Analyzing AM and FM Signals: Frequency and Data Capacity Revision Notes | A Level AQA Physics

1. Why Modulation Matters

Radio stations, mobile networks, and Wi-Fi all face the same fundamental problem: how do you send a voice, a song, or a video file through the air as an electromagnetic wave? The answer is modulation — the process of encoding information onto a carrier wave.

There are two dominant methods: amplitude modulation (AM) and frequency modulation (FM). Each leaves a distinctive fingerprint on a voltage-time graph, and AQA examiners expect you to read that fingerprint confidently. This lesson builds that skill from the ground up, then connects it to the broader question of how much data a channel can carry.

2. Reading AM and FM Signals from a Graph

Picture a voltage-time graph of a radio signal. You will see a rapid oscillation — that is the carrier wave — but its shape is being distorted by a slower pattern. That slower pattern is the information signal, sometimes called the audio frequency or modulating signal.

In an AM signal, the envelope — the outline traced by the peaks and troughs of the carrier — rises and falls in step with the information signal. The carrier frequency itself stays constant throughout. In an FM signal, the envelope stays flat (constant amplitude), but the spacing between the carrier oscillations changes: the waves bunch together when the information signal is at a high value and spread apart when it is low.

To identify the carrier frequency from a graph, count the number of complete carrier cycles within one period of the information signal, then divide by that period. Alternatively, measure the time for one carrier cycle directly and take its reciprocal: $f_c = \frac{1}{T_c}$.

A key relationship to hold onto: the carrier frequency is always much higher than the information frequency. For AM radio, a carrier might sit at 1 MHz while the audio information sits below 20 kHz — a ratio of 50:1 or more. On a graph, this means you will always see many carrier cycles for every one information cycle.

The chart above illustrates the key distinction: in AM the amplitude varies with the information signal while frequency stays fixed; in FM the frequency varies while amplitude stays fixed. Keeping this contrast sharp is essential for graph-reading questions.

Worked example: Identifying carrier and information frequencies from a voltage-time graph

A voltage-time graph shows a modulated signal. The slow envelope completes one full cycle in $5.0 \times 10^{-4}$ s. Within that envelope cycle, 200 complete carrier oscillations are visible. Identify the information frequency and the carrier frequency.

1. $f_{\text{info}} = \frac{1}{T_{\text{info}}} = \frac{1}{5.0 \times 10^{-4} \text{ s}}$ — The information frequency is the reciprocal of the period of the slow envelope variation. [M1]
2. $f_{\text{info}} = 2000 \text{ Hz} = 2.0 \text{ kHz}$ — $\frac{1}{5.0 \times 10^{-4}} = 2000$. This is a typical audio frequency. [A1]
3. $T_c = \frac{T_{\text{info}}}{200} = \frac{5.0 \times 10^{-4}}{200} = 2.5 \times 10^{-6} \text{ s}$ — 200 carrier cycles fit inside one information period, so divide the information period by 200. [M1]
4. $f_c = \frac{1}{T_c} = \frac{1}{2.5 \times 10^{-6} \text{ s}} = 4.0 \times 10^5 \text{ Hz} = 400 \text{ kHz}$ — $\frac{1}{2.5 \times 10^{-6}} = 4.0 \times 10^5$. This is a realistic AM radio carrier frequency. [A1]
5. $\frac{f_c}{f_{\text{info}}} = \frac{4.0 \times 10^5}{2.0 \times 10^3} = 200$ — The ratio equals the number of carrier cycles per information cycle — consistent with the graph description. Confirmed.

Answer: Information frequency = 2.0 kHz; Carrier frequency = 400 kHz

One more distinction worth locking in: in AM, a change in the information signal changes the height of the carrier peaks. In FM, a change in the information signal changes how tightly packed the carrier cycles are. Both methods encode the same audio information — they use different physical properties of the wave to do it.

3. Bandwidth and Data Capacity of a Channel

Now that you can read a modulated signal, the next question is: how much information can a channel carry? This is where bandwidth becomes central.

When a carrier wave is modulated, it no longer occupies a single frequency. AM and FM both spread the signal across a range of frequencies around the carrier. The width of that range is the bandwidth of the signal. A channel must have at least as much bandwidth as the signal it carries, otherwise information is lost.

For an AM signal, the modulation process creates two sidebands — one above and one below the carrier frequency — each separated from the carrier by the information frequency. So an AM station broadcasting audio up to 5 kHz occupies a bandwidth of $2 \times 5 = 10$ kHz around its carrier.

FM signals occupy a wider bandwidth than AM signals carrying the same audio content. This is because FM encodes information through frequency deviation — the carrier swings further above and below its centre frequency — which spreads the signal across a larger range. UK FM radio stations are allocated 200 kHz of bandwidth each, compared to roughly 9–10 kHz for AM stations.

The data capacity of a channel — the maximum rate at which information can be transmitted — depends directly on the available bandwidth. A wider bandwidth allows more signal variation per second, which means more data per second. This relationship is captured by the Nyquist and Shannon frameworks, though at A-level the key principle is: greater bandwidth means greater data capacity.

This is where the trade-off between AM and FM becomes clear. FM's wider bandwidth delivers higher audio fidelity — it can carry a broader range of audio frequencies with less noise — but it consumes more of the available spectrum per station. AM's narrower bandwidth allows more stations to fit within the same frequency range, but the audio quality is lower.

4. Comparing Bandwidth Availability Across Transmission Media

Different physical media — copper wire, optical fibre, radio spectrum, coaxial cable — have very different bandwidth capacities. Understanding these differences is directly assessed in AQA questions that ask you to compare or evaluate media for a given application.

Optical fibre stands apart from all other media in terms of raw bandwidth potential. The reason is physical: light has a frequency on the order of $10^{14}$ Hz, which means the carrier frequency is vastly higher than for radio or electrical signals. Since bandwidth is a fraction of the carrier frequency, a higher carrier frequency allows a proportionally larger absolute bandwidth.

Copper-based media are limited by two factors: the skin effect (at high frequencies, current flows only near the surface of the conductor, increasing resistance) and electromagnetic interference from adjacent cables. Both effects reduce the usable bandwidth as frequency increases, which is why copper telephone lines cannot match fibre for broadband delivery.

Radio spectrum is a finite, shared resource managed by national regulators. Each station or service is allocated a specific frequency band, and the width of that band determines how much information it can carry. This is why digital radio (DAB) can fit multiple stations into the bandwidth that a single analogue FM station would occupy — digital encoding is far more spectrally efficient.

5. Worked Analysis: Comparing Media for a Given Application

Exam questions often present a scenario — a hospital network, a rural broadband scheme, a live broadcast link — and ask you to evaluate which medium is most suitable. The key is to connect the data-rate requirement to the bandwidth available from each medium, then weigh up practical constraints.

Worked example: Evaluating transmission media for a high-definition video link

A broadcast company needs to transmit uncompressed HD video requiring a data rate of 1.5 Gbps over a distance of 2 km between two buildings. Evaluate whether twisted copper pair, coaxial cable, or optical fibre is most suitable. [6 marks]

1. Required data rate: 1.5 Gbps over 2 km — Always anchor your evaluation to the specific numbers given. [B1]
2. Twisted pair: maximum a few Mbps over short distances; bandwidth falls sharply beyond ~100 m at high frequencies — 1.5 Gbps far exceeds the capacity of twisted pair, especially over 2 km. Ruled out. [B1]
3. Coaxial cable: can reach hundreds of Mbps over short distances, but 1.5 Gbps over 2 km is at or beyond its practical limit; signal attenuation at high frequencies would require repeaters — Coaxial cable is marginal — it might be achievable with signal boosting but is not the preferred solution. [B1]
4. Optical fibre: supports multi-Gbps to Tbps; very low attenuation over 2 km; immune to electromagnetic interference — Optical fibre comfortably exceeds the 1.5 Gbps requirement with headroom to spare. [B1]
5. Optical fibre is the most suitable medium. — A conclusion must be explicitly stated and linked to the data-rate requirement. Mentioning cost as a trade-off adds evaluation depth. [B1 + B1 for balanced evaluation]

Answer: Optical fibre is the most suitable medium. Its bandwidth supports 1.5 Gbps over 2 km with low attenuation and no electromagnetic interference. Twisted pair is wholly inadequate at this data rate. Coaxial cable is marginal and would require repeaters. The higher installation cost of fibre is justified by the performance requirement.

6. Exam Technique and Common Mistakes

Graph-reading questions on AM and FM signals are among the most reliably tested items in this topic. The marks are there for the taking — but only if you read the graph correctly and use precise language in your answers.

When asked to identify the carrier frequency, always measure the period of the rapid oscillations, not the envelope. When asked to identify the information frequency, measure the period of the slow envelope variation (AM) or the slow variation in cycle spacing (FM). Confusing these two is the single most common error in this topic.

For six-mark evaluate questions on transmission media, the mark scheme typically rewards: a clear statement of the data-rate or bandwidth requirement, a comparison of at least two media with physical reasoning, acknowledgement of a trade-off (such as cost or installation complexity), and a justified conclusion. Hitting all four elements is what separates a 4-mark answer from a 6-mark answer.

Bringing together all three specification requirements: you can now read carrier and information frequencies from a graph, calculate AM bandwidth using $B = 2f_{\text{info}}$, and compare the data capacity of different media by linking carrier frequency to available bandwidth. These three skills connect directly — a higher carrier frequency means a wider possible bandwidth, which means a higher data capacity, which is precisely why optical fibre dominates modern high-speed networks.