1
\$\begingroup\$

Oversampling means to sample at significantly more than the Nyquist Rate. I can well understand it in signal re-construction: more samples can improve the ADC's resolution, signal-to-error ratio, etc.

However, when in a control system, I have some difficulities understanding it. Assume that the frequency of the process variable is \$f\$, so the Nyquist Rate is \$2f\$, and we sample at \$2f\$ (the controller will take action upon each sample). Maybe we don't get a stable process variable, so we decide to improve the sampling rate.

Next, we sample at \$20f\$, a sampling frequency 10 times higher (also the controller takes action upon each sample), and this time we obtain a stable process variable. However, because the controller action is also 10 times faster, the frequency of the process variable is no longer \$f\$, but much higher, say \$f'\$. Now the Nyquist Rate becomes \$2f'\$.

Comparing sampling frequency \$20f\$ and new Nyquist rate \$2f'\$, they may be very close, so we cannot call it "oversampling".

Is my understanding towards oversampling and Nyquist rate correct?

\$\endgroup\$
1
  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. Any conclusions reached should be edited back into the question and/or any answer(s). \$\endgroup\$
    – Dave Tweed
    Feb 3 '20 at 13:42
3
\$\begingroup\$
  1. Just to clean up some terminology: You're always sampling at twice the Nyquist rate. The Nyquist rate is a function of the sampling rate. You want the Nyquist rate to be greater than your maximum frequency of interest.

  2. Your loop frequency response may have a corner frequency f, which must be below the Nyquist rate. However, the loop transfer function may allow attenuated signals of higher frequency. Your requirements should tell you whether you need to address the harmonics.

  3. Your highest frequency of interest may not be limited to your loop transfer function (process variable?) if there are sources of higher frequency noise in your feedback signal. You might "oversample" with respect to your loop dynamics in order to process higher frequency noise.

  4. Oversampling can be used to perform more frequent corrections. This generally means that the corrections are smaller and more smoothed out by the loop transfer function.

  5. Oversampling also introduces inherent averaging, improving the resolution of the signal.

I hope this answers your question, as I'm not sure exactly what it was. :-)

\$\endgroup\$
2
  • \$\begingroup\$ "The Nyquist rate is a function of the sampling rate." so you mean that the Nyquist rate also changes after I increase the sampling (and correction) rate? and why do you say I always sample twice the Nyquist rate? \$\endgroup\$
    – Bloodmoon
    Jan 30 '20 at 6:57
  • \$\begingroup\$ The Nyquist rate is the maximum frequency that can be accurately represented by a sampled signal (or, more accurately, the minimum frequency that can't be accurately represented by the sampled signal). It is half the sample rate; note that a sin wave at exactly the Nyquist rate can't be reliably represented no longer how long the interval one samples). It is a function of the sampling rate. Just as the characteristics of the input signal don't change the bandwidth of a filter, they also don't change the Nyquist rate of a sampled system. \$\endgroup\$ Jan 30 '20 at 13:05
1
\$\begingroup\$

While I understand basic control loop and sample data system theory, I am by no means an expert in that field. That said, here’s the way I would look at this.

In a continuous-time (analog) control loop system, the output is measured, the error term generated, and the corrective action (voltage change, for example) all happen in a continuous, non-discrete fashion. The loop response is determined by feedback/feedforward parameters in the control loop. For sake of argument, let’s say our analog (continuous time) control loop closes at 100 Hz.

Now, replace that continuous measurement of the output with a sampled version – an ADC. If the sampling rate of the ADC is high enough compared to the loop’s response, then the entire thing can be analyzed like a traditional analog control loop. If the ADC in our example is sampling the outputs at 1 mega-sample/sec, then the discrete nature of the error samples doesn’t come into play.

As the sampling rate of the ADC is reduced, at some point you get to where you have to treat the control system as a discrete time system. This happens at a sample rate of approximately 20 times the loop closure frequency, which would be ~2 KHz in this example. Note that this is well above the 2*100 Hz sample rate you might estimate just based on the Nyquist criteria and the desired closed loop frequency response.

So as long as your ADC samples well above your loop response, you can for the most part treat the control loop in a tradition analog fashion.

\$\endgroup\$
1
  • \$\begingroup\$ There is an important concept in your answer: the loop closure frequency. And you assume that it is consistent whether the sampling rate is changed. What is the definition of it? I have googled it but only found results on frequency response. \$\endgroup\$
    – Bloodmoon
    Jan 30 '20 at 5:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.