Signals, systems and noise

Welcome to part two: the fundamentals of telecommunication, following Professor Barlabé's notes Introducció a la Telecomunicació, part one. This is where most of the short theory questions of the partial come from, so we will go definition by definition, and solve every related exam question on the way. The plan is: definitions and the general model, types of signals and their characteristic values, services and networks, then Fourier analysis, linear systems and filters, modulation, distortion, noise, decibels, and finally the Shannon capacity law.

Etymologically telecommunication means communication at a distance, but the course uses the definition of the International Telecommunication Union. According to the I T U, telecommunication is any transmission and reception of signs, signals, writing, images, sounds and information of any kind by wire, radio, optical or other electromagnetic systems. The key word is electromagnetic: an African talking drum or smoke signals are communication at a distance, but they are not telecommunication. Information is anything that increases our level of knowledge, and the message is the physical manifestation of that information. A signal is a physical magnitude whose value carries information; if it is a voltage or a current we call it an electrical signal, and we model it mathematically as a function of time, x of t. A system is a set of components with a purpose; for us it is a box that receives an input signal x of t and delivers a different output signal y of t.

Every telecommunication system fits this block diagram: a source, a transmitter, a channel, a receiver and a destination. The channel is the physical medium: a transmission line such as a two-wire cable, a coaxial cable or an optical fibre, or the air and the vacuum for radio, and that is the only big classification of telecommunication, line or radio. The two main characteristics of a channel are its attenuation and its bandwidth: every signal is attenuated as it propagates, and the attenuation depends on frequency. The transmitter conditions the signal for the channel: it modulates, codes and encrypts it, and then amplifies it to overcome the attenuation. The receiver gets a very weak signal, so it first amplifies it and then undoes the transmitter's processing: it demodulates, decodes and decrypts.

Signals are first deterministic or random. A deterministic signal is fully specified at every instant, like a cosine, while a random signal can take one of several values with a certain probability. Strictly speaking, any signal that carries information must be random, because if I already know what you will say, you are not giving me information; but random signals are hard to handle, so we model them with deterministic ones whenever we can. An analog signal is continuous in time and in value; a discrete signal has any value but only at precise instants; and a digital signal has only a finite set of values at those instants. A periodic signal repeats identically every period T zero. Converting analog to digital and back leaves a small difference, a distortion, which can be made negligible; that is why today almost everything is transmitted digitally, for a reason we will see with noise.

Next, the characteristic values of a signal, which come up in almost every partial. The peak value is the maximum absolute value, and the mean value is the time average, which for a periodic signal is the integral over one period divided by the period. The instantaneous power is x squared, which is the power dissipated in a one-ohm resistor; with another resistance it is v squared over R for a voltage, or i squared R for a current. The mean power is the average of the instantaneous power, the rms value is its square root, and the energy is the integral of x squared over all time; energy signals have finite energy, while periodic signals have finite mean power. Two exam classics follow directly: in problem 3.1 the mean of A cosine plus B sine at the same frequency is zero, because every sinusoid averages to zero over a period, and in the 2016 partial the mean of B over 2 plus A cosine is simply B over 2, the constant term.

Telecommunication services are classified by how they use the channel. Simplex means one direction only, like television, or the link that carries processed radar information to the controller's screen. Half-duplex is bidirectional but not simultaneous, on the same channel, and it needs a protocol to change direction: this is the pilot talking with the controller, on one frequency, never both at once, ending each message with the call sign. Full-duplex is bidirectional and simultaneous, and it needs two channels, one for each direction, as in the telephone. The question 'what kind of communication is established between pilot and controller' appeared in the 2016 partial and is on the question sheet: the answer is half-duplex, and you should say why.

Services are also classified by where the transmitters and receivers are. Point to point means a single transmitter and a single receiver in different places, like a radio link or a phone call. Broadcast, radiodifusió, means a few transmitters and many receivers, like television, or a satellite navigation system. In a radar, the transmitter and the receiver are in the same place. Another repeated exam question asks whether an NDB with its ADF is point to point, broadcast or radar: it is broadcast, because one non-directional beacon radiates and any number of aircraft receive it.

You cannot wire every telephone to every other telephone, and not everybody talks at once, so terminals are interconnected through networks with a few basic shapes. In a bus, all terminals hang in parallel from one single transmission line, as in Ethernet. In a ring, the line closes on itself, as in IBM's token ring. In a star, every terminal connects to one central point, a hub in a local network or an exchange in the telephone system. A tree is the intermediate case, a bus of buses, and the figure of question 7 of the 2022 partial, with small buses hanging from a main bus, is exactly a tree. The 2023 partial asked you to draw the three basic topologies, and the notes list bus, ring, star, tree and hybrid, so to be safe sketch bus, ring, star and tree, and remember that large networks use multiple stars.

Telecommunication problems are usually easier in the frequency domain than in the time domain, and it all starts with the sinusoid A cosine of two pi f zero t plus phi. A is the amplitude, f zero the frequency, the inverse of the period, and phi the phase, which shifts the peak away from the time origin. Using Euler's formula, e to the j theta equals cosine theta plus j sine theta, the sinusoid is the real part of a phasor: a vector of length A rotating counter-clockwise in the complex plane at f zero revolutions per second. Watch the projection of the rotating phasor on the real axis: it traces exactly the cosine. So the whole sinusoid is described by three numbers, and we can draw it in an amplitude versus frequency diagram: a single spectral line of height A at f zero. That diagram is the spectrum of the signal, its picture in the frequency domain.

Here is the example from the notes: w of t equals 7 minus 10 cosine of 40 pi t minus 60 degrees, plus 4 sine of 120 pi t. To read its spectrum, write every term as a positive cosine. The 7 is a cosine at zero frequency, the DC term, with amplitude 7. Minus 10 cosine becomes plus 10 cosine with 180 degrees added to the phase, so 10 at 20 Hz with phase plus 120 degrees. And a sine is a cosine delayed by 90 degrees, so 4 at 60 Hz with phase minus 90 degrees. The spectrum is therefore three lines, at 0, 20 and 60 Hz, each with an amplitude and a phase; notice also that the mean of w is 7, the DC term.

Any periodic signal can be decomposed into an infinite sum of sinusoids at multiples of the fundamental frequency, the harmonics: this is the Fourier series. In compact exponential form, x of t is the sum over n from minus to plus infinity of c n times e to the j two pi n f zero t, and each coefficient c n is the integral over one period of x of t times e to the minus j two pi n f zero t, divided by the period. Watch the square wave being rebuilt: the DC term plus the first harmonic gives a rough sinusoid, and each new odd harmonic sharpens the edges. The term for n equal to zero is the DC term, and it is just the mean value of the signal: put n equal to zero in the coefficient formula and you get the definition of the mean, which is exactly what the 2023 partial asked you to prove. Parseval's theorem gives the mean power as the sum of the squared magnitudes of the coefficients. Because the sum runs over negative frequencies too, this is a two-sided spectrum: a cosine of amplitude A shows two lines of A over 2, at plus and minus f zero, which is why the official answers draw weights of one half, as the FAQ explains.

The most important periodic signal in digital communication is the train of rectangular pulses, of duration tau and period T. Its spectrum has lines every one over T, whose heights follow the envelope A tau over T times sinc of f tau, with A the pulse amplitude, where sinc of x is sine of pi x divided by pi x; the first zero of the envelope is at one over tau. The general lesson, shown in the notes, is that the more abrupt the transitions in time, the wider the spectrum, which matters a lot when we send digital pulses through band-limited channels. Now watch what happens as the period grows: the lines get closer and closer, and in the limit, when T goes to infinity, the signal is a single pulse and the spectrum becomes continuous; that limit is the Fourier transform. And problem 3.3 asks for the mean of this train: integrate A over one pulse and divide by the period, which gives A tau over T, the DC line you see in the middle.

This derivation is short and appears in the problem collection, so let us do it once properly. By definition, the transform is the integral of x of t times e to the minus j two pi f t; the pulse is 1 between minus tau over 2 and tau over 2, and zero elsewhere, so only that interval remains. The integral of the exponential is immediate, as Barlabé insists in the FAQ: you get e to the minus j two pi f t divided by minus j two pi f, evaluated between the limits, which is e to the j pi f tau minus e to the minus j pi f tau, over j two pi f. Euler's formula says that the difference of the two exponentials is two j times the sine, so the result is sine of pi f tau divided by pi f. Multiplying and dividing by tau, we recognise tau times sinc of f tau, the official answer, whose first zeros are at plus and minus one over tau. And problem 3.4 is free once you have this: its v of t is a staircase, height 2 inside tau 1 and height 1 out to tau 2, which is exactly the sum of two centred unit pulses, and the transform is linear, so it becomes the sum of two sincs, tau 1 sinc of f tau 1 plus tau 2 sinc of f tau 2.

Two more items from the Fourier section of the notes complete the picture. Rayleigh's theorem says that the energy of a signal can be computed in either domain: the integral of x squared over time equals the integral of the squared modulus of X of f over frequency. So the squared modulus of the transform is the energy spectral density, telling you how the energy is distributed in frequency; Parseval's theorem is the same idea for periodic signals and power. A computer cannot integrate a continuous signal, so it uses the discrete Fourier transform, the DFT, defined on N samples. The fast Fourier transform, the FFT, is not a different transform but an efficient algorithm to compute the DFT, needing about N log N operations instead of N squared: if asked, the DFT is the definition and the FFT is the way to compute it.

Now systems. A system is linear if it obeys superposition: the response to a sum of inputs is the sum of the responses, and its input-output characteristic is a straight line. The fingerprint of linearity is that a sinusoid at f zero comes out as a sinusoid at the same f zero, only with a different amplitude and phase. A non-linear system has a curved characteristic, its output shape depends on the input level, and new frequencies appear at its output, as in an amplifier driven into saturation. A linear and time-invariant system is characterised by its transfer function H of f, the transform of the output divided by the transform of the input, so the output spectrum is simply the input spectrum multiplied by H; and the range of frequencies where the response is flat, usually down to the minus 3 decibel, half-power points, is the bandwidth, giving low-pass, band-pass and high-pass systems. In the time domain the same system is described by its impulse response h of t, the inverse transform of H, and the output is the convolution of the input with h, an integral that is much harder than a multiplication, which is exactly why we work in frequency.

The handwritten class notes introduce the Dirac delta, and the 2023 partial asked about it directly. The delta is the limit of a pulse of width tau and height one over tau when tau goes to zero, and its defining property is sifting: integrating a function times a delta at t zero picks the value of the function at t zero. If the input is a delta, the convolution integral collapses to h of t. So the output is the impulse response h A of t, the inverse Fourier transform of the transfer function: that is the answer to question 7, and the reason for the name impulse response. Two more identities from the notes: convolving with a delta returns the signal itself, and convolving with a delayed delta delays the signal. Finally, with deltas we can transform periodic signals: the transform of a cosine is two deltas of weight one half at plus and minus f zero, the two-sided spectrum again.

This question appears in the collection, in 2016 and in 2022, so it is worth one minute of method. The first input is a tone at W over 2: its spectrum is two lines at plus and minus W over 2, inside the pass band, where H equals one. The second input is a tone at twice W: its lines are at plus and minus two W, outside the pass band, where H equals zero. Multiplying spectra, the first output spectrum equals the input spectrum, so y 1 of t equals x 1 of t: the tone passes unchanged. The second output spectrum is zero, so y 2 of t is zero: the tone is blocked. And in the 2022 version the input is the sum of both tones; the filter is linear, so by superposition only the tone at B over 2 survives.

This is the mirror image of the rectangular pulse, and it was worth one and a half points in 2022. The impulse response is the inverse transform of H, which is the integral of e to the plus j two pi f t between minus B and B. The same algebra as before gives sine of two pi B t divided by pi t. Written with the sinc function, h of t equals 2 B sinc of 2 B t: a central peak of height 2 B at t equal to zero, and zeros at every non-zero multiple of one over 2 B. Remember the duality: a rectangle in time gives a sinc in frequency, and a rectangle in frequency gives a sinc in time; notice also that this ideal filter responds before the impulse arrives, which is why it cannot be built exactly.

Again, spectra multiply, so the output occupies only the frequencies that are both in the signal and in the pass band. The signal occupies from zero up to W, and the filter passes from B 1 to B 2. Since B 1 is below W and W is below B 2, the overlap goes from B 1 to W, shown in orange, mirrored at negative frequencies. So the output is a band-pass signal with bandwidth W minus B 1; draw it exactly like this on the exam, with both the positive and negative frequency bands.

Problem 3.7 is the reverse situation, and the collection asks you to draw the answer, not just give a number. Spectra multiply: the output spectrum is X of f times H of f, so X survives only where the filter equals one, that is between minus f zero and plus f zero. The positive band runs from f zero minus B to f zero plus B, so the filter cuts it exactly at its centre, and only the inner half, from f zero minus B to f zero, survives, shown in orange. By symmetry, at negative frequencies only the piece from minus f zero to minus f zero plus B survives. So draw each piece with the original sloping inner edge and flat top, ending in a vertical wall at plus and minus f zero, exactly as in the official solution.

Modulation is the systematic alteration of a parameter of a carrier signal according to the information carried by a modulating signal. In telecommunications the carrier is a sinusoid, because it occupies the least bandwidth, and it has only three parameters to vary: the amplitude, giving amplitude modulation, the frequency, giving frequency modulation, and the phase, giving phase modulation. The simplest example, double-sideband amplitude modulation, shows the essential effect, a translation of the spectrum: multiplying x of t by a cosine at f zero gives one half of X of f minus f zero plus X of f plus f zero; watch the baseband spectrum split and move to plus and minus f zero, now occupying 2 B; frequency and phase modulation also move the signal to the carrier, but their spectra and bandwidths follow other rules. There are five reasons to modulate: first, to radiate efficiently: an antenna must be comparable to the wavelength, and a voice signal between 300 and 3 400 Hz has wavelengths from 1 000 km down to 88 km, while at 130 MHz, in the aeronautical band, the wavelength is only 2.3 m. Second, to avoid hardware limitations; third, to reduce noise and interference, as frequency modulation does; fourth, to assign each station its own band; and fifth, to multiplex many signals on one channel.

Distortion is any difference between the shape of the output and the shape of the input. A system does not distort if the output is the input scaled by a constant K and delayed by t zero, which in frequency means a transfer function of constant modulus and a phase that is a straight line in f. If the modulus is not constant we have amplitude distortion, and if the phase is not linear we have phase distortion. Phase distortion is measured with the phase delay and the group delay; in a distortionless system they are equal, and equal to the delay of the signal. Linear distortion appears in linear systems and can be corrected: the handwritten notes show an equalizer whose transfer function is a constant divided by that of the amplifier, so their product is flat. The bandwidth is the range where amplitude and phase distortion stay below some limit, and it must fit the signal: about 5 MHz for video, 3.1 kHz for voice, and at least r over 2 for a digital signal of r symbols per second.

Non-linear distortion is produced by non-linear systems; it can be severe and it cannot be corrected with equalizers. If a single tone enters, the output contains the tone plus harmonics at integer multiples of its frequency, and the total harmonic distortion summarises them. If two close tones enter, besides their harmonics we get intermodulation products at combinations plus or minus k f 1 plus or minus l f 2; the third-order ones, like 2 f 1 minus f 2, fall right next to the wanted signals. So in the 2023 partial, a 1 kHz tone entering an amplifier and coming out with a 2 kHz tone is non-linear distortion, more precisely harmonic distortion, because a frequency appeared that was not at the input; a linear system can never create frequencies. Problem 3.6 is the same idea with algebra: squaring a cosine gives one half plus one half cosine of twice the frequency, so the output spectrum has a line of one half at zero and lines of one quarter at plus and minus 2 f zero. And a non-linear system has no transfer function at all, which is the answer to part c.

Noise is an unwanted random signal that is always present. Any resistor above absolute zero shows a random voltage between its terminals, due to the thermal agitation of its electrons: this is thermal noise, the most important kind, and other noise sources are usually treated as if they were thermal. Its spectrum is flat over the communication bands, hence the name white noise, and its power is k T B: Boltzmann's constant times the absolute temperature in kelvin times the bandwidth in hertz. Careful with the constant: the notes and the 2023 exam print 1.37 times ten to the minus 23 joules per kelvin, while the accepted value is 1.38; use whatever the exam gives you, the difference is negligible in decibels. Since noise cannot be removed, we live with it, and what matters is the signal to noise ratio S over N: with a large ratio the noise is imperceptible, with a small ratio it masks the information, and in digital systems it raises the probability of bit errors.

Since you cannot bring a formula sheet, here is the decibel toolkit you need in your head. A ratio in decibels is 10 times the base-ten logarithm of the power ratio, so doubling is about 3 dB, four times is about 6 dB, and ten times is exactly 10 dB. An absolute power in dBm is referred to one milliwatt: 0 dBm is one milliwatt, 30 dBm is one watt, minus 30 dBm is one microwatt, minus 60 is one nanowatt and minus 90 is one picowatt. In decibels, products become sums, so a received power in dBm is the transmitted power in dBm, plus both directivities in dB, minus the free-space loss, plus the loss coefficients in dB, which are zero or negative: a wall that attenuates 5 dB is a coefficient of minus 5 dB, which you add, so never subtract a negative number; and a signal to noise ratio in decibels is the signal in dBm minus the noise in dBm. One more rule, from the antenna notes: in free space, doubling the distance divides the power by four, minus 6 dB, and since the noise does not change with distance, an S over N of 20 dB at distance R becomes 14 dB at twice the distance, as in the 2022 partial. But the Friis equation itself, as Barlabé stresses in the FAQ, must be evaluated in linear units, watts and metres, so convert every decibel value first.

Two quick noise questions from recent partials. In 2022, noise power is proportional to the absolute temperature, so convert to kelvin: 17 degrees Celsius is 290 kelvin and 34 degrees Celsius is 307 kelvin; doubling the Celsius temperature does not double the noise. The ratio 307 over 290 is only plus 0.25 dB, so the noise goes from minus 114 to about minus 113.75 dBm. In 2023, compute k T B: 1.37 times ten to the minus 23, times 290, times 4 million, which is 1.59 times ten to the minus 14 watts. In dBm, divide by one milliwatt and take 10 log: minus 108 dBm. The signal is 0 dBm, so the signal to noise ratio is 0 minus minus 108, which is 108 dB.

Here is the one fundamental advantage of digital over analog, and it is a guaranteed exam question. Along a channel the signal is attenuated but the noise is not, so repeaters are installed every so often to amplify the signal. The handwritten notes show the problem: an ideal amplifier multiplies signal and noise by the same gain and keeps S over N, but a real amplifier adds its own noise, so after every hop the signal to noise ratio gets worse, and there is no way to fix it. Watch the analog chain on the top row: after each hop the waveform is noisier, which limits the maximum distance for a given transmitter power. A digital signal, however, can only take a few known levels, so if the noise is not excessive each repeater decides which level was sent and rebuilds a clean signal: that is a regenerative repeater, the bottom row. So in a digital system there is virtually no limit to the distance, and that is the answer to 'what is the main advantage of digital communication' and 'what is a regenerative repeater'.

The third limitation, after distortion and noise, is interference. Interference is the contamination of an information signal by another similar signal, normally of human origin, for example another transmitter. It is frequent and often very hard to remove in radio, and less common on lines. On lines, the typical form is crosstalk, diafonia: the conversation on one pair of a cable is heard in the background of another pair. It is caused by the partial capacitances between the conductors in the cable and is avoided by a good choice of transmission lines.

The two main limitations of a telecommunication system are bandwidth and noise, and Shannon combined them in 1948 in the Hartley-Shannon law: the capacity C equals B times the base-two logarithm of one plus S over N. The capacity is the maximum amount of information per second that the channel can carry: above it, errors are certain; below it, error-free communication is possible, at least in theory. Capacity is directly proportional to the bandwidth but grows only logarithmically with the signal to noise ratio, as the curve shows. For example, a 25 kHz aeronautical channel with an S over N of 30 dB, which is a ratio of 1 000, gives 25 000 times log base 2 of 1 001, which is 249.2 kbit/s, the answer to problem 3.11 c. And the new 8.33 kHz channels of the 2023 partial give 83.1 kbit/s at the same signal to noise ratio.

Three short capacity questions that keep coming back. Problem 3.9: white noise power is proportional to the bandwidth, so with the same signal power, dividing the bandwidth by three divides the noise by three, and the signal to noise ratio becomes 3 A; Barlabé's FAQ says it bluntly, it is much simpler than it looks. Problem 3.10, and question 7 of 2016: if you double the bandwidth you also double the noise, so the capacity becomes 2 B times log of one plus S over 2 N, which is less than twice the original capacity. With an S over N of 100, doubling the bandwidth multiplies the capacity by 1.70 and tripling it by 2.30, so the capacity increases but is not proportional; justify it exactly like this. Problem 3.12: in a radio link the received signal falls as one over R squared while the noise stays constant, so the capacity is B log 2 of one plus K 1 over K 2 R squared. When S over N is much larger than one, this becomes a constant plus a constant times log 2 of one over R: every doubling of the distance costs 2 B bits per second, which is the answer in the key.

Let us summarise part two. You know the I T U definition, the general model with the channel's attenuation and bandwidth, the duplex modes with the pilot and controller as half-duplex, and the NDB as broadcast. You can read and draw spectra, prove that the DC coefficient is the mean, transform a rectangle into a sinc and back, push tones through ideal filters, and say that a delta at the input gives the impulse response at the output. You know that modulation translates the spectrum, that linear distortion is correctable while non-linear distortion creates new frequencies, that thermal noise is k T B, and why regenerative repeaters make digital communication superior. And you can use Shannon's law and explain why doubling the bandwidth does not double the capacity. In part three we finish with data transmission, coding, analogue to digital conversion, propagation, polarization, antennas and the Friis link budget.