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