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What is the image rejection ratio when a super heterodyne receiver with quality factor of 50 is tuned with fs = 800 KHz and local oscillator frequency is 1250 KHz.
- a)81.6
- b)81.9
- c)82.1
- d)82.7
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
What is the image rejection ratio when a super heterodyne receiver wi...
IF = f1 − fs
fsi/fs - fs/fsi
= 1250 − 800 = 450KHz
fsi = fs + 2IF = 1700KHz
fsi/fs - fs/fsiIRR = 82.7
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Community Answer
What is the image rejection ratio when a super heterodyne receiver wi...
To calculate the image rejection ratio of a super heterodyne receiver, we need to consider the frequencies involved and the quality factor of the receiver.
Given:
- Intermediate frequency (IF) = fs = 800 KHz
- Local oscillator frequency (LO) = 1250 KHz
- Quality factor (Q) = 50
The image frequency is the sum of the IF and LO frequencies, and is given by the equation:
Image frequency = LO + IF
Calculating the image frequency using the given values:
Image frequency = 1250 KHz + 800 KHz = 2050 KHz
The image rejection ratio (IRR) is defined as the ratio of the desired signal power to the image signal power. It can be calculated using the quality factor (Q) of the receiver:
IRR = (Q/2)^2
Substituting the value of Q into the equation:
IRR = (50/2)^2 = 25^2 = 625
To convert this ratio into decibels (dB), we use the formula:
IRR(dB) = 10log10(IRR)
Calculating the image rejection ratio in decibels:
IRR(dB) = 10log10(625) ≈ 27.96 dB
Therefore, the image rejection ratio is approximately 27.96 dB.
However, none of the given options match this value exactly. It seems there might be an error in the options provided. The closest option is 82.7, which is not the correct answer based on the calculations.
Given:
- Intermediate frequency (IF) = fs = 800 KHz
- Local oscillator frequency (LO) = 1250 KHz
- Quality factor (Q) = 50
The image frequency is the sum of the IF and LO frequencies, and is given by the equation:
Image frequency = LO + IF
Calculating the image frequency using the given values:
Image frequency = 1250 KHz + 800 KHz = 2050 KHz
The image rejection ratio (IRR) is defined as the ratio of the desired signal power to the image signal power. It can be calculated using the quality factor (Q) of the receiver:
IRR = (Q/2)^2
Substituting the value of Q into the equation:
IRR = (50/2)^2 = 25^2 = 625
To convert this ratio into decibels (dB), we use the formula:
IRR(dB) = 10log10(IRR)
Calculating the image rejection ratio in decibels:
IRR(dB) = 10log10(625) ≈ 27.96 dB
Therefore, the image rejection ratio is approximately 27.96 dB.
However, none of the given options match this value exactly. It seems there might be an error in the options provided. The closest option is 82.7, which is not the correct answer based on the calculations.
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What is the image rejection ratio when a super heterodyne receiver with quality factor of 50 is tuned with fs = 800 KHz and local oscillator frequency is 1250 KHz.a)81.6b)81.9c)82.1d)82.7Correct answer is option 'D'. Can you explain this answer?
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