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

Choices:  
A: 3 dB  
B: 5 dB  
C: 10 dB  
D: 13 dB    

Correct Answer is  D: 13 dB  

To relate receiver noise floor to bandwidth, remember thermal noise power is proportional to bandwidth:  
$$P_n = k T B$$  
Therefore, the change in noise power (in dB) when changing bandwidth is:  
$$\Delta N_{\text{dB}} = 10 \log_{10}\!\left(\frac{B_2}{B_1}\right)$$

Applying this from 50 Hz to 1000 Hz:  
$$\Delta N_{\text{dB}} = 10 \log_{10}\!\left(\frac{1000}{50}\right) = 10 \log_{10}(20) \approx 10 \times 1.301 = 13.01\ \text{dB}$$  
So the noise floor increases by about 13 dB. Correct answer: D.

Why the other choices are incorrect:
- A: 3 dB corresponds to doubling bandwidth ($B_2/B_1 = 2$), not 20×.
- B: 5 dB corresponds to about a 3.16× bandwidth increase, not 20×.
- C: 10 dB corresponds to a 10× bandwidth increase, not 20×.

Useful rules of thumb:
- Each doubling of bandwidth adds about 3 dB of noise power.
- A 10× increase in bandwidth adds 10 dB.
- A 20× increase adds about 13 dB (10 dB + 3 dB).