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Data communication questions
A microphone's few kilohertz cannot cross a country on their own, so they ride a carrier wave, printed on its amplitude or on its frequency. Each choice has its advantages and its drawbacks, and the modern move is to send the wave as a stream of numbers instead.
17 original questions · 51 marks · the data communication notes · Electronics
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State what is meant by baseband transmission, and give two reasons a radio station modulates a carrier instead.
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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). Answering 'so it travels further' alone is the standard error: the two real reasons are sharing the spectrum and aerial length.A carrier of frequency fc is amplitude modulated by a single audio tone of frequency fM. State the frequencies transmitted.
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Three: the carrier itself at fc, a lower sideband at fc − fM and an upper sideband at fc + fM (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 instead of twice, and so quoting two frequencies, is the standard sideband slip.Write down the equation for the bandwidth of an amplitude-modulated transmission, and state what the symbol in it means.
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Bandwidth = 2fM (1), where fM is the highest audio frequency carried (1). The factor of two is there because a sideband sits on each side of the carrier. Quoting fM alone, and so halving the bandwidth, is the most common lost mark in this topic.Write down the equation for the bandwidth of a frequency-modulated transmission, and state what each symbol means.
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Bandwidth = 2(Δf + fM) (1), where Δf is the frequency deviation, the greatest swing of the carrier either side of its unmodulated frequency, and fM is the highest audio frequency carried (1). Dropping the fM term and writing 2Δf is the examiners' favourite error to punish here.State the rule that fixes the minimum sampling rate for an analogue signal, and state what goes wrong if the rule is broken.
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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). Writing 'twice the highest frequency' without the inequality, or sampling at twice the average frequency, are the standard errors.Distinguish between time-division and frequency-division multiplexing.
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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). Saying that TDM 'sends signals faster' is the standard error: it shares the same channel in time rather than in frequency.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.
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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. Mixing kHz with MHz is the trap: subtract 5.0 from 1.53 without converting and the sidebands land at −3.47 MHz and 6.53 MHz. A negative frequency is the giveaway, and neither figure is anywhere near the carrier. Sidebands always sit close to the carrier, so check that before writing the answer down.A two-way radio uses frequency modulation with a deviation of 5.0 kHz and speech limited to 3.5 kHz. Calculate its bandwidth.
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Bandwidth = 2(Δf + fM) (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; both are the standard errors, and both are marked wrong. The audio term belongs inside the bracket.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.
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.
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.
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.
Calculate the wavelength of a 3.0 kHz radio wave and of a 100 MHz carrier, and state what this says about aerial length.
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.
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.
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.
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.
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