Data, A/D, propagation, antennas

Welcome to part three, the last one. We cover digital data transmission and coding, analogue to digital conversion, radio propagation and polarization, and antennas with the Friis transmission equation, which gave the link-budget problems of the 2022 and 2023 partials. As before, every exam question on these topics is solved, and checked against the official key where one exists.

Data communication moves characters from one point to another, today almost always between computers. In parallel transmission every bit of a character travels at the same time on its own line, which is fast but needs many lines, so it is only used over very short distances; the handwritten notes show the historical example of Salvà's telegraph, one wire per letter. In serial transmission the bits travel one after another on a single line, which takes longer but is the only practical solution over long distances. A serial link must tell the receiver where each word starts and ends: in asynchronous transmission every character carries its own start and stop symbols, as drawn here, so the circuits are simple, but the extra symbols carry no user information and the gap between characters is not fixed. In synchronous transmission there are no start or stop symbols: a special sequence synchronises the receiver's clock once, and then a continuous flow of symbols is kept even when there is nothing to send, which is faster but needs more sophisticated equipment, as in the digital links between telephone exchanges.

A binary signal takes one of two values in each symbol interval, and an M-ary signal one of M values. Nothing forces data transmission to be binary: with four voltage levels, each level carries two bits, and in general M symbols represent n bits when M equals 2 to the n. In the figure, the binary signal needs twelve symbols to send twelve bits, while the 8-ary signal sends the same twelve bits with only four symbols, three bits each, so it is three times faster at the same symbol rate. Problem 3.8 is the inverse question: to code the five vowels you need at least 3 bits, because 2 bits only give four words and 3 bits give eight.

Do not confuse the two rates, it is a favourite trap. The signalling rate is the number of symbols per second, measured in baud, and a channel of bandwidth B allows at most 2 B baud, a limit only reached with special pulse shapes. The information rate is the number of bits per second, which equals the baud rate times the number of bits per symbol, n, where M equals 2 to the n levels. The two coincide only in a binary synchronous link; in a binary asynchronous link the bit rate is lower than the baud rate because of the start and stop symbols; and in an M-ary synchronous link the bit rate is higher. That is how old telephone modems worked: half a telephone channel, about 1.5 kHz, allows 3 000 baud, yet the V.34 standard reached 33 600 bits per second with multilevel symbols.

This problem, from the collection, came back in 2023 with the new 8.33 kHz channels, so let us do both. The maximum signalling rate is twice the bandwidth: 2 times 25 kHz, 50 kilobaud. With 4 levels each symbol carries 2 bits, because 2 squared is 4, so the information rate is 50 thousand times 2, 100 kilobits per second; Barlabé confirms this step in the FAQ. With 30 dB, that is a ratio of 1 000, Shannon gives 25 000 times log base 2 of 1 001, which is 249.2 kilobits per second. For the 2023 version, the 8.333 kHz channel allows 16.67 kilobaud. Eight symbols carry 3 bits each, so 50 kilobits per second, and the capacity at 30 dB is 83.1 kilobits per second.

If multilevel signals carry more bits per symbol, why not use thousands of levels? Because, for the same signal power, more levels means the levels are closer together, and noise pushes the received value across a decision threshold more easily, as the handwritten notes sketch. These curves show it: at any given signal to noise ratio, the error probability grows with the number of levels. So in the 2016 partial, with the same signal to noise ratio and the same signalling rate, the 16-level signal has a larger error probability than the binary one. In the presence of noise you cannot make a digital signal look more and more like an analog one; coding is then used to recover part of the reliability.

Modulation is processing of the signal; coding is processing of the symbols of the digital message. Source coding removes redundancy so that the source is represented with fewer bits: Morse does it by giving the shortest words to the most frequent letters, like a single dot for E, compressors like ZIP do it without loss, and JPEG, MPEG or MP3 do it with an imperceptible loss; reducing the number of bits is exactly the 'main objective of source coding' asked in the question sheet. Channel coding does the opposite: it adds bits to detect and correct transmission errors and to keep synchronism, with parity bits, block codes, Hamming codes and convolutional codes decoded with the Viterbi algorithm. A single parity bit detects one error, misses two, and corrects none. The Hamming code of the notes maps 4 data bits into 7, choosing 16 of the 128 possible words so that any two differ in at least three bits, so one or two errors are detected and a single error is corrected by picking the closest valid word; the price is redundancy. For the exam, 'name a source code and a channel code' has a ready answer: Morse or ZIP for source, parity or Hamming for channel.

The last family of channel codes shapes the electrical waveform: line codes. Binary signals can be unipolar or bipolar, and return to zero, R Z, if they go back to zero during half of the bit, or non return to zero, N R Z, if they hold the level for the whole bit. A long run of zeros in unipolar format has no transitions, so the receiver loses bit synchronism, and a long run of ones leaves a non-zero mean, a DC component, which transmission systems must avoid. A M I, alternate mark inversion, makes successive ones alternate in polarity, which removes the DC component, but a long run of zeros is still a flat line with no transitions; H D B three, not drawn here, fixes that by replacing every run of four zeros with a pattern containing pulses, so the receiver keeps synchronism. Local networks use the Manchester code, with a guaranteed transition in the middle of every bit: no DC and easy synchronism, but at the cost of a larger transmission bandwidth.

Long-distance data links modulate a sinusoidal carrier, and with a digital modulating signal amplitude, frequency and phase modulation are called ASK, FSK and PSK. Facing noise, for a given signal to noise ratio, PSK gives the lowest error probability, followed by FSK, and ASK is the worst. A modem, short for modulator-demodulator, is the equipment that converts a low-pass digital signal into a band-pass digital signal and back, which is the answer to 'what is a modem' in the 2023 partial. To go faster, modems combine amplitude and phase, and such multilevel signals are drawn as a constellation: one dot per symbol at the tip of its phasor, here twelve phases, the four inner points sharing four of them, and two amplitudes: sixteen symbols. Above the line codes, communication protocols, organised in the seven layers of the I S O O S I model, decide who talks, how errors are reported and how packets are routed.

Analogue to digital conversion is the last topic of the exam scope, and its block diagram is a sure question. First comes the anti-aliasing filter, an analogue low-pass filter that guarantees a signal band-limited to W at the input of the sampler; that is why a filter is needed as the first stage, to avoid aliasing. Then the sampler measures the signal at regular instants separated by the sampling period T s, giving a discrete signal. Then the quantizer assigns to each sample the nearest of a finite number of amplitude levels, giving a multilevel digital signal. Finally the coder replaces each level by an N-bit binary word, so the bit time is the sampling period divided by N. Filter, sampler, quantizer, coder: memorise the four names in this order, in Catalan if you like: filtre anti-aliasing, mostrejador, quantificador, codificador.

Why must the sampling frequency be at least twice the bandwidth? The spectrum of a sampled signal is periodic: the original spectrum, of width W on each side, is repeated around every multiple of the sampling frequency. As long as f s is at least twice W, the copies do not touch, and an ideal low-pass filter can recover the original exactly. Watch what happens as the sampling frequency is lowered: below twice W the replicas overlap, and the sum in red no longer looks like the original; this is aliasing, a distortion that nothing can undo afterwards. The minimum, twice W, is the Nyquist frequency, and because real filters need a transition band we sample a little faster: the compact disc samples 20 kHz audio at 44.1 kHz instead of 40.

The quantizer is a non-linear system, so it has no transfer function; it is described by its staircase input-output characteristic. Its error, the difference between each sample and the nearest level, cannot be corrected when converting back, but it becomes negligible with enough levels. Voice, in telephony or between pilot and controller, needs 256 levels, 8 bits, while high-quality music on a CD uses 65 536 levels, 16 bits, per stereo channel. The quantization steps need not be equal: non-uniform quantizers use smaller steps for the most probable values of the signal. After coding, each sample becomes N bits, so the binary signal is N times faster than the multilevel one, and the converter is characterised by its number of bits per sample together with its maximum sampling frequency.

Here is the numerical example of the notes, a perfect exam question. Sampling at 8 kHz gives one sample every 125 microseconds, and 256 levels need 8 bits per sample. Eight bits must fit in 125 microseconds, so each bit lasts 15.625 microseconds, giving 64 000 baud. In a binary synchronous link the bit rate equals the baud rate: 64 kilobits per second, the classic telephone channel. And since a channel of bandwidth B carries at most 2 B baud, the minimum bandwidth is 32 kHz, far more than the 8 kHz that the notes quote for analog transmission: digital pays in bandwidth, which is why compression algorithms such as M P 3 or JPEG were developed.

Digital to analogue conversion is much simpler. A decoder turns the binary words back into a multilevel signal, and a low-pass filter removes the spectral replicas created by sampling, leaving only the original band. If the sampling respected Nyquist, the recovery is perfect, at least in theory. The only difference between the original and the recovered signal is the small quantization error introduced in the A to D step.

Now propagation, starting with the plane wave, the simplest solution of Maxwell's equations and a good local model of any wave far from its antenna. In a plane wave the electric field, the magnetic field and the direction of propagation are mutually perpendicular, and the ratio between E and H is the intrinsic impedance of the medium, the square root of mu over epsilon, about 377 ohms in vacuum, the analogue of the characteristic impedance of a line. The wave number is omega over the propagation speed, which is one over the square root of mu epsilon, and the refractive index is the square root of the relative permittivity. At a discontinuity, part of the wave is reflected, with the angle of reflection equal to the angle of incidence, and part is refracted into the second medium following Snell's law, n 1 sine theta i equals n 2 sine theta t, with all angles measured from the normal to the surface. The reflection coefficient depends on both media, on the angle of incidence and on the polarization: decompose the electric field into a component perpendicular to the plane of incidence and one parallel to it; for a reflection from the ground, a horizontally polarized wave is the perpendicular case and a vertically polarized wave the parallel one; each has its own reflection coefficient, so an oblique linear polarization generally comes back with a different orientation, or even elliptical. Diffraction is what cannot be explained by reflection or refraction: by Huygens' principle every point of a front radiates a new wave, so waves go round obstacles, which gives some coverage behind hills.

The main propagation mechanism depends on the frequency band. Below about 100 kHz, waves are guided between the Earth's surface and the ionosphere; this was used by the old Omega and Loran navigation systems, and today only for communications with submarines. Between 100 kHz and 3 MHz the surface wave dominates, following the ground; it needs vertical polarization, which is why the NDB and ADF system uses vertically polarized antennas, and AM broadcasting lives here too. Between 3 and 30 MHz, waves bounce off the ionosphere, reaching very long distances, farther at night because the lower layer rises. Above 30 MHz propagation is optical, along the line of sight, with a direct wave and a ground-reflected wave in the troposphere: this is where aeronautical VHF voice, VOR, DME, TACAN, secondary radar and satellite systems work, and it is the regime for all the Friis problems.

Polarization is the orientation in space of the electric field of the wave, the definition asked word for word in 2016 and in the question sheet. To see it, fix a point, for example z equal to zero, and follow the tip of the electric field vector as time passes. If the field has a single component, or two components in phase, the tip moves along a line: linear polarization, vertical, horizontal or oblique. If the two components are 90 degrees apart, the tip draws an ellipse, and if they are also equal in amplitude, a circle. Watch the four cases: vertical, oblique x plus 2 y, circular x minus j y, and elliptical 2 x minus j y. Antennas take the name of the polarization they radiate far away: a vertical dipole is vertically polarized, and aeronautical voice antennas use vertical polarization.

Barlabé's method, from the FAQ: fix z equal to zero, write any j as e to the j 90 degrees, take the real part, and look at how each component varies with time. In part a, the x component is 2 E zero cosine omega t. The y component has a j in front, so it is 2 E zero cosine of omega t plus 90 degrees, which is minus 2 E zero sine omega t. Equal amplitudes and a 90 degree shift: the tip of E draws a circle, so the polarization is circular, rotating from plus x towards minus y. The answer key calls it left-hand circular, consistent with the notes' worked example, where x minus j y, rotating the other way, is right-hand circular. In part b both components are j times E zero, so both equal minus E zero sine omega t: they are in phase, the tip moves along the 45 degree diagonal, and the polarization is linear, oblique.

An antenna is the transition region, the interface, between a wave guided by a transmission line and a wave radiated into space, and it can also give the radiation a direction; opening up the end of a two-wire line is how the dipole was born. Far from the antenna, in the far-field or Fraunhofer zone, the wave arriving at a receiver can be treated locally as a plane wave. Do not confuse isotropic and omnidirectional: an isotropic radiator radiates equally in every direction of space and cannot physically exist, while an omnidirectional antenna radiates equally in all directions of one plane. Each system needs its antenna: broadcasting uses an omnidirectional transmitter and directive receivers, or omnidirectional ones if there are several transmitters; a point to point link uses two very directive antennas; and a radar uses a rotating directive antenna. By Lorentz reciprocity, the same antenna transmits and receives with the same pattern, so we simply speak of the radiation pattern. So the 2022 question about the pattern of an aeronautical voice antenna on an aircraft has this answer: omnidirectional in the horizontal plane, because the ground stations can be in any direction, with vertical polarization.

Barlabé's slides on antenna structures are a picture gallery, so for the exam just know the families and where they are used. Dipoles, usually half a wavelength long, and their arrays, as in the ILS localizer; folded dipoles; and quarter-wave monopoles over a ground plane, sometimes with a capacitive load or inside a fairing. Slots and slot arrays, common in radars, travelling-wave antennas, horns, and parabolic reflectors such as the Cassegrain, the offset and the Gregorian. Helical antennas, Yagi-Uda arrays, microstrip patches like those of the SMOS satellite, lenses, and wideband designs such as the biconical, the discone, the log-periodic and fractal antennas. On aircraft, the NAV COM antennas are monopoles or dipoles, inside fairings on fast aircraft, which is exactly your course project: a vertical quarter-wave monopole or half-wave dipole for the voice band, 118 to 137 MHz, centred at 127.5 MHz, where the wavelength is 2.35 m, so about 59 centimetres or 1.18 metres before end corrections.

Now the most important formula of the partial, derived in four steps. An isotropic radiator in a lossless medium spreads its power P T over a sphere, so at distance R the power density is P T over 4 pi R squared, falling with the square of the distance: 6 dB every time the distance doubles. A real antenna concentrates power in some directions; the directivity function D of theta and phi is the ratio between its power density and the isotropic one, so the density towards the receiver is multiplied by D T. The receiving antenna behaves like a funnel: the power it delivers to its load is the incident density times its effective area A ef. Reciprocity links the two parameters: the effective area is lambda squared over 4 pi times the directivity. Putting it together: P L equals P T times D T times D R times lambda over 4 pi R squared, times the product of loss coefficients c i, each between zero and one, for medium losses, polarization mismatch, mispointing, mismatch and efficiency; and, as the FAQ insists, everything in linear units, watts and metres.

The notes use this example to show the orders of magnitude. Convert the directivity: 3 dB is a factor of 2, and the distance is ten to the 4 metres. Power density times effective area: 1 000 watts over 4 pi times ten to the 8, times 2, times one square metre. That gives only 1.59 microwatts. From one kilowatt to one and a half microwatts is an attenuation of 88 dB, which is why received radio powers are always tiny and are expressed in dBm.

The radiation pattern draws the power density radiated at a fixed distance as a function of direction, normalised to its maximum and usually in decibels on a polar plot. The biggest lobe, at 0 dB, is the main lobe, and all the others are secondary lobes. The beamwidth is the angular width of the main lobe between the points where it falls to minus 3 dB, half power, or between its nulls. The main to secondary lobe level, N L P S in Catalan, is the difference in decibels between the main lobe and the largest secondary lobe, wherever it is. The front to back ratio compares the main lobe with the lobe 180 degrees behind it. Finally, the gain refers to the power delivered to the antenna rather than radiated, so gain equals efficiency times directivity, practically equal unless the antenna is damaged; and the radiation resistance is the resistor that would dissipate the same power the antenna radiates.

Two more antenna parameters complete the notes, and both are about connecting the antenna to a circuit. The antenna impedance is the ratio of voltage to current at its terminals; by reciprocity it is the same whether the antenna transmits or receives. To the generator, a transmitting antenna looks like a resistor that dissipates the radiated power: that resistor is the radiation resistance. The improved model adds the ohmic loss resistance of the metal, usually negligible, and a reactance, so the impedance is R rad plus R omega plus j X ant. When receiving, the same impedance sits in series with a source, the open-circuit voltage V ant that the incident wave induces at the terminals. The effective length is that open-circuit voltage divided by the incident electric field; it has units of metres but is not the physical length, and it relates to the effective area through A ef equals eta, the intrinsic impedance of free space, about 377 ohms, times l ef squared over four R ant.

Problem 2.4 gives a measured pattern, reproduced on the right, and asks three readings. The main lobe points east, to 090, and the minus 3 dB points are about 65 degrees apart, the official beamwidth; reading graphs always has some error, and Barlabé accepts anything within the reading error. The largest secondary lobe, here the back lobe at 270, is at about minus 18 dB, so the main to secondary lobe level is 18 dB, which here happens to coincide with the front to back ratio. When the antenna turns 30 degrees clockwise, the wave from the east arrives 30 degrees away from the main axis, where the pattern reads about minus 2 dB: the mispointing coefficient. So 1 microwatt, which is minus 30 dBm, becomes minus 32 dBm, which is 0.63 microwatts. Problem 2.5 is the same with a formula: sine squared theta is maximum at theta equal to 90 degrees, and turned 45 degrees it gives sine squared of 45, one half, so 0.5 microwatts.

Now the link-budget problems, starting with the two aeronautical ones of the collection. In 2.1 the wavelength is 3 times ten to the 8 over 120 MHz, 2.5 m; 5 nmi is 9 260 m; and 2 dB is a factor 1.585. Friis gives 10 watts times 1.585 squared times 2.5 over 4 pi times 9 260, squared, which is 11.6 nanowatts, the official answer. In 2.2 we solve for the transmitted power: at 127 MHz the wavelength is 2.36 m, and the product of directivities and losses is 1.585 times 1.995 times 0.251, which is 0.794, since minus 6 dB is a quarter. The minimum transmitted power is the receiver sensitivity divided by that product and by the free-space factor, which is 1.41 times ten to the minus 11 at 50 km. That gives 8.9 watts, again the official answer.

Space links push Friis to the extreme. In 2.3 we solve for R: R max equals lambda over 4 pi times the square root of P T D T D R over P L min; with lambda equal to 3 centimetres and the product under the root equal to ten to the 24, the maximum range is 2.39 million kilometres, the official answer. In the 2023 partial the distance is 20 light-minutes, 20 times 60 seconds times the speed of light, 3.6 times ten to the 11 metres, so the free-space factor is 4.4 times ten to the minus 29. The up-link uses the 3 000 watt ground transmitter, 70 dB, that is ten to the 7, and the probe's antenna, with a gain of 50 dB, ten to the 5, so the probe receives 1.3 times ten to the minus 13 watts. The down-link uses the same two antennas but only 300 watts, so the ground station receives ten times less, 1.3 times ten to the minus 14 watts. For part c, the effective area is lambda squared times D over 4 pi, 716 square metres on the ground and 7.16 on the probe, and if that is 70 % of the circular aperture, the diameters are 36 m and 3.6 m.

The 2022 partial asked a practical version, in decibels. Add up everything in decibels: 17 dBm transmitted, plus 3 and 3 dB of directivity, down to a sensitivity of minus 70 dBm, so the link can afford 93 dB of free-space loss, which is 20 log of 4 pi R over lambda. At 2.4 GHz the wavelength is 12.5 centimetres, so R equals lambda over 4 pi times ten to the 93 over 20: about 444 metres in free space. Inside the building, the loss coefficients are minus 3 dB for polarization, minus 6 for multipath and minus 5 for each of the three walls, a total of minus 24 dB, which leaves 69 dB for distance. Then R is only about 28 metres; every 6 dB of extra loss halves the range. And question 4 of the same paper is the same rule: doubling the distance costs 6 dB of signal while the noise stays the same, so an S over N of 20 dB becomes 14 dB.

Since you cannot bring anything into the exam, here is the sheet to rewrite from memory on Wednesday evening. Navigation: ground equals air plus wind, turn the wind round by 180, components with sine and cosine, the sine and cosine laws, and one minute of latitude is one nautical mile. Signals and systems: the mean is c zero, a cosine gives two lines of A over 2, rectangle and sinc, the ideal low-pass impulse response, delta in gives h out, Y equals X times H, and non-linear systems create new frequencies. Noise and data: k T B with T in kelvin, decibels and dBm, minus 6 dB per doubling of distance, regenerative repeaters, 2 B baud, bits per symbol, Shannon, and more levels mean more errors. A to D: filter, sampler, quantizer, coder, sampling at least at twice W, and 64 kilobits per second for digital voice. Antennas: Friis in linear units, the effective area formula, beamwidth and N L P S; plus the definitions that are asked again and again: half-duplex, broadcast, polarization, loxodrome and orthodrome, and the network topologies.

Let me finish with a plan for the days before the exam. On Sunday, watch parts one and two again where needed, and redo the navigation problems and the signals problems on paper without looking at the solutions. On Monday, part three, the antenna and data problems, and the list of theory questions until you can answer each one in two sentences. On Tuesday, sit the 2016 and 2022 partials as real mocks, one hour each, skipping question 5 of 2016, the envelope of a modulated signal, which is passband material and out of scope, and on Wednesday the 2023 paper, then rewrite the memory sheet from scratch. On Thursday, a light review in the morning, then room 2.3 of TR5 at 15:00 with your calculator and your ID; and remember that even if something goes wrong, this partial can be retaken on the 20th of January. Good luck with the exam.