Unfortunately the answer is not as straightforward as you might like.
Let's define to Open Loop Bandwidth as the frequency at which the loop gain is 1 (standard definition). Above this frequency the loop can not respond effectively to commands or feedback.
In an ideal system Sampling Rate compared to Open Loop bandwidth does not really matter so long as the Nyquist criterion is satisfied, but in any real system delay is usually associated with the sample rate clock (e.g. digital controller calculation time), then the relationship between Sampling Rate and Open Loop bandwidth really does matter. As a guideline the crossover frequency must be less than the reciprocal of the time delay in the loop : Wc < 1/Td .
There are several factors involved in loop delay: Time to measure the sensed value, Time to calculate the response, time to Activate the response.
The time to activate the response may be misleading. It's not just the response of the actuator. It's also the delay involved in converting from Discrete time to Continuous time (e.g. a DAC). The delay here is due to the Zero-Order-Hold effect of the conversion. All Linear Discrete Time control systems have this delay. The delay = T/2 where T is the sample period. So, assuming (as you have stated) all delays in the loop are negligible, then we only have to account for the ZOH delay (T/2) and therefore the max achievable bandwidth is Wc < 2fs (where fs is the sample rate) which gives the max achievable bandwidth in Hertz as fc < fs/pi .
Keeping the open loop bandwidth less that 1/10 sample rate is a good idea for a discrete time controller, controlling a continuous time plant. That's because you can reduce the effect of frequency warping in the model conversions from continuous to discrete.
However, despite all that, the most important aspects are not covered by any of the above discussion. You really do need to know what you are controlling because several factors may influence the bandwidth of the controller as well as the type and structure.
Is it a stable open-loop system? If not then this open loop bandwidth must be significantly greater than the frequency of the unstable mode.
Is it linear? If not then a linear controller may be unsuitable. You may need something like Receding Horizon Control (which does not have a bandwidth as such).
Is it minimum phase? If it is not the open loop bandwidth is usually required to be less than the RHP zero frequencies.
And on what factor should the cutoff of my LPF depend since the sensor
is just giving DC differential output.
Your application is a very sensitive Wheatstone bridge and, if the signal you are looking for is basically DC, then you want your filter cut-off frequency to be as low as possible in order to reduce noise from the op-amp amplifier. But, in reality you can't have a LPF with a DC cut-off frequency because nothing will ever change and, the component sizes will be infinite so you have to re-examine your requirements and possibly 10 Hz might be a good filter cut-off.
You are sampling at 19.2kHz but that is now irrelevant to your design - you could sample at 100Hz and get the same performance if 10 Hz is your low-pass filter. Remember, the LPF does two things: -
- Gets rid of unwanted self-generated noise from your op-amp amplifier (this is your main problem)
- Prevents aliasing (this won't be a problem because nothing will get through a 10 Hz filter that would cause aliasing when you sample at 19.2kHz)
In your previous question I reckoned your op-amp had a noise of 60 nV / \$\sqrt{Hz}\$ but, if you restrict your bandwidth to 10Hz, the sum of all the noises will be over a bandwidth that is 16Hz (believe it or do the math! link) therefore, your equivalent noise at the input to your op-amp will be \$\sqrt{16}\$ x 60nV = 240nV. This is then multiplied by your op-amp gain (say 10) to give you a real figure of 1.2 micro volts into the ADC.
In your previous question it was 10 micro volts because I had assumed the BW to be 16kHz.
Remember also that the op-amp noise will rise (per Hz) as frequency falls and that in the DC to 10Hz range there will be another figure in the data sheet for the op-amp that covers this area. I'm not sure about the MCP6v07 and how well it's "auto-zero" feature works well at eradicating this LF noise so you'll need to check. However, if I looked at the ADA4528 (because I use it similarly to you) it has only 97nVp-p noise in the 0.1Hz to 10Hz bandwidth and this is a really good figure for an op-amp, made so by the auto-zero feature. It appears that the MCP6v07 is 1.7 micro volts p-p for comparison.
Is this good-enough? - I can't tell you because I don't know what gain the op-amp is needed to be set at and I don't know your requirements - I can only make comparisons.
Noise Equivalent Bandwidth - for a low pass filter the NEB depends on the order of the filter: -
Noise bandwidth = 3dB cut-off frequency \$\times \dfrac{\frac{\pi}{2n}}{Sin(\frac{\pi}{2n})}\$ where n is the order of the filter. For n = 1 this reduces to Fc x pi/2
Best Answer
The required bandwidth must be great enough so faults are responded to quickly. Remember that solid state components are not as forgiving as motors when it comes to overloads.Newer devices like IGBTs have surge current ratings for about 10 microseconds and the older SCRs had surge ratings of about 10 milliseconds. When current sensing is part of the control loop which is common practice stability can be a real issue. Higher bandwidth is better from a control loop stability viewpoint.