Integration of peaks

[LabProARW]

An explanation that I had been shown for integration involves a number of vertical rectangles approximately fitting a curve. The area of each rectangle is calculated and the sum is approximately equal to the amount of response. For data rate of 20 Hz on my detector, is 0.01 minimum peak width per minute. Does this mean that there will be 20 of those imaginary vertical rectangles every second which area will be summed up? Is there a way to know how many vertical rectangles are being "integrated"/calculated? I'm trying to conceptualize peak integration for a presentation.

[sbashkyrtsev]

Those aren't really "imaginary" rectangles - they are rectangles between two neighboring points. Your signal consists of discrete points because measurements happened at some particular interval. In your example it's 20Hz, so there are 20 points per second. Each of them forms a "rectangle" with the previous point (well, except for the first one).

But rectangles are too crude of approximation. They aren't used in practice, instead there are 2 popular approaches:
* Areas of trapezoids (see Trapezoidal rule). Basically we get a reactant and a triangle and sum their areas.
* Simpson's rule - it's a little more involved method

Theoretically, Simpson's Rule should provide better approximation, but in practice this doesn't seem to be the case, or at least not always. See Comparison of integration rules in the case of very narrow chromatographic peaks, doi:10.1016/j.chemolab.2018.06.001

[lmh]

There are some areas of mathematical-analytical-chemistry that you think are going to be simple, but which turn out to be truly horrendous (calculation of intensities of isotope peaks is a good example).

There are others that you expect to be horrendously difficult, but which turn out to be very simple. Integration is one of these.

Every child knows how to work out the area of a rectangle, or a triangle. So if you have a detector measuring something at intervals, you have a chromatogram that's a series of dots. Join the dots, and you have a funny shape underneath, whose area you can find by dividing it into rectangles and triangles. It doesn't actually matter how you divide it up, the answer will be the same. Normally people divide it into rectangles with verticals at each time-point, and triangles on top of them, but you can do it any way you want. The area of a polygon doesn't depend on how you divide it up.

There are errors, because if the true signal followed convex path between two points, and your line is straight, you'll miss a bit of the area.

But fortunately, peaks start concave, become convex, then go concave again, so the errors tend to cancel out.

Old fashioned chromatographers (I include myself) like to see 12 or so points across the width of a peak, because the more points you have, the better the approximation between a join-the-dots approach and the true curve. The ref that the previous poster gave applies to "narrow" peaks, i.e. peaks where far less points have been collected.

The real problem with "narrow" peaks probably isn't a failure to reflect the true Gaussian (or whatever) shape, it's that each measurement has an error associated with it (the dot isn't in the right place!), and the less dots you have, the more likely you are to be unlucky and mis-estimate the whole thing. The points in a chromatogram are not only defining the shape of the peak, they are also "samples" creating a sort of "replication", with random errors cancelling to some extent.

Many of the old theorems for how to integrate under curves sampled at points assume the points are equally spaced, whcih isn't always true for chromatograms. The trapezoid rule, being basically rectangles-and-triangles join-the-dots geometry, doesn't care.

[sbashkyrtsev]

There are others that you expect to be horrendously difficult, but which turn out to be very simple. Integration is one of these.

Many people when they say "integration of peaks" also include determining the peak boundaries in the definition. If we include that, then suddenly it becomes one of those "horrendously difficult" calculations (: Waaaay more complicated than determining the isotopic pattern of a molecule (BTW, here we opensourced a project exactly for this: isotope-distribution in Java).