GOVERNMENT TECHNICAL INSTITUTE (INSEIN)
Department of Electronic Engineering
Electronics Communication System
CHAPTER (7)
Review Questions and Answers
Q1. Describe the radio communication system briefly with the necessary block diagram.
Answer:
 The three important blocks from the electrical communication point of view include
transmitter, receiver and channel.
 The radio transmitter is an electronic system that accepts the incoming message and
converts it into a modulated signal in the RF range by the modulation process, as
described in the analog modulation techniques case.
 The required power levels are also added to the modulated signal so that it can travel
for a longer distance.
 After adding enough power, the modulated signal is transmitted through the
communication channel towards the receiver.
 In case of free space as channel, the antenna is used as the transducer to
convert the modulating signal from guided to free space form.
 Thus, the important blocks of a radio transmitter include an oscillator to generate a
high-frequency carrier signal for modulation, modulator, power amplifier and antenna.
 The radio receiver is an electronic system designed in such a way to recover the
message from the incoming weak signal.
 The important operations of the radio receiver include converting a received signal
from free space to guided form using a receiving antenna, selecting out only the
wanted signal using the available numerous ones in the free space, demodulating the
message and delivering it to the destination in the original form.
 The two important aspects which the receiver system has to deal with, include, the
weak signal available at its input terminal due to its travel over long distance and
several signals available from many other transmitters at its input.
1
 The radio receiver should first admit only the wanted signal.
 Later, it should recover the message without distortion from the admitted weak signal.
Fig: Block diagram of radio communication system
Q2. Explain the operation of an AM transmitter with the necessary block diagram.
Answer:
 There are two types of devices in which it may be necessary to generate amplitude
modulation.
 The first of these, the AM transmitter, generates such high powers that its prime
requirement is efficiency, so quite complex means of AM generation may be used.
 The other device is the laboratory AM generator. Here, AM is produced at such a low
power level that simplicity is a more important requirement than efficiency.
 In an AM transmitter, amplitude modulation can be generated at any point after the
radio frequency source.
 As a matter of fact, even a crystal oscillator could be amplitude modulated, except that
this would be an unnecessary interference with its frequency stability.
 If the output stage in a transmitter is collector modulated in a low power transmitter,
the system is called high level modulation.
 If modulation is applied at any other point, including some other electrode of the
output amplifier, then so called low level modulation is produced.
 Naturally, the end product of both systems is the same, but the transmitter circuit
arrangements are different.
 Figure shows a typical block diagram of an AM transmitter, which may be either low
level or high level modulated.
 There are a lot of common features.
2
 Both have a stable RF source and buffer amplifiers followed by RF power amplifiers.
 In both types of transmitters, the audio voltage is processed, or filtered, so as occupy
the correct bandwidth (generally l0 kHz), and compressed somewhat to reduce the
ratio of maxi- mum to minimum amplitude.
 In both modulation systems, audio and power audio frequency (AF) amplifiers are
present, culminating in the modulator amplifies which is the highest power audio
amplifier.
Fig: Block diagram of AM transmitter
Q3. Mention the difference between AM and SSB transmitters.
Answer:
 SSB transmitter shown in the figure will be very similar to that of an AM transmitter,
except for the replacement of an amplitude modulation block with SSB modulation
block.
 The difficulty associated with the SSB is due to the suppression of a carrier
component.
 The approach followed for demodulation at the receiver is to re-insert the carrier.
 As can be appreciated, this requires excellent frequency stability on the part of both
transmitter and receiver, because, any frequency shift, anywhere along the chain of
events through which the information must pass, will cause an equal frequency shift to
the received signal.
 Imagine a 40-Hz frequency shift in a system through which three signals are being
transmitted at 200, 400 and 800 Hz.
3
 Not only will they all be shifted in frequency to 160, 360 and 760 Hz, respectively, but
their relation to one another will also stop being harmonic. The result is that it is not
possible to transmit good quality speech or music.
 There are two variants of SSB that help in mitigating this carrier stability problem,
namely, pilot carrier and independent sideband (ISB) systems.
Fig: Block diagram of SSB transmitter
Q4. Explain the operation of a pilot carrier system with the necessary block diagram.
Answer:
 The technique that is widely used to solve the frequency-stability problem is to
transmit a pilot carrier with the wanted sideband.
 The block diagram of such a transmitter is very similar to the conventional SSB
transmitter, with the one difference that an attenuated carrier signal is added to the
transmission after the unwanted sideband has been removed.
 The pilot carrier SSB system is shown in figure.
 The carrier is normally re-inserted at a level of 15 or 26 dB below the value it would
have had if it had not been suppressed in the first place, and it provides a reference
signal to help demodulation in the receiver.
 The receiver can then use an automatic frequency control (AFC) circuit to control the
frequency of a carrier signal generator inside the receiver with the help of a pilot
carrier.
4
Fig: Block diagram of an SSB pilot carrier transmitter
Q5. Draw or Explain the operation of an ISB system with the necessary block diagram.
Answer:
Fig: Block diagram of an ISB transmitter
5
Q6. Explain the operation of an FM transmitter with the necessary block diagram.
Answer:
 FM transmitters also work along the same lines as that of AM transmitters.
 Frequency modulation can be generated at any point including the radio frequency
source.
 Accordingly, we can use either direct or indirect method for the generation of FM.
Further, FM transmitters can also classify as low-level and high-level transmitters
depending on where the FM is performed.
 The crystal oscillator generates the stable carrier signal. The modulating signal and the
carrier signal are applied to the phase modulator operating in the low power level to
generate a narrowband FM wave.
 The narrowband FM wave is then passed through several stages of frequency
multipliers to increase the frequency deviation and also carrier signal frequency to the
required level.
 The several stages of frequency multiplication helps in choosing a suitable
combination for achieving the required level of multiplication factors needed for
deviation and carrier signal frequency.
 The output of the frequency multipliers stage will be a wideband FM, but at the low
power level.
 The WBFM is then passed through one or more stages of power amplifiers to add
required power levels.
 The WBFM with high power is then finally transmitted via the antenna towards the
receiver.
Fig: Block diagram of an FM transmitter
6
Q7. With the aid of the block diagram of a simple receiver, explain the basic
superheterodyne principle.
Answer:
 Of the various forms of receivers proposed at one time or another, only two have any
real practical or commercial significance - the tuned radio-frequency (TRF) receiver
and the superheterodyne receiver.
 Only the second of these is used to a large extent today, but it is convenient to explain
the operation of the TRF receiver first since it is the simpler of the two.
 The TRF receiver is a simple "logical" receiver.
 A person with just a little knowledge of communications would probably expect all
radio receivers to have this form.
 The virtues of this type, which is now not used except as a fixed-frequency receiver in
special applications, are its simplicity and high sensitivity.
 In the superheterodyne receiver, the incoming signal voltage is combined with signal
generated in the receiver.
 This local oscillator voltage is normally converted into a signal of a lower fixed
frequency.
 The signal at this intermediate frequency contains the same modulation as the original
carrier and it is now amplified and detected to reproduce the original information.
 The superhet has the same essential components as the TRF receiver, in addition to the
mixer, local oscillator and intermediate-frequency (IF) amplifier.
Fig: The TRF receiver
7
Q9. What are the advantages that the superheterodyne receiver has over the TRF
receiver? Are there any disadvantages?
Answer:
 The advantages of the superheterodyne receiver make it the most suitable type for the
great radio receiver applications; AM, FM, communications, single-sideband,
television and even radar receivers all use it, with only slight modifications in
principle.
 It may be considered as today's standard form of radio receiver, and it will now be
examined in some detail, section by section.
Q12. Define the terms sensitivity, selectivity and image Frequency.
Answer:
Sensitivity
 The sensitivity of a radio receiver is its ability to amplify weak signals.
 It is often defined in terms of the voltage that must be applied to the receiver input
terminals to give a standard output power, measured at the output terminals.
Selectivity
 The selectivity of a receiver is its ability to reject unwanted signals.
 It is expressed as a curve, which shows the attenuation that the receiver offers to
signals at frequencies near to the one to which it is tuned.
Image Frequency
 The term f si is called the image frequency and is defined as the signal frequency plus
twice the intermediate frequency.
f si =f s +2f i
8
Q21. What is simple automatic gain control? What are its functions?
Answer:
 Simple AGC is a system by means of which the overall gain of a radio receiver is
varied automatically with the changing strength of the received signal, to keep the
output substantially constant.
 A dc bias voltage, derived from the detector as shown and explained in connection
with Fig , is applied to a selected number of the RF, IF and mixer stages.
 The devices used in those stages are ones whose transconductance and hence gain
depends on the applied bias voltage or current.
Fig: Simple AGC characteristics
Q23. Describe the differences between FM and AM receivers, bearing in mind the
different frequency ranges and bandwidths over which they operate.
Answer:
The FM receiver is a superheterodyne receiver, and the block diagram of fig shows
just how similar it is to an AM receiver. The basic differences are as follows:
1. Generally much higher operating frequencies in FM
2. Need for limiting and de-emphasis in FM
3. Totally different methods of demodulation
4. Different methods of obtaining AGC
9
Fig: FM receiver block diagram
10
GOVERNMENT TECHNICAL INSTITUTE (INSEIN)
Department of Electronic Engineering
Electronics Communication System
CHAPTER (8)
Review Questions and Answers
1. Explain how television is capable of displaying complete moving pictures, instant of
time despite the fact that at any only a tiny portion of the picture tube screen is active.
Television means seeing at a distance. To be successful, a television system may be
required to reproduce faithfully:
1. The shape of each object, or structural content
2. The relative brightness of each object, or tonal content
3. Motion, or kinematic content
4. Sound
5. Color, or chromatic content
6. Perspective, or stereoscopic content
 To explain how flicker is avoided in TV it is first necessary to look at the scanning
process in a little detail.
 The moving electron beam is subjected to two motions simultaneously.
 One is fast and horizontal, and the other is vertical and slow, being 262 1/2 times
slower than the horizontal motion.
 The beam gradually moves across the screen, from left to right, while it
simultaneously descends almost imperceptibly.
 A complete frame is covered by 525 horizontal lines, which are traced out 30 times
per second.
 However, if each scene were shown traced thus from top to bottom (and left to right),
any given area of the picture tube would be scanned once every one thirtieth of a
second, too slowly to avoid flicker.
 Doubling the vertical speed, to show 60 frames per second , would do the trick but
would double the bandwidth.
11
2. Briefly describe camera and picture tubes, and explain what actually happens them
when a picture is being scanned. Why is sync transmitted?
 The scene elements at the transmitting station are produced by a mosaic of
photosensitive particles within the camera tube, onto which the scene is focused by
optical means.
 They are scanned by an electronic beam, whose intensity is modulated by the
brightness of the scene.
 A varying voltage output is thus obtained, proportional to the instantaneous brightness
of each element in turn.
 The varying voltage is amplified, impressed as modulation upon a VHF or UHF
carrier, and radiated.
 At the receiver, after amplification and demodulation, the received voltage is used to
modulate the intensity of the beam of a Cathode Ray Tube (CRT).
 If this beam is made to cover each element of the display screen area exactly in step
with the scan of the transmitter, the original scene will then be synthesized at the
receiver.
 The need for the receiver picture tube to be exactly in step with that of the transmitter
requires that appropriate information be sent.
 This is synchronizing, or sync information, which is transmitted in addition to the
picture information.
 The two sets of signals are interleaved in a kind of time-division multiplex, and the
picture carrier is amplitude-modulated by this total information.
 At the receiver, signals derived from the transmitted sync control the vertical and
horizontal scanning circuits, thus ensuring that the receiver picture tube is in step with
the transmitter camera tube.
3. Explain briefly the difference between chrominance and luminance. How is a colour
picture tube able to display white?
 Black-and-white television can be transmitted in this manner, but colour TV requires
more information.
 As well as indicating brightness or luminance, as is done in black-and-white TY
colour (or actually hue) must also be shown.
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 That is, for each picture element we must show not only how bright it is, but also
what hue this element should have, be it white, yellow, red, black or any other.
 The hue is indicated by a chrominance, or chroma, signal.
4. Explain (a) a television sound is transmitted; (b) what is meant by saying that colour
television must be compatible?
(a)
 The simplest item has been left until last; this is the sound transmission.
 A separate transmitter is used for sound, connected to the same antenna as the picture
transmitter.
 However, it is a simple matter to have a receiver with common amplification for all
signals up to a point, at which the various signals go to their respective sections for
special processing.
 This separating point is almost invariably the video detector, whose output consists of
picture, sync and sound information.
 The sound signal is amplified, applied to its own detector, amplified again and fed to a
loudspeaker.
 The modulating system used for sound in the U.S. system, and most other major
systems around the world, is wideband FM.
 It is not quite as wideband as in FM radio transmissions, but it is quite adequate for
good sound reproduction.
 The transmitting frequency for the sound transmitter is quite close to the picture
transmitting frequency.
(b)
 Frequency Division Multiplexing (FDM) is used to interleave the chrominance signal
with luminance.
 The process is quite complex.
 The chroma signal is assigned portions of the total frequency spectrum which
luminance does not use.
 The situation is complicated by the fact that color and black-and-white TV must be
compatible.
13
 That is to say, the chroma signals must be coded in such a way that a satisfactory
picture will be produced (in black and white) by a monochrome receiver tuned to
that channel.
 Conversely, color TV receivers must be designed so that they are able to reproduce
satisfactorily (in black and white) a transmitted monochrome signal.
Fig: Basic monochrome television system, (a) Transmitter (b) receiver
5. Why are television standards required? What are the major U.S TV standards?
What other TV standards are there in other parts of the world?
 It is clear that a large amount of information must be broadcast by a television
transmitter and that there are a variety of ways in which this can be done.
 Accordingly, a need exists for uniform standards for TV transmission and reception.
 Regrettably, no agreement has been reached for the adoption of worldwide standards,
and it seems unlikely in the extreme that such a standard will ever be reached.
 Thus several different systems exist, necessitating standards conversion for many
international television transmissions.
14
 The two main ones are the American [Federal Communications Commission(FCC)
system for monochrome and National Television Standards Committee (NTSC)
system for colour] and the European [Comite Consultatif International de Radio
(CCIR) system for monochrome and Phase Alternation by Line (PAL) system for
colour].
21. Explain what is meant by the Y, I and Q signals in colour TV and why they are
generated.
 White may be synthesized by the addition of blue (B), green (G) and red (R).
 By the addition of voltages that correspond to these colours in the receiver picture
tube.
 White (Y) equals 33 ½ percent each of B, G and R.
 Our eyes have a colour frequency response curve which is very similar to the response
curve of a single-tuned circuit.
 Red and blue are at the two edges, and green is right in the middle of the response
curve.
 Our eyes are most sensitive to green.
 They are about twice as sensitive to green as to red, and three times as sensitive to red
as to blue.
 The result is that "100 percent white" is given by
 Y as given by above equation, has amplitude that corresponds to 12.5 percent
modulation of the carrier, the receiver will reproduce white.
 If the amplitude of the Y video voltage yields 67.5 percent modulation, a black
image results.
 In the NTSC system, the remaining two signals are
 I stands for "in phase," and Q for “quadrature phase”.
 " Both terms are related to the manner of transmission.
15
22. With the aid of the circuit diagram of a simple matrix, show how the I, Q and Y
signals are generated in a color TV transmitter. Show typical values for the Y and I
components on your matrix.
Fig: Colour camera tube and matrix arrangement, showing typical resistor values for the
Q25. Why and how is the color burst transmitted? When is it not sent? Why not?
 The first of these is the color burst, which, as the name suggests, is a short burst of
color subcarrier. It is sent once each horizontal line and is used in the receiver as a
phase reference.
 This is required to ensure that the absolute phase of the/ and Q vectors is correct lf it
were not sent and a spurious +90° phase shift of the color subcarrier in the receiver
occurred, would be mistaken for Q, and Q for-!.
 The resulting reproduced colors would have the correct relationship to each other, but
they would be absolutely wrong.
 The (R - Y) and (B - Y) vectors are not transmitted but are often used in the receiver
16
Q26. Draw the basic block diagram of a colour television transmitter, and briefly
explain the function of each block.
Fig: Basic block diagram of colour television transmitter
17
GOVERNMENT TECHNICAL INSTITUTE (INSEIN)
Department of Electronic Engineering
Electronics Communication System
CHAPTER (10)
Review Questions and Answers
1. Electromagnetic waves are said to be transverse; what does this mean? Illustrate
with a sketch.
 Electromagnetic waves are energy propagated through free space at the velocity of
light, which is approximately 300 meters per microsecond.
 Visualize yourself standing on a bridge overlooking a calm body of water.
 If you were to drop an object (which did not float) into the pond, you would see this
energy process in action.
 Also, the direction of the electric field, the magnetic field and propagation are
mutually perpendicular in electromagnetic waves.
Fig: Transverse electromagnetic wave in free space
2. Define the term power density, and explain why it is inversely proportional to the
square of the distance from the source.
 Power density is defined as radiated power per unit area, it follows that power
density is reduced to one-quarter of its value when distance from the source has
doubled.
 It is seen that power density, is inversely proportional to the square of the distance
from the source.
18
 This is the inverse-square law, which applies universally to all forms of radiation in
free space.
 Stating this mathematically, we have
where,
= Power density at a distance r from an isotropic source
P t = Transmitted power
 An isotropic source is one that radiates uniformly in all directions in space.
3. Explain what is meant by the terms isotropic' source and isotropic medium.
 An isotropic source is one that radiates uniformly in all directions in space.
 Although no practical source has this property, the concept of the isotropic
radiator is very useful and frequently employed.
 As a matter of interest, it may be shown quite simply that the inverse-square law
applies also when the source is not isotropic.
4. Define and explain field intensity. Relate it to power density with the concept of
characteristic impedance of free space.
 Power density is defined as radiated power per unit area, it follows that power
density is reduced to one-quarter of its value when distance from the source has
doubled.
 It is seen that power density, is inversely proportional to the square of the
distance from the source.
 This is the inverse-square law, which applies universally to all forms of radiation
in free space.
 Stating this mathematically, we have
19
where,
= Power density at a distance r from an isotropic source
P t = Transmitted power
 An isotropic source is one that radiates uniformly in all directions in space.
 Field intensity is inversely proportional to the distance from the source,since it is
proportional to the square root of power density.
5. Explain fully the concept of linear polarization. Can longitudinal waves be
polarized? Explain.
 Polarization refers to the physical orientation of the radiated waves in space.
 Waves are said to be polarized (actually linearly polarized) if they all have the
same alignment in space.
 It is a characteristic of most antennas that the radiation they emit is linearly
polarized.
 A vertical antenna will radiate waves whose electric vectors will all be vertical
and will remain so in free space.
 Linearly polarized and is also
said to be vertical polarized since all the electric
intensity vectors are vertical.
 Vertical antennas radiate vertically polarized waves and similarly horizontal
antennas produce waves whose polarization is horizontal.
6. Why does the atmosphere absorb some power from waves propagating through if?
At what frequencies does this absorption become apparent?
 When propagation near the earth is examined, several factors which did not
exist in free space must be considered.
 Thus waves will be reflected by the ground, mountains and buildings.
 They will be refracted as they pass through layers of the atmosphere which
have differing densities or differing degrees of ionization.
 Also, electromagnetic waves may be diffracted around tall, massive objects.
 They may even interfere with each other, when two waves from the same
source meet after having travelled by different paths.
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7. Prove that when electromagnetic waves are reflected from a perfectly conducting
medium, the angle of reflection is equal to the angle of incidence. Hint: Bear in
mind that all parts of the wavefront travel with the same velocity, and consider what
would happen if the two angles were not equal.
 There is much similarity between the reflection of light by a mirror and the
reflection of electromagnetic waves by a conducting medium.
 In both instances the angle of reflection is equal to the angle of incidence, as
illustrated in Fig.
 Again, as with the reflection of light, the incident ray, the reflected ray and the
normal at the point of incidence are in one plane.
 The concept of images is used to advantage in both situations.
Fig: Reflection of waves, image formation.
 The incident and reflected waves travel with the same velocity.
 This is due to some absorption at each reflection; this also happens with radio
waves.
 The reflection coefficient p is defined as the ratio of the electric intensity of
the reflected wave to that of the incident wave.
 It is unity for a perfect conductor or reflector, and less than that for practical
conducting surfaces.
21
 The difference is a result of the absorption of energy (and also its
transmission) from the wave by the imperfect conductor.
 Transmission is a result of currents set up in the imperfect conductor, which
in turn permit propagation within it, accompanied by refraction.
Q8. What is refraction? Explain under what circumstances it occurs and what
causes it.
 As with light, refraction takes place when electromagnetic waves pass front
one propagating medium to a medium having a different density.
 This situation causes the wavefront to acquire a new direction in the second
medium and is brought about by a change in wave velocity.
Fig: Refraction at a plane, sharply defined boundary
 The relationship between the angle of incidence g and the angle of refractionθ,
may be calculated with the aid of simple trigonometry and geometry.
 Considering two right-angled triangles PQQ' and PP’Q we have
22
where,
v A = wave velocity in medium A
v B = wave velocity in medium B
The wave velocity in a dielectric medium is inversely proportional to the square
root of the dielectric constant of the medium.
where,
k = dielectric constant of medium I
k’=dielectric constant of medium B
µ =refractive index
Electromagnetic waves travelling from a rarer to a denser medium are refracted
toward the normal, so we see that waves travelling the other way are bent away
from the normal.
 However, if there is a linear change in density (rather than an abrupt change),
the rays will be curved away from the normal rather than bent.
The situation arises in the atmosphere just above the earth, where atmospheric
density changes (very slightly, but linearly) with height.
The top of the wave front travels in rarer atmosphere than the bottom of the
wave front and therefore travels faster, so that it is bent downward.
Fig: Refraction in a medium having linearly decreasing density (the Earth is shown flat for
simplicit
23
Q13. Describe ground-wave propagation. What is the angle of tilt? How does it affect field
strength at a distance from the transmitter?
 Ground waves progress along the surface of the earth and, as previously
mentioned, must be vertically polarized to prevent show circuiting the electric
component.
 A wave induces currents in the ground over which it passes and thus loses some
energy by absorption.
 This is made up by energy diffracted downward from the upper portions of the
wave front.
 As the wave propagates over the earth, it tilts over more and more, and the
increasing tilt causes greater short circuiting of the electric field component of
the wave and hence field strength reduction.
 Eventually, at some distance (in wavelengths) from the antenna, as partly
determined by the type of surface over which the ground wave propagates, the
wave "lies down and dies."
 Thus, in the VLF band, insufficient range of transmission can be cured by
increasing the transmitting power.
Fig: Ground-wave propagation
 If a receiving antenna is now placed at this point, the signal it will receive will
be, in volts,
24
where,
120π = characteristic impedance of free space
ht
= effective height of the transmitting antenna
hr
= effective height of the receiving antenna
I
= antenna current
d
= distance from the transmitting antenna
λ
= wavelength
Q14. Describe briefly the strata of the ionosphere and their effects on sky-wave
propagation. Why is this propagation generally better at night than during the day?
The ionosphere is the upper portion of the atmosphere, which absorbs large
quantities of radiant energy from the sun, becoming heated and ionized. There are
variations in the physical properties of the atmosphere, such as temperature, density
and composition. The overall result, as shown in figure, is a range of four main layers,
D, E, F 1 and F 2 , in ascending order.The last two combine at night to form one single
layer.
Fig: Ionospheric layers and their regular variations (F. R. East, "The Properties of the
Ionosphere Which Affect HF Transmission")
The D layer is the lowest, existing at an average height of 70 km, with an average
thickness of l0 km. The degree of its ionization depends on the altitude. Of the sun
above the horizon, and thus it disappears at night. It is the least important layer from
25
the point of view of HF propagation. It reflects some VLF and LF waves and absorbs
MF and HF waves to a certain extent.
The E layer is next in height, existing at about 100 km, with a thickness of
perhaps 25 km. Like the D layer, it all but disappears at night; the reason for these
disappearances is the recombination of the ions into molecules. This is due to the
absence of the sun (at night), when radiation is consequently no longer received. The
main effects of the E layer are to aid MF surface-wave propagation a little and to
reflect some HF waves in daytime.
The E s layer is a thin layer of very high ionization density, sometimes making
an appearance with the E layer. It is also called the sporadic E layer; when it does
occur, it often persists during the night also.On the whole, it does not have an
important part in long-distance propagation, but it sometimes permits unexpectedly
good reception. Its causes are not well understood.
The F 1 layer, as shown in figure exists at a height of 180 km in daytime and
combines with the F2 layer at night, its daytime thickness is about 20 km. Although
some HF waves are reflected from it, most pass through to be reflected from the F2
layer.Thus the main effect of the F1 layer is to provide more absorption for HF waves.
Note that the absorption effect of this and any other layer is doubled, because HF
waves are absorbed on the way up and also on the way down.
The F2 layer is by far the most important reflecting medium for high-frequency
radio waves. Its approximate thickness can be up to 200 km, and its height ranges
from 250 to 400 km in daytime. At night it falls to a height of about 300 km, where it
combines with the F1 layer.
Q18. Briefly describe the following terms connected with sky-wave propagation; virtual
height, critical frequency; maximum useable frequency, skip distance and fading?
Virtual height
The virtual height of an ionospheric layer is best understood with the aid of figure.
This figure shows that as the wave is refracted, it is bent down gradually rather than sharply.
However, below the ionized layer the incident and refracted rays follow paths that are
26
exactly the same as they would have been if refection had taken place from a surface located
at a greater height, called the virtual height of this layer.
Fig: Actual and virtual heights of an ionized layer.
Critical frequency
The critical frequency (f c ) for a given layer is the highest frequency that will be
returned down to earth by that layer after having been beamed straight up at it.
First, the higher the frequency, the shorter the wavelength, and the less likely it is that
the change in ionization density will be sufficient for refraction. Second, the closer to vertical
a given incident ray, the less likely it is to be returned to ground. When the angle of
incidence is normal the name given to this maximum frequency is critical frequency; its
value in practice ranges from 5 to 12 MHz for the F2 layer.
Maximum usable frequency
The maximum usable frequency, or MUF, is also a limiting frequency, but this time
for some specific angle of incidence other than the normal. In fact, if the angle of incidence
(between the incident ray and the normal) is θ, it follows that
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Skip distance
The skip distance is the shortest distance from a transmitter, measured along the
surface of the earth, at which a sky wave of fixed frequency (more than fc) will be
returned to earth.
Fig: Multipath sky-wave propagation.
Fading
Fading is the fluctuation in signal strength at a receiver and may be rapid or slow,
general or frequency selective. In each case it is due to interference between two
waves which left the same source but arrived at the destination by different paths.
Fading can occur because of interference between the lower and the upper rays of a
sky wave; between sky waves arriving by a different number of hops or different
paths; or even between a ground wave and a sky wave especially at the lower end of
the HF band. One of the more successful means of combating fading is to use space or
frequency diversity.
Q19. In connection with space-wave propagation, what is the radio horizon? How
does it differ from the optical horizon?
The radio horizon for space waves is about four-thirds as far as the optical
horizon. This beneficial effect is caused by the varying density of the atmosphere, and
because of diffraction around the curvature of the earth.
For transmitter,
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where,
d t = distance from transmitting antenna,km
h t = height of transmitting antenna above ground, m
For receiver,
where,
d r = distance from receving antenna,km
h r = height of receving antenna above ground, m
The total distance will be given by addition. As shown in the figure, and by the
empirical formula
Fig: Radio horizon for space waves
Q . Tropospheric Scatter Propagation
Also known as troposcatter, or forward scatter propagation, tropospheric scatter
propagation is a means of beyond-the-horizon propagation for UHF signals. It uses certain
properties of the troposphere, the nearest portion of the atmosphere (within about 15 km of
the ground).
Two directional antennas are pointed so that their beams intersect midway
between them, above the horizon. If one of these is a UHF transmitting antenna. and the
other a UHF receiving one, sufficient radio energy will be directed toward the receiving
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antenna to make this a useful communication system. The best frequencies, which are also
the most often used, are centered. On 900, 2000 and 5000 MHz. Even here the actual
proportion of forward scatter to signals incident on the scatter volume is very tiny-between 60 and -90 dB, or one- millionth to one-billionth of the incident power. High transmitting
powers are obviously needed.
Although forward scatter is subject to fading, with little signal scattered forward, it
nevertheless forms a very reliable method of over-the- horizon communication. It is not
affected by the abnormal phenomena that afflict HF sky-wave propagation. Accordingly, this
method of propagation is often used to provide long-distance telephone and other
communications links, as an alternative to microwave links or coaxial cables over rough or
inaccessible terrain. Path links are typically 300 to 500 km long.
Tropospheric scatter propagation is subject to two forms of fading. The first is fast.
occurring several times per minute at its worst, with maximum signal strength variations in
excess of 20 dB. It is often called Rayleigh fading and is caused by multipath propagation.
The second form of fading is very much slower and is caused by variations in atmospheric
conditions along the path.
It has been found in practice that the best results are obtained from troposcatter
propagation if antennas are elevated and then directed down toward the horizon.
Fig: Tropospheric scatter propagation
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