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rate computation for slow varying process analog method

curious_mind

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I am interested to compute rate of change of an input signal with the time duration of one minute. How to tune classical RC differentiator circuit to get the proposed rate of change value?
 

KlausST

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Hi,

What´s the input signal and what is the expected output signal?

I mean: You surely know that
* tau = R x C
or the step-by-step way:
* 1 minute is 60 seconds.
* and 1 Farad = 1 As / V or C = I * t / V
* and R = V/I

Is the sign important?

Klaus

Btw: I don´t recommend the analog way, because of low accuracy, high drift and high noise.
 

curious_mind

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1.If RC=60 sec , does it guarantee rate estimation /min. As I understand that Vout= RC dvin/dt. In equation RC only appears to be scaling factor. How can we say that the value of RC corresponds to rate estimation/min?

2.Since you mentioned for digital method, can we get equivalent implementation as that of analog?
 

KlausST

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Hi,

1) Math does so. Reality may differ, as already mentioned.

2) If you just use a digital differentiator, then your benefit from smaller size, higher accuracy, zero drift. But noise will be the same and you suffer from ADC stepsize.
But you are free to improve software (using filters, adding digital signal delay ...) without hardware modification.

Please do math (calculating signal currents, leakage currents, errors ..) for the analog part.

Klaus
 

curious_mind

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Simple question. In equation RC dv/dt, the term dv/dt refers to rate of change of input. How can we corelate the factor RC with rate/min computation?
 

KlausST

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Hi,

When we (physicians) speak about time, we always refer to "seconds". (Not about minutes)
So in your case: time = minutes = 60 seconds.
If you have V/s and want to transform it into V/min you basically need to use V/60s.
Now you have "multiplied" below the deviation line with 60. To correct for this you also have to multiply above the deviation line with 60.
Thus the mathematical correct formula is:
V/s = (60 × V) / (60 × s) = 60 V / min
*********

dV/dt is the mathematical deviation of Volts according time.

Simply saing: the ratio of "Volts per second" ... (at any given time)
V/t with the unit V/s

Since C = I × t / V
--> I = C × V / t
--> I = C × ( Voltage rate per time)

Example:
You have a voltage rate of 1.2 V/s on a 3uF capacitor:
--> I = 3uF × 1.2 V/s = 3.6 uA

Mind: 1 F = 1 As/V.
In the formula above you multiply "F" with "V/s",
--> "As/V" with "V/s" = AsV / sV now you may divide both with "Vs" --> remaining "A/1" = A
The "u" is the same as "a factor of 0.000001" ... and surely needs to stay.
"u" (read "micro") is no "unit".

Klaus
 

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Yes I understand now. C takes finite time to charge and it depends on R. If we analyze with respect to square waveform, it takes 5RC for C to charge. I wanted to know how much time it would take for a ramp input?
 

KlausST

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Hi,

from a mathematical vie this is not correct.
An RC takes INfinite time to charge.

Thus if you want to calclate a time than you need to use an (alloed" erro rate.

1 tau = 37% error
2 tau = 13% error
3 tau = 4.9% error
4 tau = 1.8% error
5 tau = 0.66% error
6 tau = 0.24% error
7 tau = 0.090% error (900ppm)
8 tau = 0.033% error (329 ppm)
9 tau = 120 ppm error
10 tau = 44 ppm error
and so on

5 tau is just below 1%
if you need 10 bit accuracy then you need 7 tau
if you want 16 bit accuracy than you need 9 tau

****
We don´t have an RC here (althogh the math is quite similar).
With a good active differentiator you don´t have to "wait", the output is immediate.
For sine, for ramp, for step (with limited rise rate), for any waveform

Klaus
 

danadakk

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Yes I understand now. C takes finite time to charge and it depends on R. If we analyze with respect to square waveform, it takes 5RC for C to charge. I wanted to know how much time it would take for a ramp input?

Laplace analysis of ramp function :

So just sub transform equivalent into T(s) = Vout(s) / Vin(s) then solve for Vout(s), then do inverse
transform back to f(t) and perform PFE to get at f(t) element contributions. T(s) of course the differentiator transfer F(s).



A crude sim :

1674916662429.png



Regards, Dana.
 

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