Questions on modulation and PCM

Question 1

State what is meant by baseband transmission, and give two reasons a radio station modulates a carrier instead.

Accepted Answer

Baseband transmission sends a signal in its own natural range of frequencies, as a telephone landline sends speech down a private pair of wires (1). Radio needs a carrier because every station shares the same air, so each is given its own carrier frequency and a tuned circuit at the receiver can select one and reject the rest; and because an efficient aerial must be a respectable fraction of a wavelength long, which a kilohertz signal never allows (1). 'So it travels further' alone is not one of the two reasons required.

Question 2

A carrier of frequency f_c is amplitude modulated by a single audio tone of frequency f_M. State the frequencies transmitted.

Accepted Answer

Three: the carrier itself at f_c, a lower sideband at f_c − f_M and an upper sideband at f_c + f_M (1) (1). The audio is not sent at its own frequency; it appears as the pair of new frequencies flanking the carrier. Counting the sidebands once rather than twice gives two frequencies instead of three.

Question 3

Write down the equation for the bandwidth of an amplitude-modulated transmission, and state what the symbol in it means.

Accepted Answer

Bandwidth = 2f_M (1), where f_M is the highest audio frequency carried (1). The factor of two is there because a sideband sits on each side of the carrier. Quoting f_M alone halves the bandwidth.

Question 4

Write down the equation for the bandwidth of a frequency-modulated transmission, and state what each symbol means.

Accepted Answer

Bandwidth = 2(Δf + f_M) (1), where Δf is the frequency deviation, the greatest swing of the carrier either side of its unmodulated frequency, and f_M is the highest audio frequency carried (1). Dropping the f_M term and writing 2Δf understates the bandwidth.

Question 5

State the rule that fixes the minimum sampling rate for an analogue signal, and state what goes wrong if the rule is broken.

Accepted Answer

The sampling rate must be greater than twice the highest frequency present in the signal (1). Sampled more slowly than that, a high frequency masquerades as a lower one in the reconstructed signal, and the corruption cannot be undone afterwards (1). Exactly twice the highest frequency, or twice the average frequency, does not satisfy the rule.

Question 6

Distinguish between time-division and frequency-division multiplexing.

Accepted Answer

Frequency-division multiplexing gives each user a permanent slice of the spectrum and every user transmits continuously on their own carrier, which is how the radio dial works (1). Time-division multiplexing gives each user the whole channel for a brief repeating time slot, interleaving the users in turn, which suits digital signals because samples are short pulses with dead time between them (1).

Question 7

State what is meant by the skip zone of a short-wave transmission, and explain why the highest frequency a short-wave station can use is lower at night than during the day.

Accepted Answer

The skip zone is the region between the point where the ground wave has faded out and the point where the sky wave first returns to the ground, in which the station cannot be heard at all (1). The ionosphere is produced by solar ultraviolet, so at night the ionisation falls; with fewer free electrons the layers refract a wave less strongly, and a frequency that was bent back to the ground by day passes out through them instead, so the highest usable frequency falls (1). The skip zone is a gap in coverage rather than a limit on range: beyond it the sky wave lands normally.

Question 8

An encrypted radio link is intercepted and every bit of one transmission is recorded. State what the interceptor can learn from the recording, and explain why the security of the link is said to rest on the secrecy of the key rather than on the secrecy of the method.

Accepted Answer

Nothing of the contents: the bits were combined with the key by a reversible rule, so without the key the recording carries no readable information, though the interceptor does know that a transmission took place and how long it was (1). The method has to be assumed public, because it is built into every receiver and stays secret only until somebody takes one apart, so the one thing an opponent must not hold is the key (1). Changing the key restores the security of a link, while changing the method is not something the receivers at the far end could follow.

Question 9

A station transmits by amplitude modulation on a carrier of 1.53 MHz, carrying audio up to 5.0 kHz. Calculate the frequencies of the two sidebands and the bandwidth of the transmission.

Accepted Answer

Convert the carrier first: 1.53 MHz = 1530 kHz (1). Lower sideband = 1530 − 5.0 = 1525.0 kHz and upper sideband = 1530 + 5.0 = 1535.0 kHz (1). Bandwidth = 2 × 5.0 = 10.0 kHz (1), which is also 1535.0 − 1525.0. Subtracting 5.0 from 1.53 without converting to a common unit gives −3.47 MHz and 6.53 MHz. A negative frequency is impossible, and neither figure lies close to the carrier, which is where the sidebands must fall.

Question 10

A two-way radio uses frequency modulation with a deviation of 5.0 kHz and speech limited to 3.5 kHz. Calculate its bandwidth.

Accepted Answer

Bandwidth = 2(Δf + f_M) (1) = 2 × (5.0 + 3.5) (1) = 17 kHz (1). Using 2Δf alone gives 10 kHz and dropping the factor of two gives 8.5 kHz.

Question 11

A compact disc carries two channels of audio, each limited to 20 kHz, sampled at 44.1 kHz with 16 bits per sample. Show that the sampling rate is adequate, and calculate the total bit rate.

Accepted Answer

Twice the highest frequency is 2 × 20 = 40 kHz, and 44.1 kHz exceeds it, leaving margin for imperfect filters (1). Bit rate for one channel = 44 100 × 16 = 705 600 bits per second, so for two channels (1) the total is 1 411 200 bits per second, about 1.41 Mbit s⁻¹ (1). Omitting the second channel halves the bit rate.

Question 12

An analogue-to-digital converter uses 12 bits per sample over an input range of 5.0 V. Calculate the number of quantisation levels and the size of one step.

Accepted Answer

Number of levels = 2¹² = 4096 (1). The levels span the range in 4095 steps, so one step = 5.0/4095 (1) = 1.22 × 10⁻³ V, about 1.22 mV (1). Each extra bit doubles the number of levels and halves the step, which is why quantisation error falls as bits are added. Dividing by 12 rather than by 2¹² uses the number of bits in place of the number of levels.

Question 13

Thirty digital telephone calls, each running at 64 kbit s⁻¹, are carried on one cable by time-division multiplexing. Calculate the bit rate the cable must carry, and state what each call is given by the multiplexer.

Accepted Answer

Total bit rate = 30 × 64 × 10³ (1) = 1.92 × 10⁶ bits per second, that is 1.92 Mbit s⁻¹ (1). Each call is given the whole channel for a brief repeating time slot, its samples interleaved into the gaps between everyone else's (1). Describing the calls as sharing the bandwidth simultaneously describes frequency-division multiplexing, not time-division.

Question 14

A broadcasting authority has an allocation 2.0 MHz wide for amplitude-modulated stations spaced 9 kHz apart. Calculate the number of stations that fit.

Accepted Answer

Convert the span first: 2.0 MHz = 2000 kHz (1). Number of stations = 2000/9 = 222.2 (1), so 222 stations fit, the fraction being no use to anybody (1). Dividing 2.0 by 9 without converting to a common unit gives 0.22 stations.

Question 15

Calculate the wavelength of a 3.0 kHz radio wave and of a 100 MHz carrier, and state what this says about aerial length.

Accepted Answer

λ = c/f = (3.0 × 10⁸)/(3.0 × 10³) = 1.0 × 10⁵ m, that is 100 km, against (3.0 × 10⁸)/(100 × 10⁶) = 3.0 m for the carrier (1). An efficient aerial is a respectable fraction of a wavelength, so a baseband aerial is unbuildable while a VHF one is about a metre (1). Use the speed of light, not the speed of sound: these are radio waves whatever the frequency.

Question 16

A 240 km route is to carry the same traffic either on copper cable, which needs a regenerator every 3.0 km, or on optic fibre, which needs one every 40 km. Calculate the number of regenerators each route requires, and state the property of the two media that accounts for the difference.

Accepted Answer

Copper: 240/3.0 = 80 sections, so 79 regenerators along the way, the receiver ending the last section (1). Fibre: 240/40 = 6 sections, so 5 regenerators (1). Fibre attenuates the signal far less per kilometre, so the signal can travel much further before it has to be regenerated (1).

Question 17

A communications satellite receives an uplink signal of power 2.0 μW and transmits its downlink at 20 W. Calculate the ratio of the transmitted power to the received power, and use it to explain why the uplink and the downlink must not share one frequency.

Accepted Answer

Ratio = 20/(2.0 × 10⁻⁶) (1) = 1.0 × 10⁷, that is ten million (1). On one frequency the satellite's own transmitter would therefore be ten million times stronger at its own aerial than the uplink it is listening for, so its output would swamp its receiver and the transponder would amplify itself into oscillation, as a microphone does in front of its own loudspeaker (1). Separating the two frequencies lets a filter at the satellite keep its own downlink out of its receiver.

Question 18

A broadcast band runs from 88.0 MHz to 91.0 MHz. Stations use frequency modulation carrying audio up to 15 kHz with a deviation of 75 kHz, and each channel is allowed a 20 kHz guard band on top of its own bandwidth. Calculate the bandwidth per station, the channel spacing and the number of stations the band holds. Then calculate the same three figures if the deviation is reduced to 25 kHz, and state what is lost by the change.

Accepted Answer

Span = 91.0 − 88.0 = 3.0 MHz = 3000 kHz (1). At a deviation of 75 kHz the bandwidth is 2(75 + 15) = 180 kHz (1), so the channel spacing is 180 + 20 = 200 kHz and the band holds 3000/200 = 15 stations (1). At a deviation of 25 kHz the bandwidth is 2(25 + 15) = 80 kHz (1), the spacing is 100 kHz and the band holds 30 stations (1), twice as many. What is lost is noise performance: FM's advantage over AM grows with the deviation, because a wider swing makes the wanted frequency changes large compared with the disturbance any interference can produce (1), so the narrowband stations would sound noisier and would carry less dynamic range even though the audio still reaches 15 kHz (1). Leaving out the guard band changes the count from 15 to 16.

Question 19

Telephone speech is limited to 3.4 kHz and sampled at 8.0 kHz. Calculate the bit rate with 8 bits per sample and with 12 bits per sample, and state the effect of the change on the quantisation error.

Accepted Answer

With 8 bits: 8000 × 8 = 64 000 bits per second (1), the standard worldwide figure for one digital call. With 12 bits: 8000 × 12 = 96 000 bits per second (1). The number of levels rises from 2⁸ = 256 to 2¹² = 4096 (1), a factor of 16, so each step is 16 times smaller and the quantisation error falls by the same factor (1). Note the price: half as much again in bit rate for that improvement. The error falls by 2⁴, not by 12/8, because each extra bit doubles the number of levels.

Question 20

A signal containing a 7.0 kHz tone is sampled at 10 kHz. State whether the sampling rate is adequate, calculate the frequency at which the tone appears in the reconstructed signal, and calculate the minimum acceptable sampling rate.

Accepted Answer

The rate is not adequate: it must exceed twice the highest frequency, that is more than 14 kHz, and 10 kHz falls short (1). Sampled too slowly, the 7.0 kHz tone masquerades as the difference between the sampling rate and its own frequency, 10 − 7.0 = 3.0 kHz (1) (1), and no processing afterwards can undo it, because the samples are now genuinely indistinguishable from those of a real 3.0 kHz tone. Minimum acceptable rate = 2 × 7.0 = 14 kHz, and in practice a little more (1). A filter at the receiver cannot recover the tone; it must be removed by a filter before sampling, with a fast enough sampler.

Question 21

An authority has 2.0 MHz of spectrum and wants 40 channels in it. Calculate the width available per channel. Determine whether amplitude modulation carrying audio to 4.5 kHz would fit, and whether frequency modulation with a deviation of 20 kHz carrying the same audio would fit. Recommend one scheme with a reason.

Accepted Answer

Width per channel = 2000 kHz/40 = 50 kHz (1). AM would need 2 × 4.5 = 9.0 kHz, so it fits with room to spare (1). FM would need 2(20 + 4.5) = 49 kHz, which fits, but only just, leaving 1 kHz of guard band (1). Recommend FM if the users need the noise immunity, since the amplitude clipping in an FM receiver removes most interference; recommend AM if more channels may be wanted later, since it uses a fifth of the spectrum per channel and the receivers are simpler and cheaper (1). Convert 2.0 MHz to kilohertz before dividing, and keep the f_M term: without it FM appears to need only 40 kHz.

Question 22

A communications satellite receives an uplink at 6 GHz and retransmits the same signal on a downlink at 4 GHz. Explain why both frequencies must lie far above 30 MHz, why the two frequencies must differ, and why the downlink is given the lower of the two.

Accepted Answer

Above about 30 MHz the ionosphere no longer refracts a wave back towards the ground, so both signals pass straight through it on the way to and from the satellite (1). The satellite receives microwatts and transmits watts from the same small structure, so retransmitting on the frequency it is listening to would swamp its own receiver with its own output (1). Atmospheric attenuation worsens towards higher frequency, and the satellite is the weaker, power-limited end of the link, so the downlink is given the lower and less attenuated frequency (1).

Question 23

Three transmitters send on 150 kHz, 12 MHz and 150 MHz. (a) Name the propagation mode each one relies on. (b) State the mechanism that lets the 12 MHz signal reach a receiver hundreds of kilometres beyond the horizon. (c) State why the 150 MHz transmitter's range stops at the horizon while the 12 MHz transmitter's does not.

Accepted Answer

150 kHz travels as a ground wave, 12 MHz as a sky wave and 150 MHz as a space wave, sorted by frequency alone: below about 2 MHz, between about 3 and 30 MHz, and above about 30 MHz (1). The 12 MHz wave leaves the ground and meets the ionosphere, whose free-electron density increases with height and progressively lowers the refractive index, so the wave is bent further and further until it turns back down towards the ground far beyond the horizon (1). Above about 30 MHz the ionosphere no longer bends the wave enough to return it, so the 150 MHz signal simply travels onward in a nearly straight line and its range stops at the horizon (1). The 12 MHz signal is not limited the same way, because it does not depend on a straight line at all: refraction, not line of sight, carries it over the curve of the Earth to landing points far beyond it (1).

Question 24

A weather buoy transmits its readings by radio to a shore station, and each message already carries a checksum to catch corruption caused by noise on the link. (a) State what would stop a passer-by's receiver from being able to read the buoy's readings. (b) State what would stop someone transmitting false readings of their own while pretending to be the buoy. (c) Explain why the checksum already in use protects against neither threat.

Accepted Answer

(a) Encryption: the data is combined with a secret key by a reversible rule before transmission, so an intercepted bit stream carries no information to anyone without the key (1). (b) Authentication: a check value computed from the whole message together with a secret key, sent alongside it and recomputed by the shore station; a match proves the message came from a holder of that key and that nothing in it was altered, and a forged message fails the check (1). (c) The checksum uses no key at all and is designed to be easy to recompute, so an eavesdropper reading it still sees the readings in clear, and anyone forging a message simply works out the new checksum to match their own fake data (1). A checksum only catches accidental corruption from a noisy channel; only a secret key that an attacker does not hold makes a check unforgeable (1).

Question 25

A network operator has three links to plan: (a) a trunk route between two cities carrying tens of thousands of telephone calls at once; (b) a link to a weather station on a mountain that no cable can reach; (c) the last 300 m from a street cabinet into a house, on a small budget. Name the medium best suited to each, and justify each choice by the property of that medium which decides it.

Accepted Answer

(a) Optic fibre: its bandwidth is the largest of the three, many GHz, so it carries that number of channels at once, and its attenuation is the lowest, so its regenerators stand tens of kilometres apart instead of a few (1). (b) A radio or microwave link: there is no cable to lay, which is the only property that settles this case, and in exchange the medium is shared, fades with distance and weather, and can be received by anyone in range (1). (c) Copper cable: it is cheap and already in the ground, and over 300 m its higher attenuation and its modest bandwidth of tens of MHz cost nothing that matters here, though it does pick up electrical interference and crosstalk that fibre would not (1). Fibre would serve all three technically, so say what rules it out in the other two: no cable can be laid to the mountain at all, and the short domestic run needs neither the bandwidth nor the low attenuation it would pay for (1).