Electronics Tutorial 12 - Principles of Communication and Transmission Media

 

Practical telecommunication systems have been around for about two hundred years.  The earliest electromechanical telegraph was invented in the UK by Francis Ronalds (1788 - 1873).  It was William Cooke (1806 - 1879) and Charles Wheatstone (1802 - 1875) who put the electric telegraph into commercial use.   The electric telegraph and code, were further developed by Samuel Morse (1791 - 1872) in America.  The code is still recognised internationally.  Aircraft navigation beacons transmit their identifiers in Morse Code.  Morse himself was also an artist, with a good number of portraits and other artworks to his name.  He also invented a marble-cutting machine that could cut in three dimensions.  One could argue that Morse code was an early digital transmission medium.

 

Telephony arrived in the late nineteenth century when of capturing sound wave and converting them to electrical waves were applied.  The invention of the telephone was generally credited to Alexander Bell (1847 - 1922), although others have claimed its invention as their own.

 

Wireless telegraphy started to be developed after the work of James Clerk-Maxwell (1831 - 1879) and Heinrich Hertz (1857 - 1894).  The idea was developed commercially by Guglielmo Marconi (1874 - 1937) who was the first to transmit a transatlantic message in 1901.  In the early Twentieth Century the development of speech radio spread widely and we are going to study the principles of radio and other communication technology.

 

 

General Principles

Most modern communication systems follow a general schematic like this:

Reproduced from the AQA syllabus

 

 

Important Note

If you are intending to do experiments with radio apparatus, you MUST have a licence.  It is illegal to do otherwise and you could end up getting a hefty fine, especially if your apparatus causes interference with other radio users.

 

The picture below shows a radio tuner that is separate from the audio amplifier:

 

 

In the Exam

Although these numerical calculations have not been explicitly shown in the syllabus for the Electronics option, the relationships are elsewhere on the Physics syllabus.  It is quite likely that you would be asked to do similar calculations in the exam, which is why I have included these examples.

 

You also may get new situations in which you need to apply the physics that you know.

 

 

Transmission Media

Radio wave are electromagnetic waves.  They travel at 3.0 × 108 m s-1 in a vacuum, and very close to that speed in air.  The wave equation applies:

 

All the properties of electromagnetic waves are true, and they reflect and refract in the same way as light does.  The picture shows a receiver disc for signals from a relay satellite:

 

 

Notice that the reflector is not solid, but is made of a perforated material.  This does not affect its performance as a mirror at all.

 

Question 1

The frequency from a satellite is 600 MHz.  What is the wavelength of the waves?

Answer

Question 2

A keen DIY enthusiast wants to make his satellite dish smaller, so that it's less obvious on the house.  So he makes it 20 cm in diameter.  Will it work?  Explain your answer using the physics you know.

Answer

 

Radio waves, like all electromagnetic waves are attenuated according to the inverse square law.  You have looked at the inverse square law for gamma rays and saw the equation:

 

The intensity is the power per unit area.  The equation assumes that the source is a point source which is infinitely small.   The distance x1 is the radius of the aerial, which is a distance of x1 metres from the point source.  The distance x2 is the distance between the transmitter and the receiver.  The idea is shown in the left hand diagram.

 

 

The right hand picture shows a dipole aerial that is commonly used for radio transmission.  The total length of the aerial is half a wavelength, as an electrical standing wave is set up in the dipole.  The whole thing is mounted on a bracket that is bolted to the wall of a building.

 

Question 3

A transmitter has a dipole aerial which consists of two rods of total length 1.50 m and diameter 1.0 cm.  The power from the dipole is 3.5 W.

(a) Calculate the wavelength and frequency of this transmitter.

(b) Calculate the power intensity at the surface;

(c) Calculate the power intensity at the aerial of a receiver 2.5 km away.   Give your answer to an appropriate number of significant figures.

Answer

 

From this answer you can see that the intensity of the radiation is very low indeed.  Radio transmission feeds very little energy per second to the receivers.  However you can have thousands of receivers and the total energy per second they pick up is a small fraction of that from the source.

 

 

 

Electrical Wires

Radio waves seem to be a remarkably inefficient and wasteful way of getting information about.  Most of the energy used to do so is dissipated.

 

Data can be transmitted using electric wires.  This may seen a logical way to connect a radio station with a remote transmitter, but wires present their own problems.  When we do circuits, we tend to regard the wires as perfect conductors.  They are not.  They have resistance, according to the resistivity equation:

 

Question 4

A remote transmitter is 10 km from its base station and is connected by a land line that consists of a coaxial cable which has a centre conductor of diameter 1.0 mm.  It has a thick outer sheath that has a resistance that is much lower than the centre conductor. 

Calculate the resistance of the central conductor. 

Resistivity of copper = 1.68 × 10-8 W m

Answer

 

Electric currents always have magnetic fields associated with them.  With continuously flowing direct current, there is little effect.  However with alternating currents (and direct currents that flow in pulses) there is a reverse emf due to Faraday's Law and Lenz's Law:

 

 

Question 5

Data are transmitted along a wire at a rate of 100 megabits per second.  When the current that forms a '1' flows, a magnetic field of flux density 2.5 nT is measured.  Calculate the EMF.

Answer

 

You can see the effect of this if you have an unshielded wire connected to a CRO in the physics lab.  You get a fuzzy wave along the screen which has a period of 0.02 s, even if there is no signal on the wire.  While this may seem a curiosity, it has important implications in the design of audio amplifiers.  The wave will end up being amplified as mains hum which can be intrusive.  Other spurious signals end up as noise.  Unshielded wires are often called twisted pairs.

 

 

The mess of wires here is very likely to pick up mains hum and other noise.

 

Often there are cables main of hundreds of individual conductors.  It is possible for the wires in such multi-core cables to pick up interference from signals in other conductors.

 

The wire also acts as an inductor.  The higher the frequency, the higher the reactance.

 

There is a way of preventing these problems.  The wire can be shielded easily; we simply surround the wire with an insulating layer and then copper braiding to shield the wire.    We have a coaxial cable.  Voltages induced in the braiding by changing magnetic fields are connected to earth.

 

 

The problem with coaxial cable is that:

There is another problem.  The copper core is separated from the braiding by an insulating layer.  The coaxial therefore makes a perfectly good cylindrical capacitor.  Its capacitance is not very high, but it can have a significant effect at high frequencies.  The capacitance per unit length varies from 10 to 100 pF per metre.

 

If an alternating signal is being conducted, there will be a reactance from the capacitor given by:

 

The higher the frequency, the lower the reactance.

 

Question 6

A coaxial cable is 25 metres long and has a capacitance of 56 pF m-1.  It is carrying a signal of 105 MHz.

(a) Calculate the capacitance of the cable;

(b) Calculate the reactance at this frequency.  Give the correct unit.

Answer

 

There a number of equations that deal with the capacitance of the coaxial cable.  These are NOT on the syllabus.

 

Another problem with a wire is that while a direct current is carried evenly across the conductor, this is not the case with alternating current.  High frequency alternating currents are carried on the outer layers of the conductor.  This is called the skin effect.

 

 

Optical Fibres

The most up-to-date data transmission systems use optical fibres.  These can carry large amounts of data with a very thin fibre.  They use the principle of total internal reflection, which you have studied in the section on waves.

 

The light ray makes a certain angle of incidence when it hits the boundary of an optically dense material (like glass) and an optically less dense material (like air). If this angle is greater than the critical angle, the ray is totally internally reflected. The critical angle, qc, is determined by the formula:

 

 

The rays of light should travel like this:

 

 

But instead light rays can travel several paths:

 

 

This means that the light rays can arrive at different times, resulting in dispersion or smearing. The signal that was sharp when it left the transmitter is smeared.

 

 

The picture shows us how the signal can be unacceptably distorted and even produce spurious signals that were not there.

The problem can be resolved by cladding the core with a material of slightly lower refractive index. For example the core might have a refractive index of 1.6, while the cladding has a refractive index of 1.4.


Dispersion can be reduced further by use of a graded index or multimode fibre.  Some light is passes down the middle, which has a higher refractive index, therefore slower rate of travel.  This can result in modal dispersion.  The light waves travel at different speeds, which will lead to smearing of the signals.  This is sometimes called waveguide dispersion.  With clever manipulation of the refractive indices, the ray travelling down the middle can be made to arrive at the same time as the ray that goes from side to side.  They can meet with a time difference of about 1 ns km-1


Monomode fibres are designed such that the rays pass only down the middle. If the light were perfectly monochromatic, i.e. of one wavelength only, the rays would all arrive at the same time. However even the best lasers produce a slight spread, and since refractive index varies with wavelength, there can be slight differences in arrival times, leading to smearing.

 

Optical fibres experience attenuation.

As light travels down an optical fibre, it loses intensity. This attenuation is caused by very slight impurities that you get even in the purest of glass. Also there will be defects and anomalies in the crystal structure caused by the manufacturing process. Light may even "leak out" of the fibre. Whatever the cause the light at the receiver will be dimmer than the light at the transmitter.

If the light intensity travels through 1 km of optical fibre, and its intensity is reduced to 50 % of its original, we can expect that after 2 km, then intensity is 25 % of the original, and after 3 km, it's 12.5 % (1/8) of the original. The change is exponential.

So the relationship between the intensity and the distance is going to be governed by an exponential function:

 

Where:

 

Of we have 1000 metres of optical fibre with an attenuation coefficient of 0.002 m-1, the graph is like this:

 

We can see that the intensity has dropped to 50 % of its value after 350 m.

If we plot the natural logarithm of the intensity against the distance, we get a straight line:

 

We would get a similar shape if we used logs to the base 10.

Typically attenuation coefficients are in the order of 10-5 m-1.  For long distance communication, it can be significant. Also the attenuation coefficient is different for different wavelengths in the same optical fibre.
 

Question 7

An optical fibre has an attenuation coefficient of 1.2 × 10-5 m-1.  At one end is an infra-red laser of power 250 mW.  The light passes a rectangular window that is 6.0 mm wide and 2.0 mm high.

(a) Show that the intensity of the radiation is about 21 × 106 W m-2.

(b) Calculate the intensity of the radiation at the receiver, if the optical fibre is 50 km long.

(c) What is the efficiency of the transmission? 

Answer


 

The Decibel Scale

In communication systems, the sound intensity that we hear is result of sound energy from from loudspeakers.  The rate at which this energy is given out is the power.  The human ear can just detect a doubling in power.  See Medical Physics Tutorial 3 to see how the ear behaves.  We measure the intensities of sounds, represented by powers using logarithmic scales to compress the graph into something more manageable.  In the decibel scale we use a reference power, P1,  a reference point to which we compare a second power P2.   We can write an expression for the change in power  DP.

 

 

1 Bel is the power change from 10-6 to 10-4 W.  The Bel (B) is rather a big unit and we use the decibel (dB) instead.

 

 

 A doubling of power gives a 3 dB increase.   The 3 dB increase is because if we double the power from 10 W to 20 W:

 

DP = 10 log10 (20 W ÷ 10 W)

 

DP = 10 log10 (2) = 10 × 0.3010 = 3.01 dB

 

We often describe an amplifier as having a gain of X decibels.  This means that the amplifier boosts signals by X decibels.

 

Worked example

A signal of 25 mW is boosted by 18 dB using a booster amplifier.  What is the output power of the amplifier?

Answer

DP = 18 dB = 10 lg10 (P2 ÷ 25 × 10-3 W)

1.8 = lg10 (P2 ÷ 25 × 10-3 W)

101.8 = P2 ÷ 25 × 10-3 W

P2 = 101.8 × 25 × 10-3 W = 63.1 × 25 × 10-3 W = 1.58 W = 1.6 W (2 s.f.)

 

 

The dBA scale is used to take into account the frequency dependence.  Remember that the ear is most sensitive at  frequencies between 100 to 10 000 Hz .  The table shows the levels of certain noises:

 

Level (dBA)

Noise

Effect

0

Threshold of hearing

 

20

Blood pulsing

 

30

ticking watch

 

40

Quiet conversation

 

50

Quiet street

 

70

Hoover in a room

 

90

Road drill at 7 m

Prolonged exposure can lead to hearing damage

100

Noisy factory

 

120 

Loud discothèque

Threshold of discomfort

140

Aircraft at 25 m

Threshold of pain

160

Rifle close to ear

Ear drum ruptured

 


 

Radio Transmission

Most radio transmissions travel directly from the transmitter to the receiver.  This is known as line-of sight transmission.

 

Often the transmitter may not be in direct line of sight, but the waves can diffract around objects in the way, and diffract around the curvature of the Earth. The waves are known as ground waves or surface waves

 

While the ground is opaque to light waves, it can transmit radio waves to a limited extent, especially those of a frequency lower than 3 MHz.  There is a value for refractive index for the ground, while the refractive index for air for radio waves is very close to 1.0.  Therefore the waves can be guided along the interface between the two media. 

 


This has nothing to do with total internal reflection.

 

 

 

 

 

Long distance transmission of short wave radio waves uses sky-waves which are reflected off the ionosphere.  Short wave radio waves have a frequency between 1.6 to 30 MHz (187 m down to 10 m wavelength).

Image from Wikimedia Commons.  Authors: Kf4yfd, Noldoaran, Augiasstallputzer

 

The ionosphere is a layer of the upper atmosphere from about 80 km to 1000 km above the Earth's surface.  Neutral air molecules are ionised by cosmic rays and high energy photons from the Sun.  The physics process is called back-scattering.  If the ionisation is strong enough, the radio waves are reflected back to the ground. When they strike the ground, they are diffusely reflected back to the ionosphere.  So the waves bounce off the ionosphere to the ground, enabling the waves to be received much further than would be expected.  Transmissions of just a few watts can be picked up 3500 km away.

 

The ionosphere is not a stable layer.  It undulates like a sea, leading to periodic variations in signal strength, which can be observed in the phenomenon of fading.  Fading can make broadcasts unpleasant to listen to, if not unintelligible.  (Ask your parents or grandparents who listened to Radio Luxemburg during the evening on 208 m of the Medium Wave Band when they were teens.)  Fading can be explained by considering the behaviour of standing waves.  Similar fading can be heard when listening to FM radio while driving through towns.  Waves reflected by buildings destructively interfere with the incoming radio wave.  The resultant wave has a much lower amplitude.

 

For waves below 10 MHz, sky-wave propagation is most efficient at night.  For frequencies above 10 MHz, the propagation is most efficient by day.

 

This process does not happen with high frequency waves (100 MHz), except in very unusual atmospheric conditions.

 

 

Satellite Transmission

Communication satellites are widely used in broadcasting nowadays, enabling live events to be broadcast throughout the world in real time.  Before the satellites were invented, events were recorded on film, then flown around the world.  A process that take days then is done in a fraction of a second.  The satellites have a receiver which they pick up waves from a ground station.  Then they retransmit the signals back to the ground, where they are picked up using a satellite dish like the one above.  The idea is like this:

 

 

The physics of satellite orbits is covered in the section of work on gravity fields.  Kepler III is the rule that links orbital height and periodicity.  There are three types of orbit:

Wikimedia Commons.  Author: Brandir

 

The angle of the beam leaving the parabolic dish on the satellite determines the coverage on the ground.

 

 

Consider a satellite that has a dish that is d m in diameter.  The satellite is h m above the ground and transmits with a wavelength l m.  The angle of aperture is given by the equation:

 

We can then work out the radius r by using:

 

Question 8

A geostationary satellite is 36000 km above the Earth.  It has a transmitting antenna of diameter 2.0 m.  If the frequency of the waves transmitted is 7.5 GHz, calculate the radius of coverage.

Answer

 

The UHF waves from a satellite pass straight through the ionosphere.

 

 

 

Satellite Communication Frequencies

These are in the range 1 GHz to 40 GHz (wavelength 0.3 m to 7.5 × 10-3 m).  They are in internationally agreed bands which are set out in the table below:

 

Band

Frequency /GHz

Use

L

1 - 2

Global Positioning System; satellite telephony; maritime and aeronautical communications

S

2 - 4

Weather radar; ship surface radar; some communications

C

4 - 8

Satellite communications, TV relays

X

8 - 12

Military radar, weather monitoring, air traffic control, vehicle tracking in law enforcement.

Ku

12 - 18

Satellite broadcasting downlink, e.g. Astra Satellite.

Ka

26 - 40

Satellite broadcasting uplink. high resolution radar.

 

High frequencies tend to be affected by very heavy rainfall in tropical areas (known as rain-fade).  So satellites used in these areas tend to use lower frequencies.

 

The radio signals sent from the earth are referred to as the uplink.  The signals sent from the satellite back to the Earth are called the downlink.  These two links have different frequencies.  The uplink has a higher frequency.  The reason for this is that high frequencies are needed to get radio signals through the atmosphere without being reflected.  However the high frequency signals are attenuated more by the atmosphere, so need to have a source with higher power.  This is easily achieved on a ground station; you simply use a big enough power supply.

 

Where there are several satellites in a network (sometimes called a constellation), there will be channels of communication between the satellites.  These are called cross-links.  The idea is shown below:

 

The signal from Earth Station A will not reach Satellite B, for it to relay the communication to Earth Station B.  So Earth Station A has to communicate to Satellite A.  This acts as a repeater station, sending the information on a cross-link to Satellite B, which in turn repeats the information to Earth Station B.

 

 

Satellites need to be lightweight as they are hugely expensive to launch.  The power supplies (from solar panels with back-up batteries) are limited, so a lower downlink frequency is used.  In the C-band the uplink frequency is 6 GHz, while the downlink is 4 GHz.  Another reason for this is to prevent interference that would occur if the uplink and downlink signals had the same frequency.  Another advantage of a lower frequency is less tendency for rain fade.

 

De-sensing is the a problem encountered frequently by radio engineers.  It is the effect of a strong transmitted signal on the weak received signal.  It happens if the reception and transmission frequencies are close and results in audio distortion,  and loss of range.  At worst it can lead to the complete loss of the signal.  The solutions to these problems are well-known to radio-engineers, e.g. antenna design and separation.  Frequency separation is the easiest to achieve with a satellite.

 

 

Advantages and Disadvantages of Transmission Media

 

These are summarised in the table:

 

Medium

Advantage

Disadvantage

Security

Cost

Radio

No wires needed.  Can be spread over a wide area

Energy inefficient.  Interference can make signals unintelligible.  Poor security

Poor.  Signals can be easily intercepted.

Low

Wire

Less energy needed.  Transmission and reception equipment are simpler, with no need for modulation.

Only a limited number of receivers are possible.  Degradation of quality over distance due to attenuation and noise.  High cost of infrastructure.

Better than radio.  Physical connection needed to tap wire.  Electromagnetic radiation can be detected and used to pick up signals (eavesdropping).

High

Fibre-optic

High rates of data transmission are possible.  Less susceptible to noise and interference.

Optical fibre is digital only.  Therefore conversion from analogue and back are needed.  Cracks in glass can interfere with transmission.

Better than electrical wires, but still possible to eavesdrop using sophisticated equipment.

High

 

Security is an issue with all data transmission.  With radio, it is possible for anyone with appropriate equipment to intercept radio messages.  If the data are sensitive, then encryption methods will enhance security.  During the Second World War, the German Enigma machines provided sophisticated levels of encryption, but the capture of several machines enabled allied cryptographers to crack the code.  This was helped by an early day computer called Colossus.  In an experiment conducted just a few years ago, a replica Colossus cracked the Enigma code faster than a modern-day PC.

 

Security of modern Wi-Fi systems, which are radio-based is certainly a concern.  A wired network is more secure.  An optical fibre network is even more secure, but not completely so.  Individuals and organisations that hold sensitive data on their computer systems must ensure that these data are kept secure from hackers.  The computer is a more likely target than the transmission system, so measures need to be taken to:

 

Some embassies that handle information that is sensitive towards national security have started to go back to a more low-tech approach.  They use typewriters and paper.