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sampling rate vs rate of change of input for control system

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gary36

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I am referring to http://support.ctc-control.com/customer/elearning/registered/servoBasicsForTheLayman.pdf to understand type ii servo loop. Theory provides zero velocity lag for changing velocity inputs.

To implement in hardware, I wanted to know if sampling rate is to made higher than the input rate of change? For example, If I desire to implement 12 bit measurement system (to measure velocity) and input is changing at 100 rps, then the sampling rate must be >(1/4095* 100)seconds or close to 400KHz.?

In other words, should the group delay of the control loop be lower than the input rate of change?
 

The calculation makes no sense, there's no reason why the controller shouldn't perform a large step in one sampling interval. Rate of change and group delay have different units and can't be related.

Group delay is however essential for the control loop stability. Preferably, the controller will be designed as time discrete system, considering the z-domain transfer function of the complete loop.
 

I was just thinking that by that time controller makes corrections, the input would have changed making error non-zero all the time.
 

If I desire to implement 12 bit measurement system (to measure velocity) and input is changing at 100 rps,
That doesn't make sense.
rps is a velocity, not a change in velocity.
Do you mean 100rps/s change?
How fast does the velocity change and what speed of correction you need?
Also what are you trying to correct?

Generally the sampling rate would need to be at least twice the highest frequency response of the correction loop.
 

Hi

I mean only velocity. My basic doubt arouse factoring a sampled system. Analog type II system have VCO that produces frequency proportion to error in real time. A sampled system has to wait for the delta time(reciprocal of sampling rate) before correction can be taken. In other words what could be relation ship between sampling rate vs tracking rate of a system with respect to type ii servo loop?
 

Let me give another example. if shaft speed is rotating at 360 deg/s (sawtooth wave) and if I need to quantize it to 12 bit, then sampling interval needs to be 1/4095 sec to ensure that we get 50 %duty cycle wave shape for all bits (btw, i tried this in spice with wave generator and the theory seems to be working)
 

The idea behind your examples seems to be that the sampled data should look the same as time continuous data. That's not the way time discrete systems are designed.

You need to sample the shaft position only at the calculated high speed if it can take arbitrary positions at each sample (being moved with infinite torque). But the real system has finite torque and acceleration, respectively finite bandwidth of the position signal. Time discrete system design is utilizing this knowledge.

It can be reasonable to increase the sampling rate above the theoretical minimum according to sampling theorem, may be factor 5 to 20 but not by factor > 1000.
 

Hi FvM

Quick thought process on by fundamental understanding. Suppose I sample 50Hz signal with 1KHz, all I get is 20 points (resolution is 4.5 bits), If I need 12 bits to represent the same signal, then sampling frequency must be 200KHz. What do you think.
 

Suppose I sample 50Hz signal with 1KHz, all I get is 20 points (resolution is 4.5 bits)
No.
You get 20 points with a 12-bit resolution for each point (sample).
You are confusing time resolution with amplitude resolution.
 

I meant amplitude resolution only. Consider sine cos encoder rotating at high speed. In order to detect instantaneous position with high accuracy, my sampling rate must be higher enough to do so.
 

Again, you are not understanding the concepts of time discrete signal processing. Required sample rate depends on signal bandwidth, not magnitude resolution.
 

Hi FvM

My brain is bit thicker and hard. It takes a while to understand. Please give it a last try analyzing the following example

Shaft of motor is rotating at velocity 1 rps or 360 deg/s. I have ADC (12 bit) and sampling this signal at 10Hz, that is 10 points/sec (Nyquist satisfied). Now I am talking about resolution of position. By sampling@10Hz , resolution of position becomes 360/10= 36 degrees. If I need to detect instantaneous changes say every 1 degree, then should I not sample at 360 Hz?

Sorry if I am repetitive
 

Shaft of motor is rotating at velocity 1 rps or 360 deg/s. I have ADC (12 bit) and sampling this signal at 10Hz, that is 10 points/sec (Nyquist satisfied).
Not necessarily satisfied. It depends on the speed change rate.

A more practical viewpoint could be: How is the position information processed in your system? If you are running a control algorithm at a fixed rate, it makes sense to sample the position at the same rate.
 

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