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

By observing the amplitude modulation on the carrier wave, we can identify the frequency of the test tone. The measurement of its waveform indicates that there are approximately 4 cycles within the same 10 \text{µs} period.

Thus, the time period for one cycle of the test tone is: $T_{test} = \frac{10 \, \text{µs}}{4} = 2.5 \, \text{µs}$

Using the frequency formula: $f_{test} = \frac{1}{T_{test}} = \frac{1}{2.5 \times 10^{-6}} = 400 \, \text{kHz}$

One significant advantage of using FM over AM is its improved resistance to noise and interference. FM signals are less affected by amplitude changes in the transmission medium, allowing for better sound quality and clarity.

The FM range in the UK is from 88 MHz to 108 MHz. This gives a total bandwidth of: $108 \text{ MHz} - 88 \text{ MHz} = 20 \text{ MHz} = 20,000 \text{ kHz}$

With each station separated by 200 kHz, the maximum number of stations can be calculated as follows: $\text{Number of Stations} = \frac{20,000 \, kHz}{200 \, kHz} = 100 \text{ stations}$

The FM bandwidth for a radio signal can be calculated using the formula: $B = 2(\Delta f + f_{max})$

Where:

• ( \Delta f = 75 , kHz ) (frequency deviation)
• ( f_{max} = 15 , kHz ) (maximum audio frequency)

Substituting in the values: $B = 2(75 + 15) = 2 \times 90 = 180 \, kHz$

Since the allocated bandwidth for FM in the UK is 200 kHz, and our calculated bandwidth of 180 kHz fits within this range, the radio station does fit the FM bandwidth allocation.