Figure 8 shows the first-stage filter circuit for a simple AM receiver - AQA - A-Level Physics - Question 3 - 2017 - Paper 8
Question 3
Figure 8 shows the first-stage filter circuit for a simple AM receiver. The circuit can be adjusted to resonate at 910 kHz so that it can receive a particular radio ... show full transcript
Worked Solution & Example Answer:Figure 8 shows the first-stage filter circuit for a simple AM receiver - AQA - A-Level Physics - Question 3 - 2017 - Paper 8
Step 1
Calculate the value of the capacitance when the circuit resonates at a frequency of 910 kHz.
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Answer
To find the capacitance, we use the resonance frequency formula: f = rac{1}{2 \\pi \\sqrt{LC}}
Rearranging gives us: C = rac{1}{(2 \\pi f)^2 L}
Substituting the given values:
Frequency, f=910 kHz=910×103 Hz
Inductance, L=1.1 mH=1.1×10−3 H
Thus, C=(2pi(910×103))2(1.1×10−3)1
Calculating gives: C≈27.8 pF.
Step 2
Draw on Figure 9 an ideal response curve for the resonant circuit, labelling all relevant frequency values based upon a 10 kHz bandwidth.
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Answer
The ideal response curve is typically characterized by a sharp peak at the resonant frequency. For a resonance frequency of 910 kHz and a bandwidth of 10 kHz, the relevant frequency values are:
Lower Frequency: 905 kHz
Resonant Frequency: 910 kHz
Upper Frequency: 915 kHz
The curve should show a peak gain of 1.0 at 910 kHz, gradually dropping to around 0.7 at 905 kHz and 915 kHz, creating a bell-shaped curve symmetric around the resonant frequency.
These values should be clearly labelled on Figure 9.
Step 3
Discuss the changes the listener might notice when tuning to this station due to the practical Q-factor being smaller.
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A smaller Q-factor leads to several noticeable changes for the listener:
Broader Bandwidth: The station will be broadcast over a wider frequency range, meaning the listener’s radio may pick up signals from adjacent frequencies, resulting in interference.
Less Selectivity: The circuit becomes less effective in filtering out unwanted frequencies, which could cause a muddier sound with mixed signals from nearby stations.
Increased Susceptibility to Crosstalk: With a lower Q-factor, the circuit is more prone to picking up signals from neighboring stations, leading to potential overlap and confusion in audio.
Less Gain: As a result of decreased energy transfer, the listener may notice a reduction in the clarity and strength of the signal, making it more difficult to tune in on a clear transmission.