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ECE 6130: LECTURE 2
FIELD ANALYSIS OF TRANSMISSION LINES
Text Section 2.2
Portfolio:
1) How do you calculate the RLGC parameters for a transmission line configuration that isn't in
the standard table? See Chapter 2, problem 3.
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Field Analysis of Transmission Lines
Role of E and H in R,L,G,C:
Transmission Lines in ECE 3170 used V and I as the waveform variables, R,L,G,C to represent
the properties of the line. R,L,G,C were given in tables as a function of the size and shape of
the transmission line. These tables are derived from the E and H inside the transmission line.
“Unit Length” Section of TL – E and H Field Patterns
Given a section of 2-conductor transmission line that is “one unit length long” (like 1 meter):
The two conductors have opposite charges, which makes E lines between them, and H lines
around them.
SEE FIELD PATTERNS FOR TL LINES
How do you find these fields?
Analytical (Maxwell's equations): In ECE3170, we learned how to find them for coax,
parallel plate, single wire, etc. (symmetrical systems) using Maxwell's equations.
Analytical (Boundary Value): You can also find the field distributions by applying Laplace's
equation 2 V =0 and electric field boundary conditions (tangential E is constant across a
boundary, normal D is constant across the boundary), and solve this analytically. This is called
a "boundary value problem". You can find out more about solving these in…
Computational Electromagnetics: We will solve for the fields on a microstrip and/or stripline
using the Finite Difference method, which is one way of solving boundary value problems.
For all of these methods, we assume:
The voltage between the conductors is Vo ejz and the current is Io ejz .
Z is direction of propagation, so this represents + and – traveling waves.
Inductance and Capacitance (L and C) from E and H
The time-averaged stored magnetic energy is given by (See Section 1.6 and circuit theory):
Wm 

2
 H  H ds * Wm  L | I o | / 4 W
*
L4
Io
S
2
H H
S
ds
H /m
So Inductance per unit length :
The time-averaged stored electric energy is given by (See Section 1.6 and circuit theory):
*

We   E  E ds  We  C | Vo |2 / 4 W
4 S
So Capacitance per unit length :
C

Vo
 E  E ds
*
2
S
F /m
Power Loss due to losses in the metal conductors (See equation 1.131, surface impedance
section):
Pc 
*
Rs
Rs
H

H
ds

2 S
2
  H  H dl dz
*
C
W
Power Loss PER UNIT LENGTH
*
Rs
Pc 
H  H dl W / m
2 C
Notes:
(1) 2 comes from RMS. H is given as PEAK value, so RMS value = Peak / sqrt(2), and then
you are squaring H.
(2) * represents CONJUGATE (change sign on imaginary part)
(3) C is the total contour of both conductors , C = C1 + C2
(4) Rs is the surface resistance of the conductors = 1/ (s) where s (meters) is the “skin depth”
(how far field penetrates before being reduced to 1/e,  is the conductivity (S/m)
From circuit theory: Pc = R |Io|2 / 2
So, series resistance PER UNIT LENGTH :
Rs
R
| Io | 2

*
C1  C 2
H  H dl  / m
When there is resistance between the conductors (in the substrate or insulating material) PER
UNIT LENGTH:
Pd 
 ' '
2
 E  Eds
S
unit length is
Where ’’ = imaginary part of permittivity =  / 
From circuit theory Pd = G|Vo|2 /2, so the shunt conductance per unit length is:
G
 ' '
| Vo |2
 E  E ds
S
S /m
Telegrapher’s Equations
Lumped element model represents voltage and current on transmission line. Text derives these
for field, I will derive them for voltage and current.
Kirchoff Loop Equation:
V(z,t) – R I(z,t) – L di(z,t)/dt – v(z+z,t) = 0
R = R’z, L = L’ z
Divide by z, combine V terms, and rearrange:
V(z,t) / z – R’ I(z,t) – L’ dI(z,t) /dt – V(z+z,t) / z = 0
-[ V(z+z,t) - V(z,t) ] / z = R’ I(z,t) – L’ dI(z,t)/dt
Limit as z  0 (difference goes to differential):
-dV(z,t)/dz = R’ I(z,t) – L’ dI(z,t)/dt
Kirchoff Node Equation:
I(z,t) – G v(z+z,t) – C dv(z+z,t)/dt – I(z+z,t) = 0
I(z,t) – G’z v(z+z,t) – C’z dv(z+z,t)/dt – I(z+z,t) = 0
Divide by z and rearrange :
-[ I(z+z,t) - I(z,t)] / z = G’ v(z+z,t) + C’ dv(z+z,t)/dt
Limit as z  0 :
-dI(z,t) / dt = G’ v(z+z,t) + C’ dv(z+z,t)/dt
Telegraphers equations:
-dV(z,t)/dz = R’ I(z,t) + L’ dI(z,t)/dt
-dI(z,t) / dt = G’ v(z+z,t) + C’ dv(z+z,t)/dt
Sinusoidal Steady-State Form:
Convert to Phasor form (d/dt  j).
-dV(z)/dz = (R’ + jL’) (z)
-dI(z)/dz = (G’ + jC’) V(z)
What is difference between this and “regular” circuit equations?? They are a function of z….
distance down the transmission line.
Wave Equation form:
Take d/dz of one equation, and substitute it into the other equation. (Doesn’t matter which one).
See text and copy of Chapter 2 in the IEEE room.
-d2 I(z)/dz2 –2(z) = 0
Alternate form :
-d2 V(z)/dz2 –2V(z) = 0
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