ece123

1

MAPUA INSTITUTE OF TECHNOLOGY

SCHOOL OF EECE

Lecture Notes in Transmission Media and Antenna Systems

Prepared and Compiled by:

Engr. Jennifer C.Dela Cruz Engr. Flordeliza L. Valiente

2

ECE123- Transmission Media and Antenna Systems

TRANSMISSION LINES

• A system of conductors having a precise geometry and arrangement that

is used to transfer power from source to load with minimum loss.

• Means of conveying information from one point to another.

• The conductive connections between system elements which carry signal

power.

Types of Transmission Lines

A. DIFFERENTIAL OR BALANCED LINE – where neither conductor is grounded

1. Two-Wire Open Lines are parallel lines and have uses such as power lines,

rural telephone lines, and telegraph lines. This type of line has high radiation

losses and is subject to noise pickup.

2. Twin Lead has parallel lines and is most often used to connect televisions to

their antennas.

3

3. A TWISTED PAIR consists of two insulated wires twisted together. This line

has high insulation loss.

4. A SHIELDED PAIR has parallel conductors separated by a solid dielectric and

surrounded by copper braided tubing. The conductors are balanced to ground.

Equation of the Characteristic Impedance for Parallel wire lines

B. SINGLE-ENDED OR UNBALANCED LINE – where one conductor is grounded

1. RIGID COAXIAL LINE contains two concentric conductors insulated from each

other by spacers. Some rigid coaxial lines are pressurized with an inert gas to

prevent moisture from entering. High frequency losses are less than with other

lines.

S

d

dSZo

2log276=

4

2. FLEXIBLE COAXIAL LINES consist of a flexible inner conductor and a

concentric outer conductor of metal braid. The two are separated by a

continuous insulating material.

Equation of the Characteristic Impedance for Concentric lines

COMMMON LOSSES IN A TRANSMISSION LINE

COPPER LOSSES can result from power (I2R) loss, in the form of heat, or skin

effect. These losses decrease the conductivity of a line.

DIELECTRIC LOSSES are caused by the heating of the dielectric material

between conductors, taking power from the source.

RADIATION AND INDUCTION LOSSES are caused by part of the

electromagnetic fields of a conductor being dissipated into space or nearby

objects.

A transmission line is electrically LONG if its physical length is greater than λ/16,

otherwise, the line is SHORT.

Ex. A 10m line is electrically short at 1000Hz and electrically long at 600MHz

D

ddDZ

ro log138

ε=

5

LUMPED CONSTANTS are theoretical properties (inductance, resistance, and

capacitance) of a transmission line that are lumped into a single component.

DISTRIBUTED CONSTANTS are constants of inductance, capacitance and

resistance that are distributed along the transmission line.

LEAKAGE CURRENT flows between the wires of a transmission line through the

dielectric. The dielectric acts as a resistor.

A

th

T

T

p

T

m

An ELECTRO

hrough it.

Transverse

The E and H

erpendicula

The velocity

medium with

OMAGNETIC

e Electrom

H-fields and

ar to each o

of the radi

h dielectric

C FIELD ex

magnetic W

the directi

other

io waves in

constant εr

xists along t

Wave

on of motio

free space

r:

cv =ε

transmissio

on of TEM w

e is c=3 x 1

c

r

=λε

;

on line whe

waves are m

108 m/sec, b

fv

en current f

mutually

but in a

6

flows

7

Table of Velocity Factor and Dielectric Constant of different Materials

Material Velocity

Factor (k)

Relative Dielectric

Constant (εr)

Vacuum 1.0000 1.0000

Air 0.9997 1.0006

Teflon Foam 0.8200 1.4872

Teflon 0.6901 2.1000

Polyethylene 0.6637 2.2700

Paper. praffined 0.6325 2.5000

Polysterene 0.6325 2.5000

Polyvinyl chloride 0.5505 3.3000

Rubber 0.5774 3.0000

Mica 0.4472 5.0000

Glass 0.3651 7.5000

ITERATIVE CIRCUIT

Sending

End

Receiving

End

8

TRANSMISSION LINE GENERAL EQUIVALENT CIRCUIT

R, L, G & C are all per unit length

Note: At RF R and G are ignored or line is considered lossless

Where: R = Ω / unit l G = S / unit l

L = H / unit l C = F / unit l

CHARACTERISTIC IMPEDANCE (Zo)

• Reference input impedance

• Impedance measured at the input when its length is infinite

• Also known as the surge impedance

E - dE E

dE

Z

Y

9

Where: Z = R + jωL , series impedance / section

Y = G + jωC , shunt admittance / section

By KCL: By KVL:

I + dI = I + EY E - dE = E – IZ

dI = EY (dS) --------- (1)

dE = IZ (dS) ---------- (2)

--------- (3)

--------- (4)

Differentiate I and E with respect to S:

General Solution :

But γ = √ZY = complex propagation constant

dI

dS = EY

dE dS

= IZ

IZYdS

Id=2

2

EYZdS

Ed=2

2

ZYZY eIeII −+= 21

ZYZY eEeEE −−= 21

--------- (5)

--------- (6)

10

• Propagation constant, γ, determines the variation of V or I with distance

along the line: V = Vse-S γ ; I = Ise-S γ, where VS, and IS are the voltage

and current at the source end, and S = distance from source.

where α = attenuation constant ( neper/m or dB/m)

β = phase delay constant (rad/m)

substitute (6) to (4)

Compare this to (5)

where I1 = E1/ Zo and I2 = -E2/ Zo

therefore

CjGLjRZo ω

ω++

= YZZo = --------- (7)

( )( ) βαωωγ jCjGLjR +=++=

( ) IZdS

eEeEd ZYZY

=− −

21

ZeZYEeZYEI

ZYZY −−= 21

YZ

EZZYEI 11

1 ==

11

At Radio Frequency

WAVELENGTH

- distance travelled by a point in the time required to complete one cycle.

Where:

v = velocity of propagation along the line

f = frequency of operation

C = velocity of light , 3 x 10 8 m/s

k = velocity factor, 0<k<1

SAMPLE PROBLEMS

1. Determine the wavelengths of the electromagnetic waves in free space

with the following frequencies: __ kHz, ____kHz, ____ MHz and ___GHz.

CLZo =

fvkccv

r

=== λε

;

--------- (8)

--------- (9)

12

2. For a given length of a coaxial cable with distributed capacitance C = ___

pF/m and distributed inductance L = ______ nH/m, determine the velocity

factor and velocity of propagation.

3. A very low loss cable has ___pF/ft of distributed capacitance and

_____nH/ft of distributed inductance. Calculate the following:

a. the capacitance of 4-ft length of this line

b. the characteristic impedance

c. the velocity of propagation.

d. The ratio of the shield diameter to center diameter of the coax.

13

4. Calculate the actual length in feet of a half-wavelength coax with velocity

factor of _____ at ______MHz.

5. What is the separation of two towers in feet, if the operating frequency is

_______Hz and the phase separation is _____o.

14

STANDING WAVE

- an interference pattern made by two sets of traveling waves going on

opposite direction.

From (6)

at the load; S = 0 , E = EL

EL = E1 + E2 --------- (10)

IL = I1 + I2 --------- (11)

LOSSLESS LINE TERMINATED WITH SHORT CIRCUIT

REFLECTION COEFFICIENT, Γ

- A measure of the degree of mismatch between the load and the line

ReflectedIncident

Antinode

Node Distance along the line

1

2

1

2

1

2

II

ZIZI

EE

O

O −=

−==Γ

SZYSZY eEeEE −−= 21

--------- (12)

15

From (10)

EL = E1 + E2

IL ZL = I1 Zo – I2 Zo

I1 ZL + I2 ZL = I1 ZLo – I2 Zo

I1 (ZL – Zo) = - I2 (ZL + Zo)

If Γ = 0 , ZL = Zo ( perfect match )

Γ = 1 , ZL = ∞ ( open circuit )

Γ = -1 , ZL = 0 ( short circuit )

STANDING WAVE RATIO (SWR)

- ratio of maximum to minimum I or V

φ∠Γ=+−

=−

=ΓoL

oL

ZZZZ

II1

2

Γ−

Γ+==

11

MIN

MAX

VVSWR

11

+−

=ΓSWRSWR

--------- (13)

--------- (14)

16

If ZL is purely resistive

Position of the Voltage Maxima and voltage Minima

Smax - position of the voltage maxima with respect to the load

Smin - position of the voltage minima with respect to the load

Where: m – any positive even integer

n – any positive odd integer

for the postion of the first Vmax m = 0

for the postion of the first Vmin n = 1 therefore;

O

L

ZRSWR =

L

O

RZSWR =

If ZL > ZO If ZO > ZL

Vmax

Vmin

( )β

φ2

180max

omS +=

( )β

φ2

180min

onS +=

βφ

2max =S βφ

2180

min

o

S +=

17

SAMPLE PROBEMS

1. A transmitter delivers 100W into a ____Ω lossless line that is

terminated with an antenna that has an impedance of ____Ω,

resistive. Calculate the reflection coefficient and voltage standing

wave ratio on the line.

2. A _____Ω line is terminated by a load of __________ Ω

operating at 10MHz. Find:

a. the reflection coefficient and SWR

b. the input impedance of line 6850 long

c. the position of the first voltage minimum in meters.

d. the position of the first voltage maximum in meters

⎟⎟⎠

⎞⎜⎜⎝

⎛=

+−=−

λβ

φβφ

o

oo

SS3602

1802180

2minmax

4minmaxλ

=− SS

18

3. A transmission line has a characteristic impedance of _____Ω

and a reflection coefficient equal to 0.444∠48o. Find the load and

SWR of the line.

19

4. Determine the reflection coefficient and SWR of a transmission

line with incident voltage Ei =_____V and a reflected voltage ER

=____V.

5. Using a TDR, a transmission-line impairment is located 100m

from the source. If the elapsed time from the beginning of the

pulse to the reception of the echo is ____s, determine the

velocity factor

VOLTAGE AND CURRENT AT ANY POINT ALONG THE LINE

From (10 )

[ ] 121 / EEEEL +=

Γ+=11E

EL Γ+

=11

LEE

20

Substitute to (6)

Equation of voltage at any point along the line

For current,

Equation of current at any point along the line

INPUT IMPEDANCE, ZIN

( ) ( )SZYSZY

L

OLL eeZ

ZZEE −Γ+

+=

2

( ) ( )SZYSZY

O

OLL eeZ

ZZII −Γ−+

=2

( )( )OL

OL

L

ZZZZ

EE

+−

+=

11

( )Γ

+=Γ=

L

OLL

ZZZEEE

212

--------- (15) ( )

L

OLL

ZZZEE

21+

=

--------- (16)

--------- (17)

--------- (18)

( ) ( )( ) ( )SZYSZY

O

OLL

SZYSZY

L

OLL

eeZ

ZZI

eeZ

ZZE

IE

−

−

Γ−+

Γ++

=

2

2

21

If

and

(substitute)

Recall:

[ ][ ]SZZ

SZZZZLO

OLOIN γ

γtanhtanh

++

=

OL

OL

ZZZZ

+−

=Γ ZY=γ

S

OL

OLS

S

OL

OLS

OIN

eZZZZe

eZZZZe

ZZγγ

γγ

−

−

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

+=

( ) ( )( ) ( ) S

OLOLS

SOLOL

S

OIN eZZZZeeZZZZeZZ γγ

γγ

−

−

−−+−++

=

( ) ( )[ ]( ) ( )[ ] 2/1

2/1xeeZeeZxeeZeeZZZ SS

LSS

O

SSO

SSL

OIN λγγγ

γγγγ

−−

−−

−++−++

=

2sinh

AA eeA−−

= 2cosh

AA eeA−+

= AA

AA

eeeeA −

−

+−

=tanh

[ ]( )[ ]( )SSZSZ

SSZSZZZLO

OLOIN γγγ

γγγcosh/1sinhcoshcosh/1sinhcosh

++

=

--------- (19)

22

Manipulating tanh γS:

CASE I:

α ≠ 0; β = 0 γ = α

say α = 0.1; S = 2m; γ = 0.2

tanh(0.2) = 0.1974

CASE II:

α ≠ 0; β ≠ 0 γS = αS + jβS

say α = 0.1; β = 0.2; S = 10m; γS = 1 + j2

tanh A =

e(1 +j2) = e1ej2

Recall Euler’s Identity

ejA = 1∠±A = cosA ± jsinA

CASE III:

α = 0; β ≠ 0 γS = jβS

tanhδS = tanh jβS

let βS = A

tanh jA =

e(1 + j2) – e ‐(1 + j2)

e(1 + j2) + e ‐(1 + j2)

e jA – e ‐jA

e jA + e ‐jA

23

tanh jA =

=

tanh jA = j tan A

therefore: tanh jβS =j tan βS ---- substitute in (19)

- for lossless line

Note: cosh jx = cosx

sinhjx = jsinx

LOSSLESS TRANSMISSION LINE

1. No attenuation

2. No power loss ( R=0, G=0 )

Wavelength : distance that provides a phase shift 2π radian

cosA + j sinA – (cosA ‐ j sinA)

cosA + j sinA + (cosA ‐ j sinA)

2 jsin A2 cos A

[ ][ ]SjZZ

SjZZZZ

LO

OLOIN β

βtantan

++

=

βπ

βλ 2360

===o

fv

--------- (20)

24

Complex Propagation Constant

γ = √ ( R + jωL )(G +jωC ) = α + jβ ; jβ = jω √ LC

INPUT IMPEDANCE, ZIN, for special termination cases.

CASE I: ZL = ZO (matched line)

ZIN = Zo

CASE II: ZL = 0 (short circuited line)

ZIN = Zo

[ZO + j ZL tan βS]

[ZL + j ZO tan βS]

1

ZO j ZO tan βS

βω

βπ

==fv 2

LC1

=βω

LCv 1=

OIN ZZ =

--------- (21)

--------- (22)

[ ][ ]SjZZ

SjZZZZ

LO

OLOIN β

βtantan

++

=

25

CASE III: ZL = α (open circuited line)

ZIN = Zo =

By L’Hospital’s Rule

ZIN = Zo

ZIN = Zo then

ZOCZSC = j ZO tan βS ( ZO/ j tan βS) = ZO2

ZIN for special lengths

CASE I: S = λ/4 (quarter wavelength)

ZIN = Zo = since tan(π/2) = α

[ZO + j α tan βS] [α + j ZO tan βS]

α

α

[(ZO/ ZL) + j (ZL/ ZL) tan βS]

[(ZL/ ZL) + j (ZO/ ZL) tan βS]

1 0

j tan βS

1

ZO + j ZL tan [(2π/λ)(λ/4)]

ZL + j ZO tan[(2π/λ)(λ/4)] αα

SjZZ OIN βtan=

SjZZ OIN βcot−=

SCOCO ZZZ =

--------- (23)

--------- (24)

--------- (25)

26

By L’Hospital’s Rule

ZIN = Zo = j ZO2 / jZL

CASE II: S = λ/2 (half wavelength)

ZIN = Zo

Since tan 180o = 0

ZIN = ZO

SAMPLE PROBLEMS

1. A transmission line has a characteristics impedance ZO = _____Ω, the

input impedance 0.23λ from the load is _____________Ω. Find the

load, SWR and reflection coefficient of the line.

[ZO/ tan βS + j ZL]

[ZL/ tan βS + j ZO]

ZO + j ZL tan [(2π/λ)(λ/2)]

ZL + j ZO tan[(2π/λ)(λ/2)]

ZL

ZO

L

OIN Z

ZZ2

=

LIN ZZ =

27

2. Calculate the length of a short-circuited line necessary to simulate an

inductance of ____H at _____Hz.

3. For a 50Ω lossless line operating at _______Hz, determine the input

impedance at a distance of ______λ from the open circuited

termination.

28

4. Determine the characteristic impedance of a ______λ section of a line

that simulates an inductance of ____H when short-circuited and a

capacitance of ______F when open-circuited at _____Hz.

5. A 150Ω air-filled lossless line is used to propagate a ______Hz signal.

Calculate the input impedance for a 5m length of this line when the

load is _________Ω.

COURSEWORK 2

See appendix A

End of topic discussion for Quiz 1

29

MATCHING NETWORKS

General rule: to tune out unwanted load reactance (if any) and the

transformation of the resulting impedance to the value required.

I. QUARTER-WAVE TRANFORMER

1. ZL = RL; RL ≠ZO

2. ZL = RL ± jXL (complex) ; RL ≠ZO

ZL

ZO

ZOTZIN

ZO

ZL

ZOZOT

RIN

= RL ± jXL

S

L

OTIN Z

ZZ

2

4/ =λ

LOOT RZZ =

LOOT RZZ = β

φ2

180O

MINmS +

=

βφ

2180O

MAXnS +

=

SWRZ

R OMIN =

SWRZR OMAX =

30

II. MATCHING STUBS

Condition: ZL = RL ± jXL ; RL = ZO

1. Series Short Circuit Stubs

Zins = j tan βS (Zos) = -jXL

βS = tan-1 (-XL /Zos)

Double stub

βS = tan-1 (-XL / 2Zos)

Zins = -jXL / 2 = j tan βS (Zos)

ZL = RL + jXLZos

Zins

ZO

ZL = RL + jXL

Zos

Zos

Zins

ZO

31

2. Series open Circuit Stubs

Zins = -jXL = -j cot βS (Zos)

βS = tan-1 (Zos / XL)

3. Shunt Short Circuit Stubs

Yins = 1 / Zins = -jB = 1 / (j ZOS tan βS)

βS = tan-1 (YOS / B)

ZL = RL + jXL

Zos

Zins ZO

ZL = RL + jXLZos

32

4. Shunt Open Circuit Stubs

Yins = 1 / Zins = ( j tan βS) / ZOS = -jB

βS = tan-1 (-BZOS) or βS = tan-1 (-B / YOS)

ZL = RL + jXLYL = G + jB

33

REACTANCE PROPERTIES OF SHORTED AND OPEN TRANSMISSION LINES

Impedance seen by the generator

λ/4

High R

XC

XL

Low R

XL

XC

34

SAMPLE PROBLEMS

1. It is required to match a _____Ω load to a ______Ω transmission line

to reduce the SWR along the line to unity. What must be the

characteristic impedance of the λ /4 transformer to be used for

this purpose?

2. A load ZL = 100 – j80Ω is connected to a line whose characteristic

impedance is _____Ω . Calculate the nearest point to the load at

which a λ /4 transformer maybe inserted to provide correct

matching and the characteristic impedance of the transformer.

35

3. A ____________Ω load is to be matched to a 300Ω line to give a

SWR equal to 1. Calculate the reactance of stub (shunt) and

characteristic impedance of λ /4 transformer both connected

directly to the load.

4. A transmission line has a characteristic impedance of _____Ω and a

reflection coefficient equal to 0.444∠48o. Find

a) the load and SWR of the line.

b) The nearest point to the load where a quarter wave

transformer must be inserted to reduce SWR to 1.

c) The characteristic impedance of the transformer.

d) The nearest point to the load where a series short circuit

must be placed to reduce SWR to 1.

e) The length of the stub.

37

5. A transmission line has a characteristics impedance ZO = _____Ω,

the input impedance ______λ from the load is 44.721∠63.435oΩ.

a) the load, reflection coefficient and SWR of the line, the

nearest point to the load where a quarter wave transformer

must be inserted to reduce SWR to 1.

b) the characteristic impedance of the transformer.

c) the nearest point to the load where a shunt open circuit stub


d) the length of the stub.

38

COURSE WORK 3

See appendix A

T

The Smith

The S

imped

The c

ortho

One s

the o

Smith

h Chart (

Smith chart

dance prob

coordinates

ogonal circle

set represe

ther the no

h Chart App

Determin

Finding Z

Finding st

Solution f

( Polar Im

is a graphi

blems.

on the cha

es.

nts the nor

ormalized re

plications:

ing Load im

i of a shorte

tub location

for quarter-

mpedanc

ical device

art are base

rmalized res

eactive com

mpedance Z

ed or open

n for match

-wave trans

ce Diagra

in solving t

ed on the in

sistive comp

mponent, ±

ZL, load adm

line and te

hing purpos

sformer ma

am)

transmission

ntersection

ponent, r (=

jx (= ± jX/

mittance YL

erminated li

es

atching

n-line

of two sets

= R/Zo), an

/Zo).

, SWR, |Γ|∠

ines

39

s of

nd

∠φ

40

SAMPLE PROBLEMS

See Appendix B for copies of the Smith Chart to be used in

these Exercise.

1. A transmission line has a characteristic impedance of _____Ω

and a reflection coefficient equal to 0.444∠48o. Find

a. the load and SWR of the line.

b. The input impedance 120o from the load

c. The nearest point to the load where a quarter wave


d. The characteristic impedance of the transformer.

e. The nearest point to the load where a series short circuit


f. The length of the stub.

2. A transmission line has a characteristics impedance ZO =

_____Ω, the input impedance ______λ from the load is

44.721∠63.435oΩ.

a. the load impedance, reflection coefficient and SWR of the

line

b. the nearest point to the load where a quarter wave


c. the characteristic impedance of the transformer.

d. The load admittance

e. the nearest point to the load where a shunt open circuit

stub must be placed to reduce SWR to 1.

f. the length of the stub.

41

COURSEWORK 4

See appendix A


COURSEWORK 5

Answer Questions 14-1 to 14-25 of the Textbook using 3 different book

titles from the Library

RADIOWAVE PROPAGATION

Radio Frequency Bands and Major Services

ELF – Extremely Low Frequency (30-300Hz)

Submarine Applications

VLF – Very Low Frequency (3kHz-30kHz)

• Radio waves at these frequencies are very reliable for long-range

communications. Attenuation of the ground waves is very small, and

the sky wave reflection is good.

• Radio Navigation, Aeronautical Communications and Maritime Mobile

Communications

LF – Low Frequency (30kHz - 300kHz)

• Attenuation of ground waves is higher than VLF. Sky absorption begins to

be a factor , especially at the higher end of this band.

• Radio Navigation, Aeronautical Communications and Maritime Mobile

Communications

42

MF – Medium Frequency (300kHz – 3MHz)

• This region includes the standard AM broadcast band in which it is

possible to obtain reliable ground wave coverage up to 100 miles from the

transmitting antenna.

• Amateur Communications and Maritime and Aeronautical Communications

HF – High Frequency (3MHz – 30MHz)

• Sky wave propagation is the only reliable means of communicating over

long distances especially at the upper end of this band

• Short Wave Broadcasting Point-to-Point Communications and Land,

Maritime and Aeronautical Communications

VHF – Very High Frequency (30MHz – 300MHz)

• This region includes the commercial FM and VHF TV bands. Line of sight

is the principal means of communication.

• Short Wave Broadcasting Point-to-Point Communications and Land,

Maritime and Aeronautical Communications

UHF – Ultra High Frequency (300MHz – 3GHz)

• Line of sight propagation is possible beyond the optical horizon due to the

increasing refraction effects on earth’s atmosphere

• TV Broadcasting, Radioastronomy, Aeronautical and land mobile

Communications and Satellite Communications.

SHF - Super High Frequency (3GHz – 30GHz)

• Represents the upper limit of frequencies that have any practical use in

radio-wave communication using standard method of generating and

transmitting signals.

• Microwave Relays, Satellite and Exploratory Communications

Free Space is an idealised wave environment where there are no other

transverse electromagnetic (TEM) wave, no gravity, no obstructions, no

atmosphere, no celestial events, no terrestrial events, no electrical noise, and no

43

observers. In short, the wave environment is free from everything except the

wave itself.

– space that does not interfere with the normal radiation and propagation

of waves (epitome of nothingness).

A radiated TEM wave in free space is often referred to as being in time

phase and space quadrature. This means that the E and H fields rise and fall

together in time, but are 900 apart in space.

The Isotropic model:

In free space, the TEM wave is thought as emanating from a

dimensionless source. Mathematically, such a zero-dimensional source is

obviously a point source. Moreover, the waves regarded as radiating

uniformly in all direction from this point as illustrated in Figure3. Consequently,

we call such a radiation point as isotropic source. The radiated energy of equal

intensity is required by a sphere whose surface area is given by: 4πr2

44

POLARIZATION:

The orientation of the E-field component of the TEM wave is called its

polarization. If the direction remains constant with time at a fixed point in space,

the field is said to be linearly polarized. For wave propagation near the earth’s

surface, the term vertical, horizontal and slant polarization are frequently used to

denote linear polarizations with appropriate orientations.

Direction of travel

r – any fixed distance from the source to where the intensity

is measured

45

POWER DENSITY, FIELD STRENGTH ATTENUATION

POWER DENSITY – is the total power radiated per unit area.

Isotropic Source

Field Strength – or intensity of the signal at a distance R; E is in V/m.

from,

PD – power density at any

pt. on the surface of a

spherical wavefront

Note: PT = PtGt

;

24 RPP t

D π=

Ω==

−

−

== 377120910

36

1

7104π

π

π

ε

μ

x

x

o

oZs

π120

22 EE==

SZ

P ππ 120

2

24

E=

R

tP

RPt30

=E

46

ATTENUATION

The reduction of power density with distance is equivalent to a power loss and is

commonly called wave attenuation.

SAMPLE PROBLEMS

1. Determine the power density for a radiated power of _____W at a

distance of ____km from an isotropic antenna.

2. Determine the electric field intensity for a radiated power of ____W and a

distance of ____km from a dipole antenna.

2

1log10D

Da P

P=γ

47

3. The power density at a point from a source is _____μW, and the power

density at another point is ______μW; determine the attenuation in

decibels.

OPTICAL PROPERTIES OF RADIO WAVES

a. REFLECTION – the return or change in direction of light, sound radiowaves

striking a surface or traveling from one medium to another.

Electromagnetic reflection occurs when an incident wave strikes a boundary

of two media and some or all of the incident power does not enter the second

material.

48

b. REFRACTION – the bending of a radio wave when it passes obliquely from

one medium to another in which the velocity of propagation is different.

from rare to denser medium it will be refracted towards the normal

c. DIFRACTION – the scattering of waves as it passes the edges of an object or

opening. Diffraction is defined as the modulation or redistribution of energy

within a wavefront when it passes near the edge of an opaque object.

Diffraction is the phenomena that allows light or radio waves to propagate

(peek) around corners.

by Snell’s law:

B

A

A

B

vv

nn

==2

1

sinsin

θθ

Where:

θ1 = angle of incidence

θ2 = angle of refraction

nA = refractive index of medium 1

nB = refractive index of medium 2

vA = velocity of the wave in medium 1

vB = velocity of the wave in medium 2

49

d. ABSORPTION – the dissipation of energy by radiation passing through a

medium.

e. INTERFERENCE - Radio wave interference occurs when two or more

electromagnetic waves combine in such a way that system performance is

degraded. It is subject to the principle of linear superposition of

electromagnetic waves and occurs whenever two or more waves

simultaneously occupy the same point in space.

.

Propagation of Electromagnetic Waves

50

GROUND or SURFACE WAVES

• Provide reliable 24 hour/day communication capability for frequencies of

up to about 3MHz.

• Primary mode of communication is the LF and MF bands.

Notes:

• Ground waves must be vertically polarized. This is because the electric

field in a horizontally polarized wave would be parallel to the earth’s

surface, and such waves would be short-circuited by the conductive

ground.

• Common uses:

Ship-to-Ship Communications

Ship-to-Shore Communications

Radio Navigation

Maritime Mobile Communications

• Disadvantages of Ground Wave Propagation

- requires a relatively high transmission power

- limited to VF,LF and MF bands

- ground losses vary considerably with surface material

• Advantages of Ground Wave Propagation

- given enough power, ground waves can be used to communicate

between any two locations in the world.

- Ground waves are relatively unaffected by changing atmospheric

conditions

Direction of wave travel

IncreasingTilt

Earth

Wavefront Direction of wave travel

IncreasingTilt

Earth

Wavefront

51

SKY WAVES

• Takes advantage of the ionosphere (30-250 miles above the earth’s

surface) that surrounds the earth to provide worldwide communications

with reasonably good quality, reliability and moderate power.

Notes:

• Almost all HF propagation, and night time long distance MF propagation is

by sky wave.

• Above 30MHZ, waves are more likely to penetrate the ionosphere and

continue moving out into space.

• Ionosphere is most dense during time of maximum sunlight

• In general, the lower the frequency, the more easily the signal is

refracted.

• In the UHF and SHF bands, a very small percentage of the wave’s energy

is refracted back to earth

• Under the best conditions, the maximum distance of a single hop is about

2000 miles

52

IONOSPHERIC LAYER

The ionosphere is composed of three distinct layers, designated from lowest level

to highest level (D, E, and F) as shown in figure

A

V

ap

S

b

re

S

sk

Amount of io

1. amou

2. seaso

3. sunsp

4. weath

5. local

Virtual Heigh

appears to h

Skip Zone –

ecomes too

eturned to

Skip distanc

ky wave wa

onization de

unt of sunlig

on of the ye

pots

her conditio

terrain

ht – is the h

have been r

is the zone

o weak for

earth

ce - is the d

as received

epends on

ght

ear

ons

height abov

reflected.

e of silence

reception a

distance fro

the followin

ve Earth’s s

between th

and the poin

om the tra

ng factors:

surface from

he point wh

nt where th

nsmitter to

m which a r

here the gro

he sky wave

o the point

refracted wa

ound wave

e is first

where the

53

wave

first

54

Relationship between skip zone, skip distance and ground wave

The amount of refraction depends on three (3) factors:

1. density of the ionized layer

2. frequency of the radio wave (3 – 30 MHz)

3. angle at which the wave enters the ionosphere

Critical Angle - above which the signal will not be refracted enough to return to

earth

- maximum vertical angle at which the signal can be propagated and

still be refracted back by the ionosphere.

Critical Frequency (fc) – the highest frequency returned to earth when radiated

upward in a vertical direction

effects of ionospheric density on radio wave

55

Frequency versus refraction and distance

Incidence angle of radio wave

56

Formulas For Sky Waves

From geometry (assuming flat earth):

d = 2hv tan θi

where hv = virtual height of F-layer

From theory (secant law):

MUF = fc sec θI

MAXIMUM USABLE FREQUENCY

• The higher the frequency of a radio wave, the lower the rate of refraction

by the ionosphere. Therefore, for a given angle of incidence and time of

day, there is a maximum frequency that can be used for communications

θi

hv

d

F-Layer

Earth

57

between two given locations. This frequency is known as the MAXIMUM

USABLE FREQUENCY (MUF).

• Varies between 8MHz to 30MHz with

Time of day

Distance

Direction

Season

Solar Activity

OPTIMUM WORKING FREQUENCY

The most practical operating frequency is one that you can rely onto have the

least number of problems.

FOT = 0.85 MUF

FREE SPACE LOSS

• Defined as the loss incurred by a radio wave as it travels in a straight line

through a vacuum with no absorption or reflection of energy from nearby

objects.

Lp (dB) = 92.4 + 20log f + 20log D

where : f = frequency of radio wave in GHz

D = distance in km

If f is in MHz, replace 92.4 above by 32.4

58

Fade Margin

To account for changes in atmospheric conditions, multipath loss, and terrain

sensitivity, a fade margin, Fm, must be added to total system loss:

Fm (dB) = 30log d + 10log(6ABf) - 10log(1-R) -70

where d = distance (km)

f = frequency (GHz)

R = reliability (decimal value)

A = terrain roughness factor (0.25 to 4),

B = factor to convert worst-month probability to annual probability

(0.125 to 1 depending on humidity or dryness).

A – roughness factor

= 4 over water or very smooth terrain

= 1 over average terrain

= 0.25 over very rough, mountainous terrain

B- factor to convert worst-month probability to annual probability

= 1 to convert an annual availability to a worst month basis

= 0.5 for hot humid areas

= 0.25 for average inland areas

= 0.125 for very dry mountainous areas

TROPOSPHERIC SCATTER (TROPOSCATTER)

• Is a special case of skywave propagation used for frequencies higher than

those in standard skywave propagation technique.

• Troposphere (6-10mi above the earth’s surface) is used as a reflector of

UHF signals.

59

• Is used when reliable long distance communication link is needed across

the deserts, mountain regions, off shore drilling platform and between

distant islands.

60

SPACE WAVES

Travel in a straight line from the transmitting antenna to receiving antenna.

Space-wave propagation (also called line-of-sight LOS), requires a path where

both antennas are visible to one another and there are no obstructions. VHF and

UHF communications typically use this path (frequencies above 30MHz).

LOS radio horizon for a single antenna is given as:

Therefore, for transmit and receive antennas, the distance between two antenna

is

Where:

d – total distance in km.

dt – radio horizon for transmit antenna in km

dr – radio horizon for receive antenna in km

ht –antenna height in m

hr – antenna height in m

tt hd 4=

( )RTrt hhddd +=+= 4

61

SATELLITE COMMUNICATIONS

System composed of a communications satellite in stationary orbit

approximately 22,000 miles above the earth’s surface, an earth-bound

transmitting antenna, and an earth bound receiving antenna.

Required escape velocity: 17,500 mi/hr

Lowest practical orbit: 100 miles above sea level

Satellite positions:

LEO (Low Earth Orbit)

MEO (Medium Earth Orbit)

HEO (High Earth Orbit)

SAMPLE PROBLEMS

1. Determine the MUF for a critical frequency of ____Hz and an angle of

incidence of ____degrees.

Earth Apogee Perige 30mins 30mins

62

2. Determine the radio horizon distance for a transmit antenna that is ___ft

and a receiving antenna that is ____ft.

3. Determine the fade margin for a ___km microwave hop. The RF frequency

is ___GHz, the terrain is ____ and the reliability objective is 99.9995%.

4. Determine the path loss for the following frequencies and distances:

Frequency(MHz) Distance (D)

63

TELEVISION BROADCAST BAND

Channel Frequency

(MHz)

2 54-60

Low Band

VHF

3 60-66

4 66-72

5 76-82

6 82-88

7 174-180

High Band

VHF

8 180-186

9 186-192

10 192-198

11 198-204

12 204-210

13 210-216

14 to 83 470-890 UHF Band

73,74 Govn.t & Non-Govn’t Operations/ Mobile

75 NAVI (ILS/Mbeacon)

88-108 FM Band

108-136 Aero Comm

136-174 Mobile/Marine/Air/Land

COURSEWORK 6

See Appendix A


64

ANTENNA SYSTEMS

Antenna

♣ consist of a wire or other conductor, or a collection of wires or conductors,

that converts electrical energy into electromagnetic waves for transmission, and

electromagnetic waves into electrical energy for reception

♣ An antenna is a passive reciprocal device.

♣ It acts as a transducer to convert electrical oscillations in a transmission line or

waveguide to a propagating wave in free space and vice versa.

♣ It functions as an impedance matcher between a transmission line or

waveguide and free space.

♣ All antennas have a radiation pattern which is a plot of the field strength or

power density at various angular positions relative to the antenna.

Basic Antenna Operation:

Antenna Parameters

Radiation Pattern

A polar diagram or graph representing field strengths or power densities

at various angular positions relative to an antenna.

Near and Far Fields

The term near field refers to the field pattern that is close to the antenna,

and the term far field refers to the field pattern that is at great distance.

65

Antenna Impedance and Efficiency

Za = Ra + jXa where: Ra= antenna resistance

Ra = Re + Rr Xa=antenna reactance

• Radiation resistance is the resistance that, if it replaced the antenna,

would dissipate exactly the same amount of power that the antenna

radiates.

• Feed-point impedance Ra = 73Ω (of which between 68Ω to 70Ω is the

radiation resistance). This is true for a simple dipole antenna.

Rr = radiation resistance (ohms)

P = power radiated by the antenna (watts) i = antenna current at the feedpoint (ampere)

2iPRr =

66

Antenna efficiency is the ratio of the power radiated by an antenna to the sum of

the power radiated and the power dissipated or the ratio of the power radiated

by the antenna to the total input power.

Directive gain and Power gain

Directive gain is the ratio of the power density radiated in a particular

direction to the power density radiated to the same point by a reference

antenna, assuming both antennas are radiating the same amount of power.

Directivity is the maximum directive gain; gain in the direction of one of

the major lobes of radiation pattern.

Transmitting gain (At) – If an antenna radiates A watts and a standard antenna

radiates B watts at the same locations, directions and conditions, the

transmitting gain is A/B.

Receiving gain (Ar) – If an antenna receives A watts and a standard antenna

receives B watts under the same condition, then the receiving gain is A/B.

Power gain(Ap) is given by: Ap = ηD

where:

η= antenna efficiency

Prad = radiated power

Pin = input power %100

%100%100

XRR

R

xPP

Px

PP

er

r

Drad

rad

in

rad

+=

+==

η

η

where:

D = directive gain PD = power density at some point with a given antenna

PDr = power density at the same point with a reference antenna

Dr

D

PPD =

67

If an antenna is lossless, it radiates 100% of the input power and the power gain

is equal to the directive gain. The power gain for an antenna is also given in

decibels relative to some reference antenna. Therefore, power gain is

Effective Isotropic Radiated Power

Effective Isotropic Radiated Power (EIRP) or simply ERP (effective

radiated power) is defined as an equivalent transmit power and is expressed as:

EIRP = Pt Gt (watts) = PinAp

or:

EIRP (dBW) = 10 log (Pt Gt) = 10 log (PinAp)

To determine the power density at a given point distance R from a transmit

antenna,

Received Power

where: Pt = total radiated power Gt = transmit antenna directive gain Ap = transmit antenna power gain Pin = input power

ApAp dB log10)( =

22 44 RGP

REIRPP tt

D ππ==

where:

Pr= received power PD = power density Ae = effective capture area

AePD=Pr

68

The effective capture area of an antenna can be defined as:

The received power is therefore given by the equation.

Antenna Input Impedance

Antenna input impedance is simply the ratio of the antenna’s input voltage

to input current.

Zin = Ei / Ii

Antenna input impedance is generally complex; however, if the feedpoint is at a

current maximum and there is no reactive component, the input impedance is

equal to the sum of the radiation resistance and the effective resistance.

Antenna Polarization

The polarization of the antenna refers simply to the orientation of the

electric field with respect to the ground. An antenna may be linearly, elliptically,

or circularly polarized.

where:

Ae= effective capture area(meters2) Gr = receive antenna power gain (unitless) λ=wavelength of receive signal (meters)

πλ

4

2rGAe =

2

2

)4( RGGPP rtt

r πλ

=

69

Antenna beamwidth is simply the angular separation between the two

half-power points on the major lobe of an antenna’s plane radiation pattern.

Antenna Bandwidth

This refers to the frequency range over which operation is satisfactory and

is generally taken between the half-power points.

‐3dB

‐6dB

‐9dB 0.9GHz 3GHz BW = 2.1GHz

70

SAMPLE PROBLEMS

1. For an antenna with input power Pin=___W, rms current I=__A, and

effective resistance Re=2Ω, determine:

a. Antenna’s radiation resistance

b. Antenna’s efficiency

c. Power radiated from the antenna, Prad.

2. Determine the power density at a point __km, from an antenna that has

input power Pin= 40W, efficiency η=____%, and directivity D = 16dB.

71

3. What is the dB gain of an antenna that delivers a 100 μV signal over that

of an antenna that delivers 75 μV?

4. What is the ERP if the output of a transmitter is ____ kW, the coax line

loss is _____ W, and antenna power gain is 3 dB?

BASIC ANTENNAS

Elementary Doublet

A dipole which is infinitely thin and has length l which is very short

compared to the wavelength λ・

An elementary doublet has uniform current throughout its length.

However, the current is assumed to vary sinusoidally in time and at any instant :

i(t) = I sin(2πft +θ)

With the aid of Maxwell’s equations, it can be shown that the far (radiation)field

is:

RlIEλ

θπ sin60=

72

Half-Wave Dipole

If the elements are each cut to one-quarter wavelength, the resultant

antenna is called half-wave dipole or Hertz antenna.

Grounded Dipole

A monopole (single pole) antenna one-quarter wavelength long, mounted

vertically with the lower end either connected directly to ground or grounded

through the antenna coupling network, is called a Marconi antenna.

SymbolSymbol

Balanced Feedline

λ/2

Balanced Feedline

λ/2

73

Resonant Antenna:

Corresponds to a resonant line, and the dipole antenna is a good example

whose length is a multiple of quarter wavelength.

Non-Resonant Antenna:

No standing waves and its radiation pattern is unidirectional. Usually

terminated with a load resistor

1. Long Wire Antenna

2. Rhombic Antenna

3. Vee Antenna

SAMPLE PROBLEMS

1. What is the length in feet of a half-wave dipole antenna operating at

_______Hz?

74

2. What is the frequency of operation of a dipole antenna cut to length of

_____ m?

3. Determine the radiator of a Marconi antenna cut for the frequency of

channel ___ of the TV broadcast channel.

4. A dipole is 10cm long. If the 10MHz current flowing through it is 2A, what

is the Field Strength 20km away from it in the direction of maximum

radiation?

75

Antenna Impedance Matching

♥ Antenna should be matched to their feedline for maximum power transfer

efficiency by using an LC matching network.

♥ A simple but effective technique for matching a short vertical antenna to a

feedline is to increase its electrical length by adding an inductance at its base.

♥ This inductance, called a loading coil, cancels the capacitive effect of the

antenna.

♥ Another method is to use capacitive loading.

Antenna Loading

Inductive Loading Capacitive Loading

76

Antenna Arrays• An assembly of two ore more antenna elements (often λ / 2)

situated in close proximity to each other so that their induction fields interact to

produce a radiation pattern that is a vector sum of the individual ones.

• In a phased array, all elements are fed or driven; i.e. they are connected to

the feedline.

• Some arrays have only one driven element with several parasitic elements

which act to absorb and reradiate power radiated from the driven element.

Broadside Array• one of the simplest form of antenna array consisting of a

number of dipoles having equal size, equally spaced along straight line and are

individually fed in the same phase from the same source.

•with axis placed vertically, radiation would have a narrow bidirectional

horizontal pattern

+

+ + + +

+ + + +- - - -

- - - -

ant. boom

-

λ/2

0+-

0 180° 360° 540°

f

Z

X

Y

Solid Radiation Pattern

(Bi-directional)

77

End-Fire ArrayAn array where the magnitude of the current in each element is

the same but there is a phase difference (90 degrees) between these currents

progressively from left to right.

Turnstile ArrayIt consists of two half wave dipoles mounted at right angles to

other in the same horizontal plane. When the two antennas are excited by equal

currents 90 degrees out of phase, a figure 8 radiation patterns merge to form

omnidirectional pattern.

Radiation Pattern

(Unidirectional)

+

+ +

+ +- -

- -

ant. boom

-

+

+-

-+

+-

-

0° 90°

180°

270°360°

λ/4

0°

270°

90°

180°

EACH ELEMENTIS λ/4

COAXIALFEED LINE

Radiation Pattern

(Omnidirectional)

Vector sum of the two patterns

78

Yagi-Uda Array

An array consisting of a driven element and two ore more parasitic elements.

Driven elements – elements directly connected to the transmitter output.

Parasitic Element – elements not directly connected to the transmitter output

a. reflector b. director

Characteristics of Yagi Array

• relatively narrow bandwidth since it is resonant

• 3-element array has a gain of about 7 dBi

• more directors will increase gain and reduce the beamwidth and feedpoint

impedance

• a folded dipole is generally used for the driven element to widen the bandwidth

and increase the feedpoint impedance.

+ 5% λ/2

DRIVEN POLE

DIRECTOR

- 5% λ/2

0.1λ

0.1λ

REFLECTOR

Direction of max radiatio

Radiation Pattern

(Unidirectional)

79

WIDEBAND AND SPECIAL PURPOSE ANTENNAS

Folded Dipole

•Often used alone or with other elements

- for TV and FM broadcast receiving antennas because it has a wider bandwidth

and four times the feedpoint resistance of a single dipole.

Log-Periodic Dipole Array (LPDA)

type of antenna whose array elements increase logarithmically corresponding to

a design ratio no less than 1 and the opposite ends of the array form an angle of

used in television reception including UHF range・

I/2

I/2

λ/2

λ/4

Rr= 4 x 73 = 292Ω

BW = ±10% fc a.) LC Circuit b.) Transmission Line

43

32

21

R4R3

R3R2

R2R1 τ

l

l

l

l

l

l======

l5 l6

R5

R6

DIPOLES

∝

Beam direction

80

τ= design ratio < 1 Typical values: τ= 0.7 α= 30°

Characteristics of LPDA

• feedpoint impedance is a periodic function of log f

• unidirectional radiation and wide bandwidth

• shortest element is less than or equal to λ/2 of highest frequency, while

longest element is at least λ/2 of lowest frequency

• reasonable gain, but lower than that of Yagi for the same number of

elements

• used mainly as HF, VHF, and TV antennas

Loop Antenna

Single turn coil carrying RF current; used particularly for DF(Direction Finding)

applications

Main characteristics:

•very small dimensions

•bidirectional

•greatest sensitivity in the plane of the loop

•very wide bandwidth

•efficient as RX antenna with single or multi-turn loop

feeder

SQUARE LOOP

feeder

CIRCULAR LOOP

81

Helical Antenna

・a broadband VHF and UHF antenna which is used when it is desired to provide

circular polarization characteristics.

Used for satellite and probe communication (radio telemetry)

broadband (・ + 20% of fc)

circularly polarized・

A・ p= 15 dB; θ-3dB = 20・ o are typical

( )3

215λπDNSG =

Where:

G = gain with respect to isotropic antenna N = number of turns in the helix(any positive integer) S = turn spacing ≈λ⁄4 D = diameter of the helix λ = wavelength φ = Beamwidth

NSDλ

πλφ 52

=

Helix λ/3

axial radiationλ/4

λ/8

coaxialfeed

0.8λ

Ground Plane

82

Discone Antenna

A combination of a disk and a cone in close proximity. It is a ground plane

antenna evolved from the vertical dipole and having a very similar radiation

pattern. It is characterized by an enormous bandwidth for both input impedance

and radiation pattern and behaves as though the disk were a reflector.

A wideband antenna which has usable characteristics over a frequency

range of nearly 10:1, used to radiate a vertically polarized wave in all the

horizontal directions (omnidirectional)

Coaxial

Disc

Cone

Feed

D

D

2/3D

D/25

83

UHF & MICROWAVE ANTENNAS

•highly directive and beamwidth of about 1o or less

•antenna dimensions >> wavelength of signal

•front-to-back ratio of 20 dB or more

•utilize parabolic reflector as secondary antenna for high gain

•primary feed is either a dipole or horn antenna

•use for point-to-point and satellite communications

Antenna with Parabolic Reflector

FP + PP’ = FQ + QQ’ = FR + RR’ = K

Plane waves emanating from its surface travel in a narrow beam which not only

increases gain, but also reduces susceptibility to noise.

a. b. c.

Various feed situation for a parabolic, (a) insufficient illumination (b) ideal

illumination (c) spillover

FOCUS

DIRECTRIX

A

PQ

R R'Q'P'

B

R'

DPlane waves leavinga parabolic surface:

Plane waveform

84

Remedy:

spherical reflector used to reduce back lobe radiation

Cassegrain Antenna (eliminates spill over)

Power gain and -3 dB

beamwidth are:

spherical reflector

primary feed dipoleat focus

Paraboloid

2

22

λπη DAp =

Dλφ 70

= φ0 = 2 φ

primary paraboloid reflector

vertex primary feedhorn

seconary reflectorat focus

85

for lossless,

Horn Antenna

To overcome the difficulties in radiating energy using a waveguide, the mouth of

the waveguide maybe opened out, as was done to the transmission line, but this

time an electromagnetic horn results instead of the dipole.

There are several possible horn configurations, the most common are

(a) Sectoral horn – flares out in one direction only.

(b) Pyramidal Horn – flares out in both direction and has the shape of

a truncated pyramid

(c) Conical Horn – flares out in both directions and is a logical

termination for a circular waveguide.

Special horn antennas are the Cass-horn and the Hoghorn antenna, which are

rather difficult to classify since each is a cross between a horn and a parabolic

reflector.

2

2

6λDAp =

Where: η = antenna efficiency(0.55 is typical); D = dish diameter (m); and λ= wavelength (m) φ = beamwidth between half-power points, in degrees φ0 = beamwidth between two nulls, in degrees

θ

L

86

Lens Antenna

The lens antenna is yet another example of how optical principles may be

applied to microwave antennas. It is used as a collimator at frequencies well in

excess of 3 GHz and works in the same way as a glass lens used in optics.

Principle of Wave Collimation

The function of the lens is to straighten out the wavefront ensuring that signals

are in phase after passing thru it.

Advantages: Greater design tolerances, no primary antenna to be mounted and

obstruct radiation.

Disadvantages: Greater bulk, expense and design difficulties.

Cross section ofzoned lenses

used to reduceattenuation

planewavefront

curvedwavefronts

87

SAMPLE PROBLEMS

1. A helical antenna with ___ turns is to be constructed for a frequency of

_____MHz, if the helix diameter is ___m and turn spacing of ____m find:

a) The power gain

b) The beamwidth

2. Dimension the elements of a Yagi antenna for ____MHz operation using

0.2λ inter element spacing.

88

3. Determine the gain of a 6-ft parabolic dish operating at ______MHz.

4. Design a log periodic antenna for the ____ broadcast band using design

ratio factor (τ) of 0.95 and σ = 0.08.

89

5. Design a five-element Yagi-Uda antenna for ___MHz operation with three

directors using 0.2λ inter-element spacing.

COURSE WORK 7

See Appendix A

90

WAVEGUIDE

- A specially constructed hollow metallic pipes or system of

conductors and insulators for carrying electromagnetic waves.

- They are used for microwave frequencies for the same purposes

as transmission lines were used for lower frequencies.

Reasons for using waveguide rather than coaxial cable at microwave frequency:

• easier to fabricate

• no solid dielectric and I2R losses

At microwave signal frequencies (between 100 MHz and 300 GHz), two-

conductor transmission lines of any substantial length operating in standard TEM

mode become impractical. Lines small enough in cross-sectional dimension to

maintain TEM mode signal propagation for microwave signals tend to have low

voltage ratings, and suffer from large, parasitic power losses due to conductor

"skin" and dielectric effects

Waveguides do not support TEM waves inside because of boundary conditions.

Waves travel zig-zag down the waveguide by bouncing from one side wall to the

other.

91

RECTANGULAR WAVEGUIDE

Mode of Operation

Mode Type Propagation Properties

Transverse Electric (TE) Electric field is perpendicular to the

direction of wave propagation

Transverse Magnetic (TM) Magnetic field is perpendicular to the

direction of wave propagation

DOMINANT MODE OF OPERATION

The most natural mode of operation for a waveguide, this mode is the lowest

possible frequency that can be propagated

For a waveguide’s mode of operation, the two submodes are:

1. TEmn for the transverse electric mode.

2. TMmn for the transverse magnetic mode.

Where: m – number of half-wavelength across

waveguide width (the a dimension)

n – number of half-wavelength along

waveguide height (the b dimension)

TEmn means there are m number of half-wave variations of the transverse E-

field along the “a” side and n number of half-wave variations along the “b” side.

The magnetic field forms closed loops horizontally around the E-field

92

E-Field Pattern of TE1 0 Mode

Wavelength for TE & TM Modes

Cutoff wavelength

Smallest free-space wavelength that is just unable to propagate in the

waveguide under given conditions.

The wavelength of the lowest frequency that can be accommodated in a

given waveguide.

( ) ( )22 //

2

bnamc

+=λ

a

b

λg/2

93

Any signal with λ > λc will not propagate down the waveguide.

For air-filled waveguide, cutoff freq., fc = c/λc

TE10 is called the dominant mode since λc = 2a is the longest wavelength of

any mode.

Guide wavelength

Group Velocity

The speed of transmission of a signal along a waveguide

Phase Velocity

The apparent speed of propagation along a waveguide based on the

distance between wavefronts along the walls of the waveguide.

( ) ( )22 /1/1 ffor

cc

g−−

=λ

λλ

λλ

( )2/1 cg

g corcv λλλλ

−=

( )2/1 c

gp

corcvλλλ

λ

−=

94

Waveguide Impedance

Where: ZO = 120π or 377Ω for air-filled waveguide

Circular/Cylindrical Waveguides

A waveguide having a circular cross-section, used whenever a rotating

element (radar antenna) must be attached to the transmitter/receiver.

Differences versus rectangular waveguides :

λc = 2πr/Bmn

where: r = waveguide radius

Bmn= Bessel function solution for a particular m,n

mode being propagated

=1.84 for the dominant mode of operation.

All TEmn and TMmn modes are supported since m and n subscripts are defined

differently.

Dominant mode is TE11.

( )( )2

2

/1

/1

coTM

c

oTE

ZZ

ZZ

λλ

λλ

−=

−=

95

Advantages: higher power-handling capacity, lower attenuation for a given cutoff

wavelength.

Disadvantages: polarization may rotate.

FIELD PATTERN FOR CIRCULAR WAVEGUIDE

Optical Fiber Communications

96

• Core - Thin glass center of the fiber where the light travels

• Cladding - Outer optical material surrounding the core that reflects the

light back into the core

• Buffer coating - Plastic coating that protects the fiber from damage and

moisture

Compared to conventional metal wire (copper wire), optical fibers are:

• Less expensive - Several miles of optical cable can be made cheaper

than equivalent lengths of copper wire. This saves your provider (cable

TV, Internet) and you money.

• Thinner - Optical fibers can be drawn to smaller diameters than copper

wire.

• Higher carrying capacity - Because optical fibers are thinner than

copper wires, more fibers can be bundled into a given-diameter cable than

copper wires. This allows more phone lines to go over the same cable or

more channels to come through the cable into your cable TV box.

• Less signal degradation - The loss of signal in optical fiber is less than

in copper wire.

• Light signals - Unlike electrical signals in copper wires, light signals from

one fiber do not interfere with those of other fibers in the same cable.

This means clearer phone conversations or TV reception.

• Low power - Because signals in optical fibers degrade less, lower-power

transmitters can be used instead of the high-voltage electrical transmitters

needed for copper wires. Again, this saves your provider and you money.

• Digital signals - Optical fibers are ideally suited for carrying digital

information, which is especially useful in computer networks.

• Non-flammable - Because no electricity is passed through optical fibers,

there is no fire hazard.

97

• Lightweight - An optical cable weighs less than a comparable copper

wire cable. Fiber-optic cables take up less space in the ground.

• Flexible - Because fiber optics are so flexible and can transmit and

receive light, they are used in many flexible digital cameras for the

following purposes:

Medical imaging - in bronchoscopes, endoscopes, laparoscopes

Mechanical imaging - inspecting mechanical welds in pipes and

engines (in airplanes, rockets, space shuttles, cars)

Plumbing - to inspect sewer lines

•Disadvantages:

–higher initial cost in installation & more expensive to repair/maintain

Optical Fiber Link

• Transmitter - Produces and encodes the light signals

• Optical fiber - Conducts the light signals over a distance

• Optical regenerator - May be necessary to boost the light signal (for

long distances)

• Optical receiver - Receives and decodes the light signals

InputSignal

Coder orConverter

LightSource

Source-to-fibreInterface

Fibre-to-lightInterface

LightDetector

Amplifier/ShaperDecoder

Output

Fibre-optic Cable

Transmitter

Receiver

98

Optical fibers come in two types:

• Single-mode fibers

• Multi-mode fibers

•Single-mode step-index fiber:

–minimum signal dispersion; higher TX rate possible

–difficult to couple light into fiber; highly directive light source (e.g. laser)

required; expensive to manufacture

•Multimode step-index fibers:

–inexpensive; easy to couple light into fiber

–result in higher signal distortion; lower TX rate

•Multimode graded-index fiber:

–intermediate between the other two types of fibers

Single-mode step-index fibre

Multimode step-index fibre

Multimode graded-index fibre

n1 coren2 cladding

no air

n2 claddingn1 core

Variablen

no air

Lightray

Index porfile

99

Acceptance Cone & Numerical Aperture

Acceptance angle, θc, is the maximum angle in which external light rays may

strike the air/fiber interface and still propagate down the fiber with <10 dB loss.

Numerical aperture:

Losses In Optical Fiber Cables

•The predominant losses in optic fibers are:

–absorption losses due to impurities in the fiber material

–material or Rayleigh scattering losses due to microscopic irregularities in the

fiber

–chromatic or wavelength dispersion because of the use of a non-

monochromatic source

–radiation losses caused by bends and kinks in the fiber

n2 cladding

n2 claddingn1 core

AcceptanceCone θC

22

21

1sin nnC −= −θ

22

21sin nnNA C −== θ

100

–modal dispersion or pulse spreading due to rays taking different paths down the

fiber

–coupling losses caused by misalignment & imperfect surface finishes

Absorption Losses In Optic Fiber

Fiber Alignment Impairments

Axial displacement Gap displacement

Angular displacement Imperfect surface finish

Loss

(dB

/km

)

10

0.7 0.8Wavelength (μm)

0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7

2

3

4

5

6

Peaks causedby OH- ions

Infraredabsorption

Rayleigh scattering& ultravioletabsorption

101

Light Sources

•Light-Emitting Diodes (LED)–made from material such as AlGaAs or GaAsP

–light is emitted when electrons and holes recombine

–either surface emitting or edge emitting

• Injection Laser Diodes (ILD)–similar in construction as LED except ends are

highly polished to reflect photons back & forth

ILD versus LED

•Advantages:

–more focussed radiation pattern; smaller fiber

–much higher radiant power; longer span

–faster ON, OFF time; higher bit rates possible

–monochromatic light; reduces dispersion

•Disadvantages:

–much more expensive

–higher temperature; shorter lifespan

Light Detectors

•PIN Diodes

–photons are absorbed in the intrinsic layer

–sufficient energy is added to generate carriers in the depletion layer for current

to flow through the device

•Avalanche Photodiodes (APD)

–photogenerated electrons are accelerated by relatively large reverse voltage

and collide with other atoms to produce more free electrons

102

–avalanche multiplication effect makes APD more sensitive but also more noisy

than PIN diodes.

SAMPLE PROBLEMS

1. A wave is propagated in a parallel-plane waveguide. The frequency is

___GHz, and the plane separator is ___cm. Calculate:

a) The cutoff wavelength for the dominant mode.

b) The wavelength in a waveguide.

2. A rectangular waveguide is ___cm by ___cm. Calculate the cut-off

frequency of the dominant mode.

103

3. A rectangular waveguide measures ___x___cm and has a 9GHz signal

propagated in it. Calculate the cut-off wavelength, the guide wavelength,

the group and phase velocities and the characteristics impedance for the

TE1,0 mode.

4. A waveguide has an internal width a of ___ cm, and carries the dominant

mode of a signal of unknown frequency. If the characteristic impedance is

_____Ω, what is the frequency?

104

5. Calculate the numerical aperture and the maximum angle of acceptance

for a fiber with core and cladding refraction indices of ____ and ____

respectively.

6. Determine the critical angle for a glass (n=___)/quartz (n=____)

interface. If the angle of incidence is 38o determine the angle of

refraction.


COURSEWORK 8

See Appendix A

106

ECE123 – COURSE WORK 2

Name: ___________________________ Date: _____________

Signature: _______________________ Section: ___________

Shade the letter of the corresponding answer. Strictly No Erasures.

A B C D

1. O O O O

2. O O O O

3. O O O O

4. O O O O

5. O O O O

6. O O O O

7. O O O O

8. O O O O

9. O O O O

10. O O O O

A B C D

11. O O O O

12. O O O O

13. O O O O

14. O O O O

15. O O O O

16. O O O O

17. O O O O

18. O O O O

19. O O O O

20. O O O O

GRADE

107


Name: ___________________________ Date: _____________

Signature: ________________________ Section: ___________

Show complete solution.

GRADE

110


Name: ___________________________ Date: _____________


Use the Smith Chart provided in Appendix B. Show complete solution.

GRADE

112


Name: ___________________________ Date: _____________



A B C D

1. O O O O

2. O O O O

3. O O O O

4. O O O O

5. O O O O

6. O O O O

7. O O O O

8. O O O O

9. O O O O

10. O O O O

A B C D

11. O O O O

12. O O O O

13. O O O O

14. O O O O

15. O O O O

16. O O O O

17. O O O O

18. O O O O

19. O O O O

20. O O O O

GRADE

113


Name: ___________________________ Date: _____________



A B C D

1. O O O O

2. O O O O

3. O O O O

4. O O O O

5. O O O O

6. O O O O

7. O O O O

8. O O O O

9. O O O O

10. O O O O

A B C D

11. O O O O

12. O O O O

13. O O O O

14. O O O O

15. O O O O

16. O O O O

17. O O O O

18. O O O O

19. O O O O

20. O O O O

GRADE

114


Name: ___________________________ Date: _____________



A B C D

1. O O O O

2. O O O O

3. O O O O

4. O O O O

5. O O O O

6. O O O O

7. O O O O

8. O O O O

9. O O O O

10. O O O O

A B C D

11. O O O O

12. O O O O

13. O O O O

14. O O O O

15. O O O O

16. O O O O

17. O O O O

18. O O O O

19. O O O O

20. O O O O

GRADE

115

References

Electronic Communications Systems Fundamentals Through Advanced 5th Edition

by Wayne Tomasi (textbook)

Electronic Communication Systems by George Kennedy and Bernard Davis

EE555 powerpoint presentation of Heng Chan of Mohawk College

Naval Electrical Engineering Training Series Module 10, Integrated Publishing

ece123

Documents

Transcript of ece123