Aaron3154518
commented
4 years ago Additionally in the formula y = Ae^-1/2((t-b)/c)^2, c is the standard deviation. If we were to treat each gaussian as a histogram and calculate the standard deviation from that, we could compare that to c to determine if we even need to fit the curve. (Perhaps only for single peak waveforms)
jaredwhite547
In first differencing, we calculate the FWHM by solving the equation of a Gaussian given our amplitude, amp location, and a data point on the gaussian. However we then calculate c in Gaussian Fitting by dividing the FWHM by 2log(2), which is incorrect. The relationship between c and width is as follows:
w = 2sqrt(2ln(n)) * c given w is the width when y = Amp/n.
Therefore with the FWHM, n = 2 and c = w / 2sqrt(2ln(2)).
However we do not need to calculate these two variable this way. Here is an alternative method:
Given data point (xi, yi), amplitude a, and amp location t, xi - t is half of the full width (w) at yi yi can be put in terms of A/n by noting that yi = A yi/A This means 1/n = yi/A and n = A/yi and w = 2sqrt(2ln(A/yi)) c and c = w / 2sqrt(2ln(A/yi)) Now we know that for the FWHM, n = 2 and so FWHM = 2sqrt(2ln(2)) c When we substitute c, FWHM = 2sqrt(2ln(2)) w / 2sqrt(2ln(A/yi)) = w sqrt(ln(2) / ln(A/yi)) = (xi - t) sqrt(ln(2) / ln(A/yi))
Thus we can use ANY data point to calculate the FWHM as that point gives us yi and xi and we know A and t. Then in gaussian fitting, we simply note that c = w / 2sqrt(2ln(2)). This gives us a much more accurate answer for FWHM and thus c as we can use the actual data instead of the closest point to half maximum.