Figure 5 shows the output signal from the tuner circuit of a radio receiver - AQA - A-Level Physics - Question 3 - 2022 - Paper 8

Examining the same graph, the test tone frequency can also be determined by measuring a smaller number of cycles over the same time span. We notice the test tone oscillates at a frequency much higher than the carrier wave.

Assuming there are 20 cycles in the same 60 µs, we can calculate the frequency as follows:

T for test tone is:
$T = \frac{60 \mu s}{20} = 3 \mu s = 3 \times 10^{-6} s$.

Thus, the test tone frequency is:
$f = \frac{1}{3 \times 10^{-6}} \approx 333.33 kHz$.

One significant advantage of frequency modulation (FM) over amplitude modulation (AM) is that FM is less susceptible to noise and interference. This is because changes in amplitude caused by noise do not affect the frequency of the signal, allowing for clearer audio and better quality reception.

The FM radio band in the UK spans from 88 MHz to 108 MHz, giving a total range of:
$108 MHz - 88 MHz = 20 MHz = 20000 kHz$.

Given that the FM stations are separated by 200 kHz, we can calculate the maximum number of stations using the formula:
$\text{Number of stations} = \frac{\text{Total band width}}{\text{Frequency separation}} = \frac{20000 kHz}{200 kHz} = 100$.

To determine if the radio station fits the FM bandwidth allocation, we can calculate the bandwidth required for the station. The bandwidth for FM can be approximated using Carson's Rule:

$BW = 2(f_{max} + \Delta f)$
where $f_{max}$ is the maximum audio frequency (15 kHz) and $\\Delta f$ is the frequency deviation (75 kHz).
Substituting the values we get: $BW = 2(15 + 75) = 2 \times 90 = 180 kHz$.

This means the radio station requires a bandwidth of 180 kHz, which fits within the standard allocated 200 kHz for FM stations, confirming that the station does meet the FM bandwidth allocation.