*Question E4C06:
How much does increasing a receiver’s bandwidth from 50 Hz to 1,000 Hz increase the receiver’s noise floor?*  

*Correct Answer is  D: 13 dB*  

### AI Response: (may contain errors)  

To determine how much increasing a receiver's bandwidth from 50 Hz to 1,000 Hz increases the receiver's noise floor, we can use the formula for noise power increase due to bandwidth:

$$
\Delta N = 10 \log_{10} \left( \frac{B_2}{B_1} \right)
$$

where:
- $B_1$ = initial bandwidth (50 Hz)
- $B_2$ = final bandwidth (1,000 Hz)

### Calculation:
1. Calculate the ratio of the bandwidths:
   $$ 
   \frac{B_2}{B_1} = \frac{1000 \text{ Hz}}{50 \text{ Hz}} = 20 
   $$

2. Now, apply the formula:
   $$
   \Delta N = 10 \log_{10}(20) 
   $$

3. Using the logarithm:
   - $\log_{10}(20) \approx 1.301$
   - Therefore:
   $$
   \Delta N \approx 10 \times 1.301 = 13.01 \text{ dB}
   $$

This rounds to approximately 13 dB, which is why the correct answer is **D: 13 dB**.

### Explanation of Choices:
- **A: 3 dB**: This would imply a very small increase in noise, which is not consistent with the large increase in bandwidth.
- **B: 5 dB**: This is also too low for the increase in bandwidth from 50 Hz to 1,000 Hz.
- **C: 10 dB**: While this is a common increase for a doubling of bandwidth, the increase from 50 Hz to 1,000 Hz is much larger than just a doubling.

### Summary:
Increasing the bandwidth from 50 Hz to 1,000 Hz results in a noise floor increase of approximately 13 dB due to the logarithmic relationship between bandwidth and noise power. This is a crucial concept in understanding how bandwidth affects receiver performance in amateur radio.