This may be a trivial question but I don't have a strong background in signal processing and measurements. I have a piezoelectric accelerometer with a sampling frequency of 10 kHz (sensitivity $102 \; mV/(m/s^2)$) and a 24 bit ADC. I set the ADC range to $1000 \; mV$ so the theoretical resolution of my measurement should be $$ \frac{1000}{102 \cdot 2^{24}} = 5.84 \cdot 10^{-7} \; m/s^2 $$ But this value doesn't take into account the noise.

From the datasheet I get a spectral noise in the band $10$ - $100\;Hz$ of $19.6\; (\mu m/s^2 )/\sqrt{Hz}$.

I need to calculate the RMS value of the signal in the band $40 - 50\; Hz$.

The noise power spectral density in that band is $$ P_{noise} = \left( 19.6 \cdot 10^{-6}\right)^2 \cdot 10 = 3.84 \cdot 10^{-9} \; (m/s^2)^2 $$ Thus, the RMS value of the noise is $$ RMS_{noise} = \sqrt{P_{noise}} = 6.20 \cdot 10^{-5} \; m/s^2 $$ So, correct me if I'm wrong, the actual measurement resolution depends on the noise which is the real "bottleneck". I mean, with a 24 bit ADC you should not have any problem but the real resolution should be defined given the SNR. Given a $SNR=1$ I get a RMS measurement resolution of $6.19\cdot10^{-5}\; m/s^2$. I can't find a book or a guide in which this concept is explained explicitly, I want to know if my understanding is correct or if I'm missing something.

Thank you :)


  • $\begingroup$ What you are calling ${PSD}_{noise}$ is just the noise power, not the power spectral density. PSD is the amount of power per Hz -- it's basically what you started with, after scaling the datasheet number. What you're calling ${PSD}_{noise}$ should probably be edited to say ${P}_{noise}$ $\endgroup$
    – TimWescott
    Sep 17, 2021 at 15:13
  • $\begingroup$ You are right! What a dumb error, thank you $\endgroup$
    – DavideArma
    Sep 17, 2021 at 18:31

1 Answer 1


In measurement science there is less talk of resolution and more talk of accuracy and precision.

Accuracy is how close to the true value your are; and precision is how much spread (variance) you expect a variable to have.

In my experience precision and resolution are often "mixed" together, maybe because they are a bit similar in some cases. However, accuracy and precision are well defined and general terms and are what is used to characterise any measurement (in measurement science).

Resolution is usually understood (at least in measurement science) to be the same as step-size and is a static (always the same, not dependent on time) characteristic of a device. An analogue to digital converter is a good example — there are infinitely many values in, and a limited number of values out; therefore you lose information. This is called quantisation. This behaviour can be used to determine the precision. This can be done using a common approximation due to Bennett (1948); the quantisation error is an additive white noise component with uniform distribution between $\pm\Delta/2$ and therefore has the variance: $$ \sigma_{q}^{2} = \frac{\Delta^2}{12} $$ where $\Delta$ is the step-size, or resolution. The actual error, or noise, variance is dependent on the input signal; but if your input is "busy" enough (has a lot of frequency components) then it is a good approximation. If your input is constant or almost constant, this is a very bad approximation.

From your description I have taken the liberty of drawing what I think you are working with.

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In order to make it more explicit, it looks like you are using the following measurement model $$ y = \frac{1}{K} \left( K (x + n_s) + n_q \right) $$ where $K$ is the gain/sensitivity, $x$ the "true" value of your measurement, $n_s$ is sensor noise and $n_q$ is quantisation error (or "noise").

In this case you have already determined the standard deviation of the noise from the sensor (but maybe not the environment; which is quite difficult and application specific); after the band-pass filter (40-50 Hz; and from linear system theory we know that we can treat additive signals separately) $$ \sigma_{s} = 62.0 \cdot 10^{-5} \text{ m/s}^2 $$ and from your specifications the gain (sensitivity) is $$ K = 102 \text{ mV/(m/s}^2{)} $$ the step size is $$ \Delta = \frac{1000}{2^{24}} \approx 59.6 \cdot 10^{-6} \text{ mV} $$ and the quantisation error standard deviation is $$ \sigma_{q} = 17.2 \cdot 10^{-6} \text{ mV} $$ To put a number to the accuracy and precision, we start by finding the variance of the output $y$ $$ \text{Var}(y) = \sigma_{s}^2 + \left( \frac{1}{K} \sigma_{q} \right)^2 $$ (the component $x$ is assumed to be deterministic hence $E(y) = x$). The value of the standard deviation for your output is $$ \sigma_y = 62.0 \cdot 10^{-6} \text{ m/s}^2 $$ and your output variance is almost entirely dominated by the sensor noise (as you have already noted): $$ (\sigma_q/\sigma_y)^2 \approx 0 $$ and $$ (\sigma_s/\sigma_y)^2 \approx 1 $$ If the distribution of the noise in the output is approximately Gaussian, it is common to say that the uncertainty of your measurement is $$ U = 2 \cdot \sigma_{y} = 124 \cdot 10^{-6} \text{ m/s}^2 $$ where the number 2 is called the coverage factor (and 2 is the standard value as decided by a committee of measurement scientists). If the total noise is Gaussian, a coverage factor of 2 means that 95% of your measurements will fall within the band of $\pm U$ around your measured value. This is your measure of precision and accuracy, given you have good estimates for all your error sources. If the sensor noise is Gaussian (very common in the case of physical systems), and since your sensor noise swamps the quantisation error, this is probably true in your case.

There is really no well-defined method to relate the uncertainty of your output to the concept of resolution or step-size, but if you pretend the output noise distribution is uniform (which it is not), you can use Bennett's model again (2 times output variance = pretend quantisation error variance) to obtain the effective number of bits.

In measurement science the general method used to determine uncertainty is called "Guide to the expression of uncertainty in measurement" (GUM). A hands-on introduction can be found at: https://uncertainty.nist.gov/

There are no excellent textbooks covering this subject, but the best one I have found thus far is Principles of Measurement Systems by J. Bentley: https://www.pearson.com.au/products/detail?isbn=9780130430281

  • 1
    $\begingroup$ Thank you so much for the detailed answer. You've given me a lot to work on. $\endgroup$
    – DavideArma
    Sep 21, 2021 at 8:14

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