Analog Integrator Sampling Rate

Question

What would be the effective sampling rate of an op amp analog integrator (such as the one described here ) taking in a rapidly varying signal be? Ie what time resolution would such an integrator have?

The need for this to solve a practical problem has passed, as a different method of testing was ultimately selected, and so I intend this question to be as theoretical as possible, but in order to understand situation better I shall provide the original motivation. I was attempting to use a load cell in order to measure the momentum change of an object colliding with my load cell by taking the force reading of output of the load cell over time and numerically integrating the result. Unfortunately, my data logger only sampled at 1Khz and the impulse event was estimated to take only a few ms, thus providing much too small time resolution to accurately integrate afterwards. So, My thought was to hook the load cell output directly to an analog integrator and I should get a step function output that would read off to me a value proportional to the momentum change. From the transfer function of 1/s, it would seem as though I could do it for any signal regardless of how rapidly changing the signal is, however there must be some practical limit on how fast the integrator can respond. Can I assume a transistor switching speed (on the order of many Mhz and thus much faster than needed here) for the time resolution?

Answer 1

What would be the effective sampling rate of an op amp analog integrator (such as the one described here ) taking in a rapidly varying signal be?

[...] what time resolution would such an integrator have?

[...] there must be some practical limit on how fast the integrator can respond

All of these considerations are hinting towards the Time Constant of a system which in this case, we would take to be this integrator you are considering.

The time constant is related to the physical time it takes for the response of the system to go through certain "landmarks". But, in general, it is simply a time interval kind of value (5 seconds, 0.2 seconds and so on).

So, to relate it to your problem, you can take an idealised (impulse) response of an integrator defined by the time constant you set via the external components and "test it" with a pulse input that lasts as long as you think it would in the real experiment.

I am assuming that what you are worried about is pulses so short that the integrator doesn't produce a "usable" output. But then again, that is the job of the integrator. If you work out the sum (via convolution) you can see what its output looks like for the kind of pulse you expect and in any case there are guidelines for the time constant of integrators. For example there is the guideline that says that the time constant of the integrator should be 10 times or more the period of the pulse you are trying to integrate but this number 10 is worked out from the definition of the time constant and how much you want your capacitor (i.e. the output of the integrator) to be charged at the end of your pulse.

As a side note here, what you seem to imply by:

My thought was to hook the load cell output directly to an analog integrator and I should get a step function output that would read off to me a value proportional to the momentum change.

...is not an integrator but a sample-and-hold circuit to give enough time to the rest of the equipment to read the final value. So, it is not just an RC circuit, it also needs appropriate buffering (and gating) so that reading off its value doesn't destroy it.

Can I assume a transistor switching speed (on the order of many Mhz and thus much faster than needed here) for the time resolution?

That point is slightly different. Yes, there is a parameter that measures that. It is the Slew Rate listed in datasheets but it would be more convenient to "lump" the response of the circuit along with the time constant. Otherwise, you could be looking at how far back is the integrator lagging from the actual signal and then how far back is the output of the op-amp with respect to the integration signal which does not seem to matter in your application.

Hope this helps

Answer 2

Sampling rate is only defined within discrete-time systems. So, there's really no sampling-rate equivalent that your integrator has.

What your integrator does have is finite bandwidth. It, for example, will not be sensitive for voltage variations at 2.4GHz.

The bandwidth off that circuit involves the characteristics induced by modeling the circuit based on ideal components, but also on limitations such as unintended (parasitic) inductance and capacitance of components and leads, and, most importantly: by the bandwidth of the opamp itself.

You can buy opamps with only a few Hz of bandwidth, you can buy them with multiple GHz of bandwidth, and very many values in between. Each of these have different strengths and disadvantages, so for a measurement system, choose wisely, based on an understanding of the requirements of your measurement system and signal, and at least a solid understanding of electronics and linear electrical networks; I'm not convinced designing a measurement integrator is trivial.

Generally, you might want to take a step back and actually ask over at electronics.SE how to measure a specific signal, describing exactly and in detail what you are observing, rather than assuming an integrator is the solution here. It might be - but it's equally likely you're not on the right track.