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Prof. David R. Jackson Dept. of ECE
Notes 1
ECE 5317-6351 Microwave Engineering
Fall 2017
Transmission Line Theory
1
In this set of notes we first overview different types of waveguiding systems, including transmission lines. We then develop transmission line theory in detail. We discuss several types of practical transmission lines. We discuss baluns.
Overview
2
A waveguiding structure is one that carries a signal (or power) from one point to another.
There are three common types: Transmission lines Fiber-optic guides Waveguides
Waveguiding Structures
Note: An alternative to using waveguiding structures is wireless transmission using antennas. (Antennas are discussed in ECE 5318.)
3
Transmission Line
Has two conductors running parallel (one can be inside the other) Can propagate a signal at any frequency (in theory) Becomes lossy at high frequency Can handle low or moderate amounts of power Does not have signal distortion, unless there is loss (but there always is) May or may not be immune to interference (shielding property) Does not have Ez or Hz components of the fields (TEMz)
Properties
Coaxial cable (coax) Twin lead
(shown connected to a 4:1 impedance-transforming balun)
4
Transmission Line (cont.)
CAT 5 cable (twisted pair)
The two wires of the transmission line are twisted to reduce interference and radiation from discontinuities.
5
Transmission Line (cont.)
Microstrip
h
w
εr
εr
w
Stripline
h
Transmission lines commonly met on printed-circuit boards
Coplanar strips
h εr
w w
Coplanar waveguide (CPW)
h εr
w
6
Transmission Line (cont.)
Transmission lines are commonly met on printed-circuit boards.
A microwave integrated circuit.
Microstrip line
7
Waveguides
Consists of a single hollow metal pipe Can propagate a signal only at high frequency: f > fc
The width must be at least one-half of a wavelength Has signal distortion, even in the lossless case Immune to interference Can handle large amounts of power Has low loss (compared with a transmission line) Has either Ez or Hz component of the fields (TMz or TEz)
Properties
http://en.wikipedia.org/wiki/Waveguide_(electromagnetism) 8
Fiber-Optic Guide Properties
Uses a dielectric rod Can propagate a signal at any frequency (in theory) Can be made to have very low loss Has minimal signal distortion Has a very high bandwidth Very immune to interference Not suitable for high power Has both Ez and Hz components of the fields (“hybrid mode”)
9
Fiber-Optic Guide (cont.) Two types of fiber-optic guides:
1) Single-mode fiber
2) Multi-mode fiber
Carries a single mode, as with the mode on a transmission line or waveguide. Requires the fiber diameter to be small relative to a wavelength.
Has a fiber diameter that is large relative to a wavelength. It operates on the principle of total internal reflection (critical angle effect).
Note: The carrier is at optical frequency, but the modulating (baseband) signal is often at microwave frequencies (e.g., TV, internet signal).
10
Fiber-Optic Guide (cont.)
http://en.wikipedia.org/wiki/Optical_fiber
Higher index core region
11
Multi-mode Fiber
Transmission-Line Theory
We need transmission-line theory whenever the length of a line is significant compared to a wavelength.
12
L
Transmission Line
2 conductors
4 per-unit-length parameters:
C = capacitance/length [F/m]
L = inductance/length [H/m]
R = resistance/length [Ω/m]
G = conductance/length [ /m or S/m] Ω ∆z
13
Transmission Line (cont.)
z∆
( ),i z t
+ + + + + + + - - - - - - - - - - ( ),v z tx x x B
14
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.)
+
-
+
-
( ),i z z t+ ∆( ),i z t
( ),v z z t+ ∆( ),v z t
R z∆ L z∆
G z∆ C z∆
z
( , )( , ) ( , ) ( , )
( , )( , ) ( , ) ( , )
i z tv z t v z z t i z t R z L zt
v z z ti z t i z z t v z z t G z C zt
∂= + ∆ + ∆ + ∆
∂∂ + ∆
= + ∆ + + ∆ ∆ + ∆∂
Transmission Line (cont.)
15
+
-
+
-
( ),i z z t+ ∆( ),i z t
( ),v z z t+ ∆( ),v z t
R z∆ L z∆
G z∆ C z∆
z
Hence
( , ) ( , ) ( , )( , )
( , ) ( , ) ( , )( , )
v z z t v z t i z tRi z t Lz t
i z z t i z t v z z tGv z z t Cz t
+ ∆ − ∂= − −
∆ ∂+ ∆ − ∂ + ∆
= − + ∆ −∆ ∂
Now let ∆z → 0:
v iRi Lz ti vGv Cz t
∂ ∂= − −
∂ ∂∂ ∂
= − −∂ ∂
“Telegrapher’sEquations”
TEM Transmission Line (cont.)
16
To combine these, take the derivative of the first one with respect to z:
2
2
2
2
v i iR Lz z z t
i iR Lz t z
v v vR Gv C L G Ct t t
∂ ∂ ∂ ∂ = − − ∂ ∂ ∂ ∂ ∂ ∂ ∂ = − − ∂ ∂ ∂
∂ ∂ ∂ = − − − − − − ∂ ∂ ∂
Switch the order of the derivatives.
TEM Transmission Line (cont.)
17
( )2 2
2 2( ) 0v v vRG v RC LG LC
z t t∂ ∂ ∂ − − + − = ∂ ∂ ∂
The same differential equation also holds for i.
Hence, we have:
2 2
2 2
v v v vR Gv C L G Cz t t t
∂ ∂ ∂ ∂ = − − − − − − ∂ ∂ ∂ ∂
TEM Transmission Line (cont.)
18
( )2
2
2( ) ( ) 0d V RG V RC LG j V LC V
dzω ω− − + − − =
( )2 2
2 2( ) 0v v vRG v RC LG LC
z t t∂ ∂ ∂ − − + − = ∂ ∂ ∂
TEM Transmission Line (cont.)
Time-Harmonic Waves:
19
jt
ω∂→
∂
Note that
= series impedance/length
( ) ( )2
2
2( )d V RG V j RC LG V LC V
dzω ω= + + −
2( ) ( )( )RG j RC LG LC R j L G j Cω ω ω ω+ + − = + +
Z R j LY G j C
ωω
= += + = parallel admittance/length
Then we can write: 2
2( )d V ZY V
dz=
TEM Transmission Line (cont.)
20
Define
We have:
Solution:
2 ZYγ ≡
( ) z zV z Ae Beγ γ− += +
[ ]1/2( )( )R j L G j Cγ ω ω= + +
( ) ( )2
2
2
d V zV z
dzγ=Then
TEM Transmission Line (cont.)
γ is called the “propagation constant”.
21
Question: Which sign of the square root is correct?
Principal square root:
TEM Transmission Line (cont.)
/2jz z e θ= π θ π− < ≤
22
[ ]1/2( )( )R j L G j Cγ ω ω= + +
We choose the principal square root.
( )Re 0z ≥
( )( )R j L G j Cγ ω ω= + + Re 0γ ≥
Hence
(Note the square-root (“radical”) symbol here.)
Denote:
TEM Transmission Line (cont.)
Re 0α γ= ≥
23
( )( )R j L G j Cγ ω ω= + +
[np/m]α = attenuationcontant
jγ α β= +
[ ]rad/mβ = phase constant
( )( )110 dB/m 20log 8.686e α α−= − =Attenuation in
[ ]1/mγ = propagation constant
( ) ( )10log 0.4343lnx x=
Note :
+ ( )z z j z
zAe Ae eγ α β− − −=
wave going in direction
TEM Transmission Line (cont.)
24
There are two possible locations for γ : it must be in the first quadrant (α > 0).
0 0α β≥ ⇒ ≥
Hence:
ReIm
α γβ γ
==
R j Lω+
Re
Im
G j Cω+
Re
Im
γ
γ−
Re
Im
TEM Transmission Line (cont.)
0 0( ) z z j zV z V e V e eγ α β+ + − + − −= =
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
α β ω
φ α β ω
α ω β φ
+ + − −
+ − −
+ −
=
=
= − +
gλ 0t =
z
0zV e α+ −
2
g
πβλ
=
2gβλ π=
The wave “repeats” when:
Hence:
25
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 zα ω β φ+ + −= − +
26
z
(phase velocity) pv
Phase Velocity (cont.)
0
constantω β
ω β
ωβ
− =
− =
=
t zdzdtdzdt
Set
Hence p
v ωβ
= Im )(
pv
R j L G j Cω
ω ω=
+ +
In expanded form:
27
Characteristic Impedance Z0
0( )( )
V zZI z
+
+≡
0
0
( )
( )
z
z
V z V eI z I e
γ
γ
+ + −
+ + −
=
=
so 00
0
VZI
+
+=
Assumption: A wave is traveling in the positive z direction.
(Note: Z0 is a number, not a function of z.)
28
( )I z+
( )V z++
−z
Use first Telegrapher’s Equation:
v iRi Lz t
∂ ∂= − −
∂ ∂so
dV RI j L ZIdz
ω= − − = −
Hence 0 0
z zV e ZI eγ γγ + − + −− = −
Characteristic Impedance Z0 (cont.)
29
0
0
( )
( )
z
z
V z V eI z I e
γ
γ
+ + −
+ + −
=
=Recall:
From this we have:
Use:
00
0
V Z ZZI ZYγ
+
+= = =
Characteristic Impedance Z0 (cont.)
Z R j LY G j C
ωω
= += +
30
Both are in the first quadrant
0
0
/Re 0
ZY
Z Z ZYZ
γ⇒ = ∈
⇒ = ∈⇒ ≥
first quadrant
right - half plane
0ZZY
= (principle square root)
Hence, we have
Characteristic Impedance Z0 (cont.)
31
0Z R j LZY G j C
ωω
+= =
+
0Re 0Z ≥
( )
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
eφ α
γ γ
ββ φ α−+
+ − −
−+ − − +
+
+
= +
= +
( ) ( ) ( )
( )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
ω
α
α
ω β
ω β φ
φ− +
+ +
−
−
+
=
+
+
+
−=
In the time domain:
Wave in +z direction Wave in -z
direction
General Case (Waves in Both Directions)
32
Backward-Traveling Wave
0( )( )
V z ZI z
−
− =− 0
( )( )
V z ZI z
−
− = −so
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.
33
z
( )I z−
( )V z−+
−
General Case
0 0
0 00
( )1( )
z z
z z
V z V e V e
I z V e V eZ
γ γ
γ γ
+ − − +
+ − − +
= +
= −
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.
34
z
( )I z
( )V z+
−
( )
( )
( ) ( )
0 0
0 0
0 0
0
z z
z z
V z V e V e
V VI z e eZ
j R j L G j C
R j LZG j
Z
C
γ γ
γ γ
γ α β ω ω
ωω
+ − − +
+ −− +
= + = + +
=
−
++
+
=
=
[ ]2 mgπλβ
=
[m/s]pv ωβ
=
Guided wavelength
Phase velocity
Summary of Basic TL formulas
35
dB/m 8.686α=Attenuation in
( )I z
( )V z+−
Lossless Case 0, 0R G= =
( )( )j R j L G j C
j LC
γ α β ω ω
ω
= + = + +
=
so 0
LC
α
β ω
=
=
0R j LZG j C
ωω
+=
+ 0LZC
=1
pvLC
=
pv ωβ
=
(independent of freq.) (real and independent of freq.)
36
Lossless Case (cont.) 1
pvLC
=
If the medium between the two conductors is homogeneous (uniform) and is characterized by (ε, µ), then we have that (proof omitted):
LC µε=
The speed of light in a dielectric medium is 1
dcµε
=
Hence, we have that p dv c=
The phase velocity does not depend on the frequency, and it is always the speed of light (in the material).
(proof given later)
37
( ) 0 0z zV z V e V eγ γ+ − − += +
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
38
( )( )
0
0
0
0
V V
V V
+ +
− −
=
=
( )V z+ ( )V z− Terminating impedance (load)
( )V z
( )I z
0z =z
LZ+−
( ) 00 0
zV V z e γ++ +⇒ =
( ) ( ) ( ) ( ) ( )0 00 0
z z z zV z V z e V z eγ γ− − −+ −= +
( ) 00 0
zV z V e γ+− −=
( ) 00 0
zV V z e γ−− −⇒ =
Terminated Transmission Line (cont.)
( ) 0 0z zV z V e V eγ γ+ − − += +
Hence
Can we use z = z0 as a reference plane?
39
( ) 00 0
zV z V e γ−+ +=
Terminating impedance (load)
( )V z
( )I z
0z =z
LZ+−
0z z=
Terminated Transmission Line (cont.)
( ) ( ) ( )0 0z zV z V e V eγ γ+ − − += +
Compare:
Note: This is simply a change of reference plane, from z = 0 to z = z0.
40
( ) ( ) ( ) ( ) ( )0 00 0
z z z zV z V z e V z eγ γ− − −+ −= +
Terminating impedance (load)
( )V z
( )I z
0z =z
LZ+−
0z z=
( ) 0 0z zV z V e V eγ γ+ − − += +
What is V(-d) ?
( ) 0 0d dV d V e V eγ γ+ − −− = +
( ) 0 0
0 0
d dV VI d e eZ Z
γ γ+ −
−− = −
Propagating forwards
Propagating backwards
Terminated Transmission Line (cont.)
d ≡ distance away from load
The current at z = -d is
41
(This does not necessarily have to be the length of the entire line.)
Terminating impedance (load)
( )I d−
z d= −
+−
0z =
zLZ( )V d−
( ) ( )20
0
1d dL
VI d e eZ
γ γ+
−− = − Γ
( ) 200 0 0
0
1d d d dVV d V e V e V e eV
γ γ γ γ−
+ − − + −+
− = + = +
Similarly,
ΓL ≡ Load reflection coefficient
Terminated Transmission Line (cont.)
42
( ) ( )20 1d d
LV d V e eγ γ+ −− = + Γor
0
0L
VV
−
+Γ ≡
( )I d−
z d= −
+−
0z =
zLZ( )V d−
( ) ( )
( ) ( )
20
20
0
1
1
d dL
d dL
V d V e e
VI d e eZ
γ γ
γ γ
+ −
+−
− = + Γ
− = − Γ
Z(-d) = impedance seen “looking” towards load at z = -d.
Terminated Transmission Line (cont.)
43
( ) ( )( )
2
0 2
11
dL
dL
V d eZ d ZI d e
γ
γ
−
−
− + Γ− = = − − Γ
Note: If we are at the beginning of the
line, we will call this “input impedance”.
( )I d−
z d= −
+−
0z =
zLZ( )V d−
( )Z d−
At the load (d = 0):
( ) 0101
LL
L
Z Z Z + Γ
= ≡ − Γ
Thus,
( )20
00
20
0
1
1
dL
L
dL
L
Z Z eZ Z
Z d ZZ Z eZ Z
γ
γ
−
−
−+ + − = −− +
Terminated Transmission Line (cont.)
0
0
LL
L
Z ZZ Z
−Γ =
+
( )2
0 2
11
dL
dL
eZ d Ze
γ
γ
−
−
+ Γ− = − Γ
Recall
44
⇒
Simplifying, we have:
( ) ( )( )
00
0
tanhtanh
L
L
Z Z dZ d Z
Z Z dγγ
+− = +
Terminated Transmission Line (cont.)
( ) ( ) ( )( ) ( )
( ) ( )( ) ( )
( ) ( )( ) ( )
202
0 0 00 0 2
2 0 00
0
0 00
0 0
00
0
1
1
cosh sinhcosh sinh
dLd
L L Ld
d L LL
L
d dL L
d dL L
L
L
Z Z eZ Z Z Z Z Z e
Z d Z ZZ Z Z Z eZ Z e
Z Z
Z Z e Z Z eZ
Z Z e Z Z e
Z d Z dZ
Z d Z d
γγ
γγ
γ γ
γ γ
γ γγ γ
−−
−−
+ −
+ −
−+ + + + − − = = + − − − − + + + −
= + − − +
= +
Hence, we have
45
( ) ( )
( ) ( )
20
20
0
1
1
j d j dL
j d j dL
V d V e e
VI d e eZ
β β
β β
+ −
+−
− = + Γ
− = − ΓImpedance is periodic with period λg/2:
( )
( )
2 1
2 1
2 1
2
/ 2g
g
d d
d d
d d
β ππ π
λ
λ
− =
− =
⇒ − =
Terminated Lossless Transmission Line
j jγ α β β= + =
Note: ( ) ( ) ( )tanh tanh tand j d j dγ β β= =
The tan function repeats when
( ) ( )( )
00
0
tantan
L
L
Z jZ dZ d Z
Z jZ dββ
+− = +
46
Lossless:
( )2
0 2
11
j dL
j dL
eZ d Ze
β
β
−
−
+ Γ− = − Γ
For the remainder of our transmission line discussion we will assume that the transmission line is lossless.
( )
( )( )
2
0 2
00
0
11
tantan
j dL
j dL
L
L
eZ d Ze
Z jZ dZ
Z jZ d
β
β
ββ
−
−
+ Γ− = − Γ
+= +
0
0
2
LL
L
g
p
Z ZZ Z
v
πλβ
ωβ
−Γ =
+
=
=
Summary for Lossless Transmission Line
47
( )I d−
z d= −
+−
0z =
zLZ( )V d−
( )Z d−
0
0
0LL
L
Z ZZ Z
−Γ = =
+
No reflection from the load
Matched Load (ZL=Z0)
( ) 0 Z d Z z− = for any
48
( )2
0 2
11
j dL
j dL
eZ d Ze
β
β
−
−
+ Γ− = − Γ
( )I d−
z d= −
+−
0z =
zLZ( )V d−
( )Z d−
⇒
1LΓ = −
Always imaginary!
Short-Circuit Load (ZL=0)
49
( ) ( )( )
00
0
tantan
L
L
Z jZ dZ d Z
Z jZ dββ
+− = +
( ) ( )0 tanZ d jZ dβ− =
0
0
LL
L
Z ZZ Z
−Γ =
+
( )Z d− z d= − 0z =
z
( )I d−
( )V d−+−
Note: 2g
ddβ πλ
=
( ) scZ d jX− =
S.C. can become an O.C. with a λg/4 transmission line.
Short-Circuit Load (ZL=0)
( )0 tanscX Z dβ=
50
( ) ( )0 tanZ d jZ dβ− =
scX
Inductive
Capacitive
/ gd λ0 1/ 4 1 / 2 3 / 4
0
0L
ZZ
∞ −Γ =
∞ +
Always imaginary!
Open-Circuit Load (ZL=∞)
51
1LΓ = +
0
0
LL
L
Z ZZ Z
−Γ =
+
( ) ( )( )
00
0
tantan
L
L
Z jZ dZ d Z
Z jZ dββ
+− = +
( ) ( )0 cotZ d jZ dβ− = −
( ) ( ) ( )( ) ( )
00
0
1 / tan/ tan
L
L
j Z Z dZ d Z
Z Z j dββ
+− = +
or
( )Z d− z d= − 0z =
z
( )I d−
( )V d−+−
2g
ddβ πλ
=
O.C. can become a S.C. with a λg/4 transmission line.
Open-Circuit Load (ZL=∞)
52
( ) ocZ d jX− =
( )0 cotocX Z dβ= −
( ) ( )0 cotZ d jZ dβ− = −
Note:
Inductive
Capacitive
ocX
/ gd λ0 1/ 4 1 / 2
Using Transmission Lines to Synthesize Loads
A microwave filter constructed from microstrip line.
This is very useful in microwave engineering.
53
We can obtain any reactance that we want from a short or open transmission line.
54
Find the voltage at any point on the line.
( ) ( )( )
00
0
tantan
Lin
L
Z jZ dZ Z d Z
Z jZ dββ
+= − = +
( ) inTh
in Th
ZV d VZ Z
− = +
At the input:
Voltage on a Transmission Line
0,Z β ( )V z
( )I z
+
−
0z =
LZ
d
ThZ
ThV+
−
inZ
ThVThZ
inZ( )V d−
+
−
+
−
Note: For a lossy line, replace: j jβ γ β α→ − = −
( ) ( )021j z zj
Lz V eV e β β+ +− + Γ= 0
0
LL
L
Z ZZ Z
−Γ =
+
( ) ( )20 1j d j d in
L Thin Th
ZV d V e e VZ Z
β β+ − − = + Γ = +
( ) ( )2
2
11
j zj d zin L
Th j din Th L
Z eV z V eZ Z e
ββ
β
+− +
−
+ Γ= + + Γ
At z = -d :
Hence
0 2
11
j dinTh j d
in Th L
ZV V eZ Z e
ββ
+ −−
= + + Γ
55
Voltage on a Transmission Line (cont.)
⇒
Incident wave (not the same as the initial wave!)
Let’s derive an alternative form of the previous result.
( )2
0 2
11
j dL
in j dL
eZ Z d Ze
β
β
−
−
+ Γ= − = − Γ
( )( ) ( )
( )( ) ( )
( )
2
20 20
2 220
0 2
20
20 0
20
20 0
0
111
1 111
1
1
1
j dL
j dj dLL
j d j dj dL Th LL
Thj dL
j dL
j dTh L Th
j dL
j dTh ThL
Th
in
in Th
eZ Z eeZ e Z eeZ Z
e
Z eZ Z e Z Z
eZZ
ZZ Z
Z Z ZeZ Z
Z
β
ββ
β ββ
β
β
β
β
β
−
−−
− −−
−
−
−
−
−
+ Γ + Γ− Γ ⇒ = =
+ Γ + − Γ+ Γ + − Γ
+ Γ=
+ + Γ
+
−
+ Γ = + − + Γ +
=( )2
0
20 0
0
1
1
j dL
j dTh ThL
Th
eZ Z Z Ze
Z Z
β
β
−
−
+ Γ + − − Γ +
56
Start with:
Voltage on a Transmission Line (cont.)
( ) ( )2
02
0
11
j zj d z L
Th j dTh s L
Z eV z V eZ Z e
ββ
β
+− +
−
+ Γ= + − Γ Γ
20
20
11
j din L
j din Th Th s L
Z Z eZ Z Z Z e
β
β
−
−
+ Γ= + + − Γ Γ
where 0
0
Ths
Th
Z ZZ Z
−Γ ≡
+
Therefore, we have the following alternative form for the result:
Hence, we have
57
(source reflection coefficient)
Voltage on a Transmission Line (cont.)
( ) ( )2
2
11
j zj d zin L
Th j din Th L
Z eV z V eZ Z e
ββ
β
+− +
−
+ Γ= + + Γ
Recall:
Substitute
( ) ( )( )2
02
0
11
j zj z d L
Th j dTh s L
Z eV z V eZ Z e
ββ
β
+− − −
−
+ Γ= + − Γ Γ
The “initial” voltage wave that would exist if there were no reflections from the load (a semi-infinite transmission line or a matched load).
58
Voltage on a Transmission Line (cont.)
This term accounts for the multiple (infinite) bounces.
0,Z β ( )V z
( )I z
+
−
0z =
LZ
d
ThZ
ThV+
−
inZ
( )( )
( ) ( ) ( ) ( )
2 2
2 2 2 20
0
1 j d j dL L s
j d j d j d j dTh L s L L s L s
Th
e eZV d V e e e e
Z Z
β β
β β β β
− −
− − − −
+ Γ + Γ Γ − = + Γ Γ Γ + Γ Γ Γ Γ + +
Wave-bounce method (illustrated for z = -d ):
59
Voltage on a Transmission Line (cont.)
0,Z β ( )V z
( )I z
+
−
0z =
LZ
d
ThZ
ThV+
−
inZ
We add up all of the bouncing waves.
( )
( ) ( )( ) ( )
22 2
22 2 20
0
1
1
j d j dL S L S
j d j d j dTh L L S L S
Th
e eZV d V e e e
Z Z
β β
β β β
− −
− − −
+ Γ Γ + Γ Γ +
− = + Γ + Γ Γ + Γ Γ + + +
Geometric series:
2
0
11 , 11
n
nz z z z
z
∞
=
= + + + = <−∑
( )( )
( ) ( ) ( ) ( )
2 2
2 2 2 20
0
1 j d j dL L s
j d j d j d j dTh L s L L s L s
Th
e eZV d V e e e e
Z Z
β β
β β β β
− −
− − − −
+ Γ + Γ Γ − = + Γ Γ Γ + Γ Γ Γ Γ + +
2j dL sz e β−≡ Γ Γ
60
Group together alternating terms:
Voltage on a Transmission Line (cont.)
or
( )2
0
202
11
11
j dL s
Thj dTh
L j dL s
eZV d VZ Z
ee
β
ββ
−
−−
− Γ Γ − = + + Γ − Γ Γ
( )2
02
0
11
j dL
Th j dTh L s
Z eV d VZ Z e
β
β
−
−
+ Γ− = + − Γ Γ
This agrees with the previous result (setting z = -d ).
Note: The wave-bounce method is a very tedious method – not recommended.
Hence
61
Voltage on a Transmission Line (cont.)
At a distance d from the load:
( ) ( ) ( )
( )( )*
*
2
0 2 2 * 2*0
1 Re2
1 Re 1 12
z z zL L
P z V z I z
Ve e e
Zα γ γ
+− + +
=
= + Γ − Γ
( ) ( )2
20 2 4
0
1 12
z zL
VP z e e
Zα α
+− +≈ − Γ
If Z0 ≈ real (low-loss transmission line)
Time-Average Power Flow
( ) ( )
( ) ( )
20
20
0
1
1
z zL
z zL
V z V e e
VI z e eZj
γ γ
γ γ
γ α β
+ − +
+− +
= + Γ
= − Γ
= +
( )
*2 * 2
*2 2
z zL L
z zL L
e e
e e
γ γ
γ γ
+ +
+ +
Γ − Γ
= Γ − Γ
= pure imaginary
Note:
62
( )I z
z
+−
0z =
zLZ( )V zP(z) = power flowing in + z direction
Low-loss line
( ) ( )2
20 2 4
02 2
20 02 2
0 0
1 12
1 12 2
z zL
z zL
VP z e e
Z
V Ve e
Z Z
α α
α α
+− +
+ +− +
≈ − Γ
= − Γ
Power in forward-traveling wave Power in backward-traveling wave
( ) ( )2
20
0
1 12 L
VP z
Z
+
= − Γ
Lossless line (α = 0)
Time-Average Power Flow (cont.)
63
Note: In the general lossy case (very lossy), we
cannot say that the total power is the difference of the powers in the two waves.
( )I z
z
+−
0z =
zLZ( )V z
00
0
tantan
L Tin T
T L
Z jZ dZ ZZ jZ d
ββ
+= +
24 4 2
g g
g
dλ λπ πβ β
λ= = =
00
Tin T
L
jZZ ZjZ
⇒ =
020
0
0in in
T
L
Z ZZZZ
Γ = ⇒ =
⇒ =
Quarter-Wave Transformer
20T
inL
ZZZ
=
so
0 0T LZ Z Z=
Hence
(This requires ZL to be real.)
64
Matching condition
0TZ LZ0Z
inZ d
Lossless line
Match a 100 Ω load to a 50 Ω transmission Line at a given frequency.
0 100 5070.7
TZ = ×=
0
0
2 2 2g
r rk kπ π π λλβ ε ε
= = = =
Example
0 70.7TZ = Ω
65
50Ω
/ 4gλ
[ ]100LZ = Ω0TZ0
cf
λ =
( ) 20 1 Lj j z
LV z V e eφ β+ += + Γ
( ) ( )( )
20
20
1
1 L
j z j zL
jj z j zL
V z V e e
V e e e
β β
φβ β
+ − +
+ − +
= + Γ
= + Γ
( )( )
max 0
min 0
1
1
L
L
V V
V V
+
+
= + Γ
= − Γ
Voltage Standing Wave
z
1+ LΓ
1
1- LΓ
0
( )V zV +
/ 2g
z λ∆ =0z =
66
2 2 / 2gz zβ π λ∆ = ⇒ ∆ =
( )I z
z
+−
0z =
zLZ( )V z
( ) max
min
VV
≡Voltage Standing Wave Ratio VSWR
Voltage Standing Wave Ratio
11
L
L
+ Γ=
− ΓVSWR z
1+ LΓ
1
1- LΓ
0
( )V zV +
/ 2gz λ∆ =0z =
67
( )( )
max 0
min 0
1
1
L
L
V V
V V
+
+
= + Γ
= − Γ
( )I z
z
+−
0z =
zLZ( )V z
Coaxial Cable Here we present a “case study” of one particular transmission line, the coaxial cable.
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.
68
For a TEMz mode, the shape of the fields is independent of frequency, and hence we can perform the calculation using electrostatics and magnetostatics.
,rε σ
a
b
Coaxial Cable (cont.)
-ρl0
ρl0
a
b
rε0 0
0
ˆ ˆ2 2 r
E ρ ρρ ρπ ε ρ π ε ε ρ
= =
Find C (capacitance / length)
Coaxial cable
h = 1 [m]
rε
From Gauss’s law:
0
0
ln2
B
ABA
b
ra
V V E dr
bE daρ
ρρπ ε ε
= = ⋅
= =
∫
∫
69
Coaxial cable
( )0
0
0
1
ln2 r
QCV b
a
ρρ
π ε ε
= =
Hence
We then have:
0 F/m2 [ ]ln
rCba
π ε ε=
Coaxial Cable (cont.)
70
rε
[ ]1 mh =
rε0lρ
0lρ−
a
b
ˆ2
IH φπ ρ
=
Find L (inductance / length)
From Ampere’s law:
Coaxial cable
0ˆ
2 rIB φ µ µ
π ρ
=
(1)b
a
B dφψ ρ= ∫
Magnetic flux:
Coaxial Cable (cont.)
71
Note: We ignore “internal inductance” here, and only look at the magnetic field between the two
conductors (accurate for high frequency.
rµ
I[ ]1 mh =
zS
Center conductor
I
I
1 [ ]h m=
( ) 0
0
0
1
2
ln2
b
ra
b
ra
r
H d
I d
I ba
φψ µ µ ρ
µ µ ρπρ
µ µπ
=
=
=
∫
∫
01 ln
2rbL
I aψ µ µ
π = =
0 H/mln [ ]2
r bLa
µ µπ
=
Hence
Coaxial Cable (cont.)
72
rµ
I[ ]1 mh =
0 H/mln [ ]2
r bLa
µ µπ
=
Observations:
0 F/m2 [ ]ln
rCba
π ε ε=
( )0 0 r rLC µε µ ε µ ε= =
This result actually holds for any transmission line (proof omitted).
Coaxial Cable (cont.)
73
(independent of frequency)
(independent of frequency)
0 H/mln [ ]2
r bLa
µ µπ
=
For a lossless cable:
0 F/m2 [ ]ln
rCba
π ε ε=
0LZC
=
0 01 ln [ ]
2r
r
bZa
µηε π
= Ω
00
0
376.7303 [ ]µηε
= = Ω
Coaxial Cable (cont.)
74
0 0
0
ˆ ˆ2 2 r
E ρ ρρ ρπ ε ρ π ε ε ρ
= =
Find G (conductance / length)
Coaxial cable
From Gauss’s law:
0
0
ln2
B
ABA
b
ra
V V E dr
bE daρ
ρρπ ε ε
= = ⋅
= =
∫
∫
Coaxial Cable (cont.)
75
σ
[ ]1 mh =
rε0lρ
0lρ−
a
b
J Eσ=
We then have leakIGV
=
[ ]
0
0
(1) 2
2
22
leak a
a
r
I J a
a E
aa
ρ ρ
ρ ρ
π
π σ
ρπ σπ ε ε
=
=
=
=
=
0
0
0
0
22
ln2
r
r
aa
Gba
ρπ σπ ε ε
ρπ ε ε
=
2 [S/m]ln
Gba
πσ=
or
Coaxial Cable (cont.)
76
σ
0lρ
0lρ−
a
b
Observation:
F/m2 [ ]
lnC
ba
π ε=
G C σε
=
This result actually holds for any transmission line (proof omitted).
2 [S/m]ln
Gba
πσ=
0 rε ε ε=
Coaxial Cable (cont.)
77
G C σε
=
Hence:
tanGC
σ δω ωε
= =
tanGC
δω
=
Coaxial Cable (cont.)
As just derived,
78
This is the loss tangent that would arise from conductivity.
Find R (resistance / length)
Coaxial cable
Coaxial Cable (cont.)
a bR R R= +
12a saR R
aπ =
12b sbR R
bπ =
1sa
a a
Rσ δ
=1
sbb b
Rσ δ
=
0
2a
ra a
δωµ µ σ
=0
2b
rb b
δωµ µ σ
=
Rs = surface resistance of metal
79
[ ]1 mh =
,b rbσ µ
σ,a raσ µ
a
b
(This is discussed later.)
General Transmission Line Formulas
tanGC
δω
=
( )0 0 r rLC µε µ ε µ ε= =
0losslessL Z
C= = characteristic impedance of line (neglecting loss)(1)
(2)
(3)
Equations (1) and (2) can be used to find L and C if we know the material properties and the characteristic impedance of the lossless line.
Equation (3) can then be used to find G if we know the material loss tangent.
a bR R R= +
tanGC
δω
=(4)
Equation (4) can be used to find R (discussed later).
,iC i a b= =contour of conductor,
22
1 ( )i
i s szC
R R J l dlI
=
∫
80
These formulas hold for any shape of line (not restricted to coax).
General Transmission Line Formulas
( ) tanG Cω δ=
0losslessL Z µε=
0/ losslessC Zµε=
R R=
All four per-unit-length parameters (R, L, G, C) can be found from
81
0 rε ε ε=Note:
0 , ; , , tanlosslessrZ R ω ε δ
Complex Permittivity
Accounting for Dielectric Loss
82
The permittivity becomes complex when there is polarization (molecular friction) loss.
( )jε ε ε′ ′′= −
Loss term due to polarization (molecular friction)
Example: Distilled water heats up in a microwave oven, even though there is essentially no conductivity!
Effective Complex Permittivity
c j σε εω
= −
Effective permittivity that accounts for conductivity
83
c
c
ε εσε εω
′ ′=
′′ ′′= +
The effective permittivity accounts for conductive loss.
( )c
c c
j j
j
σε ε εω
ε ε
′ ′′= − −
′ ′′= −
Hence We then have
Accounting for Dielectric Loss (cont.)
Most general expression for loss tangent:
c c cjε ε ε′ ′′= −
tan c
c
σεε ωδ
εε
′′ + ′′ ≡ =′′
Loss due to molecular friction Loss due to conductivity
84
c
c
ε εσε εω
′ ′=
′′ ′′= +
Accounting for Dielectric Loss (cont.)
The loss tangent accounts for both molecular friction and conductivity.
For most practical insulators (e.g., Teflon), we have
tan εδε′′
=′
85
0σ ≈
Note: The loss tangent is usually (approximately) constant for practical insulating materials, over a wide range of frequencies.
Typical microwave insulating material (e.g., Teflon): tanδ = 0.001.
Accounting for Dielectric Loss (cont.)
General Transmission Line Formulas
tanGC
δω
=
( )0 0 r rLC µε µ ε µ ε′ ′= =
0losslessL Z
C= = characteristic impedance of line (neglecting loss)(1)
(2)
(3)
a bR R R= +
tanGC
δω
=(4)
,iC i a b= =contour of conductor,
22
1 ( )i
i s szC
R R J l dlI
=
∫
86
Here we allow for dielectric polarization loss.
r rε ε′ is usally denoted as simplyNote: Note: tanδ accounts for both
polarization loss and conductivity.
These formulas hold for any shape of line (not restricted to coax).
General Transmission Line Formulas (cont.)
( ) tanG Cω δ=
0losslessL Z µε ′=
0/ losslessC Zµε ′=
R R=
87
Here we allow for dielectric polarization loss.
All four per-unit-length parameters (R, L, G, C) can be found from
0 , ; , , tanlosslessrZ R ω ε δ
0 rε ε ε′ ′=Note:
r rε ε′ is usally denoted as simply
Common Transmission Lines
0 01 ln [ ]
2lossless r
r
bZa
µηε π
= Ω
Coax
Twin-lead
100 cosh [ ]
2lossless r
r
hZa
η µπ ε
− = Ω
2
1 2
12
s
haR R
a ha
π
= −
1 12 2sa sbR R R
a bπ π = +
88
,r rε µa
b
,r rε µ
a a
h
Common Transmission Lines (cont.)
Microstrip ( / 1)w h ≤
89
Approximate CAD formula
εr
w t
h
060 8ln
4effr
h wZw hε
= +
1 1 12 2
1 12
eff r rr h
w
ε εε
+ −
= + +
Common Transmission Lines (cont.)
Microstrip
( ) ( ) ( )( )
( )( )0 0
1 00
0 1
eff effr reff effr r
fZ f Z
fε εε ε
−= −
( )( ) ( ) ( )( )0
12000 / 1.393 0.667 ln / 1.444eff
r
Zw h w h
π
ε=
′ ′+ + +
( / 1)w h ≥
21 lnt hw wtπ
′ = + +
90
Approximate CAD formula
εr
w t
h
Common Transmission Lines (cont.)
Microstrip ( / 1)w h ≥
( )2
1.5
(0)(0)
1 4
effr reff eff
r rfF
ε εε ε −
− = + +
( )( )
1 1 11 /02 2 4.6 /1 12 /
eff r r rr
t hw hh w
ε ε εε + − − = + − +
2
0
4 1 0.5 1 0.868ln 1rh wF
hε
λ
= − + + +
91
Approximate CAD formula
εr
w t
h
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.
92
LZ0Z
ThZ
+−
At high frequency, radiation effects can also 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.)
93
This is explored next.
LZ0Z
ThZ
+−
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.)
94
Coaxial Cable
rε a
bz
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!
95
+ −
The infinite twin lead will not radiate by itself, regardless of how far apart the lines are (this is true for any transmission line).
The incident and reflected waves represent an exact solution to Maxwell’s equations on the infinite line, at any frequency.
( )*1 ˆRe E H 02t
S
P dSρ = × ⋅ = ∫
S
+ -
Limitations of Transmission-Line Theory (cont.)
96
Incident
Reflected
No attenuation on an infinite lossless line
h
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.)
97
Incident wave Pipe
Obstacle
Reflected wave
h
Bend
Incident wave
Bend
Reflected wave
h
To reduce radiation effects of the twin lead at discontinuities:
1) Reduce the separation distance h (keep h << λ). 2) Twist the lines (twisted pair).
Limitations of Transmission-Line Theory (cont.)
CAT 5 cable (twisted pair)
98
h
Baluns
99
Baluns are used to connect coaxial cables to twin leads.
4:1 impedance-transforming baluns, connecting 75 Ω TV coax to 300 Ω 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.
Baluns (cont.)
100
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).
101
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.)
What does “balanced” and “unbalanced” mean?
Note: Even with coax, one can assume balanced voltages with respect to the ground plane if one
wants to, so the distinction is somewhat vague.
Baluns (cont.)
102
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
No ground plane
Baluns (cont.)
103
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
In the common mode, both conductors of the transmission line act as one net conductor, while the ground plane acts as the other conductor (return path for the current).
Baluns (cont.)
104
Differential mode
(desired)
Common mode
(undesired)
The differential and common modes on a coax over ground
+ -
+ -
Baluns (cont.)
105
Because of the asymmetry, a common mode current will get excited at the junction between the two lines, and propagate away from the junction.
Coax Twin Lead
Ground
Common mode Common mode
No Balun
Differential mode
Differential mode
Note: If the height above the ground plane goes to infinity, the characteristic impedance of the common mode goes to infinity, and there is no common mode current – we would not need a balun.
Baluns (cont.)
106
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
Baluns (cont.)
107
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 (current on outside of the coax)
2 1I I≠
Baluns (cont.)
108
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
+ - 0V
Baluns (cont.)
109
Next, we use superposition.
+ Solid tube
Twin Lead
Ground
+ -
+ - Differential mode 0 / 2V
0 / 2V
Solid tube Twin Lead
Ground
+ -
+ - Common mode 0 / 2V
0 / 2V
Baluns (cont.)
110
One type of balun uses an isolation transformer.
https://en.wikipedia.org/wiki/Balun
The input and output are isolated from each other, so the input can be grounded if desired (e.g., when using a coax feed for the primary) while the output is not (e.g., when using a twin lead feed for the secondary). The common mode must be zero at the transformer,
since a net current cannot flow off of the windings into the transformer core.
This balun uses a 4:1 autotransformer, having three taps on a single winding, on a ferrite rod. The center tap on the input can be
grounded if desired, while the balanced output is not. The common mode is choked off due to the high impedance of the coil.
Baluns (cont.)
111
Here is a more exotic type of 4:1 impedance transforming balun.
This balun uses two 1:1 transformers to achieve symmetric output voltages.
“Guanella balun”
0V
0V−
0V
0V0V
0V−
0 / 2I
0 / 2I
0I
0I00
00
0
0
0 / 2I
0 / 2I
Baluns (cont.)
112
Inside of 4:1 impedance transforming VHF/UHF balun for TV
Ferrite core
Baluns (cont.)
113
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).
Baluns (cont.)
114
Coax can be used to feed a microstrip line on a PCB without needing a balun.
Both the microstrip line (with ground plane) and the coax are two-conductor transmission line systems, so there is
no common mode.
Baluns (cont.)
115
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.
CPS line
Baluns (cont.)
116
= +
Excites desired differential mode
Excites undesired common mode
Superposition allows us to see what the problem is.
+ -
0V
+ - + -
0 / 2V 0 / 2V
+ - - +
0 / 2V 0 / 2V
Baluns (cont.)
117
Baluns can be implemented in microstrip form.
Tapered balun
Balun feeding a microstrip yagi antenna
Coax feed port
Paired-strip output
Baluns (cont.)
118
Baluns can be implemented in microstrip form.
Marchand balun
180o hybrid rat-race coupler used as a balun
Output (unbalanced)
Input #2 (balanced)
Input #1 (balanced)
Null output
Baluns (cont.)
119
If you try to feed a dipole antenna directly with coax, there will be a common mode current on the coax.
No balun
1 2I I=
3 0I ≠
Baluns (cont.)
120
Baluns are commonly used to feed dipole antennas from coax.
A balun first converts the coax to a twin lead, and then the twin lead feeds the dipole.
Baluns (cont.)
121
Baluns are commonly used to feed dipole antennas from coax.
A “sleeve balun” directly connects a coax to a dipole.
Coaxial “sleeve” region
Coaxial “sleeve balun”
The current on the outside of the sleeve is forced to be zero at the top of the sleeve. This makes it small everywhere on the outside.
Baluns (cont.)
122
/ 4λ
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 λ/4 height.
Baluns are commonly used to feed dipole antennas from coax.
Baluns (cont.)
123
A balun is not always necessary.
Dipole antenna Twin lead
Monopole antenna
Coax
Infinite ground plane
The system stays balanced.
Current cannot flow on the outside of the coax.
Baluns (cont.)
124
Baluns can be important when connecting to unbalanced devices like oscilloscopes.
OSC
Coax input Measuring probe
Coax leads
No Balun
There may be a strong common mode current on the coax leads. The coax leads will act as an antenna! You may not really
be measuring the field from the probe.
Baluns (cont.)
125
Baluns can be important when connecting to unbalanced devices like oscilloscopes.
With Balun
Now there is only a differential mode on the coax leads, and the coax leads act as a transmission line. We measure what is
coming from the probe.
OSC
Coax input Measuring probe
Coax leads Balun