Question E1C10:
[97.307] What is the maximum mean power level for a spurious emission below 30 MHz with respect to the fundamental emission?  

Choices:  
A: - 43 dB  
B: - 53 dB  
C: - 63 dB  
D: - 73 dB    

Correct Answer is  A: - 43 dB  

To answer E1C10:

1) Why -43 dB is correct
- FCC Part 97.307(d) sets the spurious emission limit below 30 MHz: the mean power of any spurious emission must be at least 43 dB below the mean power of the fundamental.
- This is expressed as −43 dBc (dB relative to the carrier/fundamental).

Useful conversion:
$$
P_{\text{spur,max}} = P_{\text{fund}} \times 10^{-43/10} \approx P_{\text{fund}} \times 5.01\times 10^{-5}
$$
Example: For a 100 W HF transmitter,
$$
P_{\text{spur,max}} \approx 100 \times 10^{-4.3} \approx 0.005\ \text{W} = 5\ \text{mW}
$$

2) Why the other choices are incorrect
- -53 dB, -63 dB, -73 dB: These are all stricter than required for HF (<30 MHz). Part 97 does not require these deeper suppression levels below 30 MHz. More stringent limits (e.g., -60 dB) apply above 30 MHz, not below.

3) Formulas and rules of thumb
- Spurious-to-fundamental ratio in dB:
$$
\Delta\text{dB} = 10\log_{10}\!\left(\frac{P_{\text{spur}}}{P_{\text{fund}}}\right)
$$
- Solve for allowed spurious power:
$$
P_{\text{spur}} = P_{\text{fund}} \times 10^{\Delta\text{dB}/10}
$$
- Rule of thumb: −43 dBc ≈ 1/20,000 of the fundamental power (mean power basis) for HF spurious emissions.