Question E4C05:
What does a receiver noise floor of -174 dBm represent?  

Choices:  
A: The receiver noise is 6 dB above the theoretical minimum  
B: The theoretical noise in a 1 Hz bandwidth at the input of a perfect receiver at room temperature  
C: The noise figure of a 1 Hz bandwidth receiver  
D: The receiver noise is 3 dB above theoretical minimum    

Correct Answer is  B: The theoretical noise in a 1 Hz bandwidth at the input of a perfect receiver at room temperature  

Correct answer: B

1) Why B is correct
- -174 dBm is the thermal noise power in a 1 Hz bandwidth at room temperature (about 290 K) for an ideal (noise-free) receiver.
- From thermal noise (Johnson-Nyquist) theory:
  $$P_N = k T B$$
  where:
  - $k = 1.38 \times 10^{-23}\ \text{J/K}$ (Boltzmann’s constant)
  - $T \approx 290\ \text{K}$ (room temperature)
  - $B = 1\ \text{Hz}$
- Converting $P_N$ to dBm yields approximately $-174\ \text{dBm/Hz}$.

2) Why the other choices are incorrect
- A: “6 dB above the theoretical minimum”  
  - 6 dB above −174 dBm would be −168 dBm (4× the power). −174 dBm itself is the theoretical minimum in 1 Hz at 290 K, not 6 dB above it.
- C: “The noise figure of a 1 Hz bandwidth receiver”  
  - Noise figure (NF) is a receiver parameter (in dB) indicating degradation above the thermal noise floor. −174 dBm is an absolute thermal noise density, not a noise figure.
- D: “3 dB above theoretical minimum”  
  - 3 dB above −174 dBm would be −171 dBm (2× the power). −174 dBm is the minimum, not 3 dB above.

3) Useful formulas and rules of thumb
- Thermal noise power:
  $$P_N = k T B$$
- In dBm (for room temperature):
  $$P_N\ (\text{dBm}) \approx -174\ \text{dBm/Hz} + 10\log_{10}(B\ \text{in Hz}) + \text{NF (dB)}$$
- Rule of thumb:
  - Thermal noise density at 290 K is $-174\ \text{dBm/Hz}$.
  - +3 dB doubles noise power; +6 dB quadruples it.