Time uncertainty of an ADC sample for a given sampling frequency

Question

I am working on a project which involves time difference of arrival calculations. I need to calculate the uncertainty of my timing measurements. I am using an ADC to sample the signals who's times of arrival need to be compared.

Right now, I am working under the assumption that sample at fs results in a timing uncertainty of 1/fs. For example, if a signal arrives 1 second after the system starts up, and the sampling frequency is 10Hz, the resulting arrival time would be 1s +- 0.1s. Is this method valid if I am interested in considering sample rate only?

Answer 1

Right now, I am working under the assumption that sample at fs results in a timing uncertainty of 1/fs.

This is not correct unless you are undersampling. Timing uncertainty in a Nyquist sampled signal is determined by SNR, not sampling rate since (per the Sampling Theorem) you have recorded the signal exactly and can reconstruct it with as small of a time step as you like. Thus what really limits you is noise in the measurement, not the sampling rate of the recording.

In general, you should expect to do much, much better than 1/fs.

Answer 2

As you do not disclose the scenario and details of your measurement system, I can only consider the ADC role as a discriminator module. Leaving aside the Nyquist, SNR, clock jitter issues that you pass over in your question, I offer you the probabilistic analysis of measurement errors in your scenario with only the information you provided.

You are interested in "the uncertainty of [your] timing measurements" as a result of "time difference of arrival calculations", TDA calculations. The TDA value is calculated as a difference of two individual time measurements, the event start time and the event stop (arrival) time. In your system, the result of each individual time measurement is an estimate of this time event as a uniformly distributed random variable with a value between \$(N-1)·(1/fs)\$ and \$N·(1/fs)\$, where \$N\$ is a registered ordinal number of the ADC readout (not the ADC code value!) that you qualify as a sign that the event happened. You know for sure that at the time indicated as $N·(1/fs)$ the event happened, and that it can happen earlier but not earlier than at the time \$(N-1)·(1/fs)\$, because otherwise it would be noticed in the previous ADC readout -- if only the events are not that volatile as to start and to end in a single ADC measurement cycle. We write down \$T_{event} \sim U((N-1)·(1/fs), N·(1/fs))\$.

Assume you register the ordinal numbers of the ADC readouts in the counter \$\text N\$. Let the start time is registered at the \$N_{start}\$ counter value, the stop (arrival) time is registered at the \$N_{stop}\$ counter value. Then the start time r.v. is \$T_{start} \sim U((N_{start}-1)·(1/fs), N_{start}·(1/fs))\$, and the stop time r.v. is \$T_{stop} \sim U((N_{stop}-1)·(1/fs), N_{stop}·(1/fs))\$. The TOA r.v. (\$T_{start} - T_{stop}\$) is $$ T_{TOA} = \begin{cases} \text {ASCLIN}((N_{stop}-N_{start}-2)·(1/fs), (N_{stop}-N_{start}-1)·(1/fs)) \\ \text {DESCLIN}((N_{stop}-N_{start}-1)·(1/fs), (N_{stop}-N_{start})·(1/fs)) \end{cases} $$ where \$\text {ASCLIN}\$ is an ascending linear distribution, \$\text {DESCLIN}\$ as a descending linear distribution, as shown in the image:

The total area under curves (total probability) is unity.

Your scenario does not specify even the start time choice of TOA measurements: is it the same for all TOAs or are all of the TOAs composed of individual pairs \$\{T_{start}, T_{stop}\}\$. Also, your question raises the possibility that your task is to compare TOAs: "to sample the signals who's times of arrival need to be compared", in which case, for the differences of TOA's pairs the TOA pdf may become a quadratic function of time (see Irwin–Hall distribution).

This is all that can be said about the estimation of "the uncertainty of [your] timing measurements" within the framework of the information you provided. It is unclear what the system you examine: with the time clock period of 0.1s the task may be the comparison of runner's times in elementary school athletic competitions or what else, in which case it is unclear why the ADC is used in your instrumentation at all.

If your task requires more advanced approaches then stopwatch recording, you may be interested in time-to-digital converters. If the TOA measurement is the secondary task to the other tasks which has to be done with the ADC measurements, pay attention to the other answer and to the comments.

Answer 3

Consider a pulse waveform with infinitely fast rise time. It is being sampled at a 10 Hz frequency. Assume that the ADC puts out a 00000... reading up to and including the sample at t = 1.0. After that, all readings are 11111...

The total measurement uncertainty is 1/fs, as you indicate, or 0.1 sec. However, the recovered arrival is best described as 1.05 seconds +/- .05 seconds.

This, of course, assumes no low-pass filtering at the input - which is an invitation to low SNR. If you do have an input filter, you should at least consider subtraction the group delay of the filter, which varies with both the cutoff frequency and the filter structure. For lowpass filters, the step response of the filter will be a mirror image of the frequency response, so the lower the cutoff frequency the greater the delay.