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Lecture 02
ECE 344
Microwave Fundamentals
Spring 2017
Transmission Line Theory
1
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2
– Transmission Line Theory
– The lumped-element circuit model for a transmission line.
– Wave Equations for TL model and their solutions.
– Propagation in lossy & lossless TL.
– Characteristic Impedance.
– Reflection & Transmission Coefficients.
– Standing Wave.
– Input Impedance.
Agenda
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Lumped circuits: resistors, capacitors, inductors
Neglects time delays (phase change)
Account for propagation and time delays
(phase change)
Transmission-Line Theory
Distributed circuit elements: transmission lines
We need transmission-line theory whenever the length of a line
is significant compared to a wavelength. 3
The key difference between circuit theory and transmission
line theory is electrical size.
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4
Transmission Lines
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Transmission Lines
2 conductors
4 per-unit-length parameters:
Dz
5
R = series resistance per unit length, for both conductors, in Ω/m.
L = series inductance per unit length, for both conductors, in H/m.
G = shunt conductance per unit length, in S/m or Ʊ/m.
C = shunt capacitance per unit length, in F/m.
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Transmission Line (cont.)
zD
,i z t
+ + + + + + + - - - - - - - - - -
,v z tx x x B
6
RDz LDz
GDz CDz
z
v(z+Dz,t)
+
-
v(z,t)
+
-
i(z,t) i(z+Dz,t)
Note: There are equal and opposite currents on the two conductors. (We only need to work with the current on the top conductor, since we have chosen to put all of the series elements there.)
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( , )( , ) ( , ) ( , )
( , )( , ) ( , ) ( , )
i z tv z t v z z t i z t R z L z
t
v z z ti z t i z z t v z z t G z C z
t
D D D
D D D D D
Transmission Line (cont.)
7
RDz LDz
GDz CDz
z
v(z+Dz,t)
+
-
v(z,t)
+
-
i(z,t) i(z+Dz,t)
Applying KVL
Applying KCL
(2.1)
(2.2)
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Hence
( , ) ( , ) ( , )( , )
( , ) ( , ) ( , )( , )
v z z t v z t i z tRi z t L
z t
i z z t i z t v z z tGv z z t C
z t
D
D
D D D
D
taking the limit as Dz 0:
v iRi L
z t
i vGv C
z t
“Telegrapher's
Equations”
TEM Transmission Line (cont.)
8 These are the time domain form of the transmission line equations,
(2.3)
(2.4)
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To combine these, take the derivative of the first one with respect to z:
TEM Transmission Line (cont.)
9
From instantaneous to phasor form
)()()(
)()()(
zVCjGz
zI
zILjRz
zV
)())(()(
)()(
)(
2
2
2
2
zVCjGLjRz
zV
z
zILjR
z
zV
(2.5)
(2.6)
The same differential equation also holds for i.
jt
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Solution: ( ) z zV z Ae Be
TEM Transmission Line (cont.)
10
0)()(
0)()(
2
2
2
2
2
2
zIz
zI
zVz
zV
(2.7)
(2.8)
This yields to give wave equations for V(z) and I (z):
))(( CjGLjRj Where
is the complex propagation constant, which is a function of frequency.
Equations (2.7) and (2.8) are 2nd order differential equations
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Denote:
TEM Transmission Line (cont.)
11
( )( )R j L G j C
[np/m] attenuationcontant
j
rad/m phaseconstant 1/m propagation constant
Traveling wave solutions to (2.7) and (2.8) can be found as
zz
zz
eIeIzI
eVeVzV
00
00
)(
)( (2.9)
(2.10)
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TEM Transmission Line (cont.)
0 0( ) z z j zV z V e V e e
Forward travelling wave (a wave traveling in the positive z direction):
0
0
0
( , ) Re
Re
cos
z j z j t
j z j z j t
z
v z t V e e e
V e e e e
V e t z
g 0t
z
0
zV e
2
g
2g
The wave “repeats” when:
Hence:
12
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Phase Velocity
Let’s track the velocity of a fixed point on the wave (a point of constant
phase), e.g., the crest of the wave.
0( , ) cos( )zv z t V e t z
z
vp (phase velocity)
13
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Phase Velocity (cont.)
0
constant
t z
dz
dt
dz
dt
Set
Hence p
v
14
For any point on the wave the term ωt-βz should stay constant,
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00
0 0
j z j j z
z z
z j zV e e
V z V e V
V e e e
e
e
0
0 cos
c
, R
os
e j t
z
z
V e t
v z t V z
z
V z
e
e t
In the time domain:
Wave in +z
direction Wave in -z
direction
General Case (Waves in Both Directions)
15
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Applying (2.5) to the voltage of (2.9) gives the current on the line:
so
Characteristic Impedance Z0 (cont.)
16
)()()(
zILjRz
zV
zz eVeVzV 00)(
)()(
)(
)()(
00
00
zz
zz
eVeVLjR
zI
zILjReVeV
(2.9)
(2.5)
Comparison with (2.10) shows that a characteristic impedance, Z0, can be
defined as
)(
)()(0
CjG
LjRLjRZ
(2.11)
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Characteristic Impedance Z0 (cont.)
17
0
00
0
0
I
VZ
I
V
zz eZ
Ve
Z
VzI
0
0
0
0)( (2.12)
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Characteristic Impedance Z0
0
( )
( )
V zZ
I z
0
0
( )
( )
z
z
V z V e
I z I e
so 00
0
VZ
I
+
V+(z)
-
I+ (z)
z
Assumption: A wave is traveling in the positive z direction.
(Note: Z0 is a number, not a function of z.)
18
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Backward-Traveling Wave
0
( )
( )
V zZ
I z
0
( )
( )
V zZ
I z
so
+
V -(z)
-
I - (z)
z
A wave is traveling in the negative z direction.
Note: The reference directions for voltage and current are
chosen the same as for the forward wave.
19
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General Case
0 0
0 0
0
( )
1( )
z z
z z
V z V e V e
I z V e V eZ
A general superposition of forward and
backward traveling waves: Most general case:
Note: The reference
directions for voltage
and current are the
same for forward and
backward waves.
20
+
V (z)
-
I (z)
z
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0 0
0 0
0 0
0
z z
z z
V z V e V e
V VI z e e
Z
j R j L G j C
R j LZ
G j
Z
C
I(z)
V(z)+
-z
2
mg
[m/s]pv
Guided wavelength
Phase velocity
Summary of Basic TL formulas
21
dB/m 8.686Attenuation in
1
10 dB/m 20log 8.686e
Attenuation in
10log 0.4343lnx x
Note :
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Lossless Case
0, 0R G
( )( )j R j L G j C
j LC
so 0
LC
0
R j LZ
G j C
0
LZ
C
1pv
LC
pv
(independent of freq.) (real and independent of freq.)
22
In many practical cases, however, the loss of the line is very small and
so can be neglected, resulting in a simplification of the results.
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23
zjzj eZ
Ve
Z
VzI
0
0
0
0)(
zjzj eVeVzV 00)(
Lossless Case
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0 0
z zV z V e V e
Where do we assign z = 0 ?
The usual choice is at the load.
Amplitude of voltage wave
propagating in negative z
direction at z = 0.
Amplitude of voltage wave
propagating in positive z
direction at z = 0.
Terminated Transmission Line
25
V (z) +
- z
ZL
z = 0
Terminating impedance (load)
I (z)
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0 0
z zV z V e V e
What is V(-d) ?
0 0
d dV d V e V e
0 0
0 0
d dV VI d e e
Z Z
Propagating
forwards
Propagating
backwards
Terminated Transmission Line (cont.)
d distance away from load
The current at z = -d is
28
V (-d) +
- z
ZL
z = 0
Terminating impedance (load)
I (-d)
z = -d
(This does not necessarily have to be the length of the entire line.)
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20
0
1d d
L
VI d e e
Z
200 0 0
0
1d d d dVV d V e V e V e e
V
Similarly,
L Load reflection coefficient
Terminated Transmission Line (cont.)
29
V (-d) +
- z
ZL
z = 0
I (-d)
z = -d
2
0 1d d
LV d V e e
or
0
0
L
V
V
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2
0
20
0
1
1
d d
L
d d
L
V d V e e
VI d e e
Z
Z(-d) = impedance seen “looking” towards load at z = -d.
Terminated Transmission Line (cont.)
30
V (-d) +
- z
ZL
z = 0
I (-d)
z = -d Z(-d)
2
0 2
1
1
d
L
d
L
V d eZ d Z
I d e
Note:
If we are at the
beginning of the
line, we will call this
“input impedance”.
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At the load (d = 0):
0
10
1
LL
L
Z Z Z
Thus,
20
0
0
20
0
1
1
dL
L
dL
L
Z Ze
Z ZZ d Z
Z Ze
Z Z
Terminated Transmission Line (cont.)
0
0
LL
L
Z Z
Z Z
2
0 2
1
1
d
L
d
L
eZ d Z
e
Recall
31
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Simplifying, we have:
0
0
0
tanh
tanh
L
L
Z Z dZ d Z
Z Z d
Terminated Transmission Line (cont.)
20
2
0 0 0
0 0 2
2 0 00
0
0 0
0
0 0
0
0
0
1
1
cosh sinh
cosh sinh
dL
d
L L L
d
d L LL
L
d d
L L
d d
L L
L
L
Z Ze
Z Z Z Z Z Z eZ d Z Z
Z Z Z Z eZ Ze
Z Z
Z Z e Z Z eZ
Z Z e Z Z e
Z d Z dZ
Z d Z d
Hence, we have
32
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2
0
20
0
1
1
j d j d
L
j d j d
L
V d V e e
VI d e e
Z
Impedance is periodic
with period g/2:
2 1
2 1
2 1
2
/ 2
g
g
d d
d d
d d
Terminated Lossless Transmission Line
j j
Note: tanh tanh tand j d j d
The tan function repeats when
0
0
0
tan
tan
L
L
Z jZ dZ d Z
Z jZ d
33
Lossless:
2
0 2
1
1
j d
L
j d
L
eZ d Z
e
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Note: For the remainder of our transmission line discussion we will
assume that the transmission line is lossless.
2
0 2
0
0
0
1
1
tan
tan
j d
L
j d
L
L
L
eZ d Z
e
Z jZ dZ
Z jZ d
0
0
2
LL
L
g
p
Z Z
Z Z
v
Summary for Lossless Transmission Line
34
V (-d) +
- z
ZL
z = 0
I (-d)
z = -d Z(-d)
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35
Reflection and transmission at the junction of two transmission lines with
different characteristic impedances.
the voltage for z < 0 is
reflection coefficient is given by
0z ),()( 0 zjzj eeVzV
01
01
ZZ
ZZ
the voltage for z > 0 is 0z ,)( 0 zjTeVzV
Equating these voltages at z = 0 gives the transmission coefficient, T , as
01
121
ZZ
ZT
The transmission coefficient between two points in a circuit is often expressed
in dB as the insertion loss, IL, dB ||log20 TIL
RL = −20 log |Γ| dB,
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0
0
0LL
L
Z Z
Z Z
No reflection from the load
Matched Load (ZL=Z0)
0 Z d Z z for any
36
V (-d) +
- z
ZL
z = 0
I (-d)
z = -d Z(-d)
2
0 2
1
1
j d
L
j d
L
eZ d Z
e
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1L
Always imaginary!
Short-Circuit Load (ZL=0)
37
V (-d) +
- z
z = 0
I (-d)
z = -d Z(-d)
0
0
0
tan
tan
L
L
Z jZ dZ d Z
Z jZ d
0 tanZ d jZ d
0
0
LL
L
Z Z
Z Z
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Note: 2g
dd
scZ d jX
S.C. can become an O.C. with a
g/4 transmission line.
Short-Circuit Load (ZL=0)
0 tanscX Z d
38
0 1 / 4 1 / 2 3 / 4
Xsc
Inductive
Capacitive
d/g
0 tanZ d jZ d
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0
0
L
Z
Z
Always imaginary!
Open-Circuit Load (ZL=)
39
V (-d) +
- z
z = 0
I (-d)
z = -d Z(-d)
1L
0
0
LL
L
Z Z
Z Z
0
0
0
tan
tan
L
L
Z jZ dZ d Z
Z jZ d
0 cotZ d jZ d
0
0
0
1 / tan
/ tan
L
L
j Z Z dZ d Z
Z Z j d
or
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2g
dd
O.C. can become a S.C. with a
g/4 transmission line.
Open-Circuit Load (ZL=)
40
ocZ d jX
0 cotocX Z d
0 cotZ d jZ d
Note:
0 1/4 1/2
Xoc Inductive
Capacitive
d/g
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2
0 1 Lj j z
LV z V e e
2
0
2
0
1
1 L
j z j z
L
jj z j z
L
V z V e e
V e e e
max 0
min 0
1
1
L
L
V V
V V
Voltage Standing Wave
z
1+ L
1
1- L
0
( )V z
V
/ 2g
z D 0z
41
2 2 / 2z z g D D
V (z) +
- z
ZL
z = 0
I (z)
z
𝜙L + 2βz = 0,− 2π,…….,− 2nπ
Maximum Occurs When
Minimum Occurs When
𝜙L + 2βz = −π, − 3π, ….,− (2n+1)π
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max
min
V
VVoltage Standing Wave Ratio VSWR
Voltage Standing Wave Ratio
1
1
L
L
VSWR
z
1+ L
1
1- L
0
( )V z
V
/ 2gz D 0z
42
max 0
min 0
1
1
L
L
V V
V V
V (z) +
- z
ZL
z = 0
I (z)
z
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Voltage Standing Wave Ratio
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Voltage Standing Wave Ratio
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45
VSWR=(1+0)/|1-0|=1
VSWR=(1+0.2)/|1-0.2|=1.5
VSWR=(1+0.5)/|1-0.5|=3 VSWR=(1+1)/|1-1|=∞
Voltage Standing Wave Ratio
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Using Transmission Lines to Synthesize Loads
A microwave filter constructed from microstrip line.
This is very useful in microwave engineering.
46
We can obtain any reactance that we want from a short or open transmission line.
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47
Find the voltage at any point on the line.
ZTh
VTh
+
Zin
+
- -
V (-d)
0
0
0
tan
tan
L
in
L
Z jZ dZ Z d Z
Z jZ d
inTh
in Th
ZV d V
Z Z
At the input:
Voltage on a Transmission Line
I (z)
+ Z L
-
Z Th
V Th
d
Zin
+
-
0 ,Z V (z)
z=0
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0
21j z zj
Lz V eV e 0
0
LL
L
Z Z
Z Z
2
0 1j d j d inL Th
in Th
ZV d V e e V
Z Z
2
2
1
1
j zj d zin L
Th j d
in Th L
Z eV z V e
Z Z e
At z = -d :
Hence
0 2
1
1
j dinTh j d
in Th L
ZV V e
Z Z e
48
Voltage on a Transmission Line (cont.)
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Let’s derive an alternative form of the result.
2
0 2
1
1
j d
Lin j d
L
eZ Z d Z
e
2
20 20
2 22
00 2
2
0
2
0 0
2
0
20 0
0
1
11
1 11
1
1
1
1
j d
Lj dj d
LL
j d j dj d
L Th LLThj d
L
j d
L
j d
Th L Th
j d
L
j dTh ThL
Th
in
in Th
eZ
Z ee
Z e Z eeZ Z
e
Z e
Z Z e Z Z
eZ
Z
Z
Z Z
Z Z Ze
Z Z
Z
2
0
20 0
0
1
1
j d
L
j dTh ThL
Th
e
Z Z Z Ze
Z Z
49
Start with:
Voltage on a Transmission Line (cont.)
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2
0
2
0
1
1
j zj d z L
Th j d
Th s L
Z eV z V e
Z Z e
2
0
2
0
1
1
j d
in L
j d
in Th Th s L
Z Z e
Z Z Z Z e
where 0
0
Ths
Th
Z Z
Z Z
Therefore, we have the following alternative form for the result:
Hence, we have
50
(source reflection coefficient)
Voltage on a Transmission Line (cont.)
2
2
1
1
j zj d zin L
Th j d
in Th L
Z eV z V e
Z Z e
Recall:
Substitute
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2
0
2
0
1
1
j zj z d L
Th j d
Th s L
Z eV z V e
Z Z e
Voltage wave that would exist if there were no reflections from
the load (a semi-infinite transmission line or a matched load). 51
Voltage on a Transmission Line (cont.)
This term accounts for the multiple (infinite) bounces.
I (z)
+ZL
-
ZTh
VTh
d
Zin
+
-
0 ,Z V (z)
z=0
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2 2
2 2 2 20
0
1 j d j d
L L s
j d j d j d j d
Th L s L L s L s
Th
e e
ZV d V e e e e
Z Z
Wave-bounce method (illustrated for z = -d ):
52
ZL
Z Th
V Th
d
+
-
0 ,Z
Voltage on a Transmission Line (cont.)
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22 2
22 2 20
0
1
1
j d j d
L S L S
j d j d j d
Th L L S L S
Th
e e
ZV d V e e e
Z Z
Geometric series:
2
0
11 , 1
1
n
n
z z z zz
2 2
2 2 2 20
0
1 j d j d
L L s
j d j d j d j d
Th L s L L s L s
Th
e e
ZV d V e e e e
Z Z
2j d
L sz e
53
Group together alternating terms:
Voltage on a Transmission Line (cont.)
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or
2
0
20
2
1
1
1
1
j d
L s
Th
j dTh
L j d
L s
eZV d V
Z Ze
e
2
0
2
0
1
1
j d
LTh j d
Th L s
Z eV d V
Z Z e
This agrees with the previous result (setting z = -d ).
Note: The wave-bounce method is a very tedious method – not recommended.
Hence
54
Voltage on a Transmission Line (cont.)
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At a distance d from the load:
*
*
2
0 2 2 * 2
*
0
1Re
2
1Re 1 1
2
z z z
L L
P z V z I z
Ve e e
Z
2
20 2 4
0
11
2
z z
L
VP z e e
Z
If Z0 real (low-loss transmission line)
Time-Average Power Flow
2
0
20
0
1
1
z z
L
z z
L
V z V e e
VI z e e
Z
j
*2 * 2
*2 2
z z
L L
z z
L L
e e
e e
pure imaginary
Note:
55
V (z) +
- z
ZL
z = 0
I (z)
z
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Low-loss line
2
20 2 4
0
2 2
20 02 2
0 0
11
2
1 1
2 2
z z
L
z z
L
VP z e e
Z
V Ve e
Z Z
Power in forward-traveling wave Power in backward-traveling wave
2
20
0
11
2L
VP z
Z
Lossless line ( = 0)
Time-Average Power Flow (cont.)
56
V (-d) +
- z
ZL
z = 0
I (-d)
z = -d
Note:
In the general lossy case, we cannot say
that the total power is the difference of
the powers in the two waves.
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00
0
tan
tan
L Tin T
T L
Z jZ dZ Z
Z jZ d
2
4 4 2
g g
g
d
00
Tin T
L
jZZ Z
jZ
0
2
00
0in in
T
L
Z Z
ZZ
Z
Quarter-Wave Transformer
2
0Tin
L
ZZ
Z
so
0 0T LZ Z Z
Hence
(This requires ZL to be real.)
57
Matching condition
ZL Z0 Z0T
Zin d
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Conjugate matching
Home work
Prove that the condition for maximum power delivered to the load is
58
I (z)
+ZL
-
ZTh
VTh
d
Zin
+
-
0 ,Z V (z)
z=0
*
thin ZZ
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Field Analysis of Transmission Lines
In this section we will derive the transmission line parameters (R, L, G, C) in terms of the electric and magnetic fields of the transmission line
59
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Coaxial Cable
Here we present a “case study” of one particular transmission line, the coaxial cable.
a
b ,r
Find C, L, G, R
We will assume no variation in the z direction, and take a length of one meter in
the z direction in order to calculate the per-unit-length parameters.
60
For a TEMz mode, the shape of the fields is independent of frequency, and
hence we can perform the calculation using electrostatics and magnetostatics.
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Coaxial Cable (cont.)
-l0
l0
a
b
r0 0
0
ˆ ˆ2 2 r
E
Find C (capacitance / length)
Coaxial cable
h = 1 [m]
r
From Gauss’s law:
0
0
ln2
B
AB
A
b
ra
V V E dr
bE d
a
61
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-l0
l0
a
b
r
Coaxial cable
h = 1 [m]
r
0
0
0
1
ln2 r
QC
V b
a
Hence
We then have:
0 F/m2
[ ]
ln
rCb
a
Coaxial Cable (cont.)
62
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ˆ2
IH
Find L (inductance / length)
From Ampere’s law:
Coaxial cable
h = 1 [m]
r
I
0ˆ
2r
IB
(1)
b
a
B d S
h
I
I z Center conductor Magnetic flux:
Coaxial Cable (cont.)
63
Note: We ignore “internal inductance” here, and
only look at the magnetic field between the two
conductors (accurate for high frequency.
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Coaxial cable
h = 1 [m]
r
I
0
0
0
1
2
ln2
b
r
a
b
r
a
r
H d
Id
I b
a
0
1ln
2r
bL
I a
0 H/mln [ ]2
r bL
a
Hence
Coaxial Cable (cont.)
64
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0 H/mln [ ]2
r bL
a
Observation:
0 F/m2
[ ]
ln
rCb
a
0 0 r rLC
This result actually holds for any transmission line (proof omitted).
Coaxial Cable (cont.)
65
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0 H/mln [ ]2
r bL
a
For a lossless cable:
0 F/m2
[ ]
ln
rCb
a
0
LZ
C
0 0
1ln [ ]
2
r
r
bZ
a
00
0
376.7303 [ ]
Coaxial Cable (cont.)
66
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-l0
l0
a
b
0 0
0
ˆ ˆ2 2 r
E
Find G (conductance / length)
Coaxial cable
h = 1 [m]
From Gauss’s law:
0
0
ln2
B
AB
A
b
ra
V V E dr
bE d
a
Coaxial Cable (cont.)
67
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-l0
l0
a
b
J E
We then have leakIG
V
0
0
(1) 2
2
22
leak a
a
r
I J a
a E
aa
0
0
0
0
22
ln2
r
r
aa
Gb
a
2[S/m]
ln
Gb
a
or
Coaxial Cable (cont.)
68
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Observation:
F/m2
[ ]
ln
Cb
a
G C
This result actually holds for any transmission line (proof omitted).
2[S/m]
ln
Gb
a
0 r
Coaxial Cable (cont.)
69
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G C
Hence:
tanG
C
tanG
C
Coaxial Cable (cont.)
As just derived,
70
This is the loss tangent that would
arise from conductivity.
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Find R (resistance / length)
Coaxial cable
h = 1 [m]
Coaxial Cable (cont.)
,b rb
a
b
,a ra
a bR R R
1
2a saR R
a
1
2b sbR R
b
1sa
a a
R
1
sb
b b
R
0
2a
ra a
0
2b
rb b
Rs = surface resistance of metal
71
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72
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At high frequency, discontinuity effects can become important.
Limitations of Transmission-Line Theory
Bend
Incident
Reflected
Transmitted
The simple TL model does not account for the bend.
ZTh
ZL Z0
+ -
73
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At high frequency, radiation effects can become important.
When will radiation occur?
We want energy to travel from the generator to the load, without radiating.
Limitations of Transmission-Line Theory (cont.)
74
ZTh
ZL Z0
+ -
This is explored next.
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r a
b
z
The coaxial cable is a perfectly shielded system – there is never any radiation at
any frequency, as long as the metal thickness is large compared with a skin depth.
The fields are confined
to the region between
the two conductors.
Limitations of Transmission-Line Theory (cont.)
75
Coaxial Cable
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The twin lead is an open type of transmission
line – the fields extend out to infinity.
The extended fields may cause interference
with nearby objects.
(This may be improved by using “twisted pair”.)
+ -
Limitations of Transmission-Line Theory (cont.)
Having fields that extend to infinity is not the same thing as having radiation, however!
76
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The infinite twin lead will not radiate by itself, regardless of how far
apart the lines are (this is true for any transmission line).
h
Incident
Reflected
The incident and reflected waves represent an exact solution to
Maxwell’s equations on the infinite line, at any frequency.
*1ˆRe E H 0
2t
S
P dS
S
+ -
Limitations of Transmission-Line Theory (cont.)
No attenuation on an infinite lossless line
77
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A discontinuity on the twin lead will cause radiation to occur.
Note: Radiation effects usually
increase as the frequency increases.
Limitations of Transmission-Line Theory (cont.)
h
Incident wave Pipe
Obstacle
Reflected wave
Bend h
Incident wave
Bend
Reflected wave 78
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To reduce radiation effects of the twin lead at discontinuities:
h
1) Reduce the separation distance h (keep h << ).
2) Twist the lines (twisted pair).
Limitations of Transmission-Line Theory (cont.)
CAT 5 cable
(twisted pair)
79
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Baluns
80
Baluns are used to connect coaxial cables to twin leads.
4:1 impedance-transforming baluns, connecting 75 TV coax to 300 TV twin lead
Coaxial cable: an “unbalanced” transmission line
Twin lead: a “balanced” transmission line
Balun: “Balanced to “unbalanced”
https://en.wikipedia.org/wiki/Balun
They suppress the common mode currents on the transmission lines.
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Baluns (cont.)
81
https://en.wikipedia.org/wiki/Balun
“Baluns are present in radars, transmitters, satellites, in every
telephone network, and probably in most wireless network
modem/routers used in homes.”
Baluns are also used to connect coax (unbalanced line)
to dipole antennas (balanced).
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82
Ground (zero volts)
+ -
Ground (zero volts)
+
0
The voltages are
unbalanced with
respect to ground.
The voltages are
balanced with respect
to ground.
Coax Twin Lead
Baluns (cont.)
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Baluns (cont.)
83
Baluns are necessary because, in practice, the two transmission lines
are always both running over a ground plane.
If there were no ground plane, and you only had the two lines connected to each
other, then a balun would not be necessary.
(But you would still want to have a matching network between the two lines if they have
different characteristic impedances.)
Coax Twin Lead
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Baluns (cont.)
84
When a ground plane is present, we really have three
conductors, forming a multiconductor transmission line, and
this system supports two different modes.
+ -
Differential mode
(desired)
+ +
Common mode
(undesired)
The differential and common modes on a twin lead over ground
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Baluns (cont.)
85
Differential mode
(desired)
Common mode
(undesired)
The differential and common modes on a coax over ground
+
- +
-
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Baluns (cont.)
86
A balun prevents common modes from being excited at the junction
between a coax and a twin lead.
Coax Twin Lead
Ground
Common mode Common mode
No Balun
Differential mode
Differential mode
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Baluns (cont.)
87
From another point of view, a balun prevents currents from flowing on the
outside of the coax.
Coax Twin Lead
Ground
No Balun
I1
I1
I2 I1
I2- I1
(direct current path to ground)
2 1I I
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Baluns (cont.)
88
A simple model for the mode excitation at the junction
The coax is replaced by a solid tube with a voltage source at the end.
Solid tube Twin Lead
Ground
+ - V0
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Baluns (cont.)
89
Next, we use superposition.
+
Solid tube Twin Lead
Ground
+ -
+ -
V0 / 2
V0 / 2
Differential mode
Solid tube Twin Lead
Ground
+ -
+ -
V0 / 2
V0 / 2
Common mode
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Baluns (cont.)
90
A balun prevents common modes from being excited at the junction
between a coax and a twin lead.
With Balun
Coax Twin Lead
Ground
Differential mode Differential mode
Balun
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Baluns (cont.)
91
One type of balun uses an isolation transformer.
https://en.wikipedia.org/wiki/Balun
The input and output are isolated
from each other, so one side can be
grounded while the other is not.
Circuit diagram of a 4:1 autotransformer
balun, using three taps on a single
winding, on a ferrite rod
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Baluns (cont.)
92
Here is a more exotic type of 4:1 impedance transforming balun.
0
V0 V0
0
0
V0
-V0
0 0 -V0
I0/2
I0 I0/2
I0 V0
This balun uses two 1:1 transformers to achieve symmetric output voltages.
“Guanella balun”
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Baluns (cont.)
93
Inside of 4:1 impedance transforming VHF/UHF balun for TV
Ferrite core
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Baluns (cont.)
94
Another type of balun uses a choke to “choke off” the common mode.
https://en.wikipedia.org/wiki/Balun
A coax is wound around a ferrite core. This creates a large inductance for the common mode,
while it does not affect the differential mode (whose fields are confined inside the coax).
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Baluns (cont.)
95
Baluns are very useful for feeding differential circuits with coax on PCBs.
No balun
Without a balun, there would be a
common mode current on the
coplanar strips (CPS) line.
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Baluns (cont.)
96
+ - + - + -
+ - - +
= +
V0 / 2 V0 / 2
V0 / 2 V0 / 2
V0
Excites desired differential mode
Excites undesired common mode
Superposition allows us to see what the problem is.
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Baluns (cont.)
97
Baluns can be implemented in microstrip form.
Tapered balun
Balun feeding a microstrip yagi antenna
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Baluns (cont.)
98
Baluns can be implemented in microstrip form.
Marchand balun
180o hybrid rat-race coupler used as a balun
Input (unbalanced)
Output #2
Output #1
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Baluns (cont.)
99
If you try to feed a dipole antenna directly with coax, there will be a
common mode current on the coax.
I1= I2
I3 0
No balun
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Baluns (cont.)
100
Baluns are commonly used to feed dipole antennas.
A balun first converts the coax to a twin lead,
and then the twin lead feeds the dipole. A “sleeve balun” directly
connects a coax to a dipole.
Coaxial “sleeve” region
Coaxial “sleeve balun”
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Baluns (cont.)
101
Feeding a dipole antenna over a ground plane
/ 4
This acts like voltage source for the dipole, which forces the differential mode.
The dipole is loaded by a short-circuited section of twin lead – an open circuit because of the /4 height.