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Numerical intergration of a set of measured data points

  1. Jun 23, 2012 #1
    Hi,
    I have faced the following question. In our lab we perform different measurements on Transistors. We program a scope and that controls the tests. For one of our tests we would like to calculate the total charge Q. Mathematically this is given by
    Q=∫ dt i(t), where i(t) is given by i(t) = v(t)/R, i(t)= current, v(t)=voltage, R= resistivity ( constant value), for a time interval [t1, t2]
    We cannot measure i(t) directly but we can measure v(t). This means
    Q=∫dt v(t)/R.
    We have noticed the following
    1. if we define Q1 = ∫dt ( v(t)/R)
    2. or if we define Q2= (∫dt v(t))/R
    Mathematically these two integrals should produce the same result. But we see completely different results. I wonder if this depends on the numerical methods used in integration or has another reason.

    Thank you for your help.
     
  2. jcsd
  3. Jun 23, 2012 #2

    marcusl

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    If R is truly a constant, then the answer cannot depend on when you divide. You must be making a mistake in your calculations.
     
  4. Jun 23, 2012 #3

    Integral

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    You have not told us how you are evaluating the intergrals. Without that bit of information there is no way we can hope to help you.
     
  5. Jun 24, 2012 #4
    Thank you for your feedbacks,
    1. R is constant and I expect that the integrals should be the same but they are not.
    2. I do not know what integration method is used. This is a tester connected to a circuit we can enter the function that we want to evaluate, with limits of integration.
    I have been thinking if we use spline then difference between the two way of integration may come from approximation using (v(t)/R) all the way instead of just multiplying (1/R) at the end.
     
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