- #1
ashah99
- 60
- 2
- 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.
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?
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?