Is this just a Detector and an Integrator? How does it work?

MrAl

Joined Jun 17, 2014
13,716
Thank you for that clarification, but why does it need to be charged and then discharged actually? Eric said it is to control the integration period, what does it mean?
Hello again,

That's because when we do a true definite integration (and not just an average) we have to specify the start and end times. This is mathematically necessary. Since an integration involves a summation over time, if there was no end time it would just keep adding and adding and adding the voltages that were presented at the input. That would lead to a very high value on the output that would exceed the limit of the power supply voltage, and that would not tell us anything useful at all.

Also, the choice of the start time and the end time is based on how the system operates and what is actually being measured. It's hard to tell what times these may be from the given information so far but it most likely has to do with the phase of the signal, but it could be something else. We'd have to study the entire system in order to know for sure what these times should be. That would most likely require more information about the sensors being used. For now, we can assume that the designers knew all of this information so that they could program in the required start and end times.
 
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