Area under gaussian peak (from H(max) and FWHM)

[humble_student]

First I try to find a way of finding area under Gaussian peak by using simple means. Questions at bottom of post.


The Gaussian distribution is

f(x)= 1/(sigma* sqrt{2\pi) } e^{ -(x-mu)^2/ 2sigma ^2} }


sigma being standard deviation and mu average.

A Gaussian peak in a chromatogram could be described by

f(x)= A/(sigma* sqrt{2\pi) } e^{ -(x-mu)^2/ 2sigma ^2} }

where A is the area of the peak.


The maximum peak height H is readily proven to be found at x =mu, hence the exponential part is equal to 1;

H=f(mu)= A/(sigma* sqrt{2\pi) } e^{ -(mu-mu)^2/ 2sigma ^2} }
=H= A/(sigma* sqrt{2\pi) } = 0.3989A/sigma

A = H*sigma/0.3989


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 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))*sigma = 2.35*sigma by inserting f(x) = H/2 , find x1 and x2 and then calc the width.

FWHM = 2.35 sigma

sigma= FWHM/2.35

A=H * sigma /0.3989 = H * FWHM / (0.3989 *2.35)

Then the area would easily be calculated from a paper with peaks using only ruler and calculator, by measuring the maximum height H and the 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 too 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? By using a lot of small integrals?


The integral with respect to x of [tex]
f(x)= A/(sigma* sqrt{2\pi) } e^{ -(x-mu)^2/ 2sigma ^2} }would equal A. Analytical integration yields A=A and does not give the value of the area... or does anyone another way?

3. If I just sit here with a bunch of peaks on a paper, can I use some method applying integrals to find A? Numerical integration seems heavy.
Or any other method at all than A= H * FWHM / (0.3989 *2.35)?

Thanks a lot in advance!

[tom jupille]

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

No, it does work that way *assuming you have a Gaussian peak* (as you pointed out). If all real-world peaks were Gaussian (and well-resolved) our jobs would be a lot easier .

2. How does a software compute such an area? By using a lot of small integrals?

Usually by brute force. A chromatogram is essentially a series of voltage measurements. Assuming constant frequency of measurement, the software decides where the peak begins and ends, and simply adds up all the voltage values in between. No assumptions about peak shape are required.

3. If I just sit here with a bunch of peaks on a paper, can I use some method applying integrals to find A?

Make a photocopy of the paper, cut out the peaks, and weigh them. Copy paper is remarkable homogeneous (it has to be to get through the copier without jamming!) and the mass is directly proportional to the area.

If you really want to get into this, get a copy of Norman Dyson's book "Chromatographic Integration Techniques". Here's a link to it on Amazon: http://tinyurl.com/2g53tjx