UNIT

Wave Propagation-I
&
Wave Propagation-II
(SYLLABUS
Wave Propagation: Definitions, Categorizations and GeneralClassifications, Different Modes of Wave
Propagation, Ray/Mode Concepts.
Cround Wave Propagation: Plane Earth Reflections, Space and Surface Waves, Wave Tilt, Curved
Earth Reflections.

Spgce Wave Propagation: Field Strength Variation with Distance and Height, Effect of Earth's Curvature,
Absorption, Super Refraction, M-Curves and Duct Propagation, Scattering Phenomena, Troposphere
Propagation.

Sky Wave Propagation:Structure of lonosphere, Refraction and Reflection of Sky Waves by lonosphere,
Ray Path, Critical Frequency, MUF, LUF, OF, Virtual Height and Skip Distance, Relation between MUF
and Skip Distance, Multihop Propagation.

LEARNING DBJECTIVES

Classification of waves and their modes of propagation

Concepts of ray and wave
Plane earth and curved earth refections, types of waves in ground wave propagation
Field strength variation with distance and height wave propagation in space
Concepts of super refraction, duct propagation, scattering phenomena and tropospheric propagation
Structure of ionosphere
Concepts of criticalfrequency, MUF, LUF, OF, virtual height and skip distance.
Relation between MUF and skip distance

Concept of multi-hop propagation

INTRODUGTION

the receiving antenna by different possible

Ihe radio waves propagate fromn transmitting antenna to
space wave or fropospheric wave propagation
odes as ground wave or surface wave propagation,
and sky wave or ionospheric propagation.

Ground wave at frequencies upto 2MHz. Space wave propagation is important

propagation is important
trequerncies above 30 MHz. Sky wave propagation is important
af frequencies between 2 MHz ond
30 MHz. Out of these modes, the sky wave propagation is important from the point of view of long

distance radio communication.

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 ANTENNAS AND WAVE
PROPAGATION
AND
UNIT-5
Non-uniform Plane WVave
168
PROPAGATION-DEFINITIONS,
CATEGORIZATIONS

PROPAGATION. GENERAL
D The wave in which equiphase Standing Wave
169

neither equal nor and

WAVE WAVE
5.1 DIFFERENT MODES OF surfaces are equiamplitude
CLASSIFICATIONS,
parallel When both forward and backward (reflected)
as plane wave, Thee is referred
non-uniform
waves simultaneously exist, then the resultant
aonetic (H), components may orelectric
MODE C O N C E P T S
and their
categorization. (E) and wave of their combination is called standing
introduction about wave propagation may not be
Q1. Give a brief orthogonal. wave. It is called a standing wave, as it does not
Ans:
curent carryìng conductor
propagates in free space in the formnat offelectromagnetic Slow Wave progress. The nagnitudes of the maximas and
The power radiated by 4 minimas may vary at different time instants, but
waveS. communication with the
help of electromagnetic waves The phase vel0city of a slow wave they appear at the sarme locations as shown in
Radio of long distance range of 3 x 10ll mto 3 x 10 min free space
e e gs the Studywavelengths in the
are known omal to the equiphase surface is lesswhich is
than the figure (3)
waves with velocity of light. It is useful in a
as radio waves. communication as they.t few microwave E,(x, t)
are useful for mine and marine very short wavelenve
devices like TWWT. -3T8
between 10- 10 m Waves of
The waves with wavelength within the earth's surface and water. Hert e
deep Forward Wave
the capability to penetrate is very high up to a few terra
as the frequency and bandwidth m are not of much
commerst2) 5.
used in optical communication wavelengths extending upto 10 Eorward waves travel as directed from the oriein -T/8
very long
of the propagating wave under
However,
and a better speed. the absence of any -0
of space, the characteristicsorganization. obstructions to result
Depending upon the nature and location
absorption and rotation of plane
of
reflections.
reflection, refraction, diffraction, into two types,
can be broadly categorized
The concept of wave propagation 6
Backward Wave
1 Guided waves
Unguided waves.
When a forward wave is obstructed it is reflected
2 back. This backward travelling wave is called as
1 Guided Waves medium or structure Thex the backward wave. However, the reflections of Figure (3)
guided with the help of a transmission
As the name suggests, these waves are waveguides parallel wires, optical fibers, coaxial cable. etc wh the backward wave depends on the conductivity 9. Surface Wave
man made structures may be strip lines,
waves in the required direction. andsurface of the reflecting substances material. Surface wave is one, that obtainsa support by
serve as a medium that guides the I AN
signal and data communication. A few of these are propagating in the directions parallel to the
Guided waves have large number of applications in interconnections. cable 7. Travelling Wave
internet services with the help of surface between the two media and decreases
(Local Area Network) using cables and wire pairs, used in applications such as trunk lines, closed Wave that continuously travels in one direction from
networks used by cable TV operators. Therefore, they are vertically. For example, strip lines implement
circuit TV, etc. the origin without any reflected waves is termed surface waves.
as travelling wave. The maxinmas and minimas 10 Trapped Wave
2. Unguided Waves
propagate, the waves travel along and over the earth's of the electric field compo-nents, E at different
Unguided waves do not require a physical medium to Surface wave is called a trapped wave when it
surface in the terrestrial atmosphere and in the outer
space from transmitter to receiver. These include large time instants occur at different locations as shown is unable to radiate e
continuously increasing further. in figure (1). The variation of 'E with x (space hat it camies within
number of applications which are the small distance interface except at
study of graphics to know about the upper
Unguided waves have been proved to be very useful for the parameter) for travelling waves is as shown in discontinuities. For example, trapped waves
and nebulae, radio astronomy, etc. In contrast these are figure (1). The variations of'E with'x' when there
atmospheric regions, activities of the sun, stars occur in ducting phenomena
telegraphy, mobile communication, radar
also useful in satellite communication, telecontrol, telephony, is attenuation is as shown in figure (2). Leaky Wave
communication, and remote sensing.
When there are more discontinuities placed in a
classification of electromagnetic waves. E,(x, t)
Q2. Define electromagnetic waves. Give the general Tot = 0 transmission line, then there is leakage of energy
ot = x/2
Ans: ot = x/4 due to the perturbations caused by discontinuities
v =o/B This results in a travelling wave called leaky
Electromagnetic Waves
space and is called the electromagnetic wave.
The energy radiated by a transmitting antenna propagates through the Q3. Explain in detail about various modes of
wave. This is nothing but the energy fed to the antenna by the transmitter.
propagation.
Electromagnetic waves can be expressed in termns of magnetic field vector, (H).
Ans:
V²H=uE Modes of Propagation
Electromagnetic waves may travel from
In terms of electric field vector (E) transmitting antenna to the receiving antenna in a number
Figure (1)
AE,(x, ) of ways.
Different propagations of electromagnetic waves
O=0 ot = x/4 are as follows,
The electromagnetic waves are generally classified into following types. Ground wave propagation
1 Plane Wave Sky wave propagation
2
If the equiphase surface is a plane then the wave is called a plane wave. 3 Space wave propagation
2 Uniform Plane Wave 4 Tropospheric scatter propagation.
Uniform plane wave is defined as one, for which the equiphase surface is the equiamplitude surtae This classification is based upon the frequency
electric, (E) and the magnetic, (), field components are orthogonal(perpendicular to each other) and p range, distance and several other factors.
is transverse.
Figure (2)
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 ANTENNAS AND WAVE
170 PROPAGATION UNIT-5

171
Ground Wave Propagation
Applications E2 GROUND WAVE PROPAGATION
Ground wave propagation is also
known as 1. Space wave propagation is used in -PLANE EARTH REFLECTIONs.
propagation is communication.
satellite SPACE AND SURFACE WAVEs.
surface wave propagation. This
practically important at frequencies up
to 2 MHz.
It controls radio trafic between a ground I WAVE TILT, CURVED EARTH
Ground wave propagation exists when
transmitting
2
and a satellite. sstaation REFLECTIONS
the earth's
and receiving antenna are very close to 4. Troposcatter Propagation EExplain about ground wave propagation.
curvature. Figure (2): Ground Wave Propagation
Troposcatter propagation is also
Ans:

Ground wave propagation suffers

attenuation known as Some of the energy of surface wave is absorbed
earth. forward scatter propagation, it is practicall In ground wave propagation, the waves propagate when it is passed over the surface of the earth. Lost
while propagating along the surface of the important at frequencies above 300 MHz along the surface of the earth. This phenomenon
into two energy is partially compensated by the downward
This propagation can be subdivided is also known as surface wave propagation.
types which are space wave and surface wave
This propagation covers long distances in H diffraction of waves present immediately above
propagation. range of 160 to l600 km. It is practically significant at broadcast and lower the ground's surface.
frequencies that means propagation is important The degree of attenuation depend upon
Applications: Ground wave propagation is generally Q4. Discuss briefly about ray concept and for medium and long waves. variety
used in TV, radio broadcasting etc. of factors ie., when the wave
mode concept. As ElectroMagnetic (EM) waves are guided by propagates
the curvature of the earth, depending along
2 Sky Wave Propagation waveguide or transmission line, ground waves upon
Ans:
are guided by surface of the earth. So that, frequency, surface irregularities, permittivity and
Sky wave propagation is practically important at conductivity, wave is attenuated.
frequencies between 2 MHz to 30 MHz. Here the The wave phenomena cân be analyzed by two propagation takes place around the curvature of
electromagnetic waves reach the receiving point methods, the earth. As the frequency increases, the propagating signal
after reflection from an atmospheric layer known This type of propagation is possible only when becomes weak. The ground wave propagation
1. Ray theory is also called as 'medium wave
as ionosphere. both receiving and transmitting antennas are close propagation' as
it is only possible for the waves having a
Hence, sky wave propagation is also known as 2 Mode theory. to the surface of the earth and at its lower edge low
supported by the earth. frequency of about 3 MHz to propagate in this
ionospheric wave propagation'. 1 Ray Theory mode.
It can provide communication over long distances. When ground waves spread out from the vertical
A ray can be arbitrarily considered as a line o antenna, the component of electric field is vertical Apart from the ground attenuation surface wave
Hence, it is also known as point-to-point is also attenuated because of tilt and diffraction
propagation or point-to-point communication. drawn perpendicular to the equiphase plane. Ray with respect to ground and any horizontal com
in the wave front as shown in figure (3).
theory is implemented for longer wavelength ponent of electric field which are in contact with
Disadvantage: Sky wave propagation suffers from earth is short circuited by the earth as shown in Tilt angles
fading due to reflections from earth surface, fading can
nr between reflecting layers. Surface wave must
be considered separately and is called Norton figure (1).
be reduced with the help of diversity reception. surface wave. The attenuation of surface wave is Earth
Applications: It can provide communication over long dependent on the conductivity of ground. in ray Transmitting
antenna
distances including global communication. theory ionospheric effects can be ignored and at
VLE, it is applicable only for upper edge.
3. Space Wave Propagation Figure (3): Titing Wave Wavefronts in Ground Wave Propagation
The calculation of the phase and amplitude Earth
Space wave propagation is practically important
at frequencies above 30 MHz. depends as rays moving from origin to the From the figure (3), it is clear that tilt angle
observation point. Assuming the earth to be increases as the wavefront of the ground wave
It is also known as tropospheric wave propagation, spherical and taking proper electron and collision travels along the earth's surface. Due to this
because the waves reach the receiving point after Figure (1): Simple EM Wave increase in tilt angle of wave, most of the electric
frequency variations along with height, the
reflections from tropospheric region. problems related to ray theory can be resolved. Due to the propagation of wave along the surface components are short circuited.
of the earth, charges are induced inthe earth. Thus strength of feld is consistently reducing.
Troposphere region in atmosphere is within 16 h These charges travel along the ground wave
km above the surface of carth. 2. Mode Theory Finally, surface wave is lost at some considerable
which leads to induced current in the earth. distance due to diffraction and tilt.
In space wave propagation, signal at the receiving Mode theory is implemented for shorter
wavelengths between reflecting surfaces. Surface Earth act as a leaky capacitor when it has Medium frequency and low frequency of radio
point is a combination of direct and indirect rays. induced current. Thus, earth can be depicted as
Itprovides communication over long distances waves need not necessarily be considered wave is in the range of upto 3 MHZ. Therefore,
separately. Mode theory is normally employed a resistance in parallel with capacitor. ground wave propagation is suitable for long
with VHE, UHF and microwave frequencies.
Space wave propagation is also known as "line in cases where the distance between transmitter A surface wave travels along the surface of earth wave and medium wave and long-distance radio
and receiver is too long leading to more number instead of travelling in straight line by virtue of Communication.
of sight propagation'".
of reflected waves. For resolving problems inducing currents in the earth, Q6. Write short notes on the characteristics
The field strength of receiver depends on: associated with mode theory, assumea flat earth's As a result, lower part of the wave slows down of Ground wave propagation.
Direct ray from transmitter surface and ionosphere be at a height h' above which inturn makes the whole wave to benda Aug/Sept.-24(R18), Q11(b)
Ground reflected ray the earth. little (tilt) forward.
OR
Reflected and re fracted rays from the Then easy calculations of reflection coeficients Thus bend in the wave is in synchronous with Discuss the salient features of ground
atmosphere for perpendicular and parallel polarization can be earth's curvature. This enables it to travel a
wave propagation.
Diffiracted rays aound the curvature of earth done at both lower edge of ionosphere and the distance even beyond the line of sight as shown
in figure (2). Ans: May-18, (R15), Q11(a)
and so on. ground.

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 ANTENNAS AND WAVE
PROPAGATION 173
UNIT-5

172 For a smooth earth, reflection coeficient For, r =V, =90°

propagation In figure (1), becomes - 1for both polarizations when
The characteristics of ground wave h, -Transmitter antenna height 9=90°-r
are as follows, incident wave is near grazing.
is very much useful T-Transmitter antenna When R < 0.1, the surface is considered cos®= cos(90°-)
1. The ground wave propagation 4
cos = siny
i.e., very long
at lower and broadcast frequencieswaves,
DW Direct wave smooth. ...(3)
waves, long waves and
medium 0-Angle of incidence measured from no When R > 10 the surface is considered And sin @= cosr
5 ... (4)
Ground wave propagates gliding along the
earth's oees d -Distance between transmitter rand rough. Substituting equation (3) and (4) in R We get,
2.
surface and is propagated along the
curvature of antennas receiver
Since earth is not a perfect dielectric, with proper
the earth. h, - Receiver height modifications in the expressions for a perfect (e) sin y-
are
3 These waves are vertically polarized and R,-Receiver antenna dielectric medium, the horizontal and vertical R, =
/2
produced by vertical antenna.
RW- Reflectedwave polarization reflection factors for earth can be (6,l? sin y ++ j -E, cos v
short
4 The horizontal component at electric ficld is LOwer frequencies are important consideration ot obtained.
circuit, when it comes in contact with the
earth
eround waves, since the ground losses increase Consider a medium with dielectric constant, and 2
that are placed close to the surface of earth. rapidly with frequency. conductivity. Maxwell's equations for the fields Sin r -
joE,
cos
AE R, =
5. The field strength of ground wave is E= d If the plane is smooth and l conducts for a perfect dielectricmedium are expressed as, ...(5)
finitelywavetheni
the amplitude and phase of the reflected •D Sin y +
and it changes with characteristics of the earth. Vx H= J+ .. () jOE)
different from that of the incident wave.
6 When these wave propagate along the earth's
surface, the horizontal component of field is On the otherhand, if the plane is rough then the Taking sinusoidal time variations into account Substituting, =8 and
shorted when it comes to contact with the surface. reflected wave tends to scatter and is attenuated [eanj modifying equation (1) with complex jo80
The current is induced on the earth surface. Thus The reflection at the surface of a perfect quantity can be written as,
carth acts as leaky capacitor. i.e., Rel= in equation (5), we get,
dielectric gives the reflection factors for vertical VxH=g'
7. It requires high power for transmission. polarization.
8 Ground wave attenuates and loses its energy due
As the earth is neither a perfect conductor nor a
Where, sin y- [e, - jN) - cos² y]2
sin y +[e, - j)- cos y]l2 .(7)
to earth's curvature, surface iregularities, diffirac R,
tion tilt in the wave front. perfect dielectric reflection factors at the earth's e' =et (2
9 The major applications of ground wave propa surface have to be modified by considering its jo
Reflection factor for vertical polarization,
gation are radio navigation maritime mobile finite conductivity. Reflection Factors Horizontal Polarization
communication, ship-to-ship and ship-to-shore The reflected wave and the incident wave, differ (E, -jx) sin s - [e,- j) - cos y]2
In horizontal polarization, the electric field is R,
communication. in phase and magnitude for a smooth and finite perpendicular to the incident plane and parallel to (6, -j) siny+[e,- jx) - cos* y]
10. This mode can be used at VLF (15 kHz) and upto conducting earth's surface. the refelcting surface. Hence, it is also known aas .. (8)
2 MHz. However, for a rough earth's surface, the perpendicular polarization. The reflection factors R, and R, are complex quantities, these can be
11 It is suitable for VLE, LF and MF communication. amplitude and phase of the reflected wave tends Ry and R, for horizontal and vertical polarizations rep-resented as,
12. The EM travelling in this mode are not affected to get reduced as it undergoes more scattering Vlais expressed as,
by change in atmospheric conditions. compared to the smooth earth reflections. R,-|R|4R, and R,=|R,|ZR, . (9)
cos 0 -
Q7. Briefly explain the plane earth reflection Thus, it becomes necessary, to measure the Where, |Rl and | R| =Amplitudes
in ground wave. roughness of the earth's surface. R,= 2
ZR, and R= Phase angles.
Ans: If R' indicates the roughness of the earth's surface cos 0 + - sin0
The variation of reflection factors R, and R, with
Plane Earth Reflection then for a standard deviation '¡ of the surface
iregularities relative to the mean surface height, (e)e cos -(E,-8, sin 0) r and different values of frequencies are as shown in
The signal that reaches a receiver antenna after angle of incidence 0' and wavelength 2'.The R, figures (2), (3), (4) and (5).
radiating from a transmitter antenna, trace a direct
roughness of surface is expressed using Rayleigh (e)l cos +(E,-8, sin'e)2
path and a reflected path when both antennas are Where,
criterion as,
in line of sight of each other as shown in figure 0.8
(1) 4nG Sin 0 E, - Dielectric constant in medium 1 X
RR=
8, -Dielectric constant in medium 2, 0.
elle
Characteristics of roughness of earth's surface Consider flat earth's surface as medium 2 with e,
DWh =' and free space as medium 1 with[,=E 0.4
with respect to the above expression are as
RW follows, R, can be modified as, 0.2
RW
For larger values of 0' the surface is more
12
1
rough. (E,) cos 9-+jo -egsin'o 0.0
30 60 90
Flat earth 2 The surface is smnooth, when the incident
R¡ 2 Angle in degrees
x, >x, >x, > x, andf, <f,<f, <f,
Figure (1): Direct Wave and Reflected Wave between T angle is almost equal to the grazing angle (EJ cos 0+| ja
(i.e., 0 < 0).
Figure (2): |R,|Vs Angle
and R.

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 WAVE

174
ANTENNAS AND
PROPAGATION UNIT-5

Diecuss how the space

0 Q9. wave and surface wave 175
are related to
Ans:
ground wave.
inR,degre s 30

60
Space and Surface Wave: Considering
Soace wave: The waves that Sommerfield theory, the ground wave
space waves or direct waves reach receiver from transmitter directly is divided into two parts as,
Surface wave: The waves that without any reflection are
(i) reach through reflection known as
-90 waves or surface waves. from the earth's surface are
Space wave is strong at larger known as ground reflected
-120
surface. distance, from the earth surface while
surface wave is dominant near the
Consider a vertical dipole located at earth
150
a height 'h)' from the
Vertical dipole DW
earth's surface as shown in figure.
180 P
60 90
Angle in degrees RW
x>x>x,>x, and f, <f<f,<{,

Figure (3): ZR, Vs Angle

1.0 |
X, f, Plane
carth
R
0.8 X, Image
Figure: Vertical Dipole and its lmage
The field expressions for electric
0.6 dipole at a long distances is the
space wave and is given by, combined equation of surface wave and
0.4
30 60 90 E,-j30ß/di|cos ({exp(-BR)R} +R, (exp(-BR,)R,))
Angle in degrees
x >x,>x,>x, andf <f<f<f,
+(1-R)(1-r+tcos' y) F{expGBR,YR,))
E, =-j30ßldl(sin y cos V({exp(-JBR,)/R} + ..(1)
Figure (4): |R, Vs Angle R,(exp(-JBR,) R,})
-200
-cos v(1 - R) uVa-u' cos' w) F(expBRYR,}
(1 +0.5 sin' )]
X, f
Where, ...(2)
B-190 X. R, and R, Distances from the dipole and its image to the
R -Reflection coeficient receiving point respectively.
X
-180 F-Attenuation function.
30 Angle in deere
90
u=
x >x, >x>x, and f <f<f<f,
e,-jx
Figure (5): /R, Vs Angle 18x10°6
Q8. Give the characteristics of plane earth reflections when the incident wave is horizontally
polarized and vertically polarized. Space Wave JMHz)otats
Ans:
The space wave propagation is significant at VHF bands of
Horizontally Polarized Incident Wave MHz, both sky wave and ground wave fails due to high above 30 MHz (30 MHz-300 MHZ). Above 30
For allincident angles, there is a 180° phase difference nearly between reflected wave and incident wave. attenuation. Space wave propagation is also known
as line of sight propagation. It is limited to LOS, curvature
The magnitude of the reflected wave is same as the incident wave and 180° out of phase at all frequencies The space wave equation is expressed as,
of earth and to few hundreds of kilometers.
and ground conductivities when the angle of incidence (e) is almost equal to grazing angle (y =0).
3 Magnitude and phase of the reflection factors changes with increase in angle of incidence but only to a E,(Space) =E(Space) +E(Space)
certain extent. The charige depends on the frequency and conductivity of ground. The change is large when =j30B/dl cosu({exp(jBR)R} +R, (exp(-JBR,)R,))
the frequencies are high and the ground conductivities are low. ..(3)
Surface Wave
Vertically Polarized Incident Wave
The magnitude of the reflected wave is same as the incident wave and 180° out of phase at all frequencies The attenuation function becomes unity as R reaches 0 and distances that are little smaller
then wavelength
and ground conductivities when the angle of incidence (e) is almost equal to grazing angle (y =0). of dipole.
2 As angle of incidence increases, the phase and magnitude of reftected wave decreases, Then the angle at The surface wave equation is expressed as,
which, the magnitude is minimum and a-90° phase change is observed is called Brewster Angle (orn) Pseudo
Brewster angle. Eu(Surface) = E (Surface) + E(Surface)
3. When the angles of incidence increases above the Brewster angle, the magnitude again increases and phase -j30ßldl (1 -R)F{exp (-BR,)YR,) 12u' +(cos y)ut(+0.5sin y)? ... (4)
reaches zero.
4. For low conductivities and high frequencies, the Brewster angle is almost similar to that of aperfect dielectric. Here, the higher order terms are neglected. If the electric dipole is at earth's surface, then the surface wave
However, for high conductivities and low frequencies, the Brewster's angle is very less. equation (4) is modified as,
5. When the incident wave falls normally (y =90°) on the reflecting surface then the reflection factors, R, and Ea(Surface) =j30 Blal (1l-R) F{exp-BR)/RI [a(1-) ta,cos V(l +0.5 sin' y)] u i-u'cos' y
Rhave same values but differs in phase by 180°. Here, R(R>> ) isadistance fromthe dipole to the field point.
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 WAVE
176
ANTENNAS AND
The general expression for
PROPAGATION
earth's
UNIT-5VAWCHs
e reflection coefihcient
for vertical
The attenuation factor is expressed
as, impedance is given by, surface given by,. polarization is The cffectof earth's 177
(E, -j) siny - [6,- j)- cos' j2 of the waves are.
curvature on the propagation
<5tan R= 1
Where, (E, -jX) sin y +[[,- jx) - cos w]l2 The curvature of the earth
changes
sin y Where, Wave Tilt
difference of the direct wave the path length
(JBR'U-o-cos'v2
Z. - Surface impedance
u-Permeability = 4n X107 H/m
For answer refer Unit-5, Q10.
Discuss the curved earth
wave from that of the flat and the reflected
fixed antenna-height values.
carth case even for
Q12.
1+siny o -Conductivity (mho/m)
reflection. 2 The reflected wave
Ans:
undergoes divergence because
of the reflection from
e-Permittivity = 8.85 x 1012 F/m Curved Earth Reflection
earth's
From the expression of surface In general, the propagation of decreases the power received atcurvature. This
impedance
the electric field strength can be calculated affected by the curvature of theground waves is antenna.
the receiving
iVo earth
When h is very small, the incident wave will
not The electric field cómponents in vertical ana extent. This effect is negligible when theto a large Q13. Calculate the wave
tilt in degrees of the
horizontal directions are respectively given as, is small i.e., upto d (miles) = distance Surface wave over an earth
be plane wave. of 6 mm
50/ conductivity and relative permittivity of
When h, is very large, the wave is a plane wave E, Hn, But as the distance of 12 at 2 MHz.
transmission
and the space wave field is a total ground wave.
E, -J, Z, ..(2) curvature of the earth decreases the increases, the
field strength
Ans:
May-18, (R15). Q11b)
Q10. What is angle of tilt? How does it affect Here, below the free space value. Given that,
the field strength at a distance from the For a surface wave,
Intrinsic impedance = 120 Q=377 o As a result, the waves travelling Conductivity, G=6mm
transmitter?
along the
OR Z, - Surface impedance of earth surface will not reach the receiver earth's Relative permitivity. E = 12
through a
What is wave tilt? How does it affect J. -Surface current density (A/m)
straight-line path. Operating frequency,f= 2MHz
the field strength at a distance from the In the case of surface waves, the The expression for wave tilt is given by.
E -Field due to vertical component wave may reach
transmitter? the receiver by,
E- Field due to horizontal component. Wave tilt =tan
Aug./Sept.-24(R18), Q10(b) (i) Diffraction, around the earth.
Ans: The ratio of horizontal to the vertical is given by, The expression for absolute permitivity is given
Angle off tilt is also known as wave tilt. It is (i1) Refraction, in the lower parts by
E,JZsZ, of the
defined as the deviation in orientation of the atmosphere at a certain height above the
vertically polarized ground wave on the surface E, Hno
ground. 8.854 x 102x 12 [ E, =8854 x109
of the earth. E, 1
In the case of space waves, there
are two paths = 1.0624 x 10-10
The conductivity and permittivity of the earth E. 377 |V Vo?+o'e2lan taken by the waves to reach the receiver.
determines the magnitude of tilt in the wave.
e=1.0624 x10° Farad m
'...(3) The wave that goes directly from the transmitter Wave tilt = tan
The vertically polarized wave, E has forward The angle tan represents the tilt angle. to the receiver is the direct wave. The wave 6x 10
tilt at the surface of carth resulting horizontal that
gets reflected due to the surface of the earth is the 2r x2x 10° x 1.0624 x 100
component of the current. Q11. Explain about reflection factors of earth =77.45°
reflected wave.
Due to tilt, horizontal and vertical components and write the wave tilt of the ground wave. :.Wave tilt =77459
of the electric field exists.
Because of the reflection of the wave from the
Ans: Dec.-19, (R16), Q10(a) 5.3 SPACE WAVE PROPAGATION -
curved surface of the earth, its energy decreases
These component do not have same magnitude Reflection Factors of Earth as it reaches the receiver. FIELD STRENGTH VARIATION
and are not in phase even for slight forward tilt.
Earth is neither a good conductor nor a good The effective heights of the antennas are also WITH DISTANCE AND HEIGHT,
Hence the surface becomes elliptically polarized dielectric. It is a partially conducting dielectric medium. EFFECT OF EARTH'S CURVA
wave.
lesser than the actual heights because of the
So it's dielectric constant is a complex quantity and is earth's curvature. TURE, ABSORPTION, SUPER
The graph of electric field strength for various given by, REFRACTION, M-CURVES AND
values ofx is as shown in figure below. They are as illustrated in figure (1).
E'=E+ DUCT PROPAGATION, SCAT
jo R TERING PHENOMENA, TROPO
Hence, the reflection coefficient for horizontal
SPHERICPROPAGATION
polarization is given by, R Q14. Explain in detail about space wave
sin y- [e, - jx)- cos ] M. propagation.
R, A' (or)
Sin y +[[,- jx)-cos² vjl2 Estimate the mechanism of space wave
Where,
R, is the distance A-B propagation over ideal flat earth with a
R, is the distance A-C-B d, is the length of arcA'-C neat sketch.
X=0.5 x=5 X=50 x= 500 h, is the height A'-A
jo[, h, is the height B"-B
d is the length of arc B"-C
Figure: Elliptical Polarization of Electric Field Vector at r-Angle of incidence measured from earth's d is the length of arc A"-B" Aug/Sept-24(R18), Qs(b)
Earth's Surface surface
Figure: (1) (or)
STUDENTS
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178 ANTENNAS AND WAVE 179
Explain about LOS propagation. In figure (2), PROPAGATION UNIT-5

/8 x 6.37 x
106 Let'sassurne we have an isotropic antenna, which
is an ideal antenna that radiates power equally in all
Ans: July-23(R18), Q11(b) h-Heighttof antenna A, from the eearth's 3
directions.
Space Wave Propagation h,-Height of antenna A, from surface, e10°x 16.96 (4h, +fh)
The transmission of microwave signals in a d -Horizontal distance from A' toearth's
C surface
The transmitter is radiating power P, (in watts).
line of sight link (network) is nothing but space wave d- Horizontal distance from A, to c
c4.12(/h, t/h) km This power spreads out uniformly in all directions
propagation. In other words, line of sight propagation :4=4.12(/h,tyh)km As a spherical wavefront.
is known as space wave propagation. Space wave d,- Total distance from Ato A, Ata distance d, the energy is spread over asphere
propagation is suitable for electromagnetic waves in Ris the radius of the earth R=6.37 x 10 km Where, with surface area
VHE, UHF and microwave frequency range, that is, for From right-angled triangle d-Line of sight distance (km)
waves in 30MHz to 300MHz range.
we have,
OAC of figure (2), h, h - Heights of transmitting and receiving an
A=4nd

When a transmitting antenna transmits a micro R This follows from the formula for the surface area
tenna (m)
wave signal to a receiving antenna which is in LOS, the R+h, R t .. (1) If theheights of transmitting and receiving anten of a sphere:
signal reaches the receiving antenna in two paths. One In practice a is very small. Then: for small nas, h, and h are in feet then, the line fo sight d A= 4
is direct path from transmitting antenna to the receiving of a. values is measured in miles. Where, r is the radius (here, dis the distance from
antenna and another is ground reflected wave or indirect
path. cos a l- bd From equation (7), /2R (/h, +h) the antenna).
2 As, r=3960 miles and 1 statute mile=5280 feet
Even though the direct and indirect waves begin On equating equations (1) and (2), we get. ...(2) Since the total power P, is distributed uniformly,
the power per unit area at distance d is:
at the same instance (at the transmitting end) they reach Then R'= R= x 3960
the receiving antenna at diferent instance. As a result, 1 Hence,
P .(1)
the strength of signal at the receiving end is the aggregate 2 Paminy 4nd
of both direct and indirect waves. The magnitude of the 2h,
resultant field is dependent on whether the two waves R
= a²
5280
+ h miles
5280
This is known as power flux density in watts per
square mneter (W/m²).
combining are in phase or out of phase.
= /2h, 4 Effective Aperture of the Receiving Antenna
Space wave propagation gets effected by earth's oF/2xVh, +/a,] miles
curvature, when the separation between the trans 2h, At this point, consider a receiving antenna that
d
mitting and receiving antenna is very large. When R = 2|/h,tyh,] miles captures some of this energy.
the separation is smallthe earth is assumed as a The amount of power received depends on the
flat surface. This is illustrated in figure (1). d= y2Rh, m ...(4) l4,=V2Vh, + /h] miles effective aperture A, of the receiving antenna. The,
R Similarly, by considering the right-angle triangle or d, =1.1412L/h, t h,]miles effective aperture is the area over which the antenna
OCA, of figure (2), we get, captures energy from the incoming wave.
Q15. Derive the expression for Free space
h d, = 2Rh,mstuadt ) propagation. The received power P, is given by:
From figure (2), d, =d, td,oish b Ans: July-23(R18), Q10(a) .. (2)
P=Pmi XA,
On substituting the values of d, and d, from Free Space Propagation
Earth Substituting P, in equation (1)into equation
equations (4) and (5) in the above expression we obtain. When an electromagnetic wave (such as radio (2), we get
Figure (1): Direct Ray Propagation d,= /2Rh, +/2Rh, waves, light waves, or microwaves) travels through
The radio horizon or line of sight distance lies d,= V2R(/h,+ /h,) ...(6) empty space (vacuum or air without any obstacles), P,=an) xA, As. (3)
beyond the optical horizon due to curvature of = 2x6.37x10° (Vh, ty/,) it follows the principles of free-space propagation. In
earth.
But, the effective aperture, A, is related to the
other words, free space propagation describes how
Therefore, line-of-sight distance, receiving antenna gain, Gas:
The effect of refraction is considered using effect electromagnetic waves travel through an idealized
radius of curvature of earth R. It is greater than the d, =3.57(/h, +\h,)kmh environment without any obstructions, reflections, or G,A
. (4)
actual radius of earth R. diffractions. 4T
Where, h and h are the heights of the antennas
Calculation of Line of Sight Distance Where.
in meters. The Free Space Propagation Model helps us
Consider two antennas 4, and A, elevated to a Since, effective radius of earth R' is taken, understand how the signal strength reduces as it moves G=Gain of receiving antenna
point Con the ground as shown in figure (2). equation (6) is modified as, farther from the source. The fundamental equation for
= Wavelength of the transmitted signal
B
...(7) free-space propagation is derived based on the spherical
d,= V2r(/h, t /h) spreading of energy from an isotropic source. Substituting this into the equation (3), we get
h Since for a standard atmospheric refraction, the
4 Derivation of Expression ...(5)
effective earth's radius is times the actual 4r
Transmitted Power and Isotropic Radiation Simplifying, the received power P, is given by
earth, equation (7) can be written as,
The formula gives us the received power P_ at a
/h) PG,A
Figure (2)
d,= V2R (4h, +yh)+ /2xRh,+
3 distance d' from the transmitting antenna. (4nd)
P= ...(6)

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180
Transmitter Gain
ANTENNAS AND WAVE
PROPAGATION UNIT-5

From AT'AB 181

transmitter is not
In practical scenarios, the d+ (h, +hy- d; (3) |E,1-E,2 cos
directs more energy M
isotropic; it has a gain G, meaning it h Similarly, from equation (3),
in a particular direction.
The powver density at distance d
now becomes: d, =d+4 th)?
2d
1E,|-2E, con (:0-a+B)
T .. (4)
PG, ...(7) The expression for phase difference 'a' is,
Piy 4rd
equation (6)
Using this power density equation, Phase difference (a) = 2r (path difference)
can be rewtitten as
2r |E,|-2E, sin
PG,G,? (8) Figure (1) i.e., a = (a,-a)
P= (4nd)
In figure (1), heights of transmitting and receivin 4nh, h,
Free Space Path Loss antennas are denoted by h, and h, Distanca
From equations (2) and (4), we get, |E,|=2E, sin 22d
Path loss is the ratio of transmitted power to between the transmitting and receiving antennas
received power, and it is often useful to express signal is d.
The energy received at the receiving point can
2 2d a-ndans
attenuation in these terms. This measurement helps
quantify how much a signal weakens as it travels through take two different paths. They are, a=
2rh h
1 Direct path 2d |E,|= 2E,sin
a medium, such as air or obstacles, making it essential
for analyzing and optimizing communication systems. 2 Indirect path. Effect of Field Strength with Distance and Height
Direct path is path from transmitting antenna to Since, the distance between the two antennas are
Path Loss (P) = .. (9) the receiving antenna denoted by T'R', whereas very large compared to the height of transmitting
indirect path is path from transmitting antennato 4rh, h, and receiving antennas, ie., d >> h, or h, then,
ground and ground to receiving antenna denoted = radians
Substituting the value of Pfrom equation (8) into by, T 'OR' . (5)
equation (9), we get the expression for Path Loss P, as, sin
(2rnh, h, 27h, h,
The heights of antennas are small compared to Since, ground is not a perfect conducting
(4nd)' the distance, direct ray suffers from negligible medium, phase difference is introduced due to
P, ..(10) the reflections.
G,G,A attenuation. 2rh, h,
Field Strength Calculation The total phase difference is the combination
or, in logarithmic (dB) form:
of phase difference due to reflection and path
To derive the expression for field strength, ground
Pn-20 log,, (4nd) 10 log,, (GG)') is assumed as a perfect conducting medium.
difference.
E, = E,4nh, h,
... (11) Here, .:. Total phase difference,
There fore, the expression for free-space d- Distance between the two antennas S=a+ß Where, E,-Field strength due to direct ray
propagation is, d,- Direct path distance 4th, h, 7/P
d,- Indirect path distance d + 180° (:B=180) E, =
PGG,1?
P= (4nd)2 h-Height of the transmitting antenna The field strength at receiving antenna (R')is, Where,
It indicates the amount of power received at a h,-Height ofthe receiving antenna. E,=E, (1 +ke ) P- Power radiated
distance d, based on the transmitting power P, the gains From AT'R'M' d- Distance.
Since, e 0= (cos -jsin0)
of the antennas Gand G, and the wavelength . d+ (h,-h)= d ...(1)
E,= E, (1 +k (cos 0-jsin 0) If E, is the free space field strength at unit
Q16. Sketch and explain the field strength d, -((h,- h +dy2 E,- E,(1+kcos 0)-j(ksin 0) diatance, then
variation of space waves, with distance E,
and antenna heights, deriving expressions |E,I-ENa+ kcos@)° +(k sin ) E,- d
for the same.
|E,l=Eol+k' cos +2k co se +k'sin' 9 Where, E, = 7/P V/m
Ans:
Then received field strength (E,) is given by,
Space wave propagation is very useful at ultra |E,l=EoJ+k'(cos 0+sin² 0) +2kcos8
high frequencies. Neglecting higher order terms, E,-E, 4Th, h,
Field Strength of Space or Tropospheric Wave |E,|=E,1+k² +2kcos e d

Consider a transmitting antenna and receiving

For a perfect ground, 7/P(4nh, h,
E,= d
antenna placed above the ground at distance of 15 k =1,ß= 180°
km above the surface of earth as shown in figure d, =d+ 4-h,)?
..(2)
|E,|= E,J1+l+2cos 0 88VPA,h, V/m
(1). 2d
|E,|= E,20+ cos0)
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182 183
ANTENNAS AND WAVE
Hence, field strength is inversely proportional to
the square of the distance.
Q18. Explain about space wave PROPAGATION UNIT-5

Scatter

and write the effect of the propagation

Variation offield strength with height is as shown
in figure (2).
the earth. curvature of Refraction
Ans: Diffraction
Vertical polarization
free space field Space Wave Propagation
NovJDec.-20, (R16), Q8(a)
For answer refer Unit-5, Q14, ,
Propagation. (3 paras only) Topic: Space Wave
A
’Earth

earaHeboveigtht -Actual earth

Effect of the Curvature of Earth
Figure(2)
For answer refer Unit-5, Q17. the carth. It can be measured using
signal towards
min 0 min =0 The refraction is defined as, bending of the transmitted
Q19. Explain briefly about the terms: value in free space is given
by,
i for o< 0 for o = o ofractive index n', whose
(0) Absorption
Perfect earth (ii) Super refractions. n=l+
Ans:
Where,
() Absorption: Absorption is defined as the Tis absolute temperature
E. 2E, transformation
E(uV/m) of some part of the transmitted
sienal Pis air pressure
energy to the components of the atmosphere
Figure (2)
energy can be retransmitted to the other
and this
t a t WiS partial pressure
of water in milli bars.
Q17. VWrite short notes on effects of earth cur components of relation between these parameters ('n' and N) is given
the atmosphere. 'n' can be replaced by modified index 'N', and the
vature.
As the atmosphere consists of by:
Ans: and particles, they absorb and retransmit various gases
the
Effects of Earth Curvature on Space Wave Propagation wavelengths of the radiation and the wavelengthsvarious
which N=nt
are not absorbed by any of the particles in the atmosnhere
The direct wave may or may not reach the receiv establishes a "atmospheric windows", which is shown
ing antenna and the reflected wave from ground Where a' is radius of the earth = 6.37 x 10 m
in figure (1).u)
diverges as shown in figure (1). As a result, the Windows through he Earth's atmospbere Gradient of 'N' is given by,
strength of the signal received by the receiving X-ay UV Vsible Inirared
80 dP 809600w \dT 80x4800 dw, 10°
Radio
antenna is very weak. dh
T dh T dh a

dN
Grund reflecte) ray
In the above expression when value goes beyond - 157 N units, super refraction occurs and is defined
Direct ay
as bending of the refracted signal morhan the curvature of the earth.
Q20. Discuss about the atmospheric effects in space wave propagation.
Ans: Dec.-19, (R16), Q11(b)

1A 10A 100 A 1000 A 10000 A O01 mm 01 mm 1 mm 1 cm 10 em 1m 10 m 100 m

32ul The atmosphere has several constituents like gas molecules and water vapour. It makes the dielectric
Wavekneth
constant of moist air greater than unity as the dielectric constant of water vapour.
Figure (1) The dielectric constant and the refractive index of air is given by,
Figure (1): Atmospheric Windows
2 Earth's curvature produces shadow zones. These n e,
The graph shown in figure (1) indicates the
are the zones into which the rays transmitted by Since the density of air and water vapour varies with height, the refractive index of air and dielectric constant
the transmitting antenna cannot enter. This is transmission oftheenergy into the atmosphere, in which varies with height.
illustrated in figure (1). Shadow zones are also the darker part indicates the transmission of energy
whereas valleys indicate the attenuation of the energy. Typically for a standard atmosphere, the dielectric constant is assumed to decrease uniformly with height.
known as diffraction zones. e However, these conditions give rise to phenomenon like reflection, refraction, scattering, fading etc.
3 The curvature limits the distance of propagation Basically, there are three gases in the atmosphere 1. Modified Refractive Index
of space waves. due to which absorption occurs i.e., water vapor, carbon
dioxide and ozone. This absorption' phenomenon is It is the sum of the refractive indices at a given height above the mean geometrical surface and the ratio of
4 As shown in figure (2) the actual and effectiVehelnful
antenna height differ and the field strength at the
in two ways to the earth i.e., the height to the mean geometrical radius. It is expressed as,
receiving antenna gets modified as in figure (2). () It stops the high-energy radiation signals
R
from reaching the earth. x10
Where,
(ii) Atmosphere acts as the heat source to the -Height above ground
earth.
(i1 Super Refraction: Consider two antennas i.e., R- Radius of earth
transmitting antenna and receiving antenna. When the It is introduced to study the characteristics of radius waves and troposhere.
signal radiates from the transmitter, it can undergo into
Figure (2)
three mechanisms as shown in figure (2). The atmospheric refraction depends on the change of modified refractive index, with height i.e., du,m

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 AND WAVE
184
ANTENNAS
PROPAGATION UNIT-5

Inorder to achieve greater coverage, the slope of 185

increases by 0.04S units/ft. 3
Athigh altitudes, u M-curve increases near the surface of the earth. Troposheric layer
earth.
uincreases linearly at a constant rate near the Greater coverage can be achieved when the rays Truposbertic layer
4
=0.036 units/ft. This condition is known as standard Over flat earth are straight and actual rays have
For typical atmosheric conditions, dh
(1),
atmosphere. the same curvature as that of the earth.
Evsted Guct
surface of earth are as shown in figure
The varations of u_with height near the Sarce ct
h h h h
5 Duct propagation occurs when the rays are curved
downward over the flat earth and the wave tends
Figure (1;: Duct propagation
dñn -0,.1181 to be guided along the duct.
dh per meter h, The rays are trapped between lower and upper
6 Ifthe inverted portion ofM-curve is elevated above walls of the duct, oscillating ground and upper
the surface of earth then the duct is an "elevated wall ef the duct in ground based duct. Whereas
duct". inelevated duct, the rays are oscillating between
(b) Refraction at (c) Simple surface (d) Elevated
(a) For standard
duct 022. Explain phenomenon of Duct propagation. the two walls in the atmosphere. This process is
atmosphere lower heights duct
(ground based)
known as duct propagation or super refraction
Ans: July-23(R18), Q10(b)
Duct is similar to leaky wave guide since some of
Figure (1) Duct Propagation the energy within the duct may escape into the space
Refraction of Waves The waves at UHF, VHF and microwave frequen as shown in figure (2).
cies are neither propagated along the surface of Energy Leakage
Due to the variations of refractive indices with height, the waves do not follow astraight path from transmiter earth nor reflected by the ionosphere. However, in
to receiver antennas. They follow a bent path ie., the curvature of earth.
the tropospheric region, the waves are transmitted
The radio waves travel to reach point located away than the optical horizon as shown in figure (2). beyond the LOS distances due to refraction.
High dielectric constant region
Transmitting
A normal or standard atmosphere is one where antenna

the dielectric constant is assumed to decrease

Earth surace
Low dielectric constant region uniformly with height to a value of unity at a Figure (2): Ducts as a Leaky Waveguide
height at which air density is considerably zero. Ducts to not allow the signals whose wavelength
Several layers or regions are formed over one are higher than max
Earth another in atmosphere with different temperatures Where,
and water vapour contents.
Figure (2) Ama =0.0584 P2
In the region where the frequency is adequately
The temperature inversion zone causes super refraction or duct propagation. max -Wavelength in cm
To correlate the earth's curvature and the curvature of MW beam path one of them is taken as a straight line. high and the value of gradient, dñ is negative, d- Duct height (in m)
dh
The modification of earth's curvature is done by using k-factor or kell factor. Q23. Write short notes on the following:
the curvature of the earth along which the waves
By multiplying k-factor to the actual earth radius, it gives the modified earth's radius which is used to replace (a) Duct propagation
propagate is slightly higher than that of earth. (b) Fading.
the curved path by a straight line.
Due to this, the energy of the wave is trapped Ans:
G,= ka = 1.33a Nov/Dec.-18, (R15). Q1
and makes it travel along the surface of earth
Q21. Nrite short notes on M-curves. in these regions. This phenomenon is known as
(a) Duct Propagation
duct propagation For answer refer Unit-5, Q22.
Ans:
(b Fading
M-curves Duct is a region where electromagnetic waves
Fading is defined as the lowering of the signal
are guided by two boundary surface, typically strength with the variation in the refractive index.
M-curves is known as modified index curves. These curves show the variations of refractive index with between layers of air with different refractive Variation in the refractive index occurs due to the
height. In order to account for the curvature of earth, the actual index of refraction is modified to another indices.
refractive index. abrupt change in the temperature, pressure and
The waves are refracted in duct and transmits EM humidity. Fading phenomenon having Rayleigh
Due to the change in the index of refraction, the straight rays are converted into curved rays above flat earth. waves entering duct region around the curvature nature is as shown in figure.
beyond LOS distances i.e., for distances with low A-40
The effects of non-standard atmospheric conditions can be estimated easily by transforming temperature loss.
data, meteorological data into M-curves.
du (PodwB)er
M-curves are used to predict the type of transmission path for propagation of electromagnetic waves. Ducts are formed in regions whose dh
is nega
Characteristics of M-curve tive. Temperature inversion is one of the impor
1
tant factor for formation of duct.
Standard propagation occurs when the modified index of refraction linearly varies with height. M-curve 1S -67L 20 25
05
a straight line having positive slope. A duct can be formed between two layers of Tme in seconds
2 troposphere or in the middle of troposphere and
The slope of M-curve decreases near the surface of earth which results in Figure: Fading Phenomenon
standard propagation. earth's surface as shown in figure (1).
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 ANTENNAS AND WAVE UNIT-5ARR VAw
186
Generally, the radio waves bend
PROPAGATION 187
Fading can be fast / slow path / single path ordifiract
the curved surface of earth. The strength around
The expression for maximum range (in miles) is
5.4 SKY WAVE PROPAGATION
multi-path / short-term /long-term. of given by,
diffracted wave in shadow zone depends on
The time duration required for the fast or multi roughness of earth. Further, the ionosphere hasthe - /2h, +/2h, 5.4.1 Structure of lonosphere, Refraction
in
path propagation is 0.01 sec. And the signal continual air turbulences. and Reflection of Skywaves by
long-term fading varies in the order of 10 dB. =J2x100 +2x50 =200 +10 lonosphere
Due to the turbulent eddies, discontinuities ~3
In general the effect of signals strength in summer rOughness of earth there are fluctuations in refran = 14.14 + 10
Q29. Explain the structure of ionosphere.
is 10 dB greater than that of the winter. tive index of atmosphere. Hence they may scatter daa =24.14
Similarly, the strength of the moming and evening in all directions when enter the atmosphere (or)
..d max = 24.14 miles Write short notes on D-Hayer and sporadic-E
signals is 5 dB greater than that of the afternoon The scattering mechanism decreases at greater
signals. 027. Calculate the maximum distance at which layer.
height. So far better intensity of scattered waves
Due to the fading phenomenon there is a change high power transmitting and receiving antennas signal from transmitting antenna with May-19, (R16), Q10(b)
of sudden interruption in the communication and of high gain with their beams intersect midway 144 m height would be received by the Refer Only D-layer and Sporadic E-layer)
the techniques used to solve such problems are between them and above the horizon are used. (or)
called diversity techniques, which are classified as, receiving antenna of 25 m height.
Scattering is random in nature and the scattering Ans: Briefly explain about D-region.
space diversity, frequency diversity, time diversity, signals has continuous amplitude and phase
modulation diversity and polarization diversity. fuctuations over a wide range of frequencies, Given that, (Refer Only D-layer)
Ans:
Q24. Explain about scattering phenomena. The scattering single or cone angle 'a'that depends Height of receiving antenna, h_ =25 m May-18, (R15), Q10)
Ans: on wavelength must be as small as possible. The different layers of ionosphere are as shown
t Height of transmitting antenna, h, =144 m in the figure (1).
Scattering Phenomena: The waves passing through Forward scattering propagation phenomenon has Maximum transmission distance, d_ =? Outer Atrmospbere
ionosphere and troposphere get scatter due to the con the following two theories. al The maximum line of sight transmission distance is GRegion
tinuous turbulence in the region. If the wavelength of 1. From 500 MHz with Troposphere as Scattering
the waves passing through these regions is more than the Medium: In this mode, scattering is caused from given by, F, Layer
turbulent eddies, then waves scatter in all the directions. either blobs or fine layers in the troposphere. It
However, if thewavelength is smaller than turbulent ed ranges from 300 km - 600 km. This scatteringis
= 2x6.4x 10 V144 +V25
250-400km

F, Layer C 250-400
dies, then wave scattering is confined only to a narow called tropospheric scattering. E Layer
(.Radius of earth, R= 4 x 10 m)
140-250km

cone in the forward direction of propagation. The scat 2. 30 to 50 MHZ with Troposphere as Scattering D Layer
tering phenomenon is as shown in figure below, Medium: It is assumed that scattering occurs in = 60821.04 m 9 0 - 1 4 0k m

Tropospbere 220
E layer of ionosphere witha maximum range of
S 0 - 9 0k m

Man energy

Scattering ang k 8
about 2000km.
The tropospheric scatter propagation is illustratéd
= 60.82 km
d=60.82 km
Earth
T7san
Figure (1): Layers of lonosphere
Scatterig sigTal in figure below. Q28. To produce a power density of 1.0 mWN/m?
in a given direct at a distance of 2 km an lonosphere is the upper most layer of earth's
TIras
No scattering antenna radiates a power of 180 W. An atmosphere. It ranges 70 km to 400 km above the
200
Receiver Lost isotropic antenna would have to radiate surface ofearth. In this region ionization takes place.
Earh
scater Scatter
volume
2400 Wto produce the same power density Characteristics of lonospheric layers
at the same distance. What is the directive D-Layer
Figure
Longest path gain of the practical antenna in dB? It is the lowermost region of the ionosphere and
Since, scattering phenomenon is a random Shortest path Forward scatter
Ans: lies in the range of 50-90 km.
process, the amplitude and phase of scattered waves is Given that, This layer exists only during the day time. It
not constant and changes continuously. Thus, in order to Receiving vanishes during night time due to high rate of
Transmitting Power density =1 mW/m? combination.
receive and transmit such scattered waves present beyond antenna TI
Back
antenna

the horizon, the gain of transmitter and receiver antenna scatter Distance = 2 km D-layer is responsible for day time attenuation
must be very high. In addition, the antennas should be Power radiated= 180 W of High Frequency (HF) waves.
Figure: Tropospheric Scatter Propagation
oriented in such away that they overlap in the forward Tropospheric Ducts h Power radiated by reference antenna=2400 W
It reflects Very Low Frequency (VLF) and Low
scattering region. Frequency (LF) waves.
Q25. Write short notes on tropo scattering For answer refer Unit-5, 022. SVSWw Directive gain =? It acts as ahuge waveguide at VLF for high power
propagation. Q26. Find the maximum range of tropospheric oiFor same power density at the distance of 2 km, transmitters.
transmission for which the height of the directive gain in dB is calculated as, The range of electron density is 104 to 1016
(or) transmitting antenna is 100 ft and that of per cm and at noon the ionization density is
Explain about tropospheric wave receiving antenna is 50 ft. Power radiated by reference antenna maximum.
propagation. tajbto F10log Power radiated by practical antenna
Ans:
Ans: D layer is formed by ionization of oxygen (0,)
Dec.-19(R16), Q11(a) Given that, 2400 molecule at its first ionization potential.
Tropospheric scatter propagation is used for te =10 log
For tropospheric transmission, 180 When radio waves are propagated in this layer,
propagation ofVHE, UHF and microwave signals Height of transmitting antenna = 100 ft they are attenuated since there is insufficient
beyond the radio horizon or line of sight distance, =10log[13.33] =11.249 electron density to cause bending (reflection) of
It is also known as forward scatter propagation Height of receiving antenna = 50 ft
:. Directive gain 11.249 dB these waves. Hence Dlayer is also referred to as
and troposcatter. Maximum range (d.)=? absorbing layer.
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 aVAW OHA 8ANWSTKA
U N I T

Troposphere region in atmosphere within 16 km

- 5

2
188 ANTENNAS AND WAVE PROPAGATIOA of Earch Abnormal 189
Maximum HF waves are penetrated throueh ahove the
surface
signall at the These are caused
E-Layer
pointis
wave propagation,
acombination of fdirect
receiving
and indirect rays. cycle, fading and due tides and winds, Sunspot
F-layer, even though some of them are
space

This layer behaves as a miror that turns the

back.
reflected In
over long distances Few significant whistles.
electromagnetic waves towards the earth's provides
communication
are as follows, factors effecting the
surface. The main function of F,-layer is to provide mors It
with
VHE UHF and microwave frequencies. Tides and Winds propagation
During day time, at a average height of 110 km, it absorption for HF waves. propagation is also known as "line
wave
has maximum density. Its range is of90km - 140 Space Tidal pull_ are
The density of F,-layer is lower in winter than eve due to the sun experienced
o fs i g h t p r o p a g a t i o n " .

by the atmosphere
mer, even though no great variations in height field strength of receiver depends on: and
km.
Its ionization becomes weak during night hours.
The

trom transmitter isolation period withmoon. The coincidence

the period leads of
F, Layer
F, layer is the upper end region of F-layer with (i)
Direct ray
Ground reflected
ray resonance phenomenon. tidal to
Range of electron density in day hours is 10 to () refracted rays from the atmo- more significant and This turm out to be
4.5 x105, during night it is 5 x 10 to 10. a height range of 250 km to 400 km.
(i)
Reflected and intricate
heating of the atmosphere through themal
Critical frequency ranges from 3 MHz to 5 MHz. Its critical frequency ranges from 5 MHz to 1) sphere has a24 bours time via the sun rays that
E-region is due to ionization of all gases by soft MHz (basically 10 MHZ) and may be even more
(iv)
Diffiracted rays around the curvature of earth the tidal period. peniod, whichis two times
X-ray waves. at low altitude stations. and so on.
The F2 layer possesses
Refecting HF signals during day hours is the The electron density of F,-layer may varies from S k y - w a v e P r o p a g a t i o n

of tidal motion with the maximum speed

lowest
main function of E-layer. 3 x10 to 2 x 10°. This type of propagation is important at medium located at the topmost level.particle density
Sporadic E-layer F-layer is formed by ionization of UV, X-rays and higher frequencies. The frequency range ofit layer experience maximumTherefore, F2.
The region where anomalous ionization occurs and corpuscular radiations and it disappears very sky-wave propagation is 2 to 30
MHz. Hence tides and cause abnormality effect from
in the ionosphere is known as sporadic E-region slowly. is also known as short wave propagation. and a small peak of maximumin its layer
The sky-wave propagation is also known as density at midnight. ionization
(Es). This layer does not follow Chapman's law of
Appearance of this layer is rather uncertain that variations. jonospheric wave propagation because the waves Fading
are reflected from the ionized layer. The waves The occurrence ofreduction of signal strength
may exist during day hours as well as night hours. This is the most imnportant refiecting medium for cannot reflect back to earth if the frequency is
The ionization density of this layer is high and at the receiving end because of
high frequency radio waves. unnecessary
is a very thin layer that is located at a height of
above 30 MHz. variations and fluctuations caused to eleva
Elayer with a difference of 5 to 10 km. The electron density variation during day and Ionospheric region is just 50 km to 400 km above tion and thickness of ionosphere in diverse
night are as shown in figure (2). the earth's surface, instead the waves penetrate atmospheric layers is called fading. It occurs
During day hours this layer appears in equatorial
zone. In case of polar regions it appears during 250 -400 km--------)r, Layer the ionized region. only at the receiver end as different signals
F-layer The sky-wave propagation is suitable for long move in diferent paths to arive at the receiver
night. As geomagnetic latitude increases, the (Appleton
distances point to point communication. Hence, interfere causing fluctuations in strength of
appearance of this layer also increases. layer) i St 220------F, Layer
This layer is useful in propagating VHF signals E(Kenellye it is also known as point to point propagation. received signal ie., fading.
heaviside) The waves reach the ionosphere and reflect back
to a long distance through scattering. Layer
Fading is a significant parameter in sky
10 Te2 50-90-D LayerE to the earth either in single or multiple hops as wave propagation and arises because ofre
F Layer flections from the earth. Fading is extremely
shown in figure below.
F-layer is located at height of 140 km - 400 km 0
The sky-wave propagation suffers from fading. small when the variation in signal strength
above the surface of earth with an average height Electron density (N) Electron density (N) is in between 20 to 30 dB. Fading can be
of 270 km. The fading can be reduced by diversity reception.
Figure (2) decreased by means of diversity reception.
Salient Features of Sky Wave Propagation
Appleton Barnett layer or F layer is split into F, G Layer (Outer Atmosphere) (Above 400 km)anA Sudden Ionospheric Disturbances
Sky wave propagation takes place due to the re With an abrupt manifestation of strong solar
layer and F, layer which combine at night. The region above 400 km is termed as G region. flected or refracted back waves to the earth from
In F-region, horization access both during day and flares there is an intense augment in D layer
Radiation belts encircling the earth occupies this the inosphere. ionization. An enhancement in the D-layer in
night with maximum electron density of 220 km Medium and short wave frequencies can be
with critical frequency of 5 MHz to 12 MHz.
region. creases the absorption of high frequency sig
refracted back to the earth. This makes medium results in a total
In this layer, charged particles are present that are nals passing through it. This
F, Layer and short wave frequencies helpful in the trans collapse ofall high frequency signals
known
F, layer is the lower end region of F-layer and at aligned similar tothe structure of magnetic lines continental transmission of the waves. Disturbances":
of force are captured byy magnetic field. as "Sudden lonospheric
an average height of 220 km. (generally, 140 km Critical frequency fluctuates as per atmospheric Q31. Explainthe process of refracion andretlecton
to 250 km) in summer and 150 to 300 km during Q30. Discuss the salient features of sky wave conditions, time of day and fire angle ofthe radio layers.
winter. propagation. Bring out the various waves by the antenna. of radio waves via ionospheric
(or)
problems associated with this mode of reflection and
The behaviour of F-layer is similar to that of Satellite communication is carried-out using sky
o Discuss the Mechanism of
E-region (normal) and obeys the Chapman's law propagation. wave propagation because it is dependent on the refraction of sky waves
by ionosphere.
FebJMarch-22(R18), Q7(b)
of variations. Ans: [Aug/Sep.-24(R18), Q10(a) | May-19, (R16), Q10(a)] upper atmosphere. Ans:
Its critical frequency ranges from 5 MHz to 7 Salient Features of Space Wave Propagation: roblems Associated with Sky Wave Propagation Refraction bending ofa wave
MHz at noon time. 1 Space wave propagation is practically important The abnomalities ofionosphere cause dificulties Refraction is defined as the into another. This
medium
The value of electron density varies from 2 x 10 at frequencies above 30 MHZ. ne propagation of radio waves through it. Usualy the When it
moves from
one
when the wave
undergoes a
to 4.5 x 10. 2 It is also known as tropospheric wave propagation abnormalities are of two types. phenomenon is observed
properties ofthe
Normal speed due to the varying
F-layer is formed by the ionization of oxygen because the waves reach the receiving point after change in its propagates.
atoms, due to an accepted view. reflections from tropospheric region. These are caused due to variations in height as which it
two media in
STUDENTS
thickness, seasonal changes, and noise.
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 AND WAVE
190
is as shown in (1
ANTENNAS

Consider a plane wave beettraveling

PROPAGATI
along
ON UNIT-5

Where, 2 For o 0, a negative dielectric

191
The general refraction mechanism X-axis. positive Ne?
-COS Ot ...(6) obtained resulting an imaginaryconstant is
figure (1). Then the electric field E of the wave hove
mo
constant. Hence the propagation
component only along the x-axis. a From Maxwell's equations, incident wave is totally
Refacted reflected.
Incident waves
i.e., E=E, a, VxH= J+ = gE + & ..(7)
3 For o >> 0,, 8, ~1. Hence the wave
waves
Where, and
propagates
through the plasma with small refraction.
E,= E sin ot Critical Frequency
Figure (1) VxE= OH
In sky-wave propagation, the wave propagates
And Eand E, are zero i.e, E, = 0, E =0. . .(8)
Critical frequency (f) is defined as the highest
in the ionosphere using the following two properties, Also, thereeis only one componenttof H Then by taking assumption (1) into consideration, frequency which can be reflected from the ionospheric
reflection and refraction. The reflection and the refraction y-axis. along the layer at vertical incidence.
•E, OH and
depend on the frequency of the radio waves. So, the effect i.., H=H a, Jt e, ... (9) When the frequency of electromagnetic wave is
of ionosphere on the waves in three different frequency dz
and H=0, H =0 greater than.the critical frequency, the wave is reflected
regions, low frequency waves, high frequency waves, •H, -OE,
(ii) The wave is propagating in the from that particular layer.
and mediumfrequency waves is as follows, consists of free electrons. region that dz ... (10)
Low frequency Wave: When waves at low On substituting equation (4) in equation (9), we Critical frequency is denoted by f (or) f. For a
Then, the force exerted on each particular layer, critical frequency is proportional
frequencies approach the ionosphere, the changes
in its electron density and ion density are very electric field E is given as,
electron by the obtain,
square root of the maximum electronic density in the
to the
- Ne?
large. So the layer through which these waves F,=-eE, -E cos ot+ Eo (E sin ot)= -OH, layer i.e.,
propagate behaves like a sudden discontinuous ..(0)
Because of this force, the electrons get acceleratei
mo
Sin i
medium. Thus, the layer behaves like a dielectric - Ne .. (1)
in the x-direction. E cos ot+ Eo 0 E cos ot = sin r
medium and 'reflects' the low frequency waves mO Where,
back to the earth's surface. Hence, these waves From Newton's law, we have,
F= ma
Ne? -OH, epsH- Refractive index
experience losses. oE cos ot =
o xei-Angle of incidencetet
(ii High Frequency Waves: At high frequencies d'x
the waves have shorter wave-lengths. Hence, the Ne? -YH - Angle of refractionnoes
&, oE cos ot =
changes in the ionization density are so small that
Equation (1) becomes, mo&o N- Electron density
they can bë neglected. Hence at high frequencies,
the ionosphere behaves like a dielectric with d'x
=-eE But, 1
Ne
...(11)
f-Frequency.
changing refractive index.
m
From the definition of critical frequency, u=0.
(at vertical incidence).
(iii) Medium Frequency Waves: At medium d'x E, E0E cos ot = OH,
frequencies, the ionosphere can be well thought m =eE sin ot
...(2) 0=
81 max
N.
of as a set of discrete thin layer having different From equation (11),
refractive indices. All these layers have constant On substituting initial zero conditions in equation
Ne? 8IN mas =l
ionization densities. And also the ionization (2), we get, 1

density of each layer differs from the other. So, mo»[,

An angular plasma frequency can be defined such f =81 Nme
as the wave propagates through the first layer, m=-eE
dt sin o r t l
it gets refracted partially and moves on to the that,
second layer. This process continues and there dx Ne?
are series of refractions as the wave propagates dt m Esin or m,En .. (12) Where,
through the layers of ionosphere. This results in Which is purely a function of electron density N. f-Critical frequency (MHz)
large attenuation of the wave and hence it is not dx N Maximum electron density (m)
..3) Then equation (11)can be written as,
used in long distance communication. dt mo 2 Q33. Derive the relation for dielectric constant
The propagation of wave in ionosphere through Let the number of electrons per cubic meter in of ionosphere layer in terms of plasma
this mechanism is as shown in figure (2). ¬, =1 ..(13)
space be N, the charge on each electron be -e and mass frequency.
lonosphere be m. Ans:
Then, the current density J is given as,
Neglecting the electron collisions, equation (12)
Refracted waves is reduced to, For answer refer Unit-5,.Q32 (Refer Upto
J= - Near Equation (11).
R dt 81N From equation (11 ),
...(14)
Figure (2): Sky-wave Propagation On substituting equation (3) in the above Ne'
Q32. Derive an expression for effective dielectric As the propagation constant of a wave travelling mo[
expression, we have, in plasma is &r ko, the following conclusions can be
constant and critical frequency of ionos phere An angular plasma frequency can be defined such
J= - Ne -o
Ecos . .(4)
made. that.
layer. mo 1,
Ans: Aug.-22, (R18), Q8(a) For o>,, 8,<1, with real propagation constant. Ne?
In order to derive the expression for refractive The current density is given as, Hence, wave gets reflected by plasma can be ...(12)
m,Eo
index of ionosphere, consider the following assumptions, J= oE ...5) made.

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 UNIT-5O
192 ANTENNAS AND WAVE PROPAGATION
Which is purely a function of electron density N. Substituting the values of f and 'n'in As angle of incidence, , decreases. 193
Then equation (11l) can be written as, equation (2), we get, etration of wave increases as ray-2, Hence,
3 and 4
pen 2 At lower frequencies, the waves reflect back to
If 6, decreases nore, the wave
does
Earth from the ionosphere, and the ionosonde
...(13) 179.6x10) n= sin not satisfy records the time taken for the signal to
Hence, it penetrates and moves out of return.
Neglecting the electron collisions, equation (12) 1-(0.5) the layer as ray-5 and 6. 3
As the frequency increases, at a certain
is reduced to, f=/32241.16 x 105
1-0.25
The distance of which the wave
return decreases waves will no longer reflect and insteadpoint, the
8IN until , =, where _ is the critical into space. escape
s, =1 - 1 - ... (14) f=32241.16x106 minimum distance of wave return angle and the
is known as 4
As the propagation constant of a wave travelling 0.75 skip distance. The highest frequency at which the
still reflected back is recorded signal is
in a plasma is e,ko, the following conclusions can be V42988.213 x105 When ,<, the distance ofwave as the critical
made.
return initially
increases and then penetrates the layer frequency.
f= 207.33 x 103 Hz as ray-4.
For o>,e,< 1, with real propagation constant. 036. VWhat information is Critical Frequency regarding the
1
Hence, wave gets reflected by plasma based on ..f =207.33 KHz conveyed by it Ionosphere
regarding the ionosphere? Describe a It helps scientists monitor the
ionosphere, which
E, variations with amplitude. 5.4.2 Raypath, Critical Frequency, MUF method to measure critical changes due to factors like the Sun's
LUF, OF, Virtual Height and Skip frequency. radiation,
2 For o < o,a negative dielectric constant is Ans: time of day, and space weather.
obtained resulting an imaginary propagation Distance, Relation between MUF July-23(R18), Q11(a)
constant. Hence the incident wave is totally and Skip Distance, Multi-hop Understanding Critical Frequency and Its Relation It is used to predict radio
communication
Propagation to the lonosphere conditions, For example, whether
radio signals can travel long distances.shortwave
reflected.
3 Q35. Explain Raypath. Critical Frequency
For o>> 0,, E,~1, Hence the wave propagates
through the plasma with small refraction. In simple terms, critical frequency It helps in understanding space
Ans: weather, which
Q34. At what frequency a wave must propagate highest frequency of a radio wave that can be) is the affects satellite operations and GPS accuracy.
Raypath is the path followed by a wave. In other back to Earth by a layer of the reflected Q37. Explain in brief about the
for the D-region to have an index of ionosphere following
refraction 0.5? Assume that the lonic
words, it is the path of the wave from transmitting
end to receiving end. straight up (vertically). If the frequency of the when sent with respect to wave propagation terms
density of D-layers is 400 electronslcm. is higher than the critical radio wave (a) Critical frequency
Ray path is classified into two types, instead of being reflected.
frequency, it escapes into space
Ans: Aug./Sep.-24(R18), Q11(a) (b) MUF
1
Give that, Actual Ray Path: The actual ray path of electro Importance of Critical Frequency (c) Skip distance
Refractive index of D-region, n = 0.5 magnetic wave in troposphere is curved due to the The ionosphere is a region of (d) Virtual height.
changes in refractive index of the transmission atmosphere, about 50 to 600 km above the Earth's
flled with charged particles (ions and freethe surface,
Ionic density of D-region, N= 400 electrons/cm' medium. May-18, R(15), Q10
(or)
The refractive index (n) of the ionosphere is given 2 Straight-line Ray Path: Here the wave is as These charged particles reflect radio waveselectrons). Write a short note on:
by sumed to be travelling in a straight line neglecting Earth, allowing long-distance communication back to () Virtual height
the effects of refractive indices of the media. satellites. without
(i) Critical frequency.
n= ... (1) The six different ray paths followed by a wave If a radio signal's frequency is below the
Where. under various cases is illustrated below. critical Nov/Dec.-20, R(16), Q8(b)
frequency, it bounces back to Earth after hitting (Refer Only Topics Virtual Height and Critical
f= Plasma frequency (Hz) n the ionosphere. Frequency)
f= Wave frequency (Hz) If the frequency is above the critical (or)
frequency,
it passes through the ionosphere into Derive the expression for skip distance.
n=Refractive index space,
meaning it cannot be used for long-distance AugJSept 21R(18), aBa)
Rearranging equation (1) for f, we get, Upper(Pedestrian) communication. (Refer only Topie: Skip Distance)
Ray
So, critical frequency helps in understanding how (or)
V1-n? ... (2) Write a short note on,
Lower Ray
radio waves interact with the ionosphere and in designing
Plasma Frequency radio communication systems. (a) Virtual height
T>
The plasma frequency ) is calculated using Skip distance 1 Method to Measure Critical Frequency (b) Critical frequency
Skip distance 2 (c) MUF.
S= 8.98yN (3)
n,<n,<n,<n, <n, <n, One of the main methods used to measure critical
Feb JMarch-22, (R18), Q8
Where N is electron density in electrons/cm, Figure: Schematic of Raypaths Under Different Values of f. frequency is iono_onde sounding,
Substituting N= 400 electrons/cm in equation and Angle of Incidence An ionOsonde is a special radar-like instrument (Refer only Topise: Virtual Height, Critical
Frequencey, MUF)
(3), we get, The effect of ionosphere is influénced by the angle used to study the ionosphere. It sends short pulses of
radio waves vertically upward at increasing frequencies. (or)
f- 8.98 /400 of incidence , whenf>f. Explain critical frequency.
For large value of the wave follows the relation The working of ionosonde is as follows
= 8.98 x 20 = 179.6 KHZ (Refer Only Critical Frequency)
n= sin ¢.Hence, when n < the wave returns 1 The ionosonde transmits radio pulses starting Dec.-19, (R16), Q10)
..f, = 179.6 KHz after slight penetration as ray lo1 from a low frequency and gradually increasing. (or)
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194
ANTENNAS AND WAVE PROPAGATION UNIT-5 195
Explain maximum usable frequency Where, When angle of incidence,ntG
(MUF) and skip distance. =sin i= hßIN

(Refer Only MUF and Skip Distance) SUF f-Critical frequency (MHz) ’ 8,>0, sky wave foils to reflect back
Ans: Dec.-19, R(16), Q10(b) N -Maximum electron density (m) ’0<0, waves are received beyond point B
1 Virtual Height Then, sini=
From equation (2) it can be observed that, as the ’ 0<0, waves return to earth at point B.
In long distance communication, as the sky wave angle of incidence decreases, the electron density Q38. Explain the terms and their significance
reaches ionospheric regions, the wave enters rarer sinj = 1 increases and reaches maximum when i= 0.
medium from denser medium. It starts bending with related diagrams:
down through the process of refraction. This is 4 Skip Distance (D)
(a) LUF
shown in figure (1). = cos²i (:1 -sin²) ...(3) Skip distance is the shortest distance measured (b) OF.
Ionosphere (D) from the point of transmission (transmitter) to
the point where it (sky-wave) first reaches back Ans:
Lower edge of dhonl..far=f seci orlt oilspo ...(4) to earth.
Projected ionosphere (a) Lowest Usable Frequency (LUF)
height Virtual height Equation (4) is known as secant law. It is the minimum distance at which the wave
The Lowest Usable Frequency (LUF) is that
True or (h) This equation is valid up to a distance of 1000 returns to the ground at a critical angle, 0.. frequency below which the entire power of the
Actual path
acutal (h) Actual km. The angle of incidence for F-layer is 74 The skip zone is defined as the zone or coverage wave gets absorbed by the ionosphere. Thus,
height path then maximum usable frequency, area from where sky-wave is almost in active, LUF is a lower limit below which there is no
negligibly present or very weak for reception to communication between two stations as the
Sar = Sec74° xf waves get lost.
the point where it first reflects back to earth.
R The Lowest Usable Frequency (LUF) depends
D Ground wave propagation and skip distance
3
on the following factors.
Critical Frequency determines the size of skip zone.
Figure (1) The antennas receiver sensitivity transmit
The vírtual height is defined as the height to which Critical frequency () is defined as the highest The skip distance can be better explained by using ter powers, surrOunding noise level of the
frequency which can be reflected from the figure (2). communication stations.
transmitting a short pulse with velocity of light ionospheric layer at vertical incidence.
reaches taking the same two rays travel time as The type of modulation used
the actual pulse reflected from the layer. When the frequency of electromagnetic wave is lonosphbere The characteristics of ionosphere along the
The value of virtual height is always greater than e greater than the critical frequency, the wave is laye transmission distance.
the actual height. reflected from that particular layer.
The absorption of the waves is because of the
The height at a point above the surface at whích Critical frequency is denoted by f (or)f. For a D-layer of the ionosphere as explained below.
wave bends down the earth is called actual height. particular layer, critical frequency is proportional When compared with that of the higher layers, the
Below the íonosphere the íncident and refracted to the square root of the maximum electronic
density in the layer i.e.,
mitte
Earth mum D-layer has higher number of neutral atoms and
rays follow the paths exactly that would have Ground waveSkip zone
surface
molecules. The wavelengths at lower frequencies
been followed by them if thereflection had taken sini 81N ange
are long. Because of this, the electrons may
from a surface at greater height (point B) known ...(1)
as virtual height.
sinr f Skip distance
travel a greater distance and collide with the
Where, Figure (2) neutral atoms resulting in loss of energy. Thus,
2 MUF - Refractive index at frequencies below LUF, the wave may get
Maximum usable frequency, fon defined as the From the figure, when a wave is transmitted at an
maximum possible frequency at which reflection i -Angle of incidence angle less than (angle of critical incidence) it is not completely absorbed and do not reach the
takes place for a gíven distance of propagation r-Angle of refraction received back. receiver. This is illustrated in figure.
r300
for a specific angle of incidence for that distance N- Electron density For a given frequency f=fmg skip distance, D (Hekm)jght
and for given ionospheric layer. is given by,
f- Frequency,
The values of MUF ranges from 8 MHz to 35 From the definition of critical frequency, u =0. 200 Fbyer
MHz. Maximum usable frequency can be applíed Ionosphere
(at vertical incidence). Elayer
for sky-wave communícation. 100
2h
For a sky-wave propagation. 81Nmaxdo pbyer
Angle of refraction, r= 90° 0-1 12

-1
N=N, -Sm 81N max=l
1000 2000 3000
Range (km)
Then,
sin i - 8 1 N , Figure: Absorption by the DLayer
sin r
S=81 Nmaxeiied Where,th e Above the D layer there are only few neutral
Su-Maximum Usable Frequency atoms. And below the D layer, there are no free
sin i 81N, electrons available. Hence, absorption is very less
sin 90° SMUF S-Critical frequency above and below the D layer.
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 197
196
(b)
Optimum Frequency (OF) (forsmall 0')
The Optimum ANTENNAS AND WAVE h e
= 2 = ( 1 - ) n =1 2
Frequency (OF) is PROPAGATION
ericlayer cos ..(11)
UNIT-5

at which
there is optimum that frequency n O
Sp

It is also Lonospheric io B
retun of wave and (11), we get,
(OWF). called Optimum Working energy. layer From equations (10)
Frequency E
b
h
In case of
to be usedionospheric propagation, the D

the Maximum should be as high as frequency 2h .. (12)

possible
Usable Frequency such as
But the factors (MUF). Rcos R
such as D2 D/2 (4) and (12), we get,
transmission distance and
ionospheric characteristics affect \Cy FIat R
From equations
D2 = 4 R 0² = 4R?x 2h 8hR
earth
optimum frequencyy which is 15% MUF, Hence, an
MUF is used for less than the Figure (1)
R
... (13)
ionospheric propagation. In figure (1), yo D 8hR
Thus OF lies is the
(13) and (4), D'
predicted MUF. between 50% to 85% of the h-Height of the ionospheric
layer
Figure (2) In view of
equations

As MUF
D
-Skip distance From the figure (2),
maximum skip distance.
changes view of
i Angle can be rewritten in
factors such as timeatofa particular location with of incidence
Angle
Arc
Then, equation (11l)
day, season-to-season - Angle of reflection. radius
months-to-months,
MUF also changes the OF which is a fraction of
and 0YSa From igure (1), 20 =
D (13) as,
D? (14)
similarly. R cos =| =1 8R?
But, it is practically cos i= OB D= 2R0 R
... (4)
impossible to vary
frequency of the rPropagated AB Arc d'=2 RO
From equations (13)
and (14),
to-hour. signal from bo cos i hgr enor
d ...5)
Thus, two frequencies
(1) Angle 20 =
R
sin
2h
P
2D? /8R
, BD R
are employed for 4 And AD=Rsin 0, OD =Rcos
propagation. One
day time while the frequency is
wave
used
From the relation between
Maximum Usable = OE + EBOD
during the
other at night time. D2 D' ... (15)
=R+h-R cos ... (6)
Apart from these two Frequency (MUF) and critical frequency i.e., V4R? 2R
frequencies a third
frequency may also be used AB = JAD)² +(BD) Then, the equation (8) becomes as,
from day dåring the transition
time to night time. SMUF(Sec 0).
cOs i =
huttin FJRsin e) +(R+h-Rcos 0) cos i =
Q39.Find the relation between JMUF ...(2) ..(1) JMUF
usable frequency and Maximum From equations (1) and (2), we [R+h-R(1 D? /8R?)1?
Skip distance. get, BD R+h-R cos
h COS
May-19, (R16), Q11(a) P o AB
JRsin e)² +(R+h-Rcos 0)?
SMUF
(or) 4 Squaring on both sides,
Squaring on both sides, (R+h-Rcos ) (h +D² /8R)²
Obtain the relation between skip distance cos ?
(Rsin 9)²+(R+h-Rcos )
and Maximum Usable Frequency. bette h? (D'/4) +(h+D'² /8R)
(8)
Ans: D SMUF Since,
Nov./Dec.-18, (R13), Q10(a) 4 D?
Je COS i 4
In order to obtain the expression for Maximum fMUF SUF 1
Usable Frequency (MUF) and skip distance, consider 4 D? D
h When is maximum, the skip distance D
the following two cases as: is maximum. MUF and skip distance are limited
D by curvature of the earth. When a wave leaves the
Case (i): fMUF Je 1 4h .(3) transmitter at a grazing angle OAB = 90°, this limit is D²(4
1W2

achieved. Thus, ... (16)

Flat Earth: For shorter distances about 400 km, the The equation (3), represents the relation between
R
surface of the earth can be considered as flat and MUF and skip distance for a flat earth case. cos A=
(9 The equation (1 6) )represents the relation between
OB R+h
ionization layer is thin. In this case, sharp reflections Case (ii) Curved Earth Since, actual value of 0° is very small, this MUF and skip distance 1for a curved earth case.
from the earth surface are obtained from thin ionized For curved earth, miror like reflections, can relation can be given as Since(Dm=V8hR
layer with sharp gradient of ionization density. be obtained by considering thin ionized layer R
with sharp gradient of ionization density and the R+h R(1+h/R)
Figure (1) illustrates the reflection of the
from a thin ionized layer on the flat earth.

is a
wave

CRIMINAL act.
reflecting region is considered concentric with
earth as shown in figure (2).
Anyone found guilty is LIABLE to face LEGAL proceedings.
=(1 + h/R) = (1 - h/R)
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 199

frequency is given by,

198 maximum
usable
The
Q40. Nrite a short
propagation.
note on Multi-hop ANTENNAS AND WAVE in figure, 600°viooer

Ans: Permissible
1
Ranges of
The frequency Frequencies
PROPAGATION From
triangle
OAB,
shown
-2:846 x 10 x sec
(Neglecting negative
Jar5.728 MHZ symbol)
Multi-hop Propagation lies between 10 range of mut-hop
The propagation 2 MHz to 30 MHz. propagation with
at transmitter reaches in which the wave originating Multi-hop propagation
established

is
the dis
the tion long distancesprovides MHz.Find
c o m m u n i c a t i o n

the ground is known as receiver without touching 3 over the

up to 4000 communica Q44. A VHF at 90
single hop
distances are hop distances. propagation. These The single hop covers the km
(:h <<<2rh,) 35 W
transmitter
LOS
c o m m u n i c a t i o n

which the
of Tx and
The concept of multi
for E-layer and 4000 distance of 2000
km for d -2rh,
-2x6.37x10° x3000 tance upto
possible, if
the Also find
heights

occurs in sky-wave hop propagation generally Q41. Compute the effective F-layer. 1, may be 40m and 25 m.
are
propagation. 0 antennas

with N=dielectric
of the E layer Rx end.
In sky-wave
propagation, the electromagnetic 5 x 10
constant 4=
vox637x1

...(1) field
strength at receiving
waves reach the sec, if the electrons! x 10 m
receiving point after multiple MHZ frequency of the wave is 25 d = 19.55
reflections from the ionosphere OBC shown in
figure (1), Ans:

However., the transmission path islayer. Ans: From triangle

4,- yh, +r-r Given that,
1 limited by the, Given that, Aug.-22, (R18), Q8(b)
Curvature of earth For a VHF
communication,

2. For Elayer, - + r +2h,,-r-r'

Skip distance. 4, Transmitted power,
P,= 35 W
Due to this, single hop
propagation is not possible. Density of electrons, N= 5 x d = yhit2h,,:r f=90 MHz
10° Operating frequency,
When the distance between Frequency, f= 25 MHz =25 x 10°electrons/sec = 40 m
transmitter
receiver is greater than the skip distance, the and
Effective Hz 4-y2h, r(:h; <«< 2rh,) transmitting antenna, h,
Height of the
atleast two hops to reach the receiver as waves takes dielectric constant, a,=? x5000 antenna, h_ = 25 m
(a) and (b). shown in figure The effective dielectric constant, [, is expressed d, = N2x6.37x10° Height of the receiving
as. .2) distance for possible
Terminator d, =2.524 x 10° m Line Of Sight (LOS)
the
possible distance along
communication, d=?
Day
Then the minimum the two aircrafts can have
over which receiving end, E, = ?
Night
surfaceof earth
microwave communication
is, Field strength at the
PLayaye
effective point to point
for
expression line of sight distance for a
(25x109)2 d =d +d, The
possible communication is given by,
=0.9999 Form equations (1) and (2), we have,
P
R
e, =0.999 d = 19.55 x 10 + 2.524 x 10 d=4.12 (Vh, +yh,)
Q42. TWo aircrafts are flying at d =(19.55 +25.24) x 10 m
la) Multiple-hop in East-West altitudes of
3000 m and 5000 m respectively. d =44.79 x 10 m -4.12 (W40 +/25)
F, Layer What is =4.12(11.325) = 46.659 km
theminimum possible distance along the :.d =447.9 km
surface of the earth over which they can Q43. Explain the effects of D and F layers of the
have effective point to point microwave ionosphere on propagation and estimate
..d= 46.659 km
communication? Radius of earth is 6.37
X 10 m.
the critical frequency and MUF for a layer
Single Earth Multiple Hop with 101Im3 electron density, and incident The field strength at the receiving end is given
Hop (Double Hops) Ans:
Earth angle of 600°, by,
Given that, Ans: 4 Nov/Dec.-18,(R15), Q10(b)
(b) Multiple-hop in South-West Height of aircraft 1(Assume it is transmit Efect of Dand FLayers of lonosphere on Propagation
ting) is, h =3000 m E, ...(1)
Figure For answer refer Unit-5, Q29, Topics: D-layers
Height ofaircraft =(Assumeits receiving) and F-layers. Where,
Features of Multiple Hop Propagation is, h =5000 m
Given that, A-Operating wavelength given by,
1 Sky-waves are reflected from ionospheric layer Radius of earth, r = 6.37 ×10° m.
and returm to earth by undergoing either single Minimum possible distance along the sur Electron density, N.= 10"/m
face of the earth, d=? 5A Angle of incidence, 0 = 600° 3x10
or multi-hop propagation. J90x10°
=3.33 m
2 Multi-hop propagation paths generally occur Critical frequency,f=?
e7 due to the semi-circumference of the earth (over MUE,fuup =? Substituting all the corresponding values in
20,000 km). d. The critical frequency interms of electron equation (1), we get,
When the transmitted ray is tangent to the earth is expressed as, density
3
surface, then there exist longest single hop propa 88/35x40x25
gation. = 81x10!1 3.33x(46.659)*
4. Multi-hop propagation generally involve two or = 2846049,894 Hz =71.813 V/m
each contributes
more transmission paths and Figure f=2.846 Hz
some energy to the receivers. Eg =71.813 V/m
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 201
200 of 25000
kms.
the MUE
distance
Q46. A150 m station at a Calculate

antenna transmitting at two earth 5 MHz.

by ground
wave has 1.2 MHz ANTENNAS AND WAVE to be
established
between

and Its
critical
f r e q u e n c y is

in the
lonosphere
laver.

What voltageanisantenna current for refractivePROPAGATION

of 8 A. The expression UNIT-5

ie link has is 200 kms electron density

ionosphere

receiving antenna 40 kmsreceived by a index is given, by he helahtof

Also
calculate the
(or)
radio link
between two points
height of 2 meters? away, with a for the
aiven path. frequency ionosphere is 200
Ans: for a high height of
frequency Consider the
usable earth.
Given that, maximum
surface of Q11 (b)
Calculate the
Where, f= 8IN.m distance of
2500 km on
the
is 5 MHz.
May-19, (R16),
Height of transmitting ata frequency

Height of receiving antenna, h= 150 m

critical
the
km and
antenna, h, =2 m 81N maK
Transmitting frequency, f= 1.2 MHz Electron
Density)

Antenna current,I=8A Refer

Excluding

Distance between
antenna, transmitting and receiving 09 8IN maX Ans:
Given that, D = 25000
km
d= 40 km. (10x10') Distance
between two
carth stations,
The field 8IN ax h= 200 k1m
transmitting antennastrength at a
is given by distance from the (0.9)1
(10x10
Height of ionosphere,
=5 MHz
equation, Criticalfrequency,
E= 120% h,h,I MUF of thegiven path,
ay?
Volt/meter N0.19x(10x10') ionosphere layer,
N?
. (1) 81 Electron densityin
of given path is,
Maximum Usable Frequency

Wave length à = 3x10` m/s N=2.3456x10m usable frequency

(MUF)
Then, the maximum
12x10° Hz H 250 m Since, f=81N mak
25000
|2 200
Substituting
(1) we get, the values of h, h, I, , din equation
120n x 150x 2 x 8
f-N8I2.3456x10) -5x 10 V1+390625
=312.54 MHz

250 x 40×10 =4.3588x 10 Hz A312.54 MHz

9047787 f-4.358S MHZ Electron Density

E Case () Refractive index of the medium is
given by,
10
E= 0.090477 V/m Skip distance for flat earth is given
by,
But, at critical frequency, u =0
E=0,090477 V/m
Dap=2 fe 81N as
Voltage received by receiving antenna is given 0
by,
10%10 (5x10
D=2 x 400 x 10 -1
V= Eh, 4.3588x10° N 81 81

= 1872.95 km -3.086 x 10!1 m

=0.090477 x2
|Dkin =l651.84 km N=3.086x 10'm
:V=0.180954 Volts
Q47. Assume that reflection takes place at a Case (ii) Q49. Itis defined to establish short wave commu-nication between two points on earths surface
(assumed flat) separated by 1200 km. Calculate MUF and angle of take off of the transmitted
height of 400 km and that the maximum Skip distance for curved earth is given by. wave from the following: Highest signal frequency returned to earth after vertically upward
density in the ionosphere corresponds propagation =7.2 MHz. Virtual height of the ionized layer =200 kms.
to a 0.9 refractive index at 10 MHz. What Ans:
willbe the range for which the MUF is 10 Daip
MHz?
Given that,
For a short wave communication,
65L76xt0'
Dag 40ox10' , (8x6370x1O) V4.3588)
Ans:
Distance between the two points on the earth surface, D= 1200 km
Given that,
-2[400x|0 + 53.538x10] x 2.0647 Highest signal frequency returned to earth after vertically upward
Height, h= 400 km
Virtual height of the ionized layer, h = 200 km propagation, f, = 7.2 MHz
Refractive index, u =0.9 =1872.95 km
MUF,S= 10 MHz Maximum Usable Frequency (MUF).f.-?
Dskin 1872.95k m t o Angle of takeoff of the transmited wave, ß=?
Frequency, f= 10 MHz
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202 ANTENNAS AND WAVE PROPAGATION
HenThen, the MUF is given by,
VERY SHORT QUESTIONS WITH
SOLUTIONS (VSQS)
bt lotn zio Q51. What is Reflection
?nln9
1200 Ans:
=7.2 x 106
24200) Reflection: Reflection is defined as a phenomenon in
which the angle of incidence of aplane wave is equal
=7.2 x 10/10=22.768 MHz to the angle of reflection of the same wave.
:s =22.768 MHz Q52. Define refraction.
Angle of take-off of the transmitted wave is, Ans:

B = tan Refraction: Refraction is defined as the bending

property of a plane wave in which change of speed results
in change in the direction of wave.
= tan |2x200
1200
= 18.43°
Q53. Write the expression for field strength 'E'
in ground wave propagation.
.B=1843 Ans:

Q50. calculate the value of frequency at which The field strength at a distance from transmitting
an em wave must propagate through the antenna due to ground wave can be measured using
d-region with an index of refraction of Maxwell's equation. It is expressed as,
0.5 and an electron density of 3.24 x 1024
electrons/m³. 120zh,h,!, V/m

Ans: Dec,-19, (R15),Q11(b) Where,

Given that, h, h - Effective heights of transmitting and
receiving currents
n =Refractive index = 0.5
I-Antenna currents
N=3.24 x 10 electrons/m
2- Wavelength
The refractive index is given by,
d-Distance between transmitting and receiv
81N ing antenna.
n-/1 Q54. What are the types of ground wave?
0.5 = 812(3.24 x10") Ans:

The different types of ground waves are,

Squaring both the sides, 1, Surface wave: This wave predominates at the
surface near to the earth.
81(3.24 x10*)
ooueadi 0.5 = 1 2. Space wave: This wave dominates at a larger
betiinenesi oto HhoxnhosOsionD mdistance above the earth.
bewao*-81(3.24 x10")
1u Q55. What is impact of imperfect earth on
surface wave?
81(3.24 x 10) Ans:
f? = 0.75
The surface waves or ground waves get attenuated
f=
/81(3.24 x10) due to, the earth imperfection. The attenuation
0.75 increases with the frequency.
f=1.87061 kHz n The tilt in the wave progresses along the curvature
of earth. As a result, the horizontal component of
electric field decreases reducing the strength of
E-field.

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UNIT-5A4OhR aAW NAANIA 203
wave.
Q56. Define wave tilt of ground
Ans:
Angle of tilt is also known as wave tilt. It is defined as the deviation in orientation of the vertically polarized
around wave on the surface of the earth. The conductivity and permittivity of the earth determines the magnitude
of tilt in the wave.r

Q57. What are M-Curves?

Ans:
Modified Refractive Index, M-curves are used to describe the influence of varying refractive index (of air)
on the Propagation of radio waves through earth's atmosphere. These curves illustrate the variations in modified
refractive index (M with height from surface of earth. Using the M-curves it is feasible to anticipate the transmission
path an EM wave passes through.
Q58. Define virtual height
Ans:
potsgetng
Virtual Height: The virtual height is defined as the height to which transnmitting a pulse with velocity of light would
reach taking the same two rays travel time as does the actual pulse reflected fromn the layer.
Q59. What is Skip distance?
Ans:

SkipDistance: The concept of skip distance is related with sky-wave propagation. It is the distance measured from
the point of transmission (transmitter) to the point where it (sky-wave) first reaches back to earth.
The skip zone is defined as the zone or coverage area from where sky-wave is almost inactive, negligibly
present or very weak for reception to the point where it first reflects back to earth.
Q60. Define MUF.
Ans:

MUF: Maximum usable frequency is defined as the frequency at which reflection takes place for a given distance
of propagation. The values of MUF ranges from 8MHz to 35 MHz, Maximum usable frequency can be applied
for sky-wave communication.
Q61. What is Critical frequency?
Ans:

Critical Frequency: Critical frequency ) is defined as the highest frequency which can be reflected from the
ionospheric layer at vertical incidence.
Q62. Define optimum usable frequency of an ionospheric laye.
Ans:
Optimum usable frequency of an ionosphere layer is also known as optimum working frequency. It is defined
as the day and night frequency and is usually taken as 85% of the maximum usable frequency.
OWF = 85% (MUF)
Typical values of day and night frequencies are 6.405 MHz and 5000 MHz respectively.
Q63. What is multi-hop propagation?
Ans:
The propagation of radio waves from transmitter to receiver with muliple refections (multiple hops) between
earth and ionosphere along the earth's surface is termed as multi-hop propagation.

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