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Thread: Definite integral divided by Definite Integral

  1. #1
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    Definite integral divided by Definite Integral

    I am unsure how to solve this. I have values for all variables for each wavelength in 10 nm increments, 380,390,400....
    My thought was to sum each value of $\displaystyle \tau$ at the given wavelength, but that is not correct.

    $\displaystyle \tau_{SB} = \frac{\int_{380}^{500}\tau(\lambda) \times W_B(\lambda) d\lambda}{\int_{380}^{500} W_B(\lambda) d\lambda}$

    Thanks for any help.
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  2. #2
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    Re: Definite integral divided by Definite Integral

    Hey bmccardle.

    Are you trying to figure out some sort of average/expectation?

    You should probably explain what you are trying to achieve here for further information.
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  3. #3
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    Re: Definite integral divided by Definite Integral

    This formula evaluates the transmission of a blue light, which is from 380 nm to 500 nm. The overall transmission for the filter I am trying to evaluate is 80%.
    $\displaystyle W_B$ is a weighting factor at a certain wavelength and $\displaystyle \tau$ is the transmission at a particular wavelength. I thought you could sum up the value at each wavelength, but that is not correct.
    Thanks for the help

    The transmission at:
    380 = 2.7
    390 = 3.9
    400 = 8.1
    410 = 14.9
    420 = 45.9
    430 = 84.9
    440 = 88.4
    450 = 89.3
    460 = 89.2
    470 = 87.0
    480 = 89.6
    490 = 89.6
    500 = 89.3

    The weighting factor:

    380 = 2
    390 = 10
    400 = 47
    410 = 269
    420 = 660
    430 = 771
    440 = 911
    450 = 946
    460 = 864
    470 = 706
    480 = 532
    490 = 266
    500 = 122
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  4. #4
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    Re: Definite integral divided by Definite Integral

    So, you have $\tau$ and $W_B$ at discrete points?

    In which case, you will want to estimate the integrals. There are various methods to do this. The Trapezium Rule provides a good balance between simplicity and accuracy.
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  5. #5
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    Re: Definite integral divided by Definite Integral

    Would it be the sum of $\displaystyle \tau \times W_B$ divided by the sum of $\displaystyle W_B$? This is close to what the answer should be

    4860.815/6106 = 0.7961 or 79.61%
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  6. #6
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    Re: Definite integral divided by Definite Integral

    If the weights add up to one [and are strictly positive] then what you are doing in the top integral is getting an expectation.

    Also - how are the weights determined?

    I assume you have a list of frequencies and you want to look at how much of a frequency is passed through relative to how much of some frequency is transmitted.

    You should elaborate on what the counts are [and how you get them] along with how you determine your weightings for more information.
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