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What is the maximum average effective input noise temperature that an amplifier can have if its average standard noise figure is to not exceed 1.7 ?
- a)203 K
- b)215 K
- c)235 K
- d)255 K
Correct answer is option 'A'. Can you explain this answer?
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To solve this problem, we need to understand the relationship between noise figure and noise temperature. Noise figure (NF) is a measure of the noise performance of an amplifier, while noise temperature (Tn) represents the temperature at which an ideal resistor would produce the same amount of noise power as the actual amplifier.
The relationship between noise figure and noise temperature is given by the formula:
NF = 10 * log10 (1 + (Tn / To))
where NF is the noise figure, Tn is the noise temperature, and To is the reference noise temperature (usually 290 K).
Given that the average standard noise figure should not exceed 1.7, we can rearrange the formula to solve for the maximum average effective input noise temperature (Tn):
NF = 10 * log10 (1 + (Tn / To))
1.7 = 10 * log10 (1 + (Tn / 290))
Now, let's solve for Tn.
1 + (Tn / 290) = 10^(1.7 / 10)
(Tn / 290) = 10^(1.7 / 10) - 1
Tn = 290 * (10^(1.7 / 10) - 1)
Calculating the value, we find:
Tn ≈ 202.86 K
Therefore, the maximum average effective input noise temperature that an amplifier can have, given that its average standard noise figure does not exceed 1.7, is approximately 203 K.
Hence, option A is the correct answer.
The relationship between noise figure and noise temperature is given by the formula:
NF = 10 * log10 (1 + (Tn / To))
where NF is the noise figure, Tn is the noise temperature, and To is the reference noise temperature (usually 290 K).
Given that the average standard noise figure should not exceed 1.7, we can rearrange the formula to solve for the maximum average effective input noise temperature (Tn):
NF = 10 * log10 (1 + (Tn / To))
1.7 = 10 * log10 (1 + (Tn / 290))
Now, let's solve for Tn.
1 + (Tn / 290) = 10^(1.7 / 10)
(Tn / 290) = 10^(1.7 / 10) - 1
Tn = 290 * (10^(1.7 / 10) - 1)
Calculating the value, we find:
Tn ≈ 202.86 K
Therefore, the maximum average effective input noise temperature that an amplifier can have, given that its average standard noise figure does not exceed 1.7, is approximately 203 K.
Hence, option A is the correct answer.
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