Filters, Noise, and the ADC
Filters, Noise, and the ADC
Once the sensor signal has been amplified by an instrumentation amp, it still contains noise from every source imaginable — thermal noise in the resistors, 1/f noise in the op-amps, mains pickup, EMI from switching regulators, and quantization noise from the ADC itself. A well-designed signal chain includes anti-alias filters, careful grounding, and just enough resolution at the ADC to make the noise floor visible but not wasteful.
Pallàs-Areny and Webster devote substantial space to Nyquist–Shannon sampling and to the interplay between analog filter cut-off and effective number of bits. In this lesson you will size a single-pole anti-alias filter, compute thermal (Johnson) noise, and convert between dB-SNR and effective number of bits.
These are the final skills you need to take a sensor signal all the way from the transducer into software — the ultimate goal of Webster and Pallàs-Areny's book.
$$f_c = \frac{1}{2\pi RC}$$
At $f_c$, the response is $-3\,\text{dB}$. For anti-aliasing, choose $f_c \le f_s/2$ with adequate margin.
$$V_n = \sqrt{4\,k_B\,T\,R\,B}$$
$k_B = 1.38\times10^{-23}\,\text{J/K}$, $T$ in kelvin, $R$ in ohms, $B$ = noise bandwidth (Hz). Rule of thumb at 300 K: $V_n \approx 4\,\text{nV}/\sqrt{\text{Hz}}$ per $\sqrt{1\,\text{k}\Omega}$.
An ADC samples at $f_s = 10\,\text{kHz}$. Design a one-pole RC filter with $f_c = 2\,\text{kHz}$ using $C = 10\,\text{nF}$.
$R = 1/(2\pi f_c C) = 1/(2\pi\cdot 2000\cdot 10^{-8}) = 7.96\,\text{k}\Omega$. Use $8.2\,\text{k}\Omega$ standard E12 value.
Noise bandwidth $B = 10\,\text{kHz}$, $T = 300\,\text{K}$, $R = 10\,\text{k}\Omega$. Find $V_n$.
$V_n = \sqrt{4\cdot 1.38\times10^{-23}\cdot 300\cdot 10^4\cdot 10^4}$
$= \sqrt{1.66\times10^{-12}} = 1.29\times10^{-6}\,\text{V} = 1.29\,\mu\text{V RMS}$.
This is the irreducible floor — no amplifier can do better unless its own noise is lower than this.
A 16-bit ADC with 5 V range shows an RMS noise floor of 300 µV. Find the ENOB.
Full-scale RMS (sine) $= V_{\text{FS}}/(2\sqrt{2}) = 5/(2.83) = 1.77\,\text{V}$.
SNR = $20\log_{10}(1.77/300\times10^{-6}) = 20\log_{10}(5888) = 75.4\,\text{dB}$.
$\text{ENOB} = (\text{SNR} - 1.76)/6.02 = (75.4 - 1.76)/6.02 = 12.2\,\text{bits}$.
So, despite 16 nominal bits, only about 12 are effective — a classic result that drives the selection of low-noise front-ends.
Practice Problems
Show Answer Key
1. $R = 1/(2\pi\cdot 1000\cdot 10^{-7}) = 1.59\,\text{k}\Omega$.
2. $V_n = \sqrt{4\cdot 1.38\times10^{-23}\cdot 300\cdot 10^3\cdot 1} = 4.07\,\text{nV/}\sqrt{\text{Hz}}$.
3. $10/2^{16} = 153\,\mu\text{V}$.
4. SNR = $6.02\cdot 12 + 1.76 - 20\log(2) = 72.24 + 1.76 - 6.02 = 67.98\,\text{dB}$; ENOB ≈ 11.0 bits.
5. Because aliasing folds signals above $f_s/2$ back into the baseband where no digital filter can remove them.
6. It is $20\log_{10}\sqrt{3/2}$, the ratio between a full-scale sine wave's RMS and the LSB quantization noise $\sqrt{1/12}$ LSB.