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 amplitude modulation envelope in the graph, the amplitude variation appears to complete approximately 2 cycles in 40 μs, suggesting a period (T) of 20 μs.

Thus, the frequency of the test tone can be calculated using:

$f = \frac{1}{T} \text{ where } T = 20 \mu s = 20 \times 10^{-6} s$

This gives:

$f = \frac{1}{20 \times 10^{-6}} = 50,000 \, \text{Hz} = 50 \, \text{kHz}$.

One major advantage of frequency modulation (FM) over amplitude modulation (AM) is that FM is less prone to interference from noise. This is due to its ability to maintain a consistent amplitude while varying the frequency, making it more robust in delivering clear sound quality.

The frequency range of FM in the UK is from 88 MHz to 108 MHz, which gives a total bandwidth of:

$ext{Bandwidth} = 108 \text{ MHz} - 88 \text{ MHz} = 20 \text{ MHz} = 20000 \text{ kHz}$.

Since each FM station is separated by 200 kHz, the maximum number of stations can be calculated by:

$\text{Maximum number of stations} = \frac{20000 \text{ kHz}}{200 \text{ kHz}} = 100 \text{ stations}.$

To find if the radio station fits within the FM bandwidth allocation, we need to calculate the bandwidth. The formula for FM bandwidth is:

$BW = 2(f_m + \Delta f)$

where:

• $f_m = 15 \, \text{kHz}$ (maximum audio frequency)
• $\Delta f = 75 \, \text{kHz}$ (frequency deviation)

Calculating gives:

$BW = 2(15 + 75) = 2 \times 90 = 180 \, \text{kHz}$.

Since the allocated FM bandwidth is 200 kHz, the station fits within the allocated bandwidth.