RF systems: propagation and modulation

Welcome to the first of three lectures for exam 1 of Sistemes de Radiofreqüència i Comunicació. This part covers the radio spectrum, how radio waves propagate, analogue and digital modulation, and the architectures of transmitters and receivers. Part two covers decibels, noise, linearity, sensitivity and dynamic range, and part three covers transmission lines, the Smith chart and matching. Every worked problem has been recomputed from scratch, and where the slides or last year's solutions give a number, the video says whether it agrees.

First, the rules of the game. According to the course calendar, Unit 1, radio-frequency propagation, was taught in the first three weeks and Unit 2, transmission lines, from week four, so exam 1 covers both, and it is worth 35 % of the final mark. The solved exam 1 that your classmates shared has ten short problems, roughly one per topic. They were: a noise-figure cascade, the local oscillator of a superheterodyne, the ratio between day and night maximum usable frequencies, the input impedance of a line, a VSWR, a coaxial cable from a datasheet, a cascade intercept point, a return-loss and cable-loss check, an FM bandwidth, and a compression and noise-floor question. None of them needs long algebra: the skill is choosing the right formula and converting quickly between decibels and linear units, which is exactly what we will practise.

The electromagnetic spectrum used for radio is divided by the I T U into bands, each one a decade of frequency. Low frequency goes from 30 to 300 kHz, medium frequency up to 3 MHz, high frequency up to 30 MHz, very high frequency up to 300 MHz, ultra high frequency up to 3 GHz and super high frequency up to 30 GHz. Aviation uses almost all of them: non-directional beacons sit between 190 and 535 kHz, long-range voice over the oceans uses HF, between 2 and 30 MHz, and the familiar VHF voice band runs from 118 to 137 MHz. Just below it, from 108 to 118 MHz, are the VOR and the ILS localizer, with the glide slope near 330 MHz. Around 1 GHz, the L band holds the DME, the secondary radar at 1030 and 1090 MHz and GPS at 1575 MHz, and higher up come the radio altimeter at 4.3 GHz and the weather radar near 9.4 GHz.

The first formula of the course links wavelength, frequency and the medium. The wavelength is the speed of light divided by the frequency and by the square root of the effective permittivity, because waves slow down in a dielectric. In air the permittivity is one, so 3 times ten to the 8 over ten to the 8 gives 3.00 m. In a polyethylene cable, with a permittivity of 2.3, the square root is 1.517, and the wavelength shrinks to 1.98 m: this is why a quarter-wave cable section is shorter than a quarter-wave antenna. And in water, with a permittivity of 80.4, the wavelength is only 0.335 m.

Radio waves travel through an atmosphere with distinct layers. The troposphere, from the ground to about 10 to 12 km, is where the weather is, and it bends VHF waves slightly, extending their range a little beyond the optical horizon. Above the stratosphere, from 50 to 80 km up to 400 or 500 km, the ionosphere is ionised by solar radiation, so its properties change between day and night and with solar activity. The D layer, present only in daylight, absorbs the lower frequencies below about 4 MHz, while the E, F1 and F2 layers reflect waves up to about 5, 10 and 30 MHz respectively. At night the D layer disappears and the other layers weaken, which is why the usable frequencies drop at night.

There are four ways for a radio wave to get from the transmitter to the receiver. The ground or surface wave follows the curvature of the Earth, from about 20 kHz up to 50 MHz; its losses grow with frequency and are larger over land than over sea, and they are smallest with vertical polarisation, which is why beacon antennas are vertical. Above about 50 MHz the ground wave dies quickly and communication relies on the space wave: a direct, line-of-sight ray, plus a ray reflected from the ground. In the HF band, the ionosphere can reflect the sky wave back to Earth, so a single hop can cover thousands of kilometres, which is how oceanic HF voice works. Finally, tropospheric waves are scattered or ducted in the lower atmosphere, occasionally carrying VHF far beyond the horizon.

This is a ground-wave problem with an empirical range formula. The square root of 5 000 watts is 70.7. Over sea the range is 3 times that, 212 nmi. Over land it is only 2 times that, 141 nmi. So the difference is 70.7 nmi, a third of the sea range, because the ground wave loses more energy over poorly conducting land.

Above 50 MHz, the range is set by the radio horizon. VHF voice, VOR and DME all need line of sight between the antennas, so the curvature of the Earth limits the distance. Binns gives the practical formula: the range in nautical miles is 1.23 times the sum of the square roots of the two antenna heights in feet. Because of the square roots, the range grows slowly: flying four times higher only doubles the range. The trap is in the units: the heights must be in feet and the result is in nautical miles, so convert metres first, dividing by 0.3048.

Two quick applications. For part a, the ground antenna is 10 m high, that is 32.8 ft, whose square root is 5.73. Then the square root of the aircraft height is 125 over 1.23 minus 5.73, which is 95.9, so the aircraft must fly at about 9 200 ft, 2.8 km. For part b the receiver is at ground level, so only the aircraft height counts: convert each altitude to feet, take the square root, multiply by 1.23, and multiply by 1.852 to get kilometres. At 150 m the range is 27 nmi, about 50 km, and at 300 m it is about 72 km. At 1 500 m it reaches 160 km, at 3 000 m 226 km and at 5 000 m 292 km: the plot is a square-root curve, steep at first and then flattening.

For sky waves, the key quantity is the electron density of the ionosphere. A wave sent straight up is reflected only if its frequency is below the critical frequency, fc, equal to 9 times the square root of the electron density, with fc in hertz and the density in electrons per cubic metre. Last year's solution used the same formula in friendlier units: fc in megahertz equals 9 times ten to the minus 3 times the square root of the density in electrons per cubic centimetre; both give the same number. A wave that hits the layer obliquely, leaving the ground at an elevation angle theta, can be reflected up to a higher frequency, the maximum usable frequency, approximately fc divided by the sine of theta. Below the MUF there is also a lowest usable frequency, the LUF, set by absorption in the D layer, so the usable window lies between the two. The MUF by day is roughly twice the MUF by night, and in practice one works at the optimum working frequency, 85 % of the MUF, for a safety margin.

Both sky-wave problems use the same two formulas. Convert the density first: 5 times ten to the 5 per cubic centimetre is 5 times ten to the 11 per cubic metre, so the critical frequency is 9 times its square root, 6.36 MHz. In part a, the sine of the angle is 6.36 over 24.58, which is 0.259, so the elevation angle is 15.0 degrees. In part b, assume a single hop with the reflection midway, and a flat Earth: half of 1 300 statute miles is 1 046 km, so the tangent of the elevation is 350 over 1 046, giving 18.5 degrees. Then the MUF is 6.36 over the sine of 18.5 degrees, 20.1 MHz, and the optimum working frequency is 85 % of that, 17.1 MHz. If the miles are read as nautical miles, the angle is 16.2 degrees and the OWF 19.4 MHz, so always state your assumption on the exam.

This came up in last year's exam, with the densities read from a graph of electron density against height. Use the critical-frequency formula in megahertz with the density per cubic centimetre. By day, 9 times ten to the minus 3 times the square root of ten to the 5 is 2.85 MHz; by night, with 2 times ten to the 4, it is 1.27 MHz. Dividing by the sine of 20 degrees, the MUF is 8.32 MHz by day and 3.72 MHz by night. The ratio is the square root of the density ratio, the square root of 5, which is 2.24, consistent with the rule that the day MUF is about twice the night MUF; notice that the angle cancels, so the ratio needs no trigonometry.

Obstacles and discontinuities modify propagation in four ways, and the size of the obstacle relative to the wavelength decides which. Reflection happens when a wave meets a flat object much larger than the wavelength, such as the ground or a hangar. Refraction happens when a wave passes from one medium into another and changes direction, as in the troposphere and the ionosphere. Diffraction happens when a wave meets an edge and bends around it, following the profile of the discontinuity, which gives some coverage behind hills. Scattering happens when the obstacles are smaller than the wavelength, like raindrops for radar waves, which fragment and re-radiate the wave in all directions.

The slides finish propagation with a graph of atmospheric absorption against frequency, and a question: which bands would you choose for secure applications? The absorption has peaks where water vapour and oxygen molecules resonate: water at 22 and 183 GHz, and oxygen at 60 and 119 GHz. Between the peaks there are windows, such as 35 and 94 GHz, where millimetre-wave links and radars can reach far. For a secure link you want the opposite: work at an absorption peak, especially the 60 GHz oxygen line, where the signal dies out within a few kilometres, so an eavesdropper far away cannot receive it.

Now the systems that send and receive the waves. A transmitter is the system that generates the signal, a receiver recovers the wanted signal from a wide spectrum full of other transmitters, interference and noise, and a transceiver is both in one box. The information starts as a baseband signal, whose spectrum is concentrated near zero frequency; the modulator converts it into a band-pass signal around the carrier, which is then upconverted and amplified by the power amplifier. At the other end, a low-noise amplifier, more amplifiers, downconversion and a demodulator recover the baseband signal with as little noise and distortion as possible; we will see in part two why the first amplifier matters most.

Any modulated carrier can be written as an amplitude A of t times the cosine of omega C t plus a phase phi of t: amplitude modulation varies A, frequency and phase modulation vary phi. In AM the envelope is the carrier amplitude times one plus m times the message, where the modulation index m is the message amplitude over the carrier amplitude, between zero and one; watch the envelope deepen as m grows. In the spectrum, the carrier is accompanied by a lower and an upper sideband, mirror images of the message, so double-sideband AM occupies twice the highest message frequency. In the slide example, an audio signal from 20 Hz to 30 kHz on a 100 kHz carrier needs 60 kHz, and each sideband is 29.98 kHz wide: the lower one runs from 70 to 99.98 kHz and the upper one from 100.02 to 130 kHz.

In frequency modulation the amplitude stays constant and the phase is the integral of the message, so the instantaneous frequency follows the message. The peak frequency excursion is the deviation, delta f, and the modulation index m f is the deviation divided by the modulating frequency. The bandwidth that contains about 98 % of the power is given by Carson's rule: twice the sum of the deviation and the modulating frequency, or twice f m times m f plus one. The spectrum is a set of lines spaced f m apart around the carrier, with amplitudes given by Bessel functions of the index, and Carson's band, shaded, contains the significant ones.

Two quick FM problems. In part a the deviation is the index times the modulating frequency, 5 times 10, which is 50 kHz. Carson's rule gives twice 50 plus 10, that is 120 kHz, centred on 250 MHz, with spectral lines every 10 kHz whose heights follow the Bessel functions of 5. In last year's exam the index was one, so the deviation equals the modulating frequency, 10 kHz. And the bandwidth is 2 times 10 times one plus one, 40 kHz, as both solved copies found.

With a digital message the three parameters of the carrier give three basic keyings. In amplitude shift keying a one sends the carrier and a zero sends nothing. In frequency shift keying a one and a zero are sent at two different frequencies. In phase shift keying the phase jumps by 180 degrees between the two symbols. The modulators on the slides are simple switches or multipliers, and the demodulators are an envelope detector for ASK, a pair of filters for FSK and a coherent multiplication with the carrier for PSK.

A band is the whole range of spectrum a standard may use, a channel is the bandwidth of a single user, and multiplexing lets several signals share one medium without interfering. In frequency-division multiplexing each user is given its own frequency slot, with guard bands between them so that real filters can separate them. In time-division multiplexing users take turns, each one using the whole band during its time slot. In the slide problem each signal carries 6 MHz of video and needs a 4 MHz guard band, so each channel occupies a 10 MHz slot, but only if each channel is single-sideband. With the upper sideband, a carrier at 700 MHz puts the video from 700 to 706 MHz, the guard band takes it to 710, the next carrier is at 710, and so on up to 790 MHz; the slide's figure marks its carriers at the slot centres, 705, 715 and so on, which is only schematic, although for a hand sketch 6 MHz blocks centred on those carriers, 702 to 708 MHz with 4 MHz guards, match the professor's own figure. With double sideband each channel would occupy 12 MHz and need a spacing of 16 MHz, so state your assumption. Either way the ten channels fill 700 to 800 MHz, and the 10 GHz upconverter moves the multiplex to 10.7 to 10.8 GHz.

In a direct-conversion transmitter the carrier frequency equals the local-oscillator frequency, and modulation and upconversion happen in the same circuit. It is usually built with in-phase and quadrature branches: the bit stream is divided into pairs of bits, one modulating a cosine and the other a sine of the same carrier, which increases robustness and spectral efficiency. A two-step transmitter instead modulates at an intermediate frequency and then upconverts, which makes filtering easier. This quadrature idea is quadrature amplitude modulation, QAM: two carriers 90 degrees apart, and with the right baseband signals it can generate AM, FM or digital constellations.

The workhorse receiver is the superheterodyne: a mixer multiplies the incoming signal by a local oscillator and produces the difference frequency, the intermediate frequency. Tuning is done by moving the local oscillator, while all the selective, high-gain filtering is done at a fixed IF, which gives excellent sensitivity and selectivity; the price is more circuitry and power consumption. Its weakness is the image: a signal on the other side of the local oscillator, at the same distance, also produces the IF. So the RF filter before the mixer must reject the image frequency, which is why the slides list the problem of the image among the design issues.

In last year's exam the receiver question was short. The architecture is the superheterodyne, which converts every channel to the same fixed IF of 455 kHz. With the local oscillator above the signal, its frequency is 5.5 plus 0.455, which is 5.955 MHz, the answer in the solved copy. The image is then on the other side of the oscillator, at 5.955 plus 0.455, which is 6.41 MHz, and the RF filter must remove it. One of the solved copies also gives the low-side choice, 5.045 MHz, whose image would be at 4.59 MHz; either is correct if you say which one you chose.

To summarise part one: the wavelength is c over f times the square root of the permittivity, and aviation uses bands from kilohertz beacons to gigahertz radars. Low frequencies travel as ground waves, VHF and above as space waves limited by the radio horizon, with the 1.23 rule in feet and nautical miles, and HF as sky waves. For sky waves, the critical frequency is 9 times the root of the electron density, the MUF is fc over the sine of the elevation, and you work at 85 % of it. AM needs twice the highest message frequency, FM follows Carson's rule, and the superheterodyne converts everything to a fixed IF, at the cost of an image frequency. In part two we turn to the numbers that dominate the exam: decibels, noise figure, intercept points, sensitivity and dynamic range.