Question E7F06:
What is the minimum number of bits required to sample a signal with a range of 1 volt at a resolution of 1 millivolt?  

Choices:  
A: 4 bits  
B: 6 bits  
C: 8 bits  
D: 10 bits    

Correct Answer is  D: 10 bits  

To resolve the required ADC resolution:

- Needed quantization levels:  
  $$\text{levels} = \frac{\text{range}}{\text{resolution}} = \frac{1\ \text{V}}{1\ \text{mV}} = 1000$$
- Bits required:  
  $$N = \lceil \log_2(1000) \rceil \approx \lceil 9.97 \rceil = 10$$

Why D (10 bits) is correct:
- A 10-bit ADC has $2^{10} = 1024$ levels, giving a step size  
  $$\Delta = \frac{1\ \text{V}}{1024} \approx 0.977\ \text{mV} \le 1\ \text{mV}$$  
  This meets the required resolution.

Why the others are incorrect:
- A: 4 bits → $2^4 = 16$ levels,  
  $$\Delta = \frac{1\ \text{V}}{16} = 62.5\ \text{mV}$$
- B: 6 bits → $2^6 = 64$ levels,  
  $$\Delta = \frac{1\ \text{V}}{64} \approx 15.6\ \text{mV}$$
- C: 8 bits → $2^8 = 256$ levels,  
  $$\Delta = \frac{1\ \text{V}}{256} \approx 3.91\ \text{mV}$$  
All exceed 1 mV per step, so they don’t meet the resolution requirement.

Useful formulas/rules of thumb:
- Step size (LSB): $$\Delta = \frac{\text{range}}{2^N}$$
- Required bits: $$N = \lceil \log_2\!\left(\frac{\text{range}}{\text{resolution}}\right) \rceil$$
- Rule of thumb: $2^{10} = 1024 \approx 1000$, so 10 bits ≈ 3 decimal digits of resolution.