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Frequency of an local oscillator in a AM standard broadcast receiver having on IF as 455kHz and tuned in 540kHz when local oscillator tracks above the frequency of received signal is;
- a)945 kHz
- b)970 kHz
- c)995 kHz
- d)1045 kHz
Correct answer is option 'C'. Can you explain this answer?
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Frequency of an local oscillator in a AM standard broadcast receiver ...
Frequency of Local Oscillator in AM Standard Broadcast Receiver
Frequency of local oscillator in an AM standard broadcast receiver can be calculated using the following formula:
Local oscillator frequency = Received signal frequency + Intermediate frequency
In this case, the received signal frequency is 540 kHz and the intermediate frequency is 455 kHz. Therefore, the local oscillator frequency can be calculated as:
Local oscillator frequency = 540 kHz + 455 kHz = 995 kHz
Therefore, option C is the correct answer.
Explanation:
An AM standard broadcast receiver works on the principle of heterodyne detection. The received signal is mixed with a local oscillator signal to produce an intermediate frequency (IF) signal. This IF signal is then amplified and demodulated to extract the original audio signal.
In order to ensure that the local oscillator signal is always above the frequency of the received signal, the local oscillator frequency is set to the sum of the received signal frequency and the intermediate frequency. This ensures that the IF signal is always at the desired frequency of 455 kHz.
In this case, the received signal frequency is 540 kHz and the intermediate frequency is 455 kHz. Therefore, the local oscillator frequency is calculated as 995 kHz. This ensures that the IF signal is at the desired frequency of 455 kHz, which can be easily amplified and demodulated to extract the original audio signal.
Conclusion:
The frequency of the local oscillator in an AM standard broadcast receiver can be calculated using the formula: Local oscillator frequency = Received signal frequency + Intermediate frequency. In this case, the local oscillator frequency is 995 kHz, which ensures that the IF signal is at the desired frequency of 455 kHz.
Frequency of local oscillator in an AM standard broadcast receiver can be calculated using the following formula:
Local oscillator frequency = Received signal frequency + Intermediate frequency
In this case, the received signal frequency is 540 kHz and the intermediate frequency is 455 kHz. Therefore, the local oscillator frequency can be calculated as:
Local oscillator frequency = 540 kHz + 455 kHz = 995 kHz
Therefore, option C is the correct answer.
Explanation:
An AM standard broadcast receiver works on the principle of heterodyne detection. The received signal is mixed with a local oscillator signal to produce an intermediate frequency (IF) signal. This IF signal is then amplified and demodulated to extract the original audio signal.
In order to ensure that the local oscillator signal is always above the frequency of the received signal, the local oscillator frequency is set to the sum of the received signal frequency and the intermediate frequency. This ensures that the IF signal is always at the desired frequency of 455 kHz.
In this case, the received signal frequency is 540 kHz and the intermediate frequency is 455 kHz. Therefore, the local oscillator frequency is calculated as 995 kHz. This ensures that the IF signal is at the desired frequency of 455 kHz, which can be easily amplified and demodulated to extract the original audio signal.
Conclusion:
The frequency of the local oscillator in an AM standard broadcast receiver can be calculated using the formula: Local oscillator frequency = Received signal frequency + Intermediate frequency. In this case, the local oscillator frequency is 995 kHz, which ensures that the IF signal is at the desired frequency of 455 kHz.
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Community Answer
Frequency of an local oscillator in a AM standard broadcast receiver ...
FLo = fIF + fC
= 455 × 103 + 540 × 103
= 995 kHz
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Frequency of an local oscillator in a AM standard broadcast receiver having on IF as 455kHz and tuned in 540kHz when local oscillator tracks above the frequency of received signal is;a)945 kHzb)970 kHzc)995 kHzd)1045 kHzCorrect answer is option 'C'. Can you explain this answer?
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