*Question G4D05:
How does a signal that reads 20 dB over S9 compare to one that reads S9 on a receiver, assuming a properly calibrated S meter?*  

*Correct Answer is  D: It is 100 times more powerful*  

### AI Response: (may contain errors)  

To understand how a signal that reads 20 dB over S9 compares to one that reads S9, we need to know how decibels (dB) relate to power levels.

1. **Understanding dB Levels**:
   - The S meter is calibrated such that S9 corresponds to a specific power level, typically 50 µV (microvolts) at the receiver input.
   - A reading of 20 dB over S9 means the signal is 20 dB stronger than the S9 reference level.

2. **Calculating Power Ratios**:
   - The formula to convert dB to a power ratio is:
     $$ 
     P_{ratio} = 10^{\frac{dB}{10}} 
     $$
   - For a signal that is 20 dB over S9:
     $$ 
     P_{ratio} = 10^{\frac{20}{10}} = 10^2 = 100 
     $$
   - This means the signal is 100 times more powerful than the S9 reference level.

3. **Why the Correct Answer is Correct**:
   - The correct answer is **D: It is 100 times more powerful** because a 20 dB increase corresponds to a power increase by a factor of 100.

4. **Why the Other Choices are Incorrect**:
   - **A: It is 10 times less powerful**: This is incorrect because a 20 dB increase indicates a stronger signal, not a weaker one.
   - **B: It is 20 times less powerful**: This is also incorrect for the same reason; the signal is stronger, not weaker.
   - **C: It is 20 times more powerful**: This is incorrect because a 20 dB increase corresponds to a factor of 100, not 20.

In summary, when comparing a signal that reads 20 dB over S9 to one that reads S9, the former is indeed 100 times more powerful due to the logarithmic nature of the decibel scale.