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Old 3rd July 2010, 07:03 AM   #1
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Thumbs up A theoretical question on a differentiator circuit

If we take the basic differentiator circuit (opamp configuration with a feedback resistor and a capacitor at the input), this circuit takes the derivative of the input voltage with respect to time:

output = d(input)/dt (let us assume that the RC scale factor is equal to 1)

--> Let's say I have two signals, V1 and V2. I use two differentiator circuits to obtain d(V1)/dt and d(V2)/dt

--> I then take an analog divider to divide d(V2)/dt by d(V1)/dt, to get d(V2)/d(V1) (the "dt" terms cancel out)

Is this the same as taking the derivative: dV2/dV1? In other words, is this a way to take the derivative of a variable with respect to another variable?
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Old 3rd July 2010, 01:34 PM   #2
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No, unfortunately. Your rationale applies using traditional algebraic (commutative, distributive) properties with sum and product, but not quotient.

At a point where v<>0, the quotient y = u/v of two differentiable functions is differentiable, and d/dx(u/v) = [v(du/dx) - u(dv/dx)] / v^2

An easier writing uses the ' single quote to represent the differential, so that (u/v)' = (vu' - uv') / v^2

Last edited by zigzagflux; 3rd July 2010 at 01:37 PM.
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Old 6th July 2010, 02:34 AM   #3
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Thanks zigzagflux - your answer makes perfect sense!
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Old 11th July 2010, 02:16 AM   #4
mirlo is offline mirlo  United States
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Hi Rtarbell,

Actually, if V2 really is a function of V1, then it does work. Look up the chain rule. I think zigzagflux might have misread your question the same way I did at first. At first I thought you were asking if the derivative of (V2/V1) with respect to t was the same as the quotient of the two derivatives. But you were asking how to find the deriviative of V2 with respect to V1. Different animal.

On the other hand, if you implement it, the method will become less and less accurate the more slowly V1(t) changes, because you will be dividing zero by zero so the actual result will be dominated by noise, and you will get nonsense at time-domain minima and maxima of V1(t).

-- Eric
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Old 11th July 2010, 02:56 AM   #5
mirlo is offline mirlo  United States
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One more thing about op-amp differentiator circuits: they can be interesting to stabilize, because you are intentionally putting a low frequency pole in the negative feedback loop, which already has a dominant pole.
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