Noise Floor/Device Temperature Problem

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ashah99
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Homework Statement
Please see an attachment below.
Relevant Equations
Pn= 10*log10(kTB)
dBm = dBW - 30 [dB]
Hi everyone, I would like to get some help with the following problem. I'm not sure if my answer is feasible. Thanks.

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My attempt at a solution:

I used the noise power equation: Pn= 10*log10(kTB) and converted the -100 dBm to dBW by subtracting 30 dB to get -130 dBW. Then I tried to find the overall system noise temperaure, T, and then I subtract the antenna temperature to find the temp of the LNA.

Pn= 10*log10(kTB) = Pn= 10*log10(k) + Pn= 10*log10(T) + Pn= 10*log10(B)
-130 dBW= 10*log10(1.38e-23 J/K) + 10*log10(T) + Pn= 10*log10(20e6 Hz)
10*log10(T) = -130 dBW - (-228.6 dB/J/K) - 73 dB/Hz = 25.6 dB/K
T = 10^(25.6/10) = 362.3 K
T_LNA = T - T_antenna = 262.3 K

This seems quite high for LNA device temperature. Could someone help to see if my work is right or if I'm totally off-track or misinterpreting the problem?
 
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Actually 100K is very cold so I think they mean Antenna Noise Temperature, not physical temperature. I think your method is OK but have not checked the numbers.
 
tech99 said:
Actually 100K is very cold so I think they mean Antenna Noise Temperature, not physical temperature. I think your method is OK but have not checked the numbers.
My assumption was antenna noise temp as well. Would you kind as to check if I’m on the right path?