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Area under Gaussian peak by easy measurements

  1. Jul 30, 2010 #1
    First I try to find a way of finding area under Gaussian peak by using simple means. Questions at bottom of post.

    This concerns peaks with Gaussian distribution, eg. peaks in a chromatogram in some chemical analysis.

    The Gaussian distribution is
    [tex]
    f(x) = \frac{1}{\sigma \sqrt{2\pi} } e^{ -\frac{(x-\mu)^2}{2\sigma ^2} }
    [/tex]

    and the function for a peak in the chromatogram is

    [tex]
    f(x) = \frac{A}{\sigma \sqrt{2\pi} } e^{ -\frac{(x-\mu)^2}{2\sigma ^2} }
    [/tex]

    where A is the area under the curve. The area is interesting since it represents the number of detected molecules and by using calibration curves one can find the concentration of a previously unknown sample.
    The maximum peak height H is easily proven to be found at x =[tex]\mu[/tex], hence the exponential part is equal to 1;

    [tex]
    H=f(\mu) = \frac{A}{\sigma \sqrt{2\pi} } e^{ -\frac{(\mu-\mu)^2}{2\sigma ^2} }=\frac{A}{\sigma \sqrt{2\pi} } = \frac{0.3989A}{\sigma}
    [/tex]

    This shows that the area (and hence concentration) is proportional to maximum peak height. But this assumes same /sigma. But I would like to consider that injection and stuff might be different etc and account for peaks having a bit different /sigma.

    The full width at half maximum (FWHM) can be shown to be
    FWHM=2 sqrt(2 ln(2))[tex]\sigma[/tex] = 2.35[tex]\sigma[/tex] by inserting f(x) = H/2 , find x1 and x2 and then calc the width.
    [tex]
    H = \frac{0.3989A}{\sigma}
    [/tex]

    [tex]
    A = \frac{H\sigma}{0.3989}
    [/tex]
    [tex]
    FWHM = 2.35\sigma
    [/tex]
    [tex]
    \sigma = \frac{FWHM}{2.35}
    [/tex]

    [tex]
    A = \frac{H* FWHM}{2.35*0.3989}
    [/tex]


    Then the area would easily be calculated from a paper with peaks using only ruler and calculator, by measuring the maximum height and FWHM, multiplying them and multiply by a constant.


    1.Did I overlook something here, or does it work this way?

    It seems very convenient to account for peaks having different /sigma this way, although I doubt it has any large practical significance in the high school lab... Also, this still is under assumption that all peaks are perfectly gaussian in shape.

    In school we usually have software to calculate peak areas, or just plot peak maximum height.

    2. How does a software compute such an area?


    The integral with respect to x of [tex]
    f(x) = \frac{A}{\sigma \sqrt{2\pi} } e^{ -\frac{(x-\mu)^2}{2\sigma ^2} }
    [/tex] would equal A. Integration yields A=A and does not give the value of the area...

    3. If I just sit here with a bunch of peaks on a paper, can I use some method applying integrals to find A?
     
  2. jcsd
  3. Sep 30, 2011 #2
    I just spent some time trying to figure this out. For an equation of the form:

    [itex]f(x)=N\cdot exp[-\frac{(x-u)^2}{2\sigma^2}][/itex]

    The area under the curve is given by:

    [itex]N\cdot\sigma\cdot\sqrt{2\pi}[/itex]
     
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