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Let consider a simple equation $$I = \frac{V_1-V_2}{R}$$ with $R = 0.001$ and $V_1, V_2$ around $10\ 000$, while $I$ is around $10$. The problem is $V_1$ and $V_2$ are measured with some small error (compared to $10\ 000$ as nominal value) of around $5$. But this small error will lead to unacceptable big error in $I$, when we get $$I = \frac{V_1-V_2}{R}. $$

How can we solve this problem? when we have $I$, $V_1$ and $V_2$ are continuously changing with time. how we can get a good estimate of $I$ with $V_1$ and $V_2$ measurement. Thank you very much!

Alain Remillard
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BangNguyen
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1 Answers1

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What you need to do is measure $V_1-V_2$ directly rather than measuring each separately and subtracting. This is not always possible, but sometimes it is. Think of two cars in neighboring lanes on the freeway traveling at almost the same speed. At one time they are next to each other. $5$ minutes later one is a car length ahead, say $20$ feet for a large one. This tells us that $V_1-V_2$ is about $4 ft/min$ regardless of what $V_1,V_2$ are individually.

Ross Millikan
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  • And looking at the symbols this can help: https://en.m.wikipedia.org/wiki/Instrumentation_amplifier – N74 Jan 18 '20 at 09:08
  • Here are some problem.1) The point 1 and 2 are too far away from each other ( ~ 100 meters). 2) Your example is true. Similarly, if I measure the current and have 1 (A), I can estimate V12 = 100 V. 3) Think of two cars, if you have the measured values of V1 and V2 individually at t=2s, how to estimate the distance between two cars after 3s. You can see that small error in V1 and V2 will lead to huge error of the estimated distance. – BangNguyen Jan 18 '20 at 21:21
  • Then, I am wondering that is there any methods that we can use to resolve this problem. Provided that we have V1 and V2 individually and trying to estimate I. – BangNguyen Jan 18 '20 at 21:23
  • The instrumentation amp does in analog exactly what I am suggesting here. It is supposed to measure the difference between the inputs and ignore the average value. You are correct that if you know V1 and V2 each with some error the difference between them may not be known very well. There is no solution except better data. Integration is a smoothing function, which is why measuring the distance between the cars at two long apart times gives you a good measure. – Ross Millikan Jan 18 '20 at 23:26