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Digital Conversion and Resolution

24 min Sensors & Measurement Math

Analog-to-Digital Conversion

Modern sensors produce analog signals that must be digitized. The math involves powers of 2, resolution, and sampling theory.

ADC Resolution

An $n$-bit ADC divides the input range into $2^n$ levels:

$$\text{Resolution} = \frac{V_{\text{range}}}{2^n}$$

A 10-bit ADC: $2^{10} = 1024$ levels.

Nyquist-Shannon Sampling Theorem

To accurately capture a signal of frequency $f$, you must sample at least at $f_s \geq 2f$. Below this rate, aliasing occurs.

Example 1

A 12-bit ADC has a 0–5 V range. What is the smallest voltage change it can detect?

Resolution $= 5/2^{12} = 5/4096 \approx 0.00122$ V ≈ 1.22 mV.

Example 2

An audio signal has frequencies up to 20 kHz. What is the minimum sampling rate?

$f_s \geq 2 \times 20{,}000 = 40{,}000$ Hz = 40 kHz.

(CD audio uses 44.1 kHz — slightly above the Nyquist minimum.)

Example 3

An 8-bit ADC reads 0–3.3 V. The reading is digital value 200. What voltage?

$V = 200 \times \frac{3.3}{256} = 200 \times 0.01289 = 2.578$ V.

1. How many levels does a 16-bit ADC have?

2. Resolution of a 10-bit ADC with 0–3.3 V range?

3. Minimum sampling rate for a 5 kHz signal?

4. An 8-bit ADC (0–5 V) reads value 128. Voltage?

5. How many more levels does 12-bit have vs 8-bit?

6. Why is CD audio sampled at 44.1 kHz instead of exactly 40 kHz?

Show Answer Key

1. $2^{16} = 65{,}536$

2. $3.3/1024 \approx 3.22$ mV

3. 10 kHz

4. $128 \times 5/256 = 2.5$ V

5. $4096/256 = 16$ times more levels

6. To provide a guard band above the Nyquist minimum, allowing practical filter design.