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Every electronic communication system has the same conceptual architecture: a source (microphone, camera, computer) generates a baseband signal; that signal is modulated onto a carrier; the modulated wave travels through a channel (free space, copper, optical fibre); a demodulator at the far end recovers the original signal; and an output transducer (loudspeaker, screen) converts it back into a physical phenomenon a human can perceive. This lesson develops the two classical analogue modulation schemes — amplitude modulation (AM) and frequency modulation (FM) — including their bandwidth requirements; introduces the family of digital modulation schemes (ASK, FSK, PSK, QAM) used in every modern wireless and wired link; surveys the three main physical channel types (copper twisted pair, coaxial cable, optical fibre) with their bandwidth capacities; and finishes with the key transducers that convert between physical phenomena and electrical signals — microphones, loudspeakers, photodiodes, thermistors and strain gauges. The lesson closes the AQA Electronics option by tying everything from Lesson 0 through Lesson 5 to the practical question of moving information from one place to another.
Spec mapping (AQA 7408 §3.13.5 — Data communication systems, option E): This lesson covers analogue modulation: amplitude modulation (AM) and the bandwidth = 2 × f_max relationship; frequency modulation (FM) and Carson's-rule bandwidth ≈ 2(Δf + f_max); digital modulation schemes (ASK, FSK, PSK, QAM) briefly; communication channels — twisted-pair copper, coaxial cable and optical fibre — with relative bandwidth and attenuation; input transducers including dynamic, capacitor and piezoelectric microphones, photodiodes and phototransistors, thermistors, and strain gauges; output transducers including loudspeakers (moving-coil and piezoelectric). (Refer to the official AQA specification document for exact wording.)
Synoptic links: Modulation requires the analogue signal-processing machinery of Lessons 2 and 3 — every AM transmitter uses an op-amp summing amplifier as a modulator, and every superheterodyne radio receiver uses a band-pass filter to select a channel. The transducers connect to thermal physics (thermistors), photoelectric effect / quantum physics (photodiodes — light absorbed in a reverse-biased p–n junction releases electron–hole pairs, exactly the band-gap picture from Lesson 0), and Hooke's law in mechanics (strain gauges — resistance changes with length). The digital modulation schemes connect back to the digital signals of Lesson 1, and the channel bandwidth limits set the practical maximum bitrate via the Shannon–Hartley capacity formula.
A baseband audio signal of 20 Hz–20 kHz cannot be radiated efficiently from any practical antenna. The antenna length needed for efficient radiation is comparable to a wavelength: λ = c/f gives λ ≈ 15 km at 20 kHz — clearly impractical. By modulating the audio onto a much higher-frequency carrier (typically 100 kHz to 100 MHz for radio, hundreds of MHz to tens of GHz for microwave and cellular), the radiated wavelength shrinks to a few metres or less, making compact antennas feasible.
Modulation also enables frequency-division multiplexing: many users transmit on different carrier frequencies simultaneously through the same physical medium. The radio dial — separate stations at 88.1, 88.3, 88.5 MHz, etc. — is one carrier per station, each modulated independently.
The two classical schemes vary one property of the carrier in proportion to the signal:
Let the carrier be a sinusoid c(t) = A_c cos(2πf_c t), and the signal m(t) be a band-limited audio waveform with maximum frequency f_max. The AM waveform is:
s(t) = [A_c + m(t)] cos(2πf_c t)
The envelope A_c + m(t) traces the audio waveform; the high-frequency carrier "fills" the envelope. A receiver only needs to recover the envelope — the simplest design (an envelope detector: a diode followed by an RC low-pass filter) does exactly this. Cheap and simple receivers are the historic reason AM became the first widespread radio modulation in the 1920s.
When a carrier at f_c is multiplied by an audio signal containing frequencies 0 to f_max, the result contains frequencies between (f_c − f_max) and (f_c + f_max). The AM bandwidth is:
B_AM = 2 × f_max
For a 5 kHz audio signal on AM broadcast: bandwidth = 10 kHz per station. This is why the AM broadcast band (540–1600 kHz) is divided into 10 kHz channels — exactly fitting one AM station per channel.
Question: A telephone-quality voice signal contains frequencies up to 3.4 kHz. (a) Calculate the AM bandwidth. (b) The AM broadcast band runs from 540 kHz to 1600 kHz. How many simultaneous voice stations can fit?
Solution:
(a) B_AM = 2 × 3.4 kHz = 6.8 kHz.
(b) Total band = 1600 − 540 = 1060 kHz. Number of channels = 1060 / 6.8 ≈ 156 stations. In practice the standard 10 kHz channel spacing leaves headroom for higher-quality audio (up to 5 kHz audio bandwidth), reducing the count to 106.
AM has several weaknesses:
These weaknesses drove the transition to FM for high-quality broadcast in the 1950s.
In FM the carrier's instantaneous frequency varies in proportion to the signal:
f_inst(t) = f_c + k_f m(t)
where k_f is the frequency deviation constant (Hz per volt of signal). The maximum deviation from the carrier is Δf = k_f × m_max. The amplitude of the FM waveform is constant.
Unlike AM, the FM bandwidth is not exactly twice the signal bandwidth. Carson's rule gives an excellent approximation:
B_FM ≈ 2(Δf + f_max)
where Δf is the peak frequency deviation and f_max is the highest signal frequency. For FM broadcast in the 88–108 MHz band, the standard is Δf = ±75 kHz and f_max = 15 kHz (full audio range), giving B_FM ≈ 2 × (75 + 15) = 180 kHz. The standard FM channel allocation is 200 kHz, leaving 20 kHz of guard band between adjacent stations.
FM trades extra bandwidth for several advantages:
These are why FM became the dominant analogue broadcast format above the medium-wave band.
Question: A car-radio FM receiver tunes to a 100 MHz broadcast carrying audio up to 15 kHz with frequency deviation Δf = 75 kHz. Calculate the bandwidth occupied by the signal and the percentage of the 88–108 MHz FM band used.
Solution:
B_FM = 2(Δf + f_max) = 2(75 + 15) = 180 kHz.
Total band = 108 − 88 = 20 MHz. Number of channels at 200 kHz spacing = 20 000 / 200 = 100 channels. The 180 kHz signal in a 200 kHz allocation gives a fill factor of 90%.
Question: A 5 kHz audio signal is transmitted by (a) AM, (b) FM with Δf = 50 kHz. Compare the bandwidths.
Solution:
(a) AM: 2 × 5 = 10 kHz.
(b) FM: 2 × (50 + 5) = 110 kHz — 11 times wider than AM.
FM trades bandwidth for noise immunity. This is acceptable in the VHF band where spectrum is plentiful, but on the AM band (with stations 10 kHz apart) FM would be too greedy.
Modern wireless and wired links almost universally use digital modulation:
The Shannon–Hartley capacity formula sets the ultimate limit:
C = B × log₂(1 + SNR)
where C is channel capacity in bits/s, B is bandwidth in Hz, and SNR is signal-to-noise ratio in linear (not dB) units. A 20 MHz Wi-Fi channel at 30 dB SNR achieves C = 20 × 10⁶ × log₂(1001) ≈ 199 Mbit/s — close to the actual 256-QAM Wi-Fi data rate.
The physical medium that carries the signal has a major effect on the achievable bandwidth and the attenuation per kilometre.
Two insulated copper wires twisted together. The twisting cancels external magnetic interference (the two wires pick up equal but opposite induced voltages that cancel at the receiver). Used for telephone lines, Ethernet (Cat-5, Cat-6, Cat-6a), and DSL broadband.
A central copper conductor surrounded by an insulating dielectric, a braided copper shield, and an outer insulating jacket. The shield carries the return current and excludes external interference. Used for cable TV, RF transmitter feedlines, and historic Ethernet (10BASE-2 and 10BASE-5).
A glass or plastic fibre carrying light from a laser or LED transmitter to a photodiode receiver. Single-mode fibre uses a 9 μm core for very long distances; multi-mode fibre uses a 50 μm or 62.5 μm core for shorter links.
| Medium | Bandwidth | Attenuation | Typical use |
|---|---|---|---|
| Twisted pair (Cat-6a) | 600 MHz | ~25 dB/100 m | LAN, telephony, DSL |
| Coaxial cable | GHz | ~10 dB/100 m | Cable TV, RF feedline |
| Optical fibre | THz | 0.2 dB/km | Backbone, FTTH, undersea |
Question: A cable TV operator uses 256-QAM (8 bits/symbol) on a 6 MHz coaxial channel at 35 dB SNR. (a) Calculate the maximum bitrate using Shannon–Hartley. (b) The actual bitrate achieved is around 40 Mbit/s. Comment on the efficiency.
Solution:
(a) SNR = 10^3.5 = 3162. C = 6 × 10⁶ × log₂(3163) = 6 × 10⁶ × 11.6 = 69.6 Mbit/s (Shannon limit).
(b) Actual 40 Mbit/s ÷ 69.6 = 58% of Shannon capacity. Real systems include framing overhead, forward error correction, and engineering margin; 50–70% of Shannon is typical commercial performance.
A transducer converts between a non-electrical quantity (sound, light, temperature, force, displacement) and an electrical signal.
A dynamic microphone has a diaphragm attached to a coil that moves in a magnetic field. Sound pressure moves the diaphragm; the coil cuts field lines; an EMF proportional to the diaphragm velocity is induced (Faraday's law). Robust, no power required, low cost; used in live sound, broadcast vocals, and instrument micing. Sensitivity ~2 mV/Pa.
A capacitor (condenser) microphone has a charged diaphragm forming one plate of a capacitor. Sound pressure varies the plate separation, varying the capacitance and (at constant charge) the voltage. Phantom power (48 V DC) is required to charge the diaphragm and bias an internal preamplifier. Excellent frequency response and sensitivity (~20 mV/Pa); used in studio recording and measurement.
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