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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
Step 1
Answer
To find the capacitance when the circuit resonates at a particular frequency, we use the formula for the resonant frequency of an RLC circuit:
Given:
Rearranging the formula to solve for capacitance:
Substituting the values:
Calculating this gives:
Step 2
Answer
In Figure 9, the ideal response curve should illustrate the relative voltage gain as a function of frequency. The curve peaks at the resonant frequency of 910 kHz, reaching a maximum relative voltage gain of 1.0. The bandwidth is given as 10 kHz, so we identify the frequency range from 905 kHz to 915 kHz. The shape of the curve is generally a bell shape, showing a gradual increase in gain as the frequency approaches 910 kHz from both sides and a decrease after passing it.
Therefore, the labels should indicate:
This is identified based on the 10 kHz bandwidth around the resonant frequency.
Step 3
Answer
A smaller Q-factor indicates that the circuit is less selective and more prone to picking up interference from adjacent frequencies. The listener may notice several changes:
Broader Bandwidth: The station may not sound as clear as it could with a higher Q-factor since the bandwidth of the response curve is wider, potentially allowing in more adjacent signals.
Signal Variability: The listening experience might change with fluctuations in signal quality and strength, leading to occasional static or noise due to interference.
Loss of Sensitivity: The sensitivity to the target frequency is reduced, resulting in lower sound quality or difficulty tuning in to the station clearly.
Crosstalk: The listener may hear overlapping signals from nearby stations, further degrading the listening experience due to less effective filtering.
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