = V0 e±jβz
(1.35)
and
where Vc (z, t) is the complex instantaneous voltage and Vs (z) is the phasor voltage. Again, from
15
the general wave equation (1.11) we have
∂ 2V
∂ 2V
∂V
=
LC
+ (LG + RC)
+ RGV.
2
2
∂z
∂t
∂t
In the phasor domain we also have the relation
∂
∂t
(1.36)
⇔ jω.
⇒
d2 Vs
= −ω 2 LCVs + jω(LG + RC)Vs + RGVs .
2
dz
(1.37)
d2 Vs
= (R + jωL)(G + jωC)Vs = γ 2 Vs
dz 2
(1.38)
⇒
where γ =
√
(R + jωL)(G + jωC) = α + jβ is the propagation constant along the TL. Now
assume that Vs (z) = V0+ e−γz + V0− e+γz for a solution to
d2 Vs
dz 2
= γ 2 Vs . Similarly, assume Is (z) =
I0+ e−γz + I0− e+γz . Substituting the assumed voltage and current solutions into the transformed
expressions of (1.4) and (1.8) gives the following expressions:
dVs
= −(R + jωL)Is
dz
(1.39)
and
⇒ −γV0+ e−γz + γV0− e+γz
dIs
= −(G + jωC)Vs .
(1.40)
dz
[
]
+ −γz
+ γz
= −Z I0 e
+ I0 e
where Z0 = R + jωL. Equating the coefficients
of e−γz and eγz gives −γV0+ = −ZI0+ and γV0− = −ZI0− . Therefore, the characteristic impedance
16
Z0 is
Z0 =
=
=
=
=
V0+
I0+
V−
− 0−
I0
Z
γ
Z
√
ZY
√
Z
Y
(1.41)
where Y = G + jωC.
⇒
√
Z0 =
R + jωL
= |Z0 |ejθ .
G + jωC
(1.42)
1.3. Examples: Transmission Lines
1.3.1. Example 1
Consider an 80 cm long lossless TL with a source connected to one end operating at 600 MHz.
The lumped element values of the TL are L = 0.25µH/m and C = 100pF/m. Find Z0 , β and νp .
Solution: Using (1.42) we have
√
√
Z0 =
R + jωL
=
G + jωC
0 + jω.25 × 10−6
= 50Ω.
0 + jω100 × 10−12
√
√
√
Also from (1.38) we have γ = (R + jωL)(G + jωC) = (0 + jωL)(0 + jωC) = 0+jβ = jω LC.
√
⇒ β = 2π(600 × 106 ) LC = 18.85rad/m. Finally, we have νp = ω/β = 2 × 108 m/s.
1.3.2. Example 2
A transmission line constructed of two parallel wires in air has a conductance of G = 0. The
two parallel wires are made of good conductors, therefore it is assumed that R = 0. If Z0 = 50Ω,
β = 20 rad/m and f = 700 MHz, find the per-unit inductance and capacitance of the TL.
Solution:
17
√
√
√
Since R = G = 0, β = ω LC = 20 = 2π700M Hz LC and Z0 = CL = 50. Then the ratio
of β and Z0 results in the following:
β
= ωC.
Z0
Solving for C then gives:
C =
=
β
ωZ0
20
2π ∗ 700M Hz ∗ 50
= 90.9pF/m.
Finally, solving for L gives:
L = Z02 C
= 502 ∗ 90.9 × 10−12
= 227nH/m.
1.4. Wave Reflections
1.4.1. Reflection coefficient and transmission coefficient
Next, expressions for describing the reflections of the wave along the TL are derived. To do
this, consider the TL load in Fig. 4. Next, denote
Vi (z) = V0i e−αz e−jβz
(1.43)
Vr (z) = V0r e+αz e+jβz .
(1.44)
VL (0) = Vi (0) + Vr (0) = V0i + V0r
(1.45)
and
⇒
18
and
IL (0) = I0i + I0r =
1
[V0i − V0r ].
Z0
(1.46)
Now define the reflection coefficient at the load as
V0r
ZL − Z0
=
= |Γ|ejϕΓ .
V0i
ZL + Z0
Γ=
(1.47)
Now using Γ we can write the voltage at the load in terms of the reflection coefficient and the
incident wave in the following manner:
VL = V0i + ΓV0i .
(1.48)
Z0
+
Vi(z)
VL
ZL=RL+jXL
Vr(z)
-
z=0
Figure 4. Voltage reflection at the load of a TL.
Now define the transmission coefficient as
τ=
VL
2ZL
=1+Γ=
= |τ |ejϕt .
V0i
Z0 + ZL
(1.49)
1.4.2. Reflected power
The time averaged power can be written as
(
)
∗
1
1
1 |V0 |2 −2αz
∗
−αz −jβz V0
−αz +jβz
e
e
e
cos θ.
< P >= Re(Vs Is ) = Re V0 e e
=
2
2
|Z0 |ejθ
2 |Z0 |
(1.50)
Note that θ in (1.50) refers to the angle on the characteristic impedance Z0 . Then for z = L we
19
have the following expression for the incident power:
< Pi >=
1 |V0 |2 −2αL
e
cos θ.
2 |Z0 |
(1.51)
Then, to find the reflected power at the load substitute in the reflected wave in (1.48):
)
(
∗
1
−αL −jβL (ΓV0 ) −αL jβL
e
e
< Pr > =
Re ΓV0 e
e
2
Z0 ejθ
1 |Γ|2 |V0 |2 −2αL
=
e
cos θ.
2 |Z0 |
(1.52)
This then leads to the following expression for the reflected power
< Pr >
= ΓΓ∗ = |Γ|2 .
< Pi >
(1.53)
< Pt >
= 1 − |Γ|2 .
< Pi >
(1.54)
Similarly, for the transmitted power :
1.5. Examples: Reflection Along the Transmission Line
1.5.1. Example 1
a) ZL = Z0 = 50Ω ⇒ Γ = 0 and τ = 1.
b) ZL = 0, Z0 = 50Ω ⇒ Γ = −1 and τ = 0.
c) ZL → ∞ (open), Z0 = 50Ω ⇒ Γ = 1 and τ = 2.
Notice from the previous derivations and examples that we have −1 ≤ |Γ| ≤ 1 (RL ≥ 0) and
0 ≤ |τ | ≤ 2 (RL ≥ 0).
1.5.2. Example 2
A TL with Z0 = 100Ω is loaded with a series connected 50 Ω resistor and a 10 pF capacitor.
Find the reflection coefficient at the load at 100 MHz.
Solution:
20
The load impedance at 100 MHz is ZL = 50 − j159Ω. Using (1.47) gives
Γ=
ZL − Z0
50 − j159 − 100
= 0.762∠ − 60.78.
=
ZL + Z0
50 − j159 + 100
1.5.3. Example 3
Show that |Γ| = 1 for a purely reactive load.
Solution:
For this example let ZL = 0 + jXL . From (1.47) the reflection coefficient at the load is
√
− Z02 + XL2 e−jθ
ZL − Z0
jXL − Z0
−(Z0 − jXL )
Γ=
=
=
= √ 2
= −e−j2θ .
2 jθ
ZL + Z0
jXL + Z0
Z0 + jXL
Z0 + XL e
This then gives |Γ| = | − e−j2θ | = 1.
1.6. The Coaxial Transmission Line
1.6.1. High frequency analysis
A cross-section of a coaxial TL is shown in Fig. 5. The radius of the inner conductor is a, the
radius of the inner wall on the outer conductor is b and the radius of the outer wall on the outer
conductor is c. It can be shown that the per unit values of the coaxial TL are:
2πε′
C=
ln( ab )
2πσ
G=
ln( ab )
Lext
µ
ln
=
2π
(
(
)
F
,
m
(1.55)
)
S
,
m
(1.56)
( ) ( )
b
H
,
a
m
(
) ( )
1 1
1
Ω
+
R=
,
2πδσc a b
m
and
21
(1.57)
(1.58)
√
Z0 =
Lext
1
=
C
2π
√
µ
ln
ε′
( )
b
(Ω)
a
(1.59)
where σ is the conductivity of the TL, ε′ is the permittivity of the TL and µ is the permeability of
the TL.
outer
conductor
(σc)
a
inner
conductor
(σc)
dielectric
(σ,ε,µ)
b
c
Figure 5. Cross-section of a coaxial TL.
1.6.2. Low frequency analysis
Also for the LF analysis of the coaxial TL we have the following expressions:
2πε′
C=
ln( ab )
2πσ
G=
ln( ab )
(
(
)
F
,
m
(1.60)
)
S
,
m
(1.61)
(
)( )
1
Ω
1
1
R=
+ 2
,
2
2
πσc a
c −b
m
(1.62)
and
Lext
( ))]( )
[ ( )
(
1
c
µ
b
1
4c4
H
2
2
ln
=
ln
+ +
b − 3c + 2
.
2
2
2
2π
a
4 4(c + b )
c −b
b
m
(1.63)
1.7. Two-wire Transmission Line
1.7.1. High frequency analysis
A cross section of the two-wire TL is shown in Fig. 6. The radius of each wire is denoted as
22
a and the center of each wire is separated by a distance d. It is assumed that the two wires are
emersed in a material with properties (σ, µ, ε′ ). It can be shown that the per unit values of the
two-wire TL are:
(
πε′
C=
ln( ad )
πσ
G=
d
)
cosh−1 ( 2a
µ
L = ln
π
)
F
,
m
(
(1.64)
)
S
,
m
(1.65)
( ) ( )
H
d
,
a
m
(1.66)
( )
Ω
1
R=
πaδσc m
(1.67)
and
√
Z0 =
L
(Ω).
C
(1.68)
dielectric
(σ,ε,µ)
σc
σc
a
a
d
Figure 6. Cross-section of a two-wire TL.
1.7.2. Low frequency analysis
Also for the LF analysis of the two-wire TL we have the following expressions:
πε′
C=
d
cosh−1 ( 2a
)
23
(
)
F
,
m
(1.69)
πσ
G=
d
cosh−1 ( 2a
)
(
)
S
,
m
(1.70)
[
( )] ( )
µ 1
d
H
−1
L=
+ cosh
,
π 4
2a
m
(1.71)
( )
2
Ω
R= 2
.
πa σc m
(1.72)
and
1.8. Voltage Standing Wave Ratio (VSWR)
For the voltage standing wave ratio (VSWR) we start with the following:
VsT (z) = V0 e−jβz + ΓV0 ejβz
[
]
−jβz
j(βz+ϕΓ )
= V0 e
+ |Γ|e
= V0 (1 − |Γ|)e−jβz + 2V0 |Γ|ejϕΓ /2 cos(βz + ϕΓ /2)
(1.73)
where ϕ is the angle on the reflection coefficient. Converting (1.73) to the time domain we get:
[
]
jωt
VsT (z, t) = Re VsT (z)e
= V0 (1 − |Γ|) cos(ωt − βz)
+2V0 |Γ| cos(βz + ϕΓ /2) cos(ωt + ϕΓ /2).
(1.74)
The first term in the last expression in (1.74) describes the traveling wave and the second term
in the last expression describes the standing wave along the TL. An example of a standing wave
is illustrated in Fig. 7.
The oscillations of the standing wave in Fig. 7 are described by the
trigonometric terms in (1.74). We can also derive expressions for the position of the voltage
maximums and minimums along the TL. The position of the voltage maximum is denoted as zmax
24
|VsT|
λ/2
(1+|Γ|)V0
(1−|Γ|)V0
−1
−1
−1
(φ+3π)
(φ+5π)
(φ+π)
2β Γ
2β Γ
2β Γ
−1
−1
−1
−φΓ
(φ+6π)
(φ+4π)
(φ+2π)
Γ
Γ
Γ
2β
2β
2β
2β
z
0
Figure 7. Standing wave along a TL. Note that ϕ in this figure is the phase of the reflection coefficient
Γ = |Γ|ejϕΓ .
and the position of the voltage minimum is denoted as zmin . Thus it can be shown that
zmin = −
1
[ϕΓ + (2m + 1)π]
2β
(1.75)
1
[ϕΓ + 2mπ]
2β
(1.76)
and
zmax = −
where m = 0, 1, 2.... Next, we define the VSWR (denoted as s) in the following manner:
s=
VsT (zmax )
1 + |Γ|
=
.
VsT (zmin )
1 − |Γ|
(1.77)
A few examples of a standing wave are illustrated in Fig. 8. To understand the maximum
and minimum values of the voltage along the TL we start with the following expression:
(
)
−jβz
j(βz+ϕΓ )
VsT (z) = V0 e
+ |Γ|e
.
25
(1.78)
We have a voltage minimum when βz = 0 and ϕΓ = π. This then reduces (1.78) to VsT (zmin ) =
V0 (1 − |Γ|). Similarly, we have a voltage maximum when βz = 0 and ϕ = 0. This then reduces
(1.78) to VsT (zmax ) = V0 (1 + |Γ|). Then, if the TL is matched (i.e., Z0 = ZL ) then Γ = 0 and
V0 (1 + |Γ|) = V0 (1 − |Γ|) = V0 . This results in the constant voltage along the TL shown in Fig.
8 a). If ZL = 0 (i.e., short circuit), then |Γ| = −1. This then gives a maximum and minimum
voltage of (1 + 1)V0 = 2V0 and (1 − 1)V0 = 0, respectively. This is illustrated in Fig. 8 b). Finally,
if ZL = ∞ (i.e., open circuit) then |Γ| = 1. This then gives a maximum and minimum voltage of
(1 + 1)V0 = 2V0 and (1 − 1)V0 = 0, respectively. This is illustrated in Fig. 8 c).
We can also see that the standing waves in Fig. 8 b) and c) have a period of λ/2. For example,
to calculate the position of the first minimums and maximums we use the expressions in (1.75) and
(1.76), respectively. First, if ϕΓ = π, then (1.75) and (1.76) reduce to −λ/2 and −λ/4, respectively.
These computations are illustrated in Fig. 8 b). Second, if ϕΓ = 0, then (1.75) and (1.76) reduce
to −λ/4 and 0, respectively. These computations are illustrated in Fig. 8 c).
|V(z)|
Matched line
−λ
+
|V0|
−3λ
4
−λ
2
−λ
4
a)
|V(z)|
+
Short circuit
−λ
2|V0|
−3λ
4
−λ
2
−λ
4
z
0
|V(z)|
b)
Open circuit
−λ
z
0
+
2|V0|
−3λ
4
−λ
2
−λ
4
0
z
c)
Figure 8. a) Standing wave along the TL for ZL = Z0 ; b) standing wave along the TL for ZL = 0; c)
standing wave along the TL for ZL = ∞.
26
1.9. Examples: Standing Wave Examples
1.9.1. Example 1
A 50 Ω transmission line is terminated with a load of ZL = 100 + j50 Ω. Find the voltage
reflection coefficient and the voltage standing wave ratio (VSWR).
Solution: From (1.47) we have
Γ=
ZL − Z0
100 + j50 − 50
=
= .45∠26.6.
ZL + Z0
100 + j50 + 50
Next, using (1.77) we get
s=
1 + |Γ|
1 + 0.45
=
= 2.6.
1 − |Γ|
1 − 0.45
1.9.2. Example 2
A 140 Ω lossless transmission line is terminated with a load impedance of ZL = 280 + j182 Ω.
If λ = 72 cm, find (a) Γ, (b) s and (c) the first zmin and zmax .
Solution: From (1.47) we have
Γ=
ZL − Z0
280 + j182 − 140
=
= .5∠29.
ZL + Z0
280 + j182 + 140
Next, using (1.77) we get
s=
1 + |Γ|
1 + 0.5
=
= 3.0.
1 − |Γ|
1 − 0.5
Finally, β = 2π/λ = 8.72 rad/m. Using (1.75) and (1.76) gives zmin =
cm and zmax =
−1
(29
2∗8.72
∗
π
)
180
−1
(29
2∗8.72
∗
π
180
+ π) = −20.9
= −2.9 cm.
1.10. Transmission Line of Finite Length
Now we need to analyze the TL while including the numerous forward and backward reflected
waves. To do this consider the TL in Fig. 9. From before, we have VsT (z) = V0+ e−jβz +V0− ejβz which
represents the total voltage along the TL. We also have IsT (z) = I0+ e−jβz + I0− ejβz which represents
27
the total current along the TL. Using these two expressions, we define the wave impedance as
VsT (z)
V0+ e−jβz + V0− ejβz
ZW (z) =
= + −jβz
IsT (z)
I0 e
+ I0− ejβz
[
]
ZL cos βz − jZ0 sin βz
= Z0
.
Z0 cos βz − jZL sin βz
(1.79)
Evaluating (1.79) at z = −l gives
[
Zin = ZW (−l) = Z0
Next, if βl =
2π mλ
λ 2
]
ZL cos βl + jZ0 sin βl
.
Z0 cos βl + jZL sin βl
(1.80)
= mπ where m = 0, 1, ... then Zin (l = mλ/2) = ZL . Also, if βl =
2π
(2m + 1) λ4
λ
=
(2m + 1)π/2 where m = 0, 1, ... then Zin (l = λ/4) = Z02 /ZL . The last expression is used for the
design of quarter wave transformers.
{
Lossless Transmission Line
Zg
+
VS
+
Z0
Vin
-
VL
-
Zin z = -l
ZL
z=0
Figure 9. General lossless TL in steady state.
1.11. Examples: Finite Length Transmission Lines
1.11.1. Example 1
Zs=40 Ω
+
VS
Vin
-
+
Z0=60+j40 Ω
Zin z = -l = 2m
VL
-
ZL=20+j50 Ω
z=0
Figure 10. TL for example 1.
A TL operating at ω = 106 rad/s has the following constants: α = 8 dB/m, β = 1 rad/m,
28
Z0 =60+j40 Ω and is 2m long. If Vs = 10∠0, Zs = 40 Ω and ZL = 20 + j50 Ω find:
a) Zin
b) Iin
Solution: Neglecting α gives
a)
]
ZL cos βl + jZ0 sin βl
Zin = ZW (−l) = Z0
Z0 cos βl + jZL sin βl
[
]
(20 + j50) cos(2) + j(60 + j40) sin(2)
= (60 + j40)
(60 + j40) cos(2) + j(20 + j50) sin(2)
[
= 57.28 − j2.11.
b)
Vs
Zin + Zs
10∠0
=
40 + 57.28 − j2.11
I(−l) =
= 102.7∠1.24mA.
1.11.2. Example 2
Now consider the TL with Z0 = 300Ω. The load is two 300 Ω resistors and one capacitor with
Zc = −j300Ω all connected in parallel. Calculate Zin , s, Γ and PL for l = 2m, νp = 2.5 × 108 m/s,
f = 100M Hz, Zg = 300Ω and Vs =60V.
Solution: ZL = 300||300||(−j300) = 150||(−j300) = 120 − j60Ω. ⇒
Γ=
120 − j60 − 300
ZL − Z0
=
= .447∠ − 153.4,
ZL + Z0
120 − j60 + 300
and
s=
1 + |Γ|
= 2.616.
1 − |Γ|
Since the TL is lossless, λ = νp /f = 2.5 × 108 /100M Hz = 2.5m. ⇒ β = 2π/λ = 2.51rad/m. ⇒
29
βl = 5.02rad = 287.6◦ . Solving for Zin gives
[
Zin = Z0
ZL cos 287.6 + jZ0 sin 287.6
Z0 cos 287.6 + jZL sin 287.6
]
= 760.1 − j127.6Ω.
⇒
Vs
Zg + Zin
60
=
300 + Zin
Iin =
= 56.1∠6.86mA.
Since the TL is lossless,
1 2
I R
2
1
(56.1mA)2 760
=
2
Pin =
= 1.199W.
⇒ PL ≈ 1.2W .
1.12. Review of Decibel (dB) and Phasors
1.12.1. Decibels
In this section we review the use of decibels. The number of decibels is the logarithmic ratio
of the powers of interest or:
(
P ower = 10 log
)
P2
.
P1
(1.81)
The unit of the ratio in (1.81) is dB and referred to as decibels. Equation (1.81) can be used to
express the power gain of a circuit. For example, if P1 = 10W and P2 = 5W in Fig. 11 then the
power gain = 10 log(P2 /P1 ) = −3.01dB.
30
I1
I2
+
+
Z1
V1
Z2
V2
-
-
Figure 11. Equivalent circuit.
1.12.2. Phasors
Now we switch our focus to phasors. In this course we use the following notation:
[
V (z, t) = Re Ve e
]
jωt
= |Vm |∠θv = Ve = phasor voltage
(1.82)
where V (z, t) represents the voltage in the time domain and Ve represents the voltage in the frequency
[
]
j30
jωt
j30
domain. For example, if Ve = 3∠30 = 3e then V (z, t) = Re 3e e
= 3 cos(ωt + 30).
Next, if Ve1 and Ie1 are rms quantities,
P1
]
[
∗
e
e
= Re V1 I1
[
]
∗
= Re (Ie1 Z1 )(Ie1 )
[
]
2
= Re |Ie1 | (R1 + jX1 )
= |Ie1 |2 R1
|Ve1 |2
=
.
R1
Now let V1 = |Ve1 | and I1 = |Ie1 |. This then gives
(
10 log
P2
P1
)
(
)
V22 R1
= 10 log
V 2 R2
( 12 )
( )
V2
R1
= 10 log
+ 10 log
2
V
R
( 1)
( 2)
V2
R1
= 20 log
+ 10 log
.
V1
R2
31
Therefore,
(
10 log
P2
P1
)
(
= 20 log
V2
V1
)
(1.83)
only when R1 = R2 . Next, we define dBm in the following manner:
(
P (dBm) = 10 log
)
P
.
1mW
(1.84)
The values written in terms of dBm simply mean that the power value of interest is referenced to
1 mW. ⇒ P = (10P (dBm)/10 )1mW .
1.12.3. Example 1
Suppose I measure 20 dB of power dissipated by a 50 Ω load. When just dB is written, that
means that the power is being referenced to 1 W. ⇒ 20dB = 10 log(P/1W ) ⇒ P = 102 = 100W .
Next, what is 25 mW written in (a) dBm, (b) dB and (c) dBµW?
(a) 10 log(25 × 10−3 /1 × 10−3 ) = 13.98dBm
(b) 10 log(25 × 10−3 /1) = 10 log(25) + 10 log(10−3 ) = −16.02dB
(c) 10 log(25 × 10−3 /1 × 10−6 ) = 13.98 + 30 = 43.98dBµW
1.12.4. Example 2
Next, as another example, express 20 dB across a 50 Ω resistor in terms of dBV, dBµV, dBA
and dBmA.
20dB = 10 log(P/1W ) ⇒ P = 100W . If we assume rms values of voltage and current,
P = V 2 /R = I 2 R. ⇒ V = 70.71V and I = 1.414A. Therefore,
)
70.71V
20 log
1V
(
)
70.71V
20 log
1µV
(
)
1.414A
20 log
1A
(
)
1.414A
20 log
1mA
(
32
= 37dBV
= 157dBµV
= 3dBA
= 63dBmA.
1.13. Scattering Matrices or S-parameters
The scattering matrices are used to describe the response of an N-port network to various
incident and reflected voltages on the network. For illustration, consider the N-port network in Fig.
12. The incident voltage (current) and reflected voltage (current) on the N th port is denoted as VN+
(IN+ ) and VN− (IN− ), respectively.
V+3,I3+
-
V3, I3-
- -
V 4, I4
V+4, I4+
- -
V2, I2
V+5,I5+
N-port
network
V+2,I2+
- -
V 5 , I5
- -
V1, I1
- -
V+1, I1+
VN,IN
V+N, IN+
Figure 12. An N-port network for illustration of the S-parameters.
The scattering matrices are then defined as:












V1−



S11
 
 
  S21
 
 

V3− 
 =  S31
 
..   ..
.   .
 
−
VN
SN 1
S12 S13 . . . S1N
V2−
..
..
.
.
...











SN N
or in a more compact form
[V − ] = [S][V + ].
Then, to find a particular element of [S], we first compute
V1− = S11 V1+ + S12 V2+ + ... + S1N VN+
33
V1+






V3+ 
)

.. 
. 

+
VN
V2+
and then “zero-out” the incident voltages that are not of interest. This then in general gives
Vi− .
Vj+ +
Vk =0 f or k̸=j
Sij =
The incident voltages can be “zeroed-out” by terminating the ports not of interest with a matched
load. This then reduces the reflection coefficient Γ to zero and minimizes the reflected or scattered
waves at that port.
1.13.1. Example 1
Consider the two port circuit (N = 2) in Fig. 13(a). Since we have a two port network we
have the following scattering matrices:




V1−
V2−



V1+
  S11 S12  

=

.
+
S21 S22
V2
Solving for the reflected voltage at port 1 gives V1− = V1+ S11 + V2+ S12 . To compute S11 we match
port two to give V2+ = 0. This then gives
S11 =
V1−
Zin − Zo
.
+ = Γ =
Zin + Zo
V1
Therefore, if a 50 Ω resistor is connected across port 2 (as shown in Fig. 13(b)), the input impedance
of port 1 can be shown to be 50 Ω also. Thus, S11 = 0.
Next, to compute S21 , the following equation is considered: V2− = V1+ S21 + V2+ S22 . For this
computation, the incident voltage V2 + on port 2 must be reduced to zero by a matched 50 Ω load.
This circuit is shown in Fig. 13(c). Then, if port 1 is driven with V1+ , the voltage on the 50 Ω load
can be computed as V2− = 0.67V1+ . This then gives
S21 =
0.67V1+
= 0.67
V1+
34
or in decibels
S21 = 20log0.67 = −3.47dB.
10Ω
10Ω
120Ω
Port 1
Port 2
120Ω
Port 1
a)
50Ω
b)
10Ω
10Ω
+
+
Port 1
10Ω
10Ω
+
120Ω
V1
-
V2
50Ω
-
-
c)
Figure 13. (a) two port attenuator circuit; (b) port 2 terminated with a match load to compute the input
impedance at port 1 and (c) port 2 terminated with a match load to compute S21 .
1.14. The Smith Chart
1.14.1. Introduction
To introduce the Smith chart we start with the reflection coefficient Γ = (ZL − Z0 )/(ZL + Z0 ).
All impedance values on the Smith chart are normalized and have the following notation:
ZL
RL + jXL
=
.
Z0
Z0
(1.85)
Γ=
zL − 1
zL + 1
(1.86)
zL =
1+Γ
.
1−Γ
(1.87)
zL = r + jx =
⇒
or
35
Next, using rectangular coordinates, we can write Γ = Γr + jΓi . ⇒
zL = r + jx =
1 + Γr + jΓi
.
1 − Γr − jΓi
(1.88)
It can be shown that
1 − Γ2r − Γ2i
r=
(1 − Γr )2 + Γ2i
(1.89)
and
x=
2Γi
.
(1 + Γr )2 + Γ2i
(1.90)
Equations (1.89) and (1.90) are the real and imaginary parts of (1.88), respectively. Rearranging
(
)2
( )2
(
r
1
2
2
gives the following family of circle expressions: Γr − 1+r + Γi = 1+r
and (Γr − 1) + Γi −
)2 ( )2
1
= x1 . The family of circles are shown in Figs. 14 a) and b). For example, if x = ∞, then
x
Γ = 1 + j0 and the radius of the circle is zero. If x = +1, then the circle is centered at 1 and 1.
⇒ Γ = 1 + j1 and the radius is 1. Also, if x = 2, then the circle is centered at 1 and 1/2. ⇒
Γ = 1 + j1/2 and the radius of the circle is 1/2. Combining both circles we get the smith chart
shown in Fig. 15.
Γi
Γi
r=0
r = 0.5
x=2
x = 0.5
r=1
r=2
x=1
r=
x=0
Γr
x=
Γr
x = -2
x = -0.5
x = -1
|Γ| = 1
a)
b)
Figure 14. a) r-circles in the Γr and Γi plane and b) x-circles in the Γr and Γi plane.
1.14.2. Reflection coefficient example
To demonstrate the Smith chart we consider the following example: ZL = 25 + j50 and
36
The Smith Chart
0.11
70
45
1.0
0.9
0.8
1.6
1.8
25
0.4
jX
/
Z
2.0
6
0.0
4
0.4
14
0
0.4
5
2
EN
75
T
(+
20
3.0
EC
OM
6
15
0
0.0
4
PO
N
0.6
4.0
AC
5.0
1.0
RE
0.28
0.2
10
0.8
IN D
0.25
0.26
0.24
0.27
0.23
0.25
0.24
0.26
0.23
0.27
R E FL E C T I O N C O E FFI C I E N T I N D E G
REES
L E OF
ANG
I SSI O N C O E F F I C I E N T I N
T R A N SM
DEGR
L E OF
EES
UCT
IV E
20
85
15
1.0
0.22
TA
NC
80
0.4
0.3
0.8
ANG
0.6
10
0.1
0.4
20
0.2
50
20
10
5.0
4.0
3.0
2.0
1.8
1.6
1.4
1.2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
50
50
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
0.2
20
0.4
0.1
CE
1.0
T
UC
7
0.6
0.0
3
0.4
-12
0
0.0
8
2
0.4
0.4
1
0.4
0.39
0.38
1
0.9
0.8
15
2
0.7
3
0.6
2.5
2
1.8
1.6
1.4
8
6
5
4
3
10
3
4
0.5
0.4
5
6
0.3
7
8
0.2
9
10
0.1
12
14
0.05
1.2 1.1 1
2
1
15
T O W A R D L O A D —>
10
7
5
1 1
1.1
20
30∞ 0
0.01
0 0
1.1
0 1
0.99
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
CENTER
1
ORIGIN
Figure 15. The Smith chart.
37
0.1
0.2
1.2
1.3
0.95
4
1.2
1.3 1.4
0.4
0.6
1.4
0.8
1.5
0.9
<— T O W A R D G E N E R A T O R
2
1
3
1.6
1
1.5
2
1.6 1.7 1.8 1.9 2
0.8
0.7
3
4
3
4
5
0.6
0.5
10
5
2.5
3
0.4
∞
10 15∞
4
0.3
20
6
0.2
5
10 ∞
0.1
0
SM
.C
20
4
S R
S A
AN .W. FL. .W. TTE
SM PEA LO LO N.
. C K SS [ SS [dB
OE (CO dB CO ]
EF
FF
FF N ]
F
,E
, P ST
.P
or
)
I
-70
6
4
30
-1
0
1
5
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
AN
40
-1
( -j
0.5
2.0
0.12
OM
-65
1.8
1.6
1.4
1.2
5
-4
0.13
0.37
EC
0.0
9
-110
0.1
0.11
-100
-90
0.14
0.36
0.8
1.0
-80
0.35
0.9
-4
0
0.15
0
-70
6
-5
0.1
4
0.3
-55
-35
3
0.3
R
0.7
CA
IV E
-60
-60
7
2
0.1
PA C
IT
NC
0.0
X/
0.2
-30
2
TA
T
5
),
Zo
O
R
-75
IN
D
0.3
EA
C
EN
0.4
8
0.1
0
-5
-25
PO
N
0.0
0.4
5
0.4
0.6
3.0
-20
∞ 40 30
10
TR
IV
0.8
1
0.3
R BS ]
SW d [dB F, P or I
F
SS OE F, E
LO . C EF
N. RFL CO
L.
RF
RT
20
OE
U
ES
SC
<—
T
EP
4.0
0.3
0.1
9
1.8
RADIALLY SCALED PARAMETERS
∞ 100 40
TR
0
-15 -80
WA
AN
-85
4
0.0
1.0
-15
6
5.0
0.2
-4
0
0.4
0.2
9
0.3
0.28
0.2
1
-30
0.4
0.8
0.22
0.2
-20
V EL
-160
( -jB
)
/ Yo
0.6
-10
-90
10
0.0 —> W A V E L E
0.49
N GTH
S TOW
ARD
0.0
0.49
A D <—
GEN
ARD LO
ERA
0.48
± 180
TO
170
R—
-170
0.47
>
160
90
8
0.3
50
9
0.2
30
S TOW
0.1
30
0.2
1
0.2
0.48
3
0.2
TH
EN G
7
0.3
60
0.3
0.47
0.1
35
o)
40
0.0
TI
CE
0.3
4
1
0.4
5
CI
AN
0.1
6
70
40
0.3
O
PA
PT
0.15
0.35
80
9
,
o)
CA
R
SU
VE
E
SC
/Y
( +jB
0.14
0.36
90
0.1
0.5
65
3
0.4
0
13
55
0.6 60
7
0.37
1.4
0
12
0.7
2
0.4
0.0
110
1
0.4
8
0.0
0.13
0.38
50
0.4
9
0.0
0.12
0.39
100
1.2
0.1
Z0 = 50Ω. Find ΓL .
Solution: First, normalize the load impedance in the following manner:
zl =
25 + j50
ZL
= .5 + j1.
=
Z0
50
1) Next, plot zl on the Smith chart (point A in Fig. 16).
2) Then draw a line from the origin through the point and to the outer circle.
3) Then read ϕ from the circle labeled “ANGLE OF REFLECTION COEFFICIENT IN
DEGREES”. It looks to be about 82.5 degrees.
4) Next, measure the distance from the origin to the plotted point A.
5) Finally, determine |Γ| on the scales at the bottom of Fig. 16. It looks to be about 0.62.
Therefore from the Smith chart we have Γ = .62∠82.5o .
38
The Smith Chart
0.11
70
0.4
2
45
1.0
50
0.8
25
0.4
jX
/
Z
2.0
6
0 .0
4
0.4
8
0.3
50
20
T
(+
A
EN
3.0
0.6
4
6
15
0
EC
OM
TA
NC
80
0.4
AC
1.0
RE
0.2
0.8
IN D
0.25
0.26
0.24
0.27
0.23
0.25
0.24
0.26
0.23
0.27
R E FL E C T I O N C O E FFI C I E N T I N D E G
REES
L E OF
ANG
I SSI O N C O E F F I C I E N T I N
T R A N SM
DE
L E OF
UCT
IV E
20
85
0.28
10
0.6
10
0.1
0.4
20
0.2
50
20
10
5.0
4.0
3.0
2.0
1.8
1.6
1.4
1.2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
50
Origin
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
0.2
20
0.4
10
( -jB
CE
EP
1.0
N
TA
I
CT
VE
0.8
IN
DU
2.0
1.8
1.6
1.2
5
-4
0.15
0.14
-80
0
-70
4
-1
6
3
0.4
0
-12
0.0
8
0.4
2
0.4
1
0.1
0.4
0.39
0.38
0
1
0.9
5
0.8
0.9
2
0.7
4
3
15
0.6
0.8
3
4
0.5
0.4
0.7
2.5
2
1.8
1.6
1.4
8
6
5
4
3
10
5
6
0.3
0.6
7
8
0.2
0.5
9
10
0.1
0.4
12
14
0.05
0.3
0.2
1.2 1.1 1
2
1
15
T O W A R D L O A D —>
10
7
5
1 1
1.1
20
30∞ 0
0.1
0.01
0 0
1.1
0.1
0 1
0.99
0.2
1.2
1.3
0.95
4
1.2
1.3 1.4
0.4
0.6
1.4
0.8
1.5
0.9
<— T O W A R D G E N E R A T O R
2
1
3
1.6
1
1.5
2
1.6 1.7 1.8 1.9 2
0.8
0.7
3
4
3
4
5
2.5
0.6
0.5
10
5
3
0.4
∞
10 15∞
4
0.3
20
6
0.2
5
10 ∞
0.1
0
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Figure 16. The Smith chart for example 1.
39
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
AN
0.1
ORIGIN
TR
0.0
CENTER
1
SM
.C
1
10
20
TR
0.37
0.0
7
-1
30
20
∞ 40 30
S R
S A
AN .W. FL. .W. TTE
SM PEA LO LO N.
. C K SS [ SS [dB
OE (CO dB CO ]
EF
FF
FF N ]
F
,E
, P ST
.P
or
)
I
0.36
0.12
T
0.0
9
-110
0.11
-100
-90
0.13
AC
( -j
R BS ]
SW d [dB F, P or I
F
SS OE F, E
LO . C EF
N. RFL CO
L.
RF
RT
∞ 100 40
OE
1.0
-4
0
0.35
0.9
-70
6
0
0.1
4
0.3
AC
-5
2
3
0.3
0.8
1.4
CAP
0.7
1.8
-35
-55
-60
7
0.1
RE
P
0.6
2
ITI
VE
OM
T
0
-65 .5
0.2
-30
-60
8
EC
EN
0.0
Z
X/
0.1
0.3
AN
C
ON
4
0
-5
-25
1
0.3
0.4
9
0.1
0.0
0.4
5
,
o)
O
R
-75
0.6
3.0
-20
0.3
5
0.4
4.0
4
0.0
0
-15 -80
SU
SC
1.0
-15
0.2
-4
0
0.4
0.2
9
0.2
1
-30
0.3
0.28
0.22
5.0
0.2
-20
-85
0.8
-10
0.48
)
/ Yo
0.6
RADIALLY SCALED PARAMETERS
1
0.22
5.0
GREE
S
6
15
1.0
0.1
0.4
4.0
50
0.0
0.3
0.8
ANG
0.0 —> W A V E L E
0.49
N GTH
S TOW
ARD
0.0
0.49
GEN
L O A D <—
D
R
A
ERA
TOW
0.48
S
7
180
±
H
.4
TO
T
0
170
70
NG
R—
-1
E
L
VE
0.47
>
A
W
1
0
6
<—
6
0
-90
90
-1
1
0.2
9
0.2
30
PO
N
0.3
75
0.1
30
0.2
0.2
14
0
( +jB
40
0.0
5
TI
CE
1
0.4
5
CI
AN
0.3
3
60
0.3
O
PA
PT
0.1
7
35
)
/ Yo
9
,
o)
CA
R
SU
VE
E
SC
0.3
4
0.1
0.5
65
3
0.4
0
13
0.1
6
70
40
1.8
7
0.0
0.35
1.6
0
12
0.15
0.36
80
1.4
2
0.6 60
0.4
0.7
0.0
Angle of reflection
coefficient
0.14
90
0.9
110
55
8
0.37
0.39
100
0.4
1
0.4
0.13
0.38
1.2
0.1
9
0.0
0.12
1.14.3. Input impedance example
We can also determine the input impedance of the TL. If the length of the TL is l = 60 cm,
λ = 2 m, ZL = 25 + j50 and Z0 = 50Ω, find zin .
Solution: Again, normalize the load impedance in the following manner:
zl =
ZL
25 + j50
=
= .5 + j1.
Z0
50
Next, we need to determine the length of the TL in terms of the wavelength of the source. ⇒
l/λ = 0.6/2 = 0.3. ⇒ l = 0.3λ. This means that we have to move 0.3λ down the TL from the load
to the source. This then places us at the port of the TL. To do this on the Smith chart, we need
to extend the line through point A to intersect the circle labeled “WAVELENGTHS TOWARD
GENERATOR”. This is shown in Fig. 17. This is then our reference value. This intersection
happens at l1 = 0.1345λ. Next, we need move along the circle labeled “WAVELENGTHS TOWARD
GENERATOR” a distance of 0.3λ (which is the length of the TL). Adding the reference value to
the length of the TL gives 0.3λ + 0.1345λ = 0.4345λ. This added value tells us what value we
need to move to along the “WAVELENGTHS TOWARD GENERATOR” circle to be at the port
of the TL. Next, rotate the line going through A around the Smith chart in Fig. 17 in the clockwise
direction. We are now at point B. Notice that a line extending from the origin through B intersects
the “WAVELENGTHS TOWARD GENERATOR” circle at 0.4345λ. Reading the normalized input
impedance value at point B gives zin = 0.28 − j0.4. ⇒ Zin = 50(zin ) = 14 − j20Ω. Note that the
analytical solution using (1.80) gives 13.7-j20.2.
40
The Smith Chart
0.1345 λ
0.11
0.0
4
70
0.4
0.4
45
1.0
0.9
0.8
55
1.6
1.8
0.4
A
EN
75
EC
O
4.0
TA
TO
NC
80
0.8
AC
IV E
85
1.0
RE
0.47
160
0.2
UCT
10
IN D
0.8
0.6
90
5.0
0.25
0.2
6
0.24
0.27
0.23
0.25
0.24
0.26
0.23
0.27
R E FL E C T I O N C O E FFI C I E N T I N D E G
REES
L E OF
ANG
I SSI O N C O E F F I C I E N T I N
T R A N SM
DEGR
L E OF
EES
ARD
15
1.0
0.28
ERA
M
15
0
6
0.4
R—
>
0.3
0.0
4
PO
N
0.6
0.22
GEN
20
3.0
T
(+
jX
/
Z
25
ANG
0.48
50
20
170
10
0.1
0.49
0.4
TOW
2
1
0.2
9
0.2
30
GTH S
8
0.3
0.3
0.0 —> W A V E L E
N
0.1
30
0.2
0.2
14
0
)
/ Yo
( +jB
3
40
0.0
5
S
VE
7
0.3
60
1
R
TI
CE
0.1
35
0.3
O
CI
AN
0.3
4
9
,
o)
C
A
AP
C
US
T
EP
0.1
6
70
40
2.0
0.5
65
3
0.4
0
13
0.15
0.35
80
0.1
0.4
5
0
12
7
6
0.0
2
0.14
0.36
90
1.4
0.4
0.37
0.7
8
0.6 60
0.0
110
1
0.4
0.13
0.38
50
0.4
9
0.0
0.12
0.39
100
1.2
0.1
20
0.2
50
20
10
5.0
4.0
3.0
2.0
1.8
1.6
1.4
1.2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
± 180
L O A D <—
0.2
ARD
S TOW
GTH
-170
LEN
20
1.0
R
0.0
40
-70
-1
6
0.0
0.5
0.6
1.8
1.6
0.37
-65
2.0
0.7
1.4
1.2
0.12
3
0.4
-12
0
8
0.0
2
0.4
0.0
9
-110
0.4
1
0.1
0.11
-100
-90
0.13
0.36
0.8
0.9
5
0.14
-80
0.35
0.0
7
-1
30
0.4
0.39
0.38
0
1
20
1
0.9
0.8
0.9
3
15
2
0.7
4
0.6
0.8
2
1.8
1.6
1.4
8
6
5
4
3
10
3
4
0.5
0.4
0.7
2.5
5
6
0.3
0.6
7
8
0.2
0.5
9
10
0.1
0.4
0.3
12
14
0.05
0.2
1.2 1.1 1
2
1
15
T O W A R D L O A D —>
10
7
5
1 1
1.1
20
30∞ 0
0.1
0.01
0 0
1.1
0.1
0 1
0.99
0.2
1.2
1.3
0.95
4
1.2
1.3 1.4
0.4
0.6
1.4
0.8
1.5
0.9
<— T O W A R D G E N E R A T O R
2
1
3
1.6
1
1.5
2
1.6 1.7 1.8 1.9 2
0.8
0.7
3
4
3
4
5
2.5
0.6
0.5
10
5
3
0.4
∞
10 15∞
4
0.3
20
6
0.2
5
10 ∞
0.1
0
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Figure 17. The Smith chart for example 2.
41
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
AN
0.1
ORIGIN
TR
0.0
CENTER
1
SM
.C
1
5
S R
S A
AN .W. FL. .W. TTE
SM PEA LO LO N.
. C K SS [ SS [dB
OE (CO dB CO ]
EF
FF
FF N ]
F
,E
, P ST
.P
or
)
I
IN
0
AC
P
∞ 40 30
10
TR
EP
TA
1.0
-4
0.15
-4
4
0.3
0
-70
6
-5
0.1
-55
-35
3
A
RE
-60
2
0.3
-85
SC
U
ES
V
TI
CAP
<
-75
O
-60
7
1.8
0.1
VE
M
5
),
Zo
2
C IT
I
CO
( -j
R BS ]
SW d [dB F, P or I
F
SS OE F, E
LO . C EF
N. RFL CO
L.
RF
RT
20
OE
)
/ Yo
0.2
-30
CE
T
5
X/
8
-90
( -jB
C
DU
0.1
0.3
TA
N
EN
4
0.4
0
-5
-25
ON
-160
E
NC
0.4
0.4
0.6
3.0
-20
4
0.0
0
-15 -80
0.8
6
0.4
-15
A
—W
1.0
0.3
1
0.3
0.3 λ rotation
RADIALLY SCALED PARAMETERS
∞ 100 40
0.28
0.2
0.1
9
0.4345 λ
-4
0
0.4
0.2
9
0.2
1
-30
0.3
0.22
B
4.0
0.2
5.0
VE
0.8
-10
0.48
0.6
-20
0.47
0.4
0.1
10
0.49
Origin
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
50
0.0
50
1.14.4. Voltage minimum and maximum example
We can also determine the voltage minimum and maximum values along the TL using the
Smith chart. Again, let ZL = 25 + j50 and Z0 = 50Ω and determine zmax or zmin closest to the
load. Note that a zmax occurs when r > 1 and zmin occurs when r < 1.
Solution: For our example we have
zl =
ZL
25 + j50
=
= .5 + j1.
Z0
50
Again, we plot point A (i.e., zl ) on the Smith chart in Fig. 18. Using the line drawn from the origin
through point A as a reference, we move from the load to the source along the “WAVELENGTHS
TOWARD GENERATOR” circle. When the line has rotated to point B we have encountered
our first voltage extremum. Since r = 4.3 > 1 at point B, then we have a voltage maximum.
This maximum occurs at 0.25λ − 0.1345λ = 0.1155λ from the load. This implies that the voltage
minimum occurs λ/4 further down the TL.
42
The Smith Chart
0.1345 λ
0.11
0.0
6
70
0.4
4
0
14
45
1.0
1.8
25
0.4
A
EN
75
T
0.4
N
PO
M
0.0
4
0.4
6
15
0
EC
O
4.0
TA
TO
NC
80
R—
>
0.3
AC
1.0
1.0
RE
IV E
UCT
85
160
0.2
IN D
0.6
90
10
0.8
ARD
5.0
0.25
0.26
0.24
0.27
0.23
0.25
0.24
0.26
0.23
C
0.27
R E FL E C T IO N O E FFI C I E N T I N D E G
REES
L E OF
ANG
I SSI O N C O E F F I C I E N T I N
T R A N SM
DEGR
L E OF
EES
0.48
15
20
GEN
0.47
ERA
0.8
0.28
170
10
0.1
0.49
0.4
S T OW
20
3.0
0.6
0.22
N GT H
2
ANG
20
0.2
L O A D <—
50
20
10
5.0
4.0
3.0
2.0
1.8
1.6
1.4
1.2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
± 180
Origin
B
x=0
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
0.2
ARD
S T OW
GTH
-170
LEN
20
o)
/Y
( -jB
A
PT
1.0
E
SC
U
ES
IV
D
IN
R
-70
0.6
3
8
0.0
0.4
2
0.0
0.4
1
0.1
0.4
0.39
0.38
∞ 40 30
0
20
1
0.9
5
0.8
3
15
2
0.7
4
0.6
2.5
2
1.8
1.6
1.4
8
6
5
4
3
10
3
4
0.5
0.4
5
6
0.3
7
8
0.2
9
10
0.1
12
14
0.05
1.2 1.1 1
2
1
15
T O W A R D L O A D —>
10
7
5
1 1
20
30∞ 0
1.1
0.1
0.01
0 0
1.1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0 1
0.99
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.2
1.2
1.3
0.95
4
1.2
1.3 1.4
0.4
0.6
1.4
0.8
1.5
0.9
<— T O W A R D G E N E R A T O R
2
1
3
1.6
1
1.5
2
1.6 1.7 1.8 1.9 2
0.8
0.7
3
4
3
4
5
2.5
0.6
0.5
10
5
3
0.4
∞
10 15∞
4
0.3
20
6
0.2
5
10 ∞
0.1
0
ORIGIN
Figure 18. The Smith chart for example 3.
43
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
AN
1.1
TR
CENTER
1
SM
.C
1
10
S R
S A
AN .W. FL. .W. TTE
SM PEA LO LO N.
. C K SS [ SS [dB
OE (CO dB CO ]
EF
FF
FF N ]
F
,E
, P ST
.P
or
)
I
0
4
-1
6
0.0
0.5
2.0
0.37
-65
1.8
1.6
5
0.12
0.0
7
-1
30
0.4
0
-12
9
-110
0.11
-100
-90
0.13
0.36
0.7
1.4
1.0
0.14
-80
0.35
0.9
1.2
0.15
-4
4
0.3
0
-4
0
-70
-5
6
0.1
0.8
3
A
-55
-35
0.3
RE
-60
CA
IV E
M
0.0
X/
-60
7
2
0.1
PA C
IT
EC
O
( -j
R BS ]
SW d [dB F, P or I
F
SS OE F, E
LO . C EF
N. RFL CO
L.
RF
RT
20
TR
-75
,O
0.2
-30
2
NC
T
5
)
Zo
0.3
1.8
RADIALLY SCALED PARAMETERS
∞ 100 40
OE
-85
<
4
0.0
0
-15 -80
T
UC
1
CT
A
EN
4
8
0.1
0
-5
-25
0.3
0.4
0.1
9
PO
N
-160
E
NC
0.4
5
0.6
0.4
0.3
0
-4
0.4
0.2
-20
6
0.4
0.8
9
3.0
0.48
A
—W
1.0
0.2
1
-30
0.2
0.3
0.28
0.22
4.0
VE
0.8
-15
0.2
5.0
-90
10
0.6
-20
0.47
0.4
0.1
-10
0.0
50
50
0.0 —> W A V E L E
8
0.3
50
1
0.2
9
0.2
30
0.49
0.1
30
0.2
(+
jX
/
Z
0.4
5
E
0.2
0 .0
S
VE
NC
0.3
5
TI
TA
7
40
,O
CI
EP
)
/ Yo
( +jB
3
1
0.3
o)
CA
R
PA
C
US
0.3
60
9
0.1
0.5
65
3
0.4
0
13
0.1
35
2.0
7
0.0
0.3
4
1.6
0
12
0.1
6
70
40
1.4
2
0.6 60
0.4
0.15
0.35
80
0.7
0.0
0.14
0.36
90
0.8
55
8
0.37
0.9
110
1
0.4
0.13
0.38
50
0.4
9
0.0
0.12
0.39
100
1.2
0.1
1.14.5. Stub matching example
Next we want to match a load to a 50 Ω TL by placing a short-circuit stub a distance of d
from the load. This is shown in Fig. 19. Assume the length of the short circuit stub is d1 and
has a characteristic impedance of Z0s = 50 Ω (same as the TL). We need to determine d and d1
in Fig. 19. To do this we work with admittances. For this example let zL = 2.1 + j.8. Next, plot
zL on the Smith chart in Fig 20 (point A). Then draw a line from A through the origin to the
intersection of the circle about the origin with a radius of point A. This intersection with the circle
is shown as point B in Fig. 20. The intersection is the admittance of zL and has the normalized
value yL = .42 − j.16. Also note that we use .47 λ as our reference value on the “WAVELENGTHS
TOWARDS GENERATOR” scale. We will use this value to measure how far we need to move
towards the generator to observe an input admittance value of 1 ± jb. Next, we rotate towards the
generator from point B around the circle with a radius of point A to point C on the Smith chart in
Fig. 20. Point C has an input admittance of 1 + j.95. We can also rotate around to point D which
has an input admittance of 1 − j.95. The points C and D are important because both points have
a real value of unity and a reactance that can be canceled with a reactive load.
d
d1
Z0
ZL
Zin = Z0+j0
(with stub)
short-circuit stub
Figure 19. Short-circuit stub matching example.
Next, the distance from point B to point C along the TL is .16λ + (.5 − .47)λ = .19λ and the
distance from point B to point D along the TL is .34λ + (.5 − .47)λ = .37λ. Choosing the shortest
44
distance gives d=.19 λ. At this point (i.e., point C) the input admittance is 1 + j.95. Therefore, we
need to add a stub with an admittance of −j.95 at this point. Since the stub in Fig. 19 is a short
circuit then zs = 0 ⇒ ys = ∞ which is denoted as point E on the Smith chart in Fig. 20. Next, we
need to determine how long the stub needs to be to have an input admittance of −j.95. Plotting
−j.95 on the Smith chart results in the point denoted as F in Fig. 20. Next, we move from point
E to point F towards the generator and this distance is the length of the stub. This gives a value
of d1 = .379λ − .25λ = .129λ. Another way to think about d1 is to view d1 as the distance needed
to travel from the short-circuit load down the TL to observe an input admittance of −j.95. This
distance traveled is then the length of the short-circuit stub. Now adding the admittances at the
junction of the TL and short-circuit stub in Fig. 19 gives yin = 1 + j.95 − j.95 = 1. ⇒ zin = 1. ⇒
Zin = 50 ∗ 1 = 50Ω = Z0 .
45
The Smith Chart
0.11
70
45
1.0
0.8
55
20
EN
75
EC
O
AC
IV E
UCT
85
1.0
RE
0.47
160
0.2
GEN
10
IN D
0.8
0.6
90
5.0
0.25
0.2
6
0.24
0.27
0.23
0.25
0.2
4
0.26
0.23
0.27
FL E C T IO N C O E FFI C I E N T
E
I
N
R
D
E
GREE
L E OF
S
ANG
I SSI O N C O E F F I C I E N T I N
T R A N SM
DEGR
L E OF
EES
ARD
C
0.28
1.0
ANG
0.48
15
4.0
TA
TO
NC
0.8
0.22
ERA
M
0
15
80
R—
>
0.4
6
0.3
0.0
4
PO
N
0.6
20
170
10
0.1
0.4
S T OW
3.0
T
0.4
(+
jX
/
Z
2.0
6
0.0
0.4
4
0
14
0.4
5
25
0.4
1
0.2
9
0.2
30
N GT H
50
2
0.2
0.49
A
20
0.2
L O A D <—
50
20
10
5.0
4.0
3.0
2.0
1.8
1.6
1.4
1.2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
± 180
Origin
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
0.2
)
/ Yo
jB
E (-
1.0
NC
E
SC
U
ES
IV
0.8
T
UC
ND
I
,O
R
-75
o)
Z
X/
1.2
0.15
5
0
-4
4
0.3
-4
1.0
-70
0.34 λ
0.14
-80
0.35
0.9
6
0.1
0.36
0.12
0.37
0.0
40
-70
3
8
0.4
2
0.4
1
0.1
0.11
-100
-90
0.13
0.0
7
30
-1
0.4
0
-12
0.0
9
-110
0
3
-5
-35
0.3
A
AC
( -j
0.5
2.0
1.8
1.6
1.4
CAP
0.8
-60
0.7
7
RE
-55
0.1
VE
M
T
0.6
2
C IT
I
CO
CE
EN
-65
0.2
-30
-60
0.3
TA
N
PO
N
6
0.0
8
0.1
0
-5
-25
1
0.3
F
0.1
9
4
0.4
0.4
-1
-20
0.3
5
0.0
0.2
0
0.4
-4
5
0.4
0.6
3.0
4
0.0
0
-15 -80
-15
-90
A
PT
1.0
4.0
0.3
0.2
9
-30
6
0.4
D
0.28
VE
WA
<—
-160
-85
0.8
5.0
0.2
0.22
0.47
s = 2.5 circle
-10
0.48
E
0.6
-20
ARD
S T OW
GTH
-170
LEN
20
0.4
0.2
1
0.47 λ
B
0.1
10
0.0
50
50
0.0 —> W A V E L E
8
0.3
0.3
0.49
0.1
30
0.2
40
0.0
S
VE
7
1
5
TI
CE
)
/ Yo
( +jB
3
0.3
CA
R
CI
AN
0.1
0.3
60
9
o)
,O
PA
C
US
T
EP
0.3
4
35
0.1
0.5
65
3
0.4
0
13
70
40
1.8
7
0.0
0.1
6
1.6
0
0.16 λ
0.15
0.35
1.4
12
0.6 60
0
0.14
0.36
80
90
0.7
0.0
.42
0.37
0.9
110
1
0.4
8
0.13
0.38
50
0.4
9
0.0
0.12
0.39
100
1.2
0.1
0.4
0.39
0.38
0.379 λ
20
∞ 40 30
0
1
1
0.9
5
0.8
0.9
2
0.7
4
3
15
0.6
0.8
3
4
0.5
0.4
0.7
2.5
2
1.8
1.6
1.4
8
6
5
4
3
10
5
6
0.3
0.6
7
8
0.2
0.5
9
10
0.1
0.4
12
14
0.05
0.3
0.2
1.2 1.1 1
2
1
15
T O W A R D L O A D —>
10
7
5
1 1
1.1
20
30∞ 0
0.1
0.01
0 0
1.1
0.1
0 1
0.99
0.2
1.2
1.3
0.95
4
1.2
1.3 1.4
0.4
0.6
1.4
0.8
1.5
0.9
1
1.8
1.5
2
2
1.6 1.7 1.8 1.9 2
0.8
0
2
<— T O W A R D G E N E R A T O R
2
1
3
1.6
0.7
3
4
3
4
5
2.5
0.6
0.5
10
5
3
0.4
∞
10 15∞
4
0.3
20
6
0.2
5
10 ∞
0.1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
CENTER
1
ORIGIN
Figure 20. The Smith chart for example 4.
46
SM
.C
1
10
20
TR S.W RF S.W AT
AN . L.
. T
SM PEA LO LO EN.
. C K SS [ SS [dB
OE
OE (CO dB CO ]
EF
FF
FF N ]
F
,E
, P ST
.P
or
)
I
R BS ]
SW d [dB F, P or I
F
SS OE F, E
LO . C EF
N. FL O
RT R FL. C
R
∞ 100 40
TR
AN
RADIALLY SCALED PARAMETERS
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.14.6. Load impedance example 1
Let s = 3 on a 50 Ω TL. We have zmin = −5cm from the load and the distance between the
minimum values is 20cm. Find ZL .
Solution: We know the value between the minimum values is λ/2 = 20cm. ⇒ λ = 40cm. This
then gives zmin = −5cm/40cm = −0.125λ. Point A on the Smith chart in Fig. 21 corresponds to
the s = 3 circle and point B corresponds to the zmin value (i.e., r<1). Now, move towards the load
0.125λ. ⇒ λ′ = (0+0.125)λ = 0.125λ. Point C in Fig. 21 is zL = .6−j.8. ⇒ ZL = 50zL = 30−j40.
47
The Smith Chart
0.11
0.0
6
70
0.4
4
0
14
45
1.0
0.9
0.8
55
1.8
0.4
20
EN
75
EC
OM
TA
TO
4.0
AC
1.0
1.0
RE
IV E
UCT
85
160
0.2
IN D
0.6
90
10
0.8
ARD
5.0
170
10
0.1
0.49
0.4
20
50
20
10
5.0
4.0
3.0
2.0
1.8
1.6
1.4
A
Origin
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
A D <—
ARD LO
1.2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
± 180
50
50
0.2
/ Yo
)
-90
jB
E (-
1.0
C
AN
T
P
CE
VE
I
CT
DU
IN
,
o)
OR
-75
5
X
/Z
0.0
-20
0.14
-80
0.35
0.36
0.12
0.37
0.0
40
-70
6
-1
0.0
7
30
3
8
0.4
0.4
0.1
-1
( -j
0.4
0
-12
0.0
9
-110
0.11
-100
-90
0.13
AC
M
T
0.5
2.0
1.8
1.6
1.2
0.15
5
-4
0
-4
4
0.3
1.0
-70
0.9
6
0.1
0
-35
3
A
-5
2
0.3
0.8
1.4
CAP
RE
0.7
-60
7
1.8
0.1
VE
-55
2
C IT
I
CO
CE
EN
0.6
0.2
-30
-60
0.3
TA
N
PO
N
-65
8
0.1
0
-5
-25
0.3
1
4
9
0.0
0.4
0.1
0.4
0.3
5
0.4
0.6
3.0
4
0.0
0
-15 -80
0.8
0
-15
SU
S
1.0
0.2
-4
0
0.4
0.2
9
.46
V
WA
<—
-160
-85
0.28
0.22
C
0.2
1
-30
0.3
4.0
T OW
TH S
NG
ELE
0.8
5.0
0.2
-20
0.47
E
s = 3.0 circle
0.6
-10
-170
20
0.4
0.1
10
0.0
0.2
B
R EES
0.48
0.25
0.26
0.24
0.27
0.23
0.25
0.24
0.26
0.23
C
0.27
R E FL E C T IO N O E FFI C I E N T I N D E G
REES
L E OF
ANG
I SSI O N C O E F F I C I E N T I N
T R A N SM
DEG
L E OF
0.48
15
20
GEN
0.47
ERA
0.8
NC
80
0.3
R—
>
0.4
6
15
0
0.0
4
PO
N
0.6
0.28
S T OW
3.0
T
0.4
(+
jX
/
Z
0.4
5
25
0.22
N GT H
50
2
ANG
0.0 —> W A V E L E
8
0.3
1
0.2
9
0.2
30
0.49
0.1
30
0.2
0.2
0.0
S
VE
E
0.3
5
R
TI
NC
)
/ Yo
( +jB
3
40
,O
CI
TA
7
0.3
60
1
0.3
o)
CA
PA
US
P
CE
0.1
35
9
0.1
0.5
65
3
0.4
0
13
0.3
4
2.0
7
0.0
0.1
6
70
40
1.6
0
12
0.15
0.35
1.4
2
0.14
0.36
80
0.7
0.4
0.6 60
0
0.37
90
110
1
0.4
.08
0.13
0.38
50
0.4
9
0.0
0.12
0.39
100
1.2
0.1
2
1
0.4
0.39
0.38
0.125λ
0
1
1
0.9
5
0.8
0.9
2
0.7
4
3
15
0.6
0.8
3
4
0.5
0.4
0.7
2.5
2
1.8
1.6
1.4
8
6
5
4
3
10
5
6
0.3
0.6
7
8
0.2
0.5
9
10
0.1
0.4
12
14
0.05
0.3
0.2
1.2 1.1 1
2
1
15
T O W A R D L O A D —>
10
7
5
1 1
1.1
20
30∞ 0
0.1
0.01
0 0
1.1
0.1
0 1
0.99
0.2
1.2
1.3
0.95
4
1.2
1.3 1.4
0.4
0.6
1.4
0.8
1.5
0.9
1
1.5
2
1.6 1.7 1.8 1.9 2
0.8
0
2
<— T O W A R D G E N E R A T O R
2
1
3
1.6
0.7
3
4
3
4
5
2.5
0.6
0.5
10
5
3
0.4
∞
10 15∞
4
0.3
20
6
0.2
5
10 ∞
0.1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
ORIGIN
Figure 21. The Smith chart for example 5.
48
NS
CENTER
1
M
.C
1
10
20
TR S.W RF S.W AT
AN . L.
. T
SM PEA LO LO EN.
. C K SS [ SS [dB
OE (CO dB CO ]
EF
FF
FF N ]
F
,E
, P ST
.P
or
)
I
20
∞ 40 30
OE
R BS ]
SW d [dB F, P or I
F
SS OE F, E
LO . C EF
N. RFL CO
L.
RF
RT
∞ 100 40
TR
A
RADIALLY SCALED PARAMETERS
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.14.7. Load impedance example 2
A lossless 100Ω TL of length 3λ/8 is terminated with an unknown impedance. If Zin = −j2.5
then calculate
a) ZL
b) What length of open and short circuit stubs could be used to replace ZL ?
Solution:
a) To calculate the unknown load we first need to normalize the input impedance. ⇒ zin =
Zin /Z0 = −j.025. This point is plotted as point A in Fig. 22. Next, we need to express the length
of the TL in terms of a wavelength. ⇒ L = 3λ/8 = .375λ. This means we need to move from the
source on the TL to the load that is .375λ down the TL. The scale we use for this rotation is labeled
“WAVELENGTHS TOWARDS LOAD”. Rotating .375λ from point A at .004λ puts us at point B
on the Smith chart in Fig. 22. Point B is at .379λ. The load impedance at point B is zL = +j.95.
⇒ ZL = +j95Ω.
b) Next, to find the length of the the open circuit stub that has an input impedance of ZL
we start by plotting point C (zL = ∞) on the Smith chart in Fig. 22. Next, moving towards the
generator on the “WAVELENGTHS TOWARDS GENERATOR” scale to point B we get an input
impedance for the stub of zL = +j.95. The length of the stub is then do = .25λ + .121λ = .371λ.
Next, we plot the short circuit load (zl = 0) on the stub at point D on the Smith chart in Fig. 22.
Then moving towards the generator to point B gives a stub length of ds = .121λ.
What we have shown in part b) is that an open circuit stub of .371λ or a short circuit stub of
.121λ could be used as an equivalent load to the TL as the lumped load ZL determined in part a).
49
Finding the
length of the
short circuit
stub
The Smith Chart
0.11
0.0
70
0.4
4
0
14
45
1.0
1.4
0.8
7
1.6
o)
3
0.1
30
8
0.3
2
1.8
0.2
50
25
0.4
(+
jX
/
3.0
EN
75
T
0.4
20
0.6
EC
OM
NC
TA
AC
1.0
RE
0.8
IN D
0.25
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6
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0.23
0.25
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C
0.27
R E FL E C T IO N O E FFI C I E N T I N D E G
REES
L E OF
ANG
I SSI O N C O E F F I C I E N T I N
T R A N SM
D
L E OF
UCT
10
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0.4
20
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50
20
10
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1.8
1.6
1.4
1.2
0.9
1.0
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0.7
0.6
0.5
0.4
0.3
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50
Origin
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
50
0.2
20
0.4
0.1
/ Yo
( -jB
CE
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0.37
0
-70
4
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6
8
0.4
2
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0
0.1
0.11
-100
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0.38
R BS ]
SW d dB F, P or I
[
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N. RFL CO
L.
RF
RT
20
∞ 40 30
0
10
20
1
0.8
3
15
2
0.7
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2
1.8
1.6
1.4
8
6
5
4
3
10
3
4
0.5
0.4
5
6
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1 1
20
30∞ 0
1.1
0.1
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0 0
1.1
1
0.9
0.8
0.7
0.6
0.5
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0.3
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1
3
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0.7
3
4
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4
5
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10
5
3
0.4
∞
10 15∞
4
0.3
20
6
0.2
5
10 ∞
0.1
ORIGIN
Figure 22. The Smith chart for example 6.
50
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
AN
CENTER
1
SM
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0.9
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TR S.W RF S.W AT
AN . L.
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SM PEA LO LO EN.
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or
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5
-4
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3
0.4
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0
9
-110
0.0
7
30
-1
RADIALLY SCALED PARAMETERS
∞ 100 40
TR
1.2
1.0
-4
0
0.35
0.9
-70
0
6
0.1
Finding the
length of the
open circuit
stub
4
0.3
-5
-35
3
A
M
( -j
0.5
2.0
1.8
1.6
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0
80
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ANG
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S T OW
ARD
0.0
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0.49
GEN
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W
ERA
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± 180
TO
TH S
0.47
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17
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0
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-170
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V
0.47
>
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W
160
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-160
0.4
0.3
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1
9
0.2
30
0.8
10
D
1
0.2
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N
0.3
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0.1
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60
0.2
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E
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40
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NC
70
40
1
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R
PA
TA
0.1
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o)
CA
S
SU
P
CE
B
Distance
to the load
0.15
0.35
2.0
0.5
65
3
0.4
0
13
/Y
( +jB
0.14
0.36
80
9
0.1
5
0.6 60
7
6
0.0
0
12
0.7
2
0.4
55
0.4
8
0.37
90
0.9
110
1
0.0
0.13
0.38
50
0.4
9
0.0
0.12
0.39
100
1.2
0.1
1.15. Transient Analysis
1.15.1. Analytical solution
In this section we study the transient characteristics of a TL. To do this we consider the TL in
Fig. 23 a). The problem has a forward traveling wave with amplitude V + . The wave is propagating
at a wave velocity of ν (group velocity). At time t = 0, the DC battery is connected to the TL
through the switch. At this point a wave front starts propagating down the TL. The closing of the
switch can be modeled as the step function in Fig 23 b). If the TL is lossless, then the current
can be written as IL = V0 /RL . Now, if RL ̸= Z0 , then the reflected wave adds with the forward
traveling wave giving a total wave along the TL as V1+ + V1− were V1+ is the first forward traveling
wave and V1− is the first reflected wave. ⇒
ΓL =
RL − Z0
V−
= 1+
RL + Z0
V1
(1.91)
at the load. Since, Rg = 0, we have a reflected wave at the source. This then gives
Γg =
Zg − Z0
Z0
V+
=−
= −1 = 2− .
Zg + Z0
Z0
V1
(1.92)
Now V2+ propagates to the load. ⇒ V2− = ΓL V2+ . This returns to the battery and V3+ = −V2− .
Since |ΓL | < 1, the TL will eventually reach a steady state. ⇒
VL = V1+ + V1− + V2+ + V2− + V3+ + V3− + ...
= V1+ (1 + ΓL + Γg ΓL + Γg Γ2L + ...)
)
(
1 + ΓL
+
.
= V1
1 − Γg ΓL
(1.93)
If Rg ̸= 0 then
V1+ =
V0 Z0
.
Rg + Z0
51
(1.94)
+
V
t=0
I
+
+
+
V0
Z0
-
z=0
RL = Z0
VL
-
wave propagating at wave
velocity ν (group velocity).
z=l
a)
VL
V0
t
t’ = l/ν
b)
Figure 23. a) A general transmission line with a transient source and b) the step function used to model
the switch
1.15.2. Voltage reflection diagrams
A graphical method can be used to study the behavior of wave reflections on the TL. To do
this, consider the voltage reflection diagram in Fig. 24 a). On the abscissa (x-axis) we measure
the length of the TL and on the ordinate (y-axis) we measure time. As the wave is launched down
the TL at time zero and at position zero. This point is shown in the bottom left-hand corner of
Fig. 24 a). The first forward wave propagating down the TL reaches the end of the TL at time
l/ν. Then the wave reflects back to the source and arrives at the source at time 2l/ν...and so on.
This behavior results in the back and forth illustration in Fig. 24 a). If we consider the voltage at
z = 3l/4 along the TL then we have the voltage response w.r.t. time shown in Fig. 24 b). We can
see that the voltage slowly converges to V0 RL /(Rg + RL ).
Now say the line was initially charged. This then results in the problem described in Fig. 25.
We know the voltage to the left of the dotted line is less than V0 . ⇒ V1+ has a sign change. Also,
52
t
4l/ν
V2- = Γ
2
Γ V1+
g L
13l/4ν
3l/ν
11l/4ν
+
+ ΓgΓLV1
=
V2
2l/ν
V1- = Γ +
LV
1
5l/4ν
l/ν
3l/4ν
+
V1
z
l
3l/4
0
a)
V3/4
V0RL
Rg+RL
+ - + + - + V1+V1+V2+V2
V1+V1+V2
+ V1+V1
+
V1
3l/4ν 5l/4ν
t
11l/4ν 13l/4ν 19l/4ν 21l/4ν
b)
Figure 24. a) Voltage reflections on the transmission line and b) line voltage on the transmission line over
time
+
V1
t=0
+
Rg
V0+V+1
V0
+
I1
VR
Z0
(open)
{
IR
{
-
charge not in motion
charge in motion
Figure 25. Transient response of a charged transmission line.
53
IR = −I1+ = −V1+ /Z0 . ⇒ VR = V0 + V1+ = IR Rg = −I1+ Rg = −V1+ Rg /Z0 . ⇒
V1+ =
−V0 Z0
.
Z0 + Rg
54
(1.95)
CHAPTER 2. ELECTROSTATIC FIELDS
2.1. Vector Analysis
2.1.1. Vector notation
In this section we introduce the notation used for the remainder of the notes. Vectors are
denoted as Ā. In rectangular coordinates, Ā = Ax âx + Ay ây + Az âz . From this we define the
√
magnitude of Ā as |Ā| = A2x + A2y + A2z . The unit vector of Ā is defined as
âA =
Ā
Ax
Ay
Az
=
âx +
ây +
âz .
|Ā|
|Ā|
|Ā|
|Ā|
Ā and âA are drawn in Fig. 26.
z
z
az
A
ay
y
ax
x
x
Figure 26. Vector definitions.
2.1.2. Dot product
For this course, the dot product (Fig. 27) has the following notation:
Ā · B̄ = |Ā||B̄| cos(θAB )
= Ax Bx + Ay By + Az Bz
Also note that Ā · B̄ = B̄ · Ā and Ā · Ā = |Ā|2 .
55
y
y
A
B
θΑΒ
x
Figure 27. Dot product.
2.1.3. Cross product
For this course, the cross product (Fig. 28) has the following notation:
Ā × B̄ = âN |Ā||B̄| sin(θAB )
âx ây âz = Ax Ay Az Bx By Bz = (Ay Bz − Az By )âx − (Ax Bz − Az Bx )ây + (Ax By − Ay Bx )âz .
z
AXB
B
y
A
x
Figure 28. Cross product.
2.1.4. Gradient
Next, we discuss the gradient of a scalar function. If T (x, y, z) is our scalar function. Then
the gradient of T is defined as:
∇T =
∂T
∂T
∂T
âx +
ây +
âz
∂x
∂y
∂z
56
where the gradient operation is referred to as “del” and the operation is in cartesian coordinates.
After the operation of del on a scalar function T we have a vector whose magnitude is equal to the
maximum rate of change of the physical quantity (i.e., volt) per unit distance and whose direction
is along the direction of maximum increase. Now dot ∇T with a unit vector âl , ∇T · âl , then we
have a directional derivative.
2.1.5. Coordinate conversion example 1
Convert P1 (x = 1, y = 2, z = 3) to cylindrical coordinates.
√
Solution: P1 ( 12 + 22 , ϕ = tan−1 (2/1), z)
2.1.6. Coordinate conversion example 2
Convert P2 (ρ = 2, ϕ = 30, z = 4) to cartesian coordinates.
Solution: P2 (2 cos 30, 2 sin 30, 4)
2.1.7. Coordinate conversion example 3
Let F̄ = zâx − xây + yâz , determine F̄ in cylindrical coordinates.
Solution: We know x = ρ cos ϕ and y = ρ sin ϕ. ⇒



 Aρ   cos ϕ sin ϕ 0

 
 A  =  − sin ϕ cos ϕ 0
 ϕ  

 
Az
0
0
1


Ax = z


  A = −x = −ρ cos ϕ
 y

Az = y = ρ sin ϕ


.


⇒
F̄ = âρ [z cos ϕ − ρ sin ϕ cos ϕ]
+ âϕ [−z sin ϕ − ρ cos2 (ϕ)]
+ âz [ρ sin ϕ].
2.2. Coulomb’s Law
The following relation between two charges and the force between those two charges was
57
determined by Colonel Charles Coulomb. The result was the following:
F =k
Q1 Q2
R2
(2.1)
where Q1 and Q2 are the positive or negative charges, R is the separation distance between the
charges, F is the force between the charges and k is a constant. An illustration of the problem is
shown in Fig. 29. We also have k = 1/(4πε0 ) where ε0 = 8.854 × 10−12 F/m which is called the
free-space permittivity. Using (2.1) gives
F =
Q1 Q2
.
4πε0 R2
(2.2)
The units for Q are Coulombs (C), for R are meters (m) and F are Newtons (N). In vector form
we have
F̄1 = −F̄2 =
Q1
Q1 Q2
Q1 Q2
â21 = −
â12 .
2
2
4πε0 R12
4πε0 R12
a12
R12 =
r1
r2 - r1
(2.3)
Q2
r2
origin
Figure 29. The force between two charges as described by Coulomb’s equation.
2.3. Electric Field Intensity
2.3.1. Field from a single point charge
Now, fix Q1 and move a test charge Qt in the space around Q1 . This force per-unit charge is
Q1
F̄t
=
â .
2 1t
Qt
4πε0 R1t
58
(2.4)
Equation (2.4) describes a vector field. This field is called the electric field intensity. Therefore,
F̄t
Q1
=
â .
2 1t
Qt
4πε0 R1t
Ē =
(2.5)
The units of (2.5) is Volts/meter or V/m. Equation (2.5) describes the electric field from a single
point charge at a point anywhere in space at vector R1t . This is described in Fig. 30. This then
gives
r̄ − r̄′
Q
.
4πε0 |r̄ − r̄′ |2 |r̄ − r̄′ |
Ē(r̄) =
Therefore, in general we have
Ē(r̄) =
(2.6)
n
∑
Qm
â .
2 m
4πε
|r̄
−
r̄
|
0
m
m=1
P(x‘,y‘,z‘)
Q
R= r -
r‘
r‘
(2.7)
P(x,y,z)
E
r
origin
Figure 30. Electric field due to a single point charge.
2.3.2. Field from a charge distribution
In this section we will write an expression for the electric field intensity due to a continuous
∫
volume charge distribution. If we denote the volume charge as ρv (C/m3 ) then Q = v ρv dV where
Q is the total charge in Coulombs. This then gives
∫
Ē(r̄) =
V
ρv (r̄′ )dV ′ r̄ − r̄′
4πε0 |r̄ − r̄′ |2 |r̄ − r̄′ |
(2.8)
where the units are V/m.
2.3.3. Total charge example
Calculate Q in the volume: 0 ≤ ρ ≤ 0.1m, 0 ≤ ϕ ≤ π and 2m≤ z ≤ 4m where ρv = ρ2 z 2 sin .6ϕ.
59
Solution: Using a volume integral we have the following:
∫
π
∫
0.1
∫
4
ρ2 z 2 sin(0.6ϕ)ρdzdρdϕ
0
0
2
4
∫ π ∫ 0.1
1 3 3
=
ρ sin(0.6ϕ) z dρdϕ
3 2
0
0
0.1
∫ π
56
1 4 sin(0.6ϕ)dϕ ρ =
3 0
4 0
π
−6
56 100 × 10 − cos(0.6ϕ) =
3
4
0.6
0
Q =
= −777.7 × 10−6 (cos(0.6π) − 1)
= 1.0181mC.
2.3.4. Field from a line charge
Next, consider the infinite line of electric charge on the z-axis in Fig. 31. An incremental unit
of the electric field in the region around the line can be written as
ρl dz ′ (r̄ − r̄′ )
dĒ =
4πε0 |r̄ − r̄′ |3
(2.9)
where r̄ = yây = ρâρ , r̄′ = z ′ âz and r̄ − r̄′ = ρâρ − z ′ âz . This then gives
dĒ =
ρl dz ′ (ρâρ − z ′ âz )
.
4πε0 (ρ2 + (z ′ )2 )3/2
(2.10)
dEρ =
ρl ρdz ′
.
4πε0 (ρ2 + (z ′ )2 )3/2
(2.11)
⇒
⇒
∫
Eρ =
∞
−∞
ρl ρdz ′
.
4πε0 (ρ2 + (z ′ )2 )3/2
(2.12)
⇒
Eρ =
ρl
.
2πε0 ρ
60
(2.13)
z
ρl
dQ = ρldz’
aR
(0,0,z’)
R = r - r’
r’
θ
P
r
dEz
dEρ
y
dE
x
Figure 31. Electric field due to a line charge.
2.3.5. Field from a sheet of charge
Consider the infinite uniform sheet of charge in the y-z plane in Fig. 32. It can be shown that
the electric field anywhere above the sheet of charge can be written as
Ē =
ρs
âN .
2ε0
(2.14)
Notice that (2.14) has no dependence on ρ. Think of this problem as an infinite uniform light with
no attenuation.
z
dy’
Infinite sheet of
charge in the y-z plane.
R
y
ρs
x
Figure 32. Electric field due to an infinite sheet of charge in the y-z plane.
61
2.3.6. Streamlines and sketches of fields
Stream lines are used to help us visualize the field lines from a source. To do this consider
Ē =
ρl
âρ .
2πε0 ρ
The field lines from a single point charge are sketched in Fig. 33.
a)
b)
c)
d)
Figure 33. a) A stream line sketch that is not good; b) and c) two good stream line sketches that show
the intensity of the field and d) the traditional stream line sketch indicating the fields in the region around
the charge.
2.4. Electric Flux Density
Next, we define the electric flux density. To do this consider the problem defined in Fig. 34.
The electric flux density, D̄ (C/m2 ), exists between the inner and outer conductors. The electric
flux density is another way to describe the field and it should be noted that the flux density is not
a force field like the electric field, but it describes the density of the field. Therefore, from a point
62
charge we have the following electric flux density:
Q
âr .
4πr2
D̄ = ε0 Ē =
⇒
∫
D̄ =
V
Metal
conducting
spheres
(2.15)
ρv dV
âr .
4πR2
Insulating
or dielectric
material
-Q
r=b
(2.16)
+Q
r=a
Electric
flux
lines
Figure 34. Electric field density between two conducting spheres with a dielectric insulator.
2.5. Gauss’s Law
2.5.1. Introduction
Gauss’s law states: The electric flux passing through any closed surface is equal to the total
charge enclosed by that surface. To understand this law, consider the image in Fig. 35. The total
flux Ψ passing through the surface is
I
D̄s · dS̄.
Ψ=
(2.17)
S
Equation (2.17) is adding up all of the flux passing through the Gaussian surface S. Then Gauss’s
law states Ψ = Qenc or
I
D̄s · dS̄ = Qenc .
(2.18)
S
If the charge is a volume then:
I
∫
D̄s · dS̄ =
S
ρv dV.
V
63
(2.19)
∆S
Gaussian
surface S
Ds,normal
∆S
θ
Q
P
Figure 35. Definition of Gauss’s law with a Gaussian surface.
2.5.2. Example of Gauss’s law
To demonstrate Gauss’s law, consider a point charge Q in Fig. 36. We know the field from
that point charge is Ē =
Q
â
4πε0 r2 r
and D̄ = ε0 Ē. ⇒ D̄ =
Q
â
4πr2 r
and at the surface, D̄s =
know that dS̄ = r2 sin θdθdϕâr = a2 sin θdθdϕâr . This then gives D̄s · dS̄ =
Q
4π
Q
â
4πa2 r
Q
â .
4πa2 r
We
· a2 sin θdθdϕâr =
sin θdθdϕ. This then gives
∫
0
2π
∫
π
0
Q
sin θdθdϕ =
4π
∫
2π
0
π
∫ 2π
Q
Q
(− cos θ) dϕ =
dϕ = Q.
4π
2π
0
0
z
Ds
dS
r=
a
θ
Q
y
φ
x
Figure 36. Example of Gauss’s law.
2.5.3. Application of Gauss’s law
The main application of Gauss’s law is to solve for the electric flux density as a result of some
charge distribution. We can usually do this in the problem if:
64
1) D̄s is either normal or tangential to the closed surface everywhere, so that D̄s · dS̄ becomes
either Ds dS or zero.
2) On the portion of the closed surface for which D̄s · dS̄ is not zero, Ds = constant.
If D̄s · dS̄ = Ds dS then, Ds dS is a scalar and Ds can be moved outside of the integration.
2.5.4. Point charge example
Again, consider the point charge Q at the origin in Fig. 36. For this example, we choose the
closed surface S to be a sphere. This then gives D̄s = Ds as a constant over the sphere because
the only component of the electric flux density is the r component. Therefore, if the radius of the
sphere is r then
I
I
D̄s · dS̄ =
Q=
I
dS = 4πr2 Ds .
Ds dS = Ds
S
S
S
Solving for Ds gives
Ds =
Q
.
4πr2
(2.20)
This expression is similar to (2.5) multiplied by ε0 . In general we have
D̄s =
Q
âr .
4πr2
(2.21)
2.5.5. Line charge example
Next, consider the line charge ρl extending to infinity in the ±z-direction in Fig. 37. On the
surface of the cylinder, we have D̄ = Dρ âρ normal to the surface. This then gives
I
D̄s · dS̄
Q =
∫
∫
cylinder
Dρ · dS̄ +
=
∫
sides
= Dρ
0dS̄ +
top
1dS̄
sides
∫ L ∫ 2π
ρdϕdz
= Dρ
0
0
= Dρ 2πρL.
65
∫
0dS̄
bottom
Therefore,
Ds = Dρ =
Q
.
2πρL
z
a
ρl
Gaussian
surface
(line charge)
L
Figure 37. Calculating the electric field due to a line charge using Gauss’s law.
2.6. Divergence
H
From the previous section we have that Gauss’s law states that S D̄ · dS̄ = Qenc . To illustrate
H
divergence, consider the illustration in Fig. 38. To evaluate S D̄ · dS̄ in Fig. 38 we have
I
∫
∫
D̄ · dS̄ =
S
∫
+
f ront
∫
+
back
Since the surface area is very small we have
+
lef t
∫
f ront
∫
+
right
∫
+
top
.
bottom
= D̄f ront · ∆S̄f ront = Dx,f ront ∆y∆z. Since P is
in the middle of the cube in Fig. 38 we have
Dx,f ront − Dx0
∂Dx
=
.
∆x/2
∂x
⇒
Dx,f ront =
⇒
∆x ∂Dx
+ Dx0 .
2 ∂x
]
∆x ∂Dx
Dx,f ront dS =
+ Dx0 ∆y∆z.
2 ∂x
f ront
I
[
66
Similarly for the back of the cube in Fig. 38 we have
(
∫
=
back
⇒
)
∆x ∂Dx
− Dx0 +
∆y∆z.
2 ∂x
∫
∫
+
f ront
=
∂Dx
∆x∆y∆z,
∂x
=
∂Dy
∆x∆y∆z
∂y
back
∫
∫
+
lef t
right
and
z
Flux density in
the cube
D=Dx0 ax+Dy0 ay+Dz0 az
P(x,y,z) (middle of
the cube)
∆z
∆x
∆y
y
x
Figure 38. Illustration of divergence.
∫
∫
+
top
=
bottom
∂Dz
∆x∆y∆z.
∂z
This then gives
(
)
∂Dx ∂Dy ∂Dz
D̄ · dS̄ =
+
+
∆x∆y∆z
∂x
∂y
∂z
S
(
)
∂Dx ∂Dy ∂Dz
+
+
=
∆V
∂x
∂y
∂z
I
(2.22)
= Qenc .
67
(
What (2.22) says is that the charged enclosed in a unit volume ∆V is equal to
y
∂Dx
z
+ ∂D
+ ∂D
∂x
∂y
∂z
)
∆V .
Now let ∆V → 0 and divide both sides of (2.22) by ∆V gives
Q
lim
=
∆V →0 ∆V
(
∂Dx ∂Dy ∂Dz
+
+
∂x
∂y
∂z
This leads to the following notation: ∇ · D̄ = div D̄ =
∂Dx
∂x
+
)
∂Dy
∂y
= ρV .
+
∂Dz
∂z
which gives
∇ · D̄ = ρV .
(2.23)
Equation (2.23) is known as one of Maxwell’s equations. Equation (2.23) is very useful because it
is a nice relation between the charge in a region and the fields being radiated from that charge in
the region. As long as (2.23) is satisfied, then the solution to (2.23) is unique in that region.
2.6.1. Example of computing the divergence in a region
Find ∇ · D̄ at the origin if D̄ = e−x sin yâx − e−x cos yây + 2zâz . This then gives ∇ · D̄ =
∂Dx
∂x
+
∂Dy
∂y
+
∂Dz
∂z
= −e−x sin y + e−x sin y + 2 = 2. This says that the divergence of D̄ is constant
everywhere.
2.6.2. Solving for volume charge in a region
Find ρV for D̄ =
4y
(z 2
z3
4xy
âx
z
2
2
+ 2xz ây − 2xz2 y âz . Solving gives ∇ · D̄ =
4y
z
+ 0 − 2x
2 y(−2)
z3
=
4y
z
2
+ 4xz3 y =
+ x2 ).
2.7. The Divergence Theorem
∂
∂
∂
âx + ∂y
ây + ∂z
âz . Also, from Gauss’s law we have
From our previous definition ∇ = ∂x
H
∫
D̄ · dS̄ = Q = vol ρV dV . This then gives the following Divergence Theorem
I
∫
D̄ · dS̄ =
S
∇ · D̄dV.
(2.24)
V
2.8. Energy and Potential
2.8.1. Work
Suppose we want to move a charge Q a distance dL in an Ē-field. We know that Ē = F̄ /Q
68
⇒ F̄ = ĒQ where F̄ is the force on Q arising from the electric field. Thus, to move the charge we
have the applied force Fapplied = −QĒ · âL . We know our work is force times distance = energy
expended. ⇒ the differential work dW is dW = −QĒ · dL̄ = −QĒ · âL dL. ⇒
∫
f inal
W = −Q
Ē · dL̄.
(2.25)
initial
Note that âL dL = dL̄.
2.8.2. Differential work example
Let Ē =
1
(8xyzâx
z2
+ 4x2 zây − 4x2 yâz ) V/m. Find the differential amount of work done to
move a 6 nC charge a distance of 2 µm from P(2,-2,3) in the âL = − 67 âx + 37 ây + 27 âz direction.
Solution: The electric field at P is Ē(2, −2, 3) =
112
9
−32
âx
3
+
16
â
3 y
+
32
â .
9 z
⇒ Ē · âL =
64
7
+
16
7
+
64
63
=
= 12.44. ⇒ dW = −QĒ · âL · dL = −6nC · 12.44 · 2µm = −149 fJ.
2.8.3. Line integral example
A line integral is used if the path is continuous. For this example let the electric field be
Ē = yâx + xây + 2âz . Find the work to carry 2 C from B(1,0,1) to A(.8,.6,1) along the arc of the
circle x2 + y 2 = 1 and z = 1.
Solution: We will work in rectangular coordinates for the problem in Fig. 39. Note that dL̄ =
dxâx + dyây + dzâz . This then gives
∫
A
W = −Q
Ē · dL̄
B
∫ A
= −2
(yâx + xây + 2âz )(dxâx + dyây + dzâz )
B
∫ .8
∫ .6
∫ 1
= −2
ydx − 2
xdy − 4
dz.
1
0
1
Now we need to move along a circular path. This then gives y =
This then gives
∫
W = −2
1
.8
∫
√
2
1 − x dx − 2
.6
0
69
√
√
1 − x2 and x = 1 − y 2 .
√
1 − y 2 dy − 0 = −.96J.
y
A(.8, .6, 1)
x
B(1, 0, 1)
Figure 39. Work along a continuous line example.
Next, integrating along a straight line from point B to A gives
∫
∫
.8
∫
.6
1
ydx − 2
xdy − 4
dz
1
0
1
∫ .8
∫ .6
= −2
(−3x + 3)dx − 2
(1 − y/3)dy − 0
W = −2
1
0
= −.96J.
This shows that the work done is independent of the path taken in any electrostatic field.
2.8.4. Potential
Now define the potential difference V as the work done in moving a unit positive charge from
one point to another in an Ē-field in the following manner:
∫
f inal
V =−
Ē · dL̄.
(2.26)
Ē · dL̄.
(2.27)
initial
A commonly used notation is also
∫
A
VAB = −
B
2.8.5. Point charge example
For this example, consider the point charge Q. Find the potential between the radial distance
rA and rB from the point charge.
70
Solution: From before, we know the electric field from a point charge is
Ē = Er âr =
Q
âr .
4πε0 r2
We also know that dL̄ = drâr . This then gives
∫
∫
A
VAB = −
Ē · dL̄ = −
B
rA
rB
(
)
Q
1
Q
1
dr =
−
.
4πε0 r2
4πε0 rA rB
Note that when we use the potential notation, we have a reference point. Usually “ground” or
“infinity”.
2.8.6. Potential field of a line charge example
From before, we know that the potential from a point charge is
∫
∫
A
VAB = −
Ē · dL̄ = −
B
(
)
Q
1
Q
1
dr =
−
= VA − VB .
4πε0 r2
4πε0 rA rB
rA
rB
We know work is independent of the path between the two points. This implies VAB is independent
of the path. Therefore, if we define rB → ∞ to be our reference point, then VAB →
Q 1
.
4πε0 rA
Therefore, in general the potential field of a point charge Q is
V =
Q
.
4πε0 r
Therefore, for a line of charges, the potential field is
∫
V (r̄) =
l
ρL (r̄′ )
dL′ .
4πε0 |r̄ − r̄′ |
(2.28)
2.8.7. Potential fields due to surface and volume charges
The previous example can be generalized to the following expressions involving surface and
volume charges:
∫
V (r̄) =
S
ρs (r̄′ )
dS ′
4πε0 |r̄ − r̄′ |
71
(2.29)
and
∫
V (r̄) =
V
ρv (r̄′ )
dV ′ .
′
4πε0 |r̄ − r̄ |
(2.30)
2.8.8. Potential due to a ring of charge
Next, consider the ring of charge in Fig. 40. The charge is denoted as ρl with a radius a on
the x-y plane and centered about the z-axis. Since this is a line charge, we need to evaluate a line
integral over the charge. This then gives the following:
∫
2π
V (r̄) =
0
ρL a′
dϕ′
4πε0 |r̄ − r̄′ |
where dL′ = a′ dϕ, r̄ = zâz and r̄′ = a′ âρ . This then gives |r̄ − r̄′ | =
∫
V (r̄) =
0
2π
√
a′2 + z 2 . ⇒
ρ a′
ρ a′
√L
√L
dϕ′ =
.
4πε0 a′2 + z 2
2ε0 a′2 + z 2
This is the potential at infinity. For a potential difference in general we have VAB = VA − VB =
∫A
− B Ē · dL̄. Now say A → B then VAB → 0. ⇒
∫
A
−
Ē · dL̄ → 0
B
and is independent of the path. Therefore,
I
Ē · dL̄ = 0
(2.31)
around a closed path.
2.8.9. Potential gradient
From the previous section, we know V = −
∫
l
Ē·dL̄. What if we know V and want to determine
Ē. To do this start with the incremental differential element ∆L̄. This then says ∆V = −Ē · ∆L̄.
Next, let θ be the angle between Ē and L̄ (Fig. 41). Now lim∆L→0 ∆V
= dV
= −E cos θ. Note that
∆L
dL
dV = E when θ = π or when ∆L̄ is opposite to Ē. Therefore, the magnitude of Ē is given as
dL max
72
z
(0,0,z)
r
y
a’
ρl
φ’
dL’ = a’ dφ’
x
Figure 40. Potential due to a ring of charge.
∆L
θ
Figure 41. Potential gradient.
73
the max rate of change of V . To illustrate this consider the equipotential surfaces in Fig. 42. We
want the electric field at point P. The direction of max change is in the direction of âN pointed
towards the higher potentials. ⇒
dV dV
Ē = − âN = −
âN
dL max
dN
where the last term is the derivative in the normal direction. ⇒
−
dV
dV
dV
dV
âN = − âx,N −
ây,N −
âz,N = −∇V
dN
dx
dy
dz
which is also called “the gradient of V”. Therefore,
Ē = −∇V.
(2.32)
+50
+60
+40
+70
+30
+80
Direction of
max change
Figure 42. Equipotential surfaces.
2.8.10. Gradient example
For this example let V = 2x2 y − 5z and P(-4,3,6) be given. Find Ē(x, y, z).
Solution: We know that Ē = −∇V . ⇒ ∇V =
∂
∂
∂
(2x2 y −5z)âx + ∂y
(2x2 y −5z)ây + ∂z
(2x2 y −5z)âz
∂x
4xyâx + 2x2 ây − 5âz V/m. ⇒ Ē = −4xyâx − 2x2 ây + 5âz = 48âx − 32ây + 5âz .
74
=
2.8.11. The dipole
The electric dipole has the geometry shown in Fig. 43 where d << r. Now, the potential at
(
)
Q
Q
Q
R2 −R1
P w.r.t. ∞ is V = 4πε0 R1 − 4πε0 R2 = 4πε0 R1 R2 . We want to simplify this expression. To do
this, choose P such that R1 and R2 are much much greater than d. This is shown in Fig. 44. This
results in the following approximations: R1 ≈ R2 ≈ r and R2 − R1 = d cos θ. This then simplifies
the expression for the voltage to the following:
(
)
Q
d cos θ
Q d cos θ
V =
=
.
4πε0
rr
4πε0 r2
z
P
R1
+Q
r
θ
R2
d
y
x
-Q
Figure 43. A dipole.
This then gives
Ē = −∇V
)
(
∂
Q d cos θ
âr
= −
∂r 4πε0 r2
)
(
1 ∂
Q d cos θ
−
âθ
r ∂θ 4πε0 r2
)
(
1 ∂
Q d cos θ
−
âϕ
r sin θ ∂ϕ 4πε0 r2
Q d cos θ
Q d sin θ
=
âr +
âθ + 0.
3
2πε0 r
4πε0 r3
75
⇒
Qd
[2 cos θâr + sin θâθ ].
4πε0 r3
Ē =
The equipotential surfaces are shown in Fig. 45.
z
θ
R1
P
+Q
r
d
y
{
R2
R2 - R1 = d cos θ
x
-Q
Figure 44. Far-field approximation of a dipole.
z
0.4
0.6
0.8
1
0
0
-1
-0.8
-0.6
-0.4
Figure 45. Equipotential surfaces of a dipole.
76
CHAPTER 3. PROPERTIES OF DIELECTRICS AND
CONDUCTORS
3.1. Current
3.1.1. Definition
To introduce current we first consider the illustration in Fig. 46. Current is defined as the
rate of movement of charge passing a given reference point of 1 C/s. ⇒
I=
dQ
.
dt
(3.1)
The unit for current is Amperes or A. The current density J¯ describes the current distribution over
the reference plane. The unit for current density is A/m2 and ∆I = J¯ · ∆S̄. Thus, in general we
have
∫
J¯ · dS̄.
I=
(3.2)
S
∆S
Q
Q
-Q
Q
Reference plane
Figure 46. Current definition.
3.1.2. Convection current
The next type of current we introduce is convection current. To do this, consider the illustration in Fig. 47. In this case, we move ∆Q which is an element of charge ∆Q = ρV ∆V = ρV ∆S∆L
a distance ∆x. This means we have moved a charge ∆Q = ρV ∆S∆x through a reference plane. ⇒
∆I =
∆x
∆Q
= ρV ∆S
.
∆t
∆t
77
⇒
∆x
= ρV ∆SVx = ∆I
∆t
lim ρV ∆S
∆t→0
where Vx is the velocity in the x-direction of the charge. ⇒
J¯ = ρV V̄
(3.3)
where V̄ is the general velocity vector. Note that (3.3) shows that charge in motion results in a
current called the convection current.
z
∆Q = ρv∆V
z
∆Q = ρv∆V
∆S
∆S
y
y
∆L
∆x
∆L
x
x
Figure 47. Convection current.
3.1.3. Continuity of current
Consider a region bounded by a closed surface S. The current through the surface S is
I
J¯ · dS̄
I=
S
where this outward flow of positive charge must be balanced by a decrease of positive charge. ⇒
I
I=
dQ
J¯ · dS̄ = −
dt
S
where Q is the charge inside a closed surface and the negative sign represents an outward traveling
78
current. Using the divergence theorem
I
∫
D̄ · dS̄ =
S
gives
I
∫
J¯ · dS̄ =
S
V
∇ · D̄dV
V
¯ = − dQ = −
∇ · JdV
dt
∫
V
∂ρv
dV.
∂t
(3.4)
This then gives the current continuity equation in point form as:
∂ρv
∇ · J¯ = −
.
∂t
(3.5)
3.1.4. Metallic conductors (resistance)
In a conductor with an electric field, a charge will feel a force from the Ē field as F̄ = −eĒ
where Q = −e. The charges then collide with each other and the conductor and eventually results
in a drift velocity vd . ⇒ vd = −µe Ē where −µe is the mobility of an electron. Substituting this
into (3.3) gives J¯ = −ρv µe Ē. This then gives (Fig. 48)
J¯ = σ Ē
(3.6)
where σ is the conductivity of the conductor and has units S/m.
σ
J
Area = S
L
Figure 48. Current in metallic conductors.
Now, assume the current is uniform throughout the cylindrical region in Fig. 48. We know
79
that I =
⇒V =
∫
J¯ ·dS̄ = JS and VAB = −
S V
L
I
σS
∫a
b
Ē ·d¯l = −ELba = ELab or V = EL. ⇒ J =
I
S
= σE = σ VL
= RI where
L
.
σS
R=
(3.7)
3.1.5. Resistance example
For this example, consider a copper wire with L = 1mi and a diameter of 0.05808 in. Find
Req which is the equivalent resistance of the wire.
Solution: We know that the diameter = 0.05808 in 2.54 cm/1 in = 1.4752 mm. ⇒ S = πr2 =
1.7092µm2 . We also have L = 5280 ft = 1.6093 km. ⇒
Req =
1.6093km
= 16.23Ω
5.8 × 107 S/m1.7092µm2
where 5.8 × 107 S/m is the conductivity of copper.
3.2. Boundary Conditions for Perfect Conductors
Now we want to relate the field around a conductor to the charge flowing on a perfect
conductor. A perfect conductor has a conductivity of infinity or σ = ∞. In our case we assume
static charges. To do this, consider the problem defined in Fig. 49. We have Dt = 0 otherwise
the flux would change the charge at the surface and then the problem would not be static. For the
normal component, consider the pill box in Fig. 49. For an incremental ∆S we have from Gauss’s
law:
I
D̄ · dS̄.
Qenc =
S
⇒
∫
∫
∫
+
top
+
bottom
= Qenc .
sides
As ∆S → 0 and ∆h → 0, then D through the sides → 0 and D = 0 in the conductor. ⇒
DN ∆S = Q = ρs ∆S ⇒
DN = ρs .
80
For the Ē-field we consider the loop in Fig. 49 and we know:
∫
∫
b
+
a
∫
c
∫
d
+
c
In the conductor we have Ē = 0 which gives
∫d
c
Ē · dL̄ = 0. ⇒
a
+
b
H
= 0.
d
= 0. Now let ∆W → 0 and ∆h → 0. ⇒
Et ∆W − EN 12 ∆h + EN 12 ∆h = 0. ⇒ Et ∆W = 0. ⇒
Et = 0.
Therefore, we have the following B.C. on the conductor:
Dt = E t = 0
(3.8)
DN = ε 0 E N = ρ s .
(3.9)
âN × Ē = 0
(3.10)
âN · D̄ = ρs .
(3.11)
and
Or in general we have
and
3.3. Boundary Conditions for Perfect Dielectrics
In this section, we derive the boundary conditions for perfect dielectrics. A perfect dielectric
has a conductivity of zero or σ = 0. To derive the B.C. we consider the problem of two perfect
H
dielectrics illustrated in Fig. 50. From before, we know that Ē · dL̄ = 0 around the loop in Fig.
50. Therefore, as ∆h → 0 and ∆W → 0 we have
I
Ē · dL̄ → Etan1 ∆W − Etan2 ∆W = 0.
81
Free-space
EN
a
E
N
∆S
DN
Dt
ρenc
{
D
a
∆W
∆h
∆h
d
b
∆h
∆W
Et
c
Loop
Pill box
(Gaussian surface)
Conductor
Figure 49. Boundary conditions on a conductor.
Dielectric for
region 1
ε1
EN
a
E
N
∆S
DN
Dt
ρenc
{
D
a
∆h
∆W
∆h
d
b
∆h
∆W
Pill box
(Gaussian surface)
Et
c
Loop
Dielectric for
region 2
ε2
Figure 50. Boundary conditions between two perfect dielectrics.
82
This then gives
Etan1 = Etan2 .
(3.12)
Also, as ∆S → 0 and ∆h → 0 on the gaussian surface in Fig. 50 we have
I
εĒ · dS̄ → DN 1 ∆S − DN 2 ∆S = ∆Q = ∆Sρs .
Qenc =
This then gives
DN 1 − DN 2 = ρ s .
(3.13)
âN × (Ē1 − Ē2 ) = 0
(3.14)
âN · (D̄1 − D̄2 ) = ρs .
(3.15)
Or in general we have
and
3.4. Method of Images
To introduce the method of images or image theory we first consider the illustration in Fig. 51.
We need to satisfy the B.C. at the conducting plane on the left of Fig. 51. We know that V =
Q
4πε0 r
is the absolute potential of a charge. Therefore, at the boundary of the conducting plane on the
right side of Fig. 51 we have V =
+Q
4πε0 r+
+
−Q
.
4πε0 r−
But r+ = r− = r giving V =
along the boundary. Other examples of image theory are shown in Fig. 52.
z
z
+Q
+Q
d
d
r+
(Image Theory)
V=0
y
Conducting plane
with V = 0.
-d r
-Q
Figure 51. Illustration of image theory.
83
y
+Q
4πε0 r
+
−Q
4πε0 r
=0V
z
z
ρ
ρ
L
L
d
d
(Image Theory)
V=0
V=0
y
y
-d
−ρ
L
z
z
IL
IL
IL’
d
IL’
d
(Image Theory)
V=0
V=0
y
y
-d
IL’
IL
Figure 52. Other examples of image theory.
84
3.5. Capacitance
3.5.1. Definition
In this section we introduce capacitance. To do this, we first consider the problem defined in
Fig. 53. Two conductors are embedded in a uniform dielectric. Conductor M2 carries a total positive
charge +Q and conductor M1 caries an equal negative charge −Q. The electric flux travels from
M2 to M1 . ⇒ there exists a potential Vo between the two conductors. Now define the capacitance
between the conductors as
Q
,
Vo
(3.16)
εi Ē · dS̄
.
∫+
− − Ē · dL̄
(3.17)
C=
or in general
H
C=
We can see in equations (3.16) and (3.17) that C is independent of Vo and Q. C is only dependent
on the physical dimensions of the problem.
+ +
+
+
+ M2 +
+
+
D
+
+
+ ++
- - - M1
ε
- - - - (dielectric)
Figure 53. Definition of capacitance.
3.5.2. Capacitance between finite parallel plates
Next, we compute the capacitance between the infinite parallel plates shown in Fig. 54. From
before, we know that Ē = 2 ρεsi âz on an infinite plate. This then gives D̄ = ρs âz . ⇒ DN = Dz = ρs
which is the B.C. on the lower plane. On the upper plane we have the B.C. DN = −Dz . This then
85
gives
∫
∫
lower
Vo = −
Ē · dL̄ = −
upper
d
0
ρs
ρs
dz = d.
εi
εi
If the surface area of the plane is finite with area S then Q = ρs S. This then gives
C=
Q
εS
=
.
V
d
(3.18)
Note that equation (3.18) neglects fringing. The energy stored in a capacitor is
WE =
=
=
=
∫
1
εE 2 dV
2 vol
1
CV 2
2 o
1
QVo
2
1 Q2
.
2C
(3.19)
z
−ρs
εi
Top plate
z=d
Bottom plate
z=0
E
+ρs
y
Figure 54. Capacitance between two parallel plates.
3.5.3. Capacitance between two parallel plates with two dielectrics example
In this example, we want to determine C for the structure shown in Fig. 55. We know that
C=
Q
.
Vo
Assume a potential of Vo between the plates and that E1 and E2 are uniform. This then
gives Vo = E1 d1 + E2 d2 . At the dielectric interface we know DN 1 = DN 2 (ρs = 0). This then gives
86
ε1 E1 = ε2 E2 . ⇒ Vo = E1 d1 +
ε1 E1 d2
ε2
= E1 [d1 +
E1 =
ε1
d ].
ε2 2
This then gives
Vo
.
d1 + εε12 d2
We know that on the plate we have
ρs1 = ε1 E1 = ε1
Vo
=
d1 + εε12 d2
Vo
.
d1
d2
+
ε1
ε2
This then gives
C=
ρs S
S
Q
=
=
Vo
Vo
Vo
d1
ε1
Vo
=
+ dε22
d1
ε1 S
1
+
d2
ε2 S
=
1
C1
1
+
1
C2
=
C1 C2
.
C1 + C2
z
Upper plate
Surface area S
Vo
+
-
E2
d
E1
d2
ε2
d1
ε1
Dielectric between
the plates
Lower plate
Figure 55. Capacitance between two parallel plates with two dielectrics.
3.6. Poisson’s and Laplace’s Equations
In this section we introduce Poisson’s and Laplace’s equations. These equations relate the
potential in a region to the sources in that region in a general manner. From the previous sections,
we know ∇· D̄ = ρv and D̄ = εĒ. We also know Ē = −∇V . ⇒ ∇· D̄ = ∇·(εĒ) = −∇·(ε∇V ) = ρv .
87
This then gives
∇ · ∇V
ρv
ε
∂ 2V
∂2V
∂ 2V
=
+
+
∂x2
∂y 2
∂z 2
= −
= ∇2 V.
this then gives
∇2 V = −
ρv
.
ε
(3.20)
Equation (3.20) is called Poisson’s equation. Thus, if ρv = 0 then we have Laplace’s equation:
∇2 V = 0.
(3.21)
3.7. Uniqueness Theorem
In this section we show that the solutions to (3.20) and (3.21) are unique. Assume that V1
and V2 are solutions to Laplace’s equation or are a solution to (3.21). This then gives ∇2 V1 = 0 and
∇2 V2 = 0. Therefore, ∇2 (V1 − V2 ) = 0. The uniqueness theorem states that V1 = V2 and that V1
and V2 satisfy the B.C. of a problem. Therefore, there exists only one unique solution to the space.
3.8. Parallel Plate Capacitance Example Using Laplace’s Equation
For this example, we will consider the problem defined in Fig. 56. We have two parallel plates
each with a surface area S separated by a distance d. For this example, assume that the plates
are close enough such that if a voltage is applied between the top and bottom plate the resulting
electric field is in the z-direction only. To compute the equivalent capacitance we need to evaluate
C = Q/Vo where Vo is the voltage across the plates and Q is the resulting charge on each plate from
this voltage. For the dielectric region between the plates, we need to satisfy Laplace’s equation.
This then means we have a solution in the following form: V (z) = Az + B. We will determine the
coefficients A and B by enforcing the B.C. on each plate. At z = 0 the voltage is V = 0 and at z =
d the voltage is V = Vo . Therefore, V (0) = B = 0 and V (d) = Ad = Vo . This then gives A = V o/d.
88
This then gives the potential between the plates as V (z) =
Vo
z.
d
Now we know the potential in
∂ Vo
the dielectric. Taking the gradient of V (z) gives Ē = −∇V = − ∂z
zâz = − Vdo âz , which is the
d
electric field in the dielectric. ⇒ D̄ = εĒ = −ε Vdo âz = DN . At the plate at z = 0 we have the B.C.
∫
= εS
.
ρs = DN = −ε Vdo âz . This then gives Q = S − εVdo dS = −ε VodS . Therefore, C = VQo = |Q|
Vo
d
z
Upper plate
Surface area S
Vo
+
-
ε
d
Dielectric between
the plates
Lower plate
Figure 56. Computing the capacitance between two parallel plates of finite size.
3.9. Non-Parallel Plate Capacitance Example Using Laplace’s Equation
In this problem we calculate the electric field between the two plates shown in Fig. 57 in the
cylindrical coordinate system. We know that ρs = 0 which then gives ∇2 V = 0. In cylindrical
coordinates we have
(
)
1 ∂
∂V
1 ∂2V
∂ 2V
1 ∂ 2V
ρ
+ 2 2 +
=
0
+
+ 0 = 0.
ρ ∂ρ
∂ρ
ρ ∂ϕ
∂z 2
ρ2 ∂ϕ2
This then gives
1 ∂ 2V
= 0.
ρ2 ∂ϕ2
This then gives
∂ 2V
= 0.
∂ϕ2
For this second partial differential equation we have the following solution V = Aϕ + B. Applying
the B.C. we have V = 0 at ϕ = 0 ⇒ B = 0. Also, at ϕ = α we have Aα = Vo . ⇒ A =
gives the potential between the plates as V =
Vo
ϕ.
α
89
Vo
.
α
This then
Therefore, Ē = −∇V = − ρ1 ∂V
â = − ρ1 Vαo âϕ .
∂ϕ ϕ
V = Vo
φ=α
V=0
φ=0
Insulating
gap
φ
α
z
Infinite plates
(both)
Figure 57. Computing the electric field between two plates of finite size in cylindrical coordinates.
90
CHAPTER 4. MAGNETOSTATIC FIELDS
4.1. Biot-Savart Law
4.1.1. Introduction
In this section, we introduce Magnetostatic (or magnetic) fields. To illustrate magnetic fields
we need to consider the differential DC current element in free-space shown in Fig. 58. To relate
the magnetic field to the DC current we start with the Biot-Savart law. The Biot-Savart law is
written in the following manner:
dH̄2 =
I1 dL̄1 × âR12
2
4πR12
(4.1)
where H̄ is the magnetic field intensity with units A/m and R12 = |R̄12 |. Thus, in general and in
integral form we have
I
H̄ =
I1 dL̄1 × âR
.
4πR2
(4.2)
{
Point 1
dL1
aR
I1
12
R12
P
(Point 2)
Figure 58. Illustration of the Biot-Savart law.
Surface
{
Js
I
Figure 59. Illustration of the Biot-Savart law using current density.
We can also describe the Biot-Savart law in terms of current density J¯s . The total current
91
flowing on the surface in Fig. 59 is IdL̄ = J¯s dS = J¯v dV . ⇒
∫
H̄ =
S
and
∫
H̄ =
vol
J¯s × âR
dS
4πR2
(4.3)
J¯v × âR
dV.
4πR2
(4.4)
4.1.2. Biot-Savart law example of an infinite line of current.
To illustrate the Biot-Savart law we will consider the infinite line of current on the z-axis and
differential current segment in the z-direction in Fig. 60. We have r̄ = ρâρ and r̄′ = z ′ âz . ⇒
R̄12 = r̄ − r̄′ = ρâρ − z ′ âz . This then gives
ρâρ − z ′ âz
.
âR12 = √
ρ2 + z ′ 2
Now let dL̄ = dz ′ âz . This then gives
∫
∞
Idz ′ âz × (ρâρ − z ′ âz )
4π(ρ2 + z ′ 2 )3/2
−∞
∫ ∞
ρâϕ
I
dz ′
=
2
4π −∞ (ρ + z ′ 2 )3/2
∫
Iρâϕ ∞
dz ′
=
4π −∞ (ρ2 + z ′ 2 )3/2
I
=
âϕ .
2πρ
H̄2 =
(4.5)
Equation (4.5) is the magnetic field due to an infinite line of current segment.
4.2. Magnetic Flux Density
Next, we define the magnetic flux density as
B̄ = µ0 H̄
(4.6)
where µ0 = 4π × 10−7 H/m. µ0 is referred to as the permeability of free-space. The units for B̄ are
92
z
Infinite
current I
Point 1
Differential current element
{
dL
aR
z’ az
R12
y
ρ aρ
P
(Point 2)
x
Figure 60. Field from an infinite line of current.
W b/m2 or T (Tesla).
4.3. Ampere’s Law
In this section we introduce Ampere’s law. To do this consider the problem defined in Fig. 61.
The problem is a current segment with infinite length. The current has a magnitude Ienc . Ampere’s
law states the following:
I
H̄ · dL̄.
Ienc =
(4.7)
l
Or in words, the integral around a closed path of the magnetic field is equal to the total current
enclosed by the closed path of integration.
path of integration
with radius a.
I enc
Figure 61. Magnetic field from an infinite line current.
4.3.1. Magnetic field from an infinitely long current
For this example we consider the infinitely long current in Fig. 61. If we assume the current in
93
Fig. 61 is on the z-axis and define our path of integration to be along a circle enclosing the current
with radius a, then we have the following expression for the magnetic field from the current:
I
∫
H̄ · dL̄ =
∫
2π
2π
Hϕ ρdϕ = Hϕ ρ
dϕ = Hϕ 2πρ = I.
0
0
Therefore,
Hϕ =
I
.
2πρ
4.3.2. Magnetic field in a coax
In this example, we calculate the magnetic field in and around the coax in Fig. 62. For this
example, we assume that I is uniformly distributed in the conductors. +I is on the center conductor
and −I is on the outer conductor. Notice H is not a function of ϕ or z. For a ≤ ρ < b we have
Hϕ =
I
.
2πρ
2
2
2
πρ
ρ
For 0 < ρ ≤ a we have Ienc = I aρ2 = I πa
2 and 2πρHϕ = I a2 where 2πρ is the path of
integration in the inner conductor. This then gives H =
(
πρ2 − πb2
2πρHϕ = I − I
πc2 − πb2
)
Iρ
.
2πa2
Next, for b ≤ ρ ≤ c we have
(
)
ρ2 − b2
=I −I 2
.
c − b2
This then gives
Hϕ =
I c2 − ρ2
.
2πρ c2 − b2
Finally, for ρ >C we have Hϕ = 0 because Ienc = 0. A plot of the magnetic field is shown in Fig.
63.
4.4. Curl
In this section we define the curl operation. To do this consider the illustration in Fig. 64.
H
For this, we want to evaluate H̄ · dL̄ = I = Jz ∆x∆y. On a side we have (H̄ · ∆L̄)1−2 = Hy,1−2 ∆y.
We want to write this expression in terms of Hyo . We know that
∂Hy
Hy,1−2 − Hyo
=
.
1
∂x
∆x
2
94
c
a
b
I
Infinitely long
I
Figure 62. Magnetic field from an infinite coax.
I/2πa
I/4πa
a
0
2a
Figure 63. Plot of the magnetic field along the coax.
95
3a = b 4a = c
This then implies
∂Hy
= Hyo +
∂x
Hy,1−2
This then gives
(H̄ · ∆L̄)1−2
(
)
1
∆x .
2
(
)
1 ∂Hy
= Hyo +
∆x ∆y.
2 ∂x
After a few steps we have
(
I
H̄ · dL̄ =
)
∂Hy ∂Hx
∆x∆y = Jz ∆x∆y.
−
∂x
∂y
z
H = Ho = Hxo ax + Hyo ay + Hzo az
4
3
∆x
1
∆y
2
y
x
Figure 64. Curl of the magnetic field in a region.
Next, evaluating the following limits give
H
lim
∆x→0,∆y→0
H̄ · dL̄
∂Hy ∂Hx
=
−
= Jz .
∆x∆y
∂x
∂y
Similarly, in other planes we have
H
lim
∆y→0,∆z→0
H̄ · dL̄
∂Hz ∂Hy
=
−
= Jx
∆y∆z
∂y
∂z
96
(4.8)
(4.9)
and
H
lim
∆z→0,∆x→0
H̄ · dL̄
∂Hx ∂Hz
=
−
= Jy .
∆z∆x
∂z
∂x
This then leads to the following definition:
H
(CurlH̄)N = lim
SN →0
H̄ · dL̄
= ∇ × H̄.
∆SN
(4.10)
This then gives the point form of Ampere’s law as:
Also,
H
¯
∇ × H̄ = J.
(4.11)
∇ × Ē = 0.
(4.12)
Ē · dL̄ = 0 or
4.5. Stokes Theorem
Similar to the divergence theorem. From before we have ∇ × H̄ = J¯ and
This then gives
I
∫ ∫
H̄ · dL̄ = Ienc =
Finally,
H̄ · dL̄ = Ienc .
∫ ∫
J¯ · dS̄ =
l
H
(∇ × H̄) · dS̄.
S
I
∫ ∫
H̄ · dL̄ =
(∇ × H̄) · dS̄.
l
(4.13)
S
4.6. Magnetic Flux
From before we defined the magnetic flux density as B̄ = µH̄. The magnetic flux is defined
as
∫
B̄ · dS̄.
Φ=
S
The units for Φ are Wb. We also have
I
B̄ · dS̄ = 0.
S
97
(4.14)
Using the divergence theorem we have
∇ · B̄ = 0.
(4.15)
∇ × Ē = 0
(4.16)
∇ × H̄ = J¯
(4.17)
∇ · D̄ = ρv
(4.18)
∇ · B̄ = 0.
(4.19)
Therefore, in summary we have:
Equations (4.16)-(4.19) are referred to as Maxwell’s static equations. The integral form of Maxwell’s
static equations are written in the following manner:
I
I
I
Ē · dL̄ = 0
l
H̄ · dL̄ = I =
l
IS
(4.20)
∫
∫
J¯ · dS̄
(4.21)
ρV dV
(4.22)
S
D̄ · dS̄ = Q =
V
B̄ · dS̄ = 0.
(4.23)
S
4.7. Solid Conductor Example
A solid conductor of circular cross-section is made of a homogeneous nonmagnetic material.
If the radius is a = 1.0 mm, the conductor is on the z-axis and Iz = 20 A, find:
a) Hϕ at ρ = 0.5 mm
b) Bϕ at ρ = 0.8 mm
c) Total magnetic flux per unit length inside the wire
d) Total flux for ρ < 0.5 mm
e) Total flux for ρ > 1.0 mm.
Solution:
98
Ienc
.
2πρ
a) From before we had Hϕ =
This then gives
(0.5)2
20
= 1591.5A/m.
2
(1.0) 2π ∗ 0.5mm
Hϕ =
Ienc
.
2πρ
b) Again from before we had Hϕ =
This then gives
20
(0.8)2
= 2546.4A/m.
(1.0)2 2π ∗ 0.8mm
Hϕ =
Computing the magnetic flux density then gives Bϕ = µ0 1Hϕ = 3.2 mT.
c) In general the magnetic flux density in the wire can be written as
Bϕ =
ρ2 I
µ0 Iρ
µ0 =
.
2
a 2πρ
2πa2
Then the total flux can be computed as:
∫
∫
1m
1mm
Φ=
0
0
µ0 Iρ
µ0 20 1
20µ0
5µ0
dρdz =
(1mm)2 =
=
= 2 µW b/m.
2
2
2πa
2π(1mm) 2
2π2
π
d) The total flux for ρ < 0.5 mm can be computed as:
∫
1m
∫
Φ=
0
0
.5mm
20µ0 1
5µ0 (.5mm)2
µ0 Iρ
2
dρdz
=
(.5mm)
=
= 0.5 µW b/m.
2πa2
2π(1mm)2 2
(1mm)2
Next, to compute the TOTAL flux, multiply the previous result by 1 m. This then gives Φ =
0.5 µW b.
e) Computing the total flux for ρ > 1 mm leads to the following result:
∫
Φ = lim
′
ρ →∞
0
1.0m
∫
ρ′
1.0mm
ρ′
10µ0
10µ0
= ∞.
dρdz = lim
ln(ρ)
ρ′ →∞ π
πρ
1mm
4.8. Scalar and Vector Magnetic Potentials
In a similar manner to the electric scalar potentials, we define scalar vector and magnetic
99
potentials. To do this, assume there exists a scalar magnetic potential denoted as Vm where
H̄ = −∇Vm .
(4.24)
¯ It can be shown that the curl of
Using Maxwell’s equation then gives ∇ × H̄ = ∇ × (−∇Vm ) = J.
¯ Therefore,
the gradient of any scalar is zero. This then gives ∇ × (−∇Vm ) = 0 = J.
H̄ = −∇Vm
(4.25)
where J¯ = 0. Vm also satisfies Laplace’s equation:
∇ · B̄ = µ0 ∇ · H̄
= 0
= µ0 ∇ · (−∇Vm ).
This then gives
∇2 Vm = 0
(4.26)
where again J¯ = 0.
4.9. Coaxial Scalar Potential Example
Consider the Coax shown in Fig. 65. From equation (4.5) we know that the magnetic field
from a coax can be written as Hϕ = I/2πρ. This then gives
I
= −∇Vm 2πρ
ϕ
1 ∂Vm
= −
.
ρ ∂ϕ
This then gives
−I
−I
∂Vm
=
⇐⇒ Vm =
ϕ.
∂ϕ
2π
2π
100
Therefore, when ϕ = 0 then Vm = 0. However, if ϕ = 2π then Vm = −I. This illustrates that
Vm is not a single valued function. This is because as ϕ rotates around the z-axis a different value
of current is enclosed after each rotation. In summary, if Ienc = 0 then we have a single valued
function. This then gives
∫
Vm,ab = −
a
H̄ · dL̄
(4.27)
b
which is path specific.
y
P(ρ,π/4,0)
φ
x
Iout
ρ=a
ρ=b
ρ=c
Figure 65. Coax for the scalar potential example.
4.10. Vector Magnetic Potential
From Maxwell’s equation we have ∇ · B̄ = 0. It can be shown that ∇ · (∇ × Ā) = 0 where
Ā is the vector magnetic potential. Using this property results in the following expression for the
magnetic flux density
B̄ = ∇ × Ā.
(4.28)
Next, computing the magnetic field then gives
H̄ =
1
∇ × Ā
µ0
(4.29)
and
1
∇ × J¯ = ∇ × ∇ × Ā.
µ0
Finally, it can be shown that
I
Ā =
l
µ0 IdL̄
.
4πR
101
(4.30)
(4.31)
Equation (4.31) can be used in general to compute the magnetic vector potential from any current
distribution. This expression results in a purely mathematical result and cannot be measured.
However, once Ā is computed, the magnetic field can then be computed using (4.29). Finally, for
surface and volume currents we have the following:
∫
Ā =
S
and
∫
Ā =
V
µ0 J¯s dS
4πR
(4.32)
µ0 J¯v dV
.
4πR
(4.33)
4.11. Magnetic Forces
In an electric field problem we have F̄ = QĒ (which was experimentally determined). Now,
if the charge Q is moving with velocity v̄ in a region with flux density B̄, then it has been
experimentally determined that the force on the charge Q can be computed as F̄ = Q(v̄ × B̄).
This then gives a total force of
F̄ = Q(Ē + v̄ × B̄).
(4.34)
Equation (4.34) is known as the Lorentz Force Equation.
4.12. Force on a Charge Example
Consider the charge Q = 18 nC traveling with velocity v̄ = 5 × 106 (0.6âx + 0.75ây + 0.3âz )
m/s in the presence of a magnetic flux density B̄ = −3.0âx + 4.0ây + 6.0âz mT. Compute |F̄ |.
Solution:
Using (4.34) with Ē = 0 we first compute v̄ × B̄ in the following manner:
0.6 0.75 0.3
v̄ × B̄ = −3.0 4.0 6.0
∗ 5 × 106 ∗ 10−3
= 16.5 × 103 âx + 22.5 × 103 ây + 23.25 × 103 âz .
102
This then gives
|F̄ | = Q|v̄ × B̄| = 653µN.
4.13. Force on a Differential Current Element
In this section the expression for the force on a differential current segment from a magnetic
flux density is derived. To compute this, the force equation (4.34) with Ē = 0 is written in
differential form in the following manner: dF̄ = dQ(v̄×B̄). Next, we define J¯ = ρv v̄ and dQ = ρv dV .
This then gives
dF̄ = ρv dV (v̄ × B̄)
= (J¯ × B̄)dV.
We also know that J¯v dV = J¯s dS = IdL̄. This then gives
dF̄ = (J¯s × B̄)dS
and
dF̄ = IdL̄ × B̄.
In integral form, the previous three equations can be written as
∫
F̄ =
J¯v × B̄dV,
(4.35)
J¯s × B̄dS
(4.36)
IdL̄ × B̄
(4.37)
V
∫
F̄ =
S
and
I
F̄ =
or
I
F̄ = −I
103
B̄ × dL̄
(4.38)
Therefore, for a straight conductor we have
F̄ = I L̄ × B̄.
(4.39)
4.14. Force on a Square Conducting Loop Example
For this example the force from the current on the conducting square loop in Fig. 66 is
computed. Using (4.5), the magnetic field from the current on the wire can be computed as
H̄ = 15/2πxâz A/m. This then gives B̄ = µ0 H̄ = 3/xâz µT . Next, using (4.38) we get
I
F̄ = −I
B̄ × dL̄
]
∫ 2
∫ 1
∫ 0
1
1
1
1
âz × dxâx +
âz × dyây +
âz × dxâx +
âz × dyây
= −2 × 10−3 ∗ 3 × 10−6
0 3
3 x
2 1
1 x
3
2
1
0 ]
[
y
−9
= −6 × 10
ln(x)ây + (−âx ) + ln(x)ây + y(−âx )
3
[∫
3
1
0
3
2
= −8âx nN.
(4.40)
Therefore |F̄ | = 8 nN.
z
Free space
15 A
(1,0,0)
(3,0,0)
y
(1,2,0)
2 mA
(3,2,0)
x
Figure 66. Force on a square loop example.
4.15. Force Between Differential Current Elements
In this section, the force between the two differential current segments in Fig. 67 is determined.
To do this, the differential magnetic field at point 2 due to a differential current element at point 1
104
is written using (4.1) in the following manner:
dH̄2 =
IdL̄1 × âR12
.
2
4πR12
(4.41)
Also, the differential force on a differential current element is dF̄ = IdL̄ × B̄. Now let dB̄2 be
the differential flux density at point 2 caused by the current element at point 1. This then gives
IdL̄ = I2 dL̄2 . This then gives
d(dF̄2 ) = I2 (dL̄2 × dB̄2 )
I1 I2
dL̄2 × (dL̄1 × âR12 ).
2
4πR12
= µ0
(4.42)
The last term in (4.42) represents the force between two differential current segments. Therefore,
in general if we have two long parallel wires with equal constant currents, the integration of (4.42)
gives the following resultant force:
F̄ =
µ0 I 2
N/m.
2πd
(4.43)
The resulting force is illustrated in Fig. 67.
d
F
F
I
I
Figure 67. Force between two infinite parallel wires.
4.16. Magnetic Boundary Conditions
In this section, the magnetic boundary conditions illustrated in Fig. 68 are derived. First
start by applying Gauss’s law to the surface. This then gives
I
B̄ · dS̄ = 0.
S
105
⇒
∆sBN 1 − ∆sBN 2 = 0.
⇒
BN 1 = BN 2 .
Therefore, in general we have
âN12 · (B̄2 − B̄1 ) = 0.
(4.44)
Next, applying Ampere’s law to the line integration in Fig. 68 we get
I
H̄ · dL̄ = Ienc .
L
This then gives
Ht1 ∆L − Ht2 ∆L = Js ∆L = I.
⇒
Ht1 − Ht2 = Js .
Therefore, in general we have
âN12 × (H̄2 − H̄1 ) = J¯s .
(4.45)
Dielectric for
region 1
µ1
BN
Ht
∆S
B
{
BN
Bt
HN
∆h
1
a
∆h
∆L
d
b
∆h
N12
BN
2
Figure 68. Magnetic boundary conditions.
106
Ht
c
Ht
2
Pill box
(Gaussian surface)
a
H
1
Loop
Dielectric for
region 2
µ2
4.17. Inductance
In this section we define the inductance (or self-inductance) as the ratio of the total flux
linkages to the current which they link as
L=
NΦ
I
(4.46)
where N is the number of turns in the coil and I is the current flowing in the coil.
4.18. Inductance of a Coax Example
In this example, we calculate the per-unit inductance of the coaxial cable in Fig. 69. First,
using the expression in (4.5) the field from the current on the inner conductor can be written as
H̄ = Hϕ =
I
.
2πρ
Next, to compute the total flux, the surface S (also shown in Fig. 69) must be defined. Then, using
this surface the total flux passing through it can be computed as
∫
d
∫
b
I
dρdz
µ0
2πρ
0
a
( )
µ0 Id
b
=
ln
.
2π
a
Φ =
(top view)
a
inner
conductor
(σc)
b
dielectric
(σ,ε,µ)
Z=d
H
I
Z=0
Surface S
Figure 69. Coax for computing the per-unit inductance.
107
Therefore,
Φ
I
( )
µ0 d
b
=
ln
.
2π
a
L =
Or in per-unit notation:
µ0
L=
ln
2π
( )
b
H/m.
a
108
(4.47)
CHAPTER 5. TIME-VARYING FIELDS
5.1. Faraday’s Law
In this section, the first law for time-varying fields is presented and illustrated in Fig. 70.
This law is called Faraday’s law and it states that a time varying flux passing through a closed loop
induces a voltage on that loop. In differential form this can be written as
Binc(t)
I
Bind(t)
closed path
Figure 70. Illustration of Faraday’s law.
Vemf = −
dΦ
dt
(5.1)
where Vemf is the induced voltage. Then, for an N -turn loop,
dΦ
.
dt
(5.2)
Ē · dL̄.
(5.3)
Vemf = −N
This then leads to the following expression
I
Vemf =
l
Note that (5.3) is closed and path specific. Next, the induced emf can be written as
I
Vemf
d
= Ē · dL̄ = −
dt
l
109
∫
B̄ · dS̄.
S
(5.4)
Now applying Stokes Theorem
I
∫
H̄ · dL̄ =
(∇ × H̄) · dS̄
l
to (5.4) gives
(5.5)
S
∫
∫
(∇ × Ē) · dS̄ =
Vemf =
−
S
S
dB̄
· dS̄.
dt
(5.6)
Therefore, (5.6) is satisfied if and only if
∇ × Ē = −
∂ B̄
.
∂t
(5.7)
Equation (5.7) is known as Maxwell’s first equation for time-varying fields.
5.2. Induced Voltage on a Coil Example
In this example, the induced voltage from a time-varying field on the coil of N-turns shown in
Fig. 71 is computed. The loop has a radius of a and is in the x-y plane. A resistor R is connected
to the port and B̄ = Bo (2ây + 3âz ) sin(ωt) is defined in the region around the antenna. For this
example, compute
a) The total flux Φ for N = 1.
b) Vemf for N = 10, Bo = 0.2 T, a = 10 cm and ω = 103 rad/s.
c) Polarity of Vemf for t = 0.
d) I for R = 1 KΩ.
z
N-turns
I
B
1
R
+
V
emf
2
Figure 71. N-turn coil.
110
a
y
Solution:
a) The total flux can be computed as
∫
Φ = 1 B̄ · dS̄
∫ S
=
Bo (2ây + 3âz ) sin ωtâz dS
S
= 3πa2 Bo sin ωt.
b) The induced voltage can be computed as
Vemf = −N
dΦ
dt
= −3πωN a2 Bo cos ωt
= −188.5 cos(103 t).
c) At time t = 0,
dΦ
dt
> 0 with amplitude Vemf = −188.5 V. Since Φ is increasing , I is from 2 to 1
in the figure (Lenz’s law). This then gives Vemf = V1 − V2 = −188.5 V.
d) Finally, for the current
V2 − V1
R
188.5
=
cos(103 t)
1000
I =
= 0.19 cos(103 t).
5.3. Displacement Current
In this section we define displacement current through a discussion using Maxwell’s equations.
Using Maxwell’s first equation we can notice that a time-varying magnetic field produces an electric
field. Also, from before we have the boundary condition ∇ × H̄ = J¯ for the static case. This then
v
gives ∇ · (∇ × H̄) = 0 = ∇ · J¯ = − ∂ρ
̸= 0. This is a contradiction. Therefore, to solve this issue
∂t
111
we add a Ḡ term to both sides. This results in
∇ × H̄ = J¯ + Ḡ.
This then gives
∇ · J¯ + ∇ · Ḡ = 0.
⇒
∂ρv
∂t
∂
=
(∇ · D̄)
∂t (
)
∂ D̄
= ∇·
.
∂t
∇ · Ḡ =
Therefore,
Ḡ =
∂ D̄
.
∂t
This finally leads to the following relation
∂ D̄
∇ × H̄ = J¯ +
.
∂t
Equation (5.8) is called Maxwell’s second equation for time-varying fields and the
(5.8)
∂ D̄
∂t
term is denoted
as the displacement current density.
5.4. Maxwell’s Equations for Time-Varying Fields
In this section we summarize Maxwell’s equations for time-varying fields in both the differential
and integral forms. First the differential form:
∇ × Ē = −
∂ B̄
∂t
∂ D̄
∇ × H̄ = J¯ +
∂t
112
(5.9)
(5.10)
∇ · D̄ = ρv
(5.11)
∇ · B̄ = 0.
(5.12)
J¯ = σ Ē.
(5.13)
We also have
Finally, in integral form we have
I
∫
Ē · dL̄ = −
L
S
I
∂ B̄
· dS̄
∂t
∫
∂ D̄
· dS̄
∂t
(5.15)
ρv dV
(5.16)
H̄ · dL̄ = I +
L
I
∫
S
D̄ · dS̄ =
S
(5.14)
V
I
B̄ · dS̄ = 0.
(5.17)
S
A few comments can be made about the differential form of Maxwell’s equations. Equation
(5.9) states that if the H̄-field is changing with respect to time at some point, then the Ē-field has
a curl at that point. Next, equation (5.10) states that a time-varying Ē-field generates an H̄-field.
5.5. Wave Propagation in Free Space (Lossless)
This section derives expressions that describe an electromagnetic wave propagating in free
space. This is done by deriving the Transverse Electromagnetic (TEM) wave. This means that both
the Ē- and H̄-fields are orthogonal to the direction of propagation. Now, first assume Ē = Ex âx
and that the wave is travelling in the z-direction. This then gives
∂Ex
ây
∂z
∂ H̄
= −µ0
∂t
∂Hy
= −µ0
ây .
∂t
∇ × Ē =
113
Notice that the previous result is a magnetic field with a y-component. Using Maxwell’s second
equation then gives
∂Hy
âx
∂z
∂ Ē
= ε0
∂t
∂Ex
= ε0
âx .
∂t
∇ × H̄ =
Notice in this case that the previous result is an electric field with an x-component. Comparing the
previous two expressions results in the following
∂Ex
∂Hy
= −µ0
∂z
∂t
(5.18)
∂Hy
∂Ex
= −ε0
.
∂z
∂t
(5.19)
and
The expressions in (5.18) and (5.19) have two unknowns (Ex and Hy ). In other words, we have two
equations with two unknowns. Next, if we differentiate (5.18) with respect to z we get
∂ 2 Ex
∂ 2 Hy
= −µ0
.
∂z 2
∂t∂z
(5.20)
Then if we differentiate (5.19) with respect to t we get
∂ 2 Hy
∂ 2 Ex
= −ε0 2 .
∂z∂t
∂t
(5.21)
Substituting (5.21) back into (5.20) gives
∂ 2 Ex
∂ 2 Ex
=
µ
ε
.
0 0
∂z 2
∂t2
(5.22)
Equation (5.22) is referred to as the wave equation and is one equation with one unknown Ex . Next,
114
we define the propagation velocity as
1
ν=√
≈ 3 × 108 = c.
µ0 ε0
(5.23)
Similarly, we have the following wave equation:
∂ 2 Hy
∂ 2 Hy
=
µ
ε
.
0 0
∂z 2
∂t2
(5.24)
Note that the expressions for the electric and magnetic fields in the a region MUST satisfy equations
(5.22) and (5.24), respectively. This will result in a solution to the electromagnetic problem in that
region. Next, to solve the wave equations we assume a solution in the following form: Ex (z, t) =
f1 (t − z/ν) + f2 (t + z/ν). It can be shown that
′
Ex (z, t) = |Exo | cos(ωt − ko z + ϕ1 ) + |Exo
| cos(ωt + ko z + ϕ2 ).
(5.25)
The first term in (5.25) represents the forward traveling wave and the second term represents the
backward traveling (or reflected) wave. We can also define the wave number in free-space as
ko =
ω
rad/m.
c
(5.26)
Next, the wavelength is the distance in space that ko z shifts by a value of 2π. This then gives
ko z = ko λ = 2π. ⇒
λ=
2π
.
ko
(5.27)
λ is called the free-space wavelength. We also have the phasor form of the propagating wave (5.25):
1
1 ′′ jϕ2 jko z jωt
|e e e .
Ex (z, t) = |Exo |ejϕ1 e−jko z ejωt + |Exo
2
2
(5.28)
Note that (5.28) is written in the time domain. Next, We can also write Maxwell’s equations in the
115
frequency domain in the following manner:
e = jωε0 E
e
∇×H
(5.29)
e = −jωµ0 H
e
∇×E
(5.30)
e = 0
∇·E
(5.31)
e = 0.
∇·H
(5.32)
Equations (5.29)-(5.32) assume a source free region and a sinusoidal steady state source. This then
leads to the generalization of (5.22) which is the wave equation in the frequency domain:
e = −k 2 E
e
∇2 E
o
(5.33)
√
ex = −ko E
ex .
where k = ω/c = ω µ0 ε0 . Next, to simplify (5.33) we consider the x-component: ∇2 E
Expanding gives
ex
∂2E
∂x2
2e
2e
ex . Next, assuming a plane wave such that E
ex does not
+ ∂∂yE2x + ∂∂zE2x = −ko2 E
vary with respect to x or y gives:
ex
∂ 2E
ex
= −ko E
∂z 2
(5.34)
ex (z) = Exo e−jko z + E ′ ejko z .
E
xo
(5.35)
which has the following solution:
e = −jωµ0 H
e we can obtain H:
e
Now, using Maxwell’s equation ∇ × E
]
[
1
−jko z
′ jko z
e
Hy = −
(−jko )Exo e
+ (jko )Exo e
jωµ0
√
√
ε0 −jko z
ε0 jko z
′
= Exo
e
− Exo
e
µ0
µ0
′ jko z
e .
= Hyo e−jko z + Hyo
116
Next, comparing Hyo to Exo gives
√
Exo =
and
′
Exo
where η0 =
√
=−
µ0
Hyo = η0 Hyo
ε0
(5.36)
µ0 ′
′
Hyo = −η0 Hyo
ε0
(5.37)
√
µ0 /ε0 = 377Ω ≈ 120πΩ. Equation (5.36) represents the forward traveling wave and
(5.37) represents the reflected wave. Therefore, in the far-field we have
η0 =
e
E
.
e
H
(5.38)
Note: be careful with vector components in (5.38). Also in instantaneous form we have
√
Hy (z, t) = Exo
ε0
′
cos(ωt − ko z) − Exo
µ0
√
ε0
cos(ωt + ko z).
µ0
(5.39)
5.6. Wave Propagation in Lossy Dielectrics
In this section we derive the expressions for a propagating wave in a lossy dielectric. To do
this we define the dielectric medium to have a constant µr and εr . This then gives
e = −k 2 E
e
∇2 E
(5.40)
√
√
where k (the wave number) in a more general form is k = ω µε = ko εr µr , ε = εr ε0 and µ = µr µ0 .
Next, considering the x-component of the electric field gives
ex
d2 E
ex .
= −k 2 E
2
dz
(5.41)
The solution to (5.41) allows for a complex solution. This then gives jk = α + jβ where α is the
117
attenuation constant and β is the phase constant. This then gives a solution of the following form
ex = Exo e−jkz = Exo e−αz e−jβz
E
(5.42)
Ex = Exo e−αz cos(ωt − βz).
(5.43)
or in the time domain:
Next, the loss in a dielectric can be represented with complex values of ε where ε = ε′ − jε′′ =
ε0 (ε′r − jε′′r ). Similarly, we have µ = µ′ − jµ′′ = µ0 (µ′r − jµ′′r ). This then results in the following
wave number:
√
√
k = ω µ(ε′ − jε′′ ) = ω µε′
√
1−
jε′′
.
ε′
Therefore, in general we have
√
α = Re[jk] = ω
and
√
β = Im[jk] = ω
µε′
2
µε′
2
[√
(
1+
[√
1+
(
ε′′
ε′
ε′′
ε′
]1/2
)2
−1
(5.44)
]1/2
)2
+1
.
(5.45)
We also have the wave velocity
νp =
ω
.
β
(5.46)
Note that (5.46) changes with material properties. This then implies λ = 2π/β also changes with
material properties. Next, the wave impedance can ge written as
√
η=
µ
=
′
ε − jε′′
√
1
µ
√
( )
′
ε
′′
1 − jεε′
(5.47)
which is a complex value. Because of this, the Electric and Magnetic fields are no longer in phase.
118
Also notice that
λ=
2π
2π
1
c
λ0
= √ ′ = √ ′ = √ ′ =
β
f µr εr
µr ε′r
ω µε
f µε
(5.48)
where λ0 is the free-space wavelength. As long as µr ε′r > 1 then λ < λ0 . Therefore, waves in
dielectrics have a shorter wavelength. This is useful for 1) printed transmission lines, 2) printed
antennas and 3) printed filters.
119
CHAPTER 6. TOPICS ON ELECTROMAGNETIC
COMPATIBILITY
6.1. Coupling Between Transmission Lines
6.1.1. General Problem
To investigate the coupling between two TLs we start with the general problem defined in
Fig. 72. The problem consists of two parallel TLs above a common ground plane. The TL with
the source is called the generator conductor and the TL with the two resistive loads is called the
receptor conductor. Rs is the source resistance, RL is the load resistance, RN E is the near-end load
on the receptor conductor and RF E is the far-end load on the receptor conductor. The receptor
conductor has a length L.
Receptor conductor
Generator conductor
RL
RFE
length L
RNE
Rs
Vs
Ground plane
Figure 72. General TL coupling problem.
Next, we can derive an equivalent circuit of the coupled TLs. By looking at one small section
of the coupled TLs we will be able to describe the interaction between the two TLs. The equivalent
circuit for one section of coupled TLs is shown in Fig. 73. We want to simplify the analysis of the
coupling between the two TLs in Fig. 73 to the following two cases:
1) capacitive coupling (occurs in high impedance circuits)
2) inductive coupling (occurs in low impedance circuits)
6.1.2. Capacitive coupling (low frequencies)
120
Rg/2
Rg/2
Lg/2
Lg/2
Cgr
Lgr/2
Lgr/2
Rs
Rr/2
Rr/2
Lr/2
Cr
Cg
RNE
Vs
RL
Lr/2
RFE
Figure 73. Equivalent circuit of the coupled transmission lines.
For this analysis, we assume that we have high impedance loads on the coupled TLs in Fig.
72. If this is the case, then the coupling between the TLs is mostly capacitive. This reduces the
equivalent circuit in Fig. 73 to the equivalent circuits shown in Figs. 74 a) and b).
Receptor conductor
Generator conductor
RL
Cgr = (cgr)L
RFE
+
+
Cg
Rs
+
cap
Rs
Cr
RNE
Cg
Vin
RL
Cr
R = RNE||RFE
cap
cap
VNE = VFE
Vs
VNE
Vs
Cgr
-
-
Ground plane
a)
b)
Figure 74. a) Capacitive coupling between the coupled transmission lines and b) the equivalent circuit of
the capacitively coupled transmission lines.
Next, we want to write the voltages induced on RN E and RF E by Vs . Evaluating the equivalent
circuit in Fig. 74 b) gives
(
VNcap
E
=
VFcap
E
= Vin
(
= Vin
121
)
1
R|| jωC
r
1
+
R|| jωC
r
1
jωCgr
gr
jω CgrC+C
r
jω +
1
R(Cgr +Cr )
)
.
(6.1)
Next, for low frequencies we have the following assumption:
ω <<
1
.
R(Cgr + Cr )
(6.2)
This then simplifies (6.1) to
cap
VNcap
E = VF E ≈ jωCgr Vin R.
(6.3)
Note again that (6.3) is for high impedance circuits. We can also write
(
Vin = Vs
and
(
Zin =
Zin
Zin + Rs
)
)
[
(
)]
1
1
1
||RL ||
+ R||
.
jωCg
jωCgr
jωCr
(6.4)
(6.5)
But, for low frequencies we can approximate a capacitor as an open. This then gives
(
Vin ≈ Vg(DC) ≈ Vs
)
RL
.
RL + Rs
(6.6)
6.1.3. Inductive coupling (low frequencies)
In this section we want to look at the low frequency inductive coupling between the TLs. For
this case we simply open all capacitors in Fig. 73. This then results in the equivalent circuit shown
in Fig. 75. The near-end voltage can be written as VNind
E = Ir RN E and the far-end voltage can be
written as VFind
E = Ir RF E . Next, KVL for the receptor gives Ir RN E + jωLr Ir + RF E Ir − jωLgr Ig = 0.
⇒
Ir =
(RN E
jωLgr Ig
.
+ RF E ) + jωLr
(6.7)
Next, KVL around the generator gives −Vs + Rs Ig + jωLg Ig + Ig RL − jωIr Lgr = 0. ⇒
Ig =
jωLgr
Vs
+
Ir .
(Rs + RL ) + jωLg (Rs + RL ) + jωLg
122
(6.8)
Receptor conductor
Generator conductor
+ ind
VFE
RL
RFE
Lr
Lg
-
Lgr
Ig
+
Rs
V
Vs
ind
NE
Ir
RNE
Ground plane
Figure 75. The equivalent circuit of the inductively coupled transmission lines.
Next, if we assume that ωLr << RN E + RF E , ωLg << (Rs + RL ) and a weak coupling condition of
(ωLgr )2 << (Rs + RL )(RN E + RF E ) then
Ig ≈
Vs
= Ig (DC)
Rs + RL
(6.9)
Ir ≈
jωLgr
Ig (DC).
RN E + RF E
(6.10)
and
Then
VNind
E ≈ jωLgr
−RN E
Ig (DC)
RN E + RF E
(6.11)
RF E
Ig (DC).
RN E + RF E
(6.12)
and
VFind
E ≈ +jωLgr
123
6.1.4. Equations for both inductive and capacitive coupling
Noting that both V ind and V cap are proportional to jω and Vs we can write the results as
and
(
)
VN E
cap
ind
= jω MN E + MN E
Vs
(6.13)
(
)
VF E
cap
ind
= jω MF E + MF E
Vs
(6.14)
where
MNind
E =
RN E
Lgr
,
RN E + RF E Rs + RL
(6.15)
MFind
E =
RF E
Lgr
RN E + RF E Rs + RL
(6.16)
and
cap
MNcap
E = MF E = (RN E ||RF E )
RL Cgr
.
Rs + RL
(6.17)
This then results in the inductive/capacitive model shown in Fig. 76 where
RL
RL + Rs
(6.18)
Vs
.
Rs + RL
(6.19)
Vg (DC) = Vs
and
Ig (DC) =
jωLgrIg(DC)
+
+
-
+
VNE RNE
RFE VFE
jωCgrVg(DC)
-
-
Figure 76. The equivalent inductive/capacitive model of the coupled transmission lines.
124
Note that the equivalent circuit model shown in Fig. 76 breaks down for TL lengths longer
than ≈ 0.2λ. Therefore, if the frequency is high enough then a smaller segment of the TL should
be analyzed or a more accurate distributive model of the coupled TL should be used.
Next, we consider the problem in Fig. 72 for the case when the ground plane has finite
conductivity. We represent the loss with a lumped resistor Ro = ro l where l is the length of the
TL and ro is the per-unit loss of the ground plane. The problem with Ro defined is shown in Fig.
77. Ro is called the common impedance and Vo is called the common impedance voltage where
Vo = ro Iref . ⇒ Iref = Ir + Ig ≈ Ig (DC) at low frequencies. ⇒
(
Vo ≈ Vs
)
Ro
.
Rs + RL
(6.20)
This voltage appears on the receptor circuit also. Therefore, this contributes to the near- and
far-end noise voltages at low frequencies. This then gives:
VNciE
= MNciE
Vs
(6.21)
VFciE
= MFciE
Vs
(6.22)
RN E
Ro
RN E + RF E Rs + RL
(6.23)
RF E
Ro
.
RN E + RF E Rs + RL
(6.24)
and
where
MNciE =
and
MFciE = −
With this notation, we have the following final coupling equations that include the common impedance
noise voltages:
)
(
VN E
cap
ind
= jω MN E + MN E + MNciE
Vs
125
(6.25)
and
(
)
VF E
cap
ind
= jω MF E + MF E + MFciE .
Vs
(6.26)
This results in the equivalent circuit show in Fig. 78.
Rs
Vs
Ig
+
+
VNE
Ir
VFE
RNE
+
-
Vo
RL
RFE
-
Iref
Ro
Figure 77. Coupled transmission lines with a lossy reference conductor.
RoIg(DC)
+
-
+
+
-
jωLgrIg(DC)
VNE RNE
+
RFE VFE
jωCgrVg(DC)
-
-
Figure 78. The equivalent inductive/capacitive model of the coupled transmission lines with a lossy
reference conductor.
6.2. Shielding
6.2.1. Reducing capacitive coupling
In this section we investigate the use of shielding to reduce the coupling between the TLs in
Fig. 72. From (6.3) we can see that we can reduce the capacitive coupling by :
1) reducing frequency
2) reducing RN E and/or RF E
3) reducing Vg (DC) by reducing Vs
4) reducing Cgr by increasing the spacing between the traces and shielding
For this section we will focus on using a shield around the receptor to reduce the capacitive
126
coupling. This problem is defined in Fig. 79 where Vs is the generator (source) open circuit voltage,
Vs,o is the output voltage of the source, Cg is the capacitance from the generator to the reference
conductor, Cgs is the capacitance from the generator to the shield, Cgr is the capacitance from
the generator to the reference conductor, Cs is the capacitance from the shield to the reference
conductor, Crs is the capacitance from the receptor conductor to the shield, Cr is the capacitance
from the receptor to the reference conductor and Vsh is the voltage between the shield and reference
conductor. If we ground the shield on the receptor conductor (at both ends), then we have the
equivalent circuit shown in Fig. 80.
Shield
Generator conductor
RFE VFE
RL
+ Cs
Cgs
Cgr
+
Rs Vs,o
Vs
+
Cg
VNE
-
Receptor conductor
+
-
Crs Vsh
Cr
RNE
-
Ground plane
Figure 79. Capacitively coupled transmission lines with a shielded receptor.
+
+
Rs
Cgr
Cg
Vs,o
Cr
RL Cgs
Crs
R = RNE||RFE
cap
cap
VNE = VFE
Vg
-
-
Figure 80. The equivalent circuit of the coupled transmission lines with a shielded receptor.
Using circuit analysis, the following expression can be evaluated from the circuit in Fig. 80:
(
VNcap
E = Vs,o
(
)
)
R|| jω(Crs1 +Cr )
(
)
R|| jω(Crs1 +Cr ) + jωC1 gr
or
127
(6.27)
(
VNcap
E
= Vs,o
jω
(
jω +
)
Cgr
Cgr +Crs +Cr
1
R(Cgr +Crs +Cr )
)
.
(6.28)
At low frequencies we have Vs,o ≈ Vs RL /(RL + Rs ) = Vg (DC) and ω << 1/(R(Cgr + Cr + Crs )).
This then simplifies (6.28) down to
cap
VNcap
E = VF E ≈ jωRCgr Vg (DC).
(6.29)
Equation (6.29) is similar to (6.3), except in (6.29) Cgr is much less than Cgr in (6.3).
6.2.2. Reducing inductive coupling
When looking at shielding for inductive coupling, we need to first consider the expressions in
(6.11) and (6.12). We can see that we can reduce the inductive coupling by :
1) reducing frequency
2) reducing Ig (DC)
3) reducing Lgr by
i) reducing the area between the trace and ground plane
ii) orientation
iii) a choke
Shield
Generator conductor
RL
Lg,sh
Lsh
Receptor conductor
Lg
Ig
Rsh
Ir
Rs
+ ind
VFE
Lsh,r
RFE
Lr
Lgr
-
Vs
+
V
ind
NE
Ir
RNE
-
Figure 81. Inductively coupled transmission lines with a shielded receptor.
To understand the best method of shielding for inductive coupling we need to consider the
problem in Fig. 81. We want to write the voltages across the near- and far-end resistors in
128
terms of the inductive coupling and the source voltage. First, KVL around the shield loop gives:
Ish Rsh + (jωLg,sh )Ig − (jωLsh )Ish − (jωLsh,r )Ir = 0. Solving for Ish gives
Ish =
+jω(Lg,sh Ig + Lsh,r Ir )
.
Rsh + jωLsh
(6.30)
Next, KVL around the receptor loop gives: Ir RN E −(jωLgr )Ig −(jωLsh,r )Ish +(jωLr )Ir +Ir RF E = 0
ind
where VNind
E = Ir RN E and VF E = −Ir RF E . Solving for Ir gives:
Ir =
jω(Lgr Ig + Lsh,r Ish )
.
RN E + RF E + jωLr
(6.31)
Next, substituting (6.30) into (6.31) gives
Ir =
RN E
[
]
Lgr Ig (Rsh + jωLsh ) − jωLsh,r (Lg,sh Ig + Lsh,r Ir )
jω
.
+ RF E + jωLr
Rsh + jωLsh
(6.32)
Rearranging then gives:
Ir =
RN E
[
]
jω
Ig (Lgr Rsh + jωLsh Lgr − jωLsh,r Lg,sh ) + jω(Lsh,r )2 Ir
.
+ RF E + jωLr
Rsh + jωLsh
(6.33)
Now assume we have a uniform current distribution along the conductors in Fig. 81. This then
gives Lgr ≈ Lg,sh and Lsh ≈ Lsh,r . Rearranging (6.33) and substituting in the above assumptions
gives:
[
Ir 1 +
[RN E
]
]
[
jωLgr Rsh
(ωLsh,r )2
= Ig
.
+ RF E + jωLr ][Rsh + jωLsh ]
[RN E + RF E + jωLr ][Rsh + jωLsh ]
(6.34)
Solving for Ir gives
Ir =
[RN E + RF E
jωIg Lgr Rsh
.
+ jωLr ][Rsh + jωLsh ] + (ωLsh,r )2
(6.35)
Next, if we denote the denominator of (6.35) at DEN = [RN E +RF E +jωLr ][Rsh +jωLsh ]+(ωLsh,r )2
129
and assume that Lr ≈ Lsh ≈ Lsh,r then DEN simplifies to DEN = (RN E +RF E )Rsh +jωLsh [RN E +
RF E + Rsh ]. Next, if we assume that RN E + RF E >> Rsh then (6.35) simplifies to
Ir ≈
Using Ig ≈ Ig (DC) =
Vs
.
Rs +RL
+jωIg Lgr Rsh
.
+ RF E )(Rsh + jωLsh )
(RN E
(6.36)
This then gives the following expressions for the near- and far-end
voltages in Fig. 81:
[
VNind
E
][
]
RN E
Rsh
= −RN E Ir ≈
(jωLgr Ig (DC))
RN E + RF E
Rsh + jωLsh
and
[
VFind
E
−RF E
= RF E Ir ≈
(jωLgr Ig (DC))
RN E + RF E
The
[
Rsh
Rsh + jωLsh
][
]
Rsh
.
Rsh + jωLsh
(6.37)
(6.38)
]
(6.39)
term in (6.38) is referred to as the Shield Factor (SF). This term is the same in both (6.37) and
(6.38). Also note that
SF =
This then implies:



SF =
1
Rsh
=
.
Lsh
Rsh + jωLsh
1 + jω R
sh
1
if
ω << Rsh /Lsh

 Rsh /jωLsh if
ω >> Rsh /Lsh .
(6.40)
(6.41)
What we see from (6.41) is that the shield has no effect on the coupling caused by the mutual
inductance at low frequencies. The voltage w.r.t. frequency caused by mutual inductance is
plotted in Fig. 82. We can see that the shield for the inductive coupling starts to reduce V ind at
approximately Rsh /Lsh . Therefore, the shield grounded at one end does not improve the inductive
crosstalk. To do this we need to ground the shield at both ends. However, shields grounded at both
ends are susceptible to ground loop problems.
130
|V IND|
Low frequency
High frequency
20 dB/decade
ω ≅ Rsh/Lsh
ω
Figure 82. Induced voltage caused by inductive coupling.
6.2.3. Summary of transmission line coupling equations
Without a shield
For capacitive coupling without a shield we have the following equations:
cap
VNcap
E = VF E ≈ jωCgr Vin R.
(6.42)
For inductive coupling without a shield we have the following equations:
VNind
E ≈ jωLgr
−RN E
Ig (DC)
RN E + RF E
(6.43)
VFind
E ≈ jωLgr
RF E
Ig (DC)
RN E + RF E
(6.44)
and
where Ig (DC) = Vs /(Rs + RL ).
Then to compute both the inductive and capacitive coupling between TLs without a shield we use
the following expressions:
(
)
VN E
cap
ind
= jω MN E + MN E
Vs
131
(6.45)
and
(
)
VF E
cap
ind
= jω MF E + MF E
Vs
(6.46)
where
MNind
E =
RN E
Lgr
,
RN E + RF E Rs + RL
(6.47)
MFind
E =
RF E
Lgr
RN E + RF E Rs + RL
(6.48)
and
cap
MNcap
E = MF E = (RN E ||RF E )
RL Cgr
.
Rs + RL
(6.49)
Next, if we include the common impedance noise voltage (these expressions are valid with or without
a shield) we have:
and
(
)
VN E
cap
ind
= jω MN E + MN E + MNciE
Vs
(6.50)
(
)
VF E
cap
ind
= jω MF E + MF E + MFciE
Vs
(6.51)
where
MNciE =
Ro
RN E
RN E + RF E Rs + RL
(6.52)
RF E
Ro
.
RN E + RF E Rs + RL
(6.53)
and
MFciE = −
With a shield
For capacitive coupling with a shield we have the following equations:
cap
VNcap
E = VF E ≈ jωRCgr Vg (DC).
(6.54)
For inductive coupling with a shield we have the following equations:
[
VNind
E
]
−RN E
=
jωLgr Ig (DC) SF
RN E + RF E
132
(6.55)
and
[
VFind
E
]
RF E
=
jωLgr Ig (DC) SF
RN E + RF E
(6.56)
where
SF =
This then implies:
Rsh
1
.
=
Lsh
Rsh + jωLsh
1 + jω R
sh



SF =
1
if
ω << Rsh /Lsh

 Rsh /jωLsh if
ω >> Rsh /Lsh .
(6.57)
(6.58)
Then to compute both the inductive and capacitive coupling between TLs with a shield we use the
following expressions:
and
(
)
VN E
cap
ind
= jω MN E + MN E
Vs
(6.59)
(
)
VF E
cap
ind
= jω MF E + MF E
Vs
(6.60)
where
MNind
E =
−RN E
Lgr
SF,
RN E + RF E Rs + RL
(6.61)
MFind
E =
RF E
Lgr
SF
RN E + RF E Rs + RL
(6.62)
and
cap
MNcap
E = MF E = (RN E ||RF E )
RL Cgr
.
Rs + RL
(6.63)
6.3. Twisted pair
The next TL coupling problem investigated is the twisted pair problem shown in Fig. 83. The
problem consists of two conductors twisted together in the presence of a single conductor carrying a
current IeG . In this problem there exists capacitive and inductive coupling between the twisted pair
and the single conductor. To represent this coupling, we define the equivalent circuit shown in Fig.
84. We will evaluate this equivalent circuit to understand how the twisted affect of the wire can
reduce the inductive coupling. For this investigation we have two possible loading configurations:
133
Coupling
magnetic
fields
IG
Figure 83. Twisted pair coupling problem.
LHT
+
-
+
-
E1 = jωlgr1LHTIg
E1 = jωlgr1LHTIg
E2 = jωlgr2LHTIg
E2 = jωlgr2LHTIg
RFE
RNE
+
I1 = jωcgr1LHTVg
-
+
I2 = jωcgr2LHTVg
I2 = jωcgr2LHTVg
-
I1 = jωcgr1LHTVg
Ground reference
Figure 84. Equivalent circuit of the twisted pair coupling problem.
1) Balanced configuration - both wires have the same impedance to ground.
2) Unbalanced configuration - both wires do not have the same impedance to ground.
The ind./cap. model used to investigate the crosstalk in the twisted pair will depend on how
the ground is connected (i.e., if a balanced or unbalanced system is used). First, if we assume
an unbalanced system we get the equivalent circuits in Figs. 85 and 86. The model in Fig. 85
represents the inductive coupling and the model in Fig. 86 represents the capacitive coupling.
6.3.1. Inductive coupling - unbalanced
When evaluating the equivalent circuit in Fig. 85 using KVL we get the following equation:
e1 + E
e2 − E
e1 − E
e2 = 0. This shows that for each unit of twisted wires along the twisted pair cable,
E
134
we have a zero-sum value for the KVL analysis of each twist. This indicates that the inductive
coupling between the single wire and the twisted pair in Fig. 83 is almost zero or negligible. This
then tells us that twisted pair is useful when it is desired to reduce the crosstalk due to inductive
coupling.
+
+
RNE
V
IND
NE
-
-
+
-
E1 = jωlgr1LHTIg
E2 = jωlgr2LHTIg
E2 = jωlgr2LHTIg
E1 = jωlgr1LHTIg
RFE
+
-
+
-
Figure 85. Equivalent circuit for the inductive coupling in the twisted pair coupling problem.
6.3.2. Capacitive coupling - unbalanced
Next, for the capacitive coupling we consider the equivalent circuit in Fig. 86. If we assume
that the twisted pair is tight, then we can assume the the following: cgr1 ≈ cgr2 . This assumption
states that the capacitance between the generator in Fig. 83 and both of the conductors in the
twisted pair are the same. This then reduces the circuit in Fig. 86 to the circuit in Fig. 87. We
can then write the near- and far-end voltage as
VNCAP
≈ jωcgr1 LHT NT Vg (DC)RN E ||RF E
E
(6.64)
T
VNCAP
≈ jωCgr1
Vg (DC)RN E ||RF E
E
(6.65)
or
T
= cgr1 LHT NT .
where LHT is the length of the full-twist, NT is the number of twists and Cgr1
Therefore, using a twisted pair with unbalanced grounding reduces V IN D but not V CAP . Finally,
in general we have
VN E ≈
]
[
RN E
T
Vg (DC)RN E ||RF E .
jω(lgr1 − lgr2 )LHT I(DC) + jωCgr1
RN E + RF E
135
(6.66)
+
RNE
CAP
RFE
VNE
-
I2 = jωcgr2LHTVg
I1 = jωcgr1LHTVg
I1 = jωcgr1LHTVg
I2 = jωcgr2LHTVg
Figure 86. Equivalent circuit for the capacitive coupling in the twisted pair coupling problem.
+
RNE
CAP
2I1
VNE
RFE
-
Figure 87. Reduced equivalent circuit for the capacitive coupling in the twisted pair coupling problem.
6.3.3. Balanced case
For this section we look at the balanced case (Fig. 88). The equivalent circuit for capacitive
coupling for this case is shown in Fig. 89. The inductive coupling is the same as for the unbalanced
case. We can see that the near-end voltage is zero for an even number of twists and close to zero
for an odd number of twists. Therefore, for the balanced case the capacitive coupling is essentially
zero. In summary, if possible it is best to use the twisted pair in a balanced case because both the
inductive and capacitive coupling is minimized.
RNE/2
RFE/2
RNE/2
RFE/2
Figure 88. The twisted pair coupling problem with a balanced load.
136
{
+
0
for an even number of twists
E1-E2
for an odd number of twists
-
+
RNE/2
RFE/2
CAP
RNE/2
VNE
RFE/2
-
N(I1+II2)
N(I1+II2)
Figure 89. The equivalent circuit for the capacitive coupling in the twisted pair coupling problem with a
balanced load.
137