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.
36
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
must be placed to reduce SWR to 1.
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
transformer must be inserted to reduce SWR to 1.
d. The characteristic impedance of the transformer.
e. The nearest point to the load where a series short circuit
must be placed to reduce SWR to 1.
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
transformer must be inserted to reduce SWR to 1.
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
End of topic discussion for Quiz 2
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
End of topic discussion for Quiz 3
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.
End of topic discussion for Quiz 4
COURSEWORK 8
See Appendix A
105
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
ECE123 – COURSE WORK 3
Name: ___________________________ Date: _____________
Signature: ________________________ Section: ___________
Show complete solution.
GRADE
108
109
110
ECE123 – COURSE WORK 4
Name: ___________________________ Date: _____________
Signature: _______________________ Section: ___________
Use the Smith Chart provided in Appendix B. Show complete solution.
GRADE
111
112
ECE123 – COURSE WORK 6
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
113
ECE123 – COURSE WORK 7
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
114
ECE123 – COURSE WORK 8
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
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