# Fitting two peaks at the same time with NonLinearFit

If I have the following data:

data={{68.029,0.0570654},{68.062,0.0571538},{68.095,0.0573135},{68.128,0.0573148},{68.161,0.0574761},{68.194,0.0576357},{68.227,0.0577004},{68.261,0.0574547},{68.293,0.0576841},{68.326,0.0576759},{68.359,0.0576677},{68.392,0.0577419},{68.426,0.0576339},{68.459,0.0577603},{68.492,0.0578867},{68.525,0.0580131},{68.558,0.0581791},{68.591,0.058209},{68.624,0.0583987},{68.657,0.0585884},{68.69,0.0587781},{68.723,0.0589489},{68.756,0.0589455},{68.788,0.0593332},{68.821,0.0594897},{68.855,0.0594118},{68.888,0.0597297},{68.921,0.0598878},{68.954,0.0600459},{68.987,0.0602039},{69.02,0.060362},{69.053,0.0605185},{69.086,0.0608349},{69.119,0.0609898},{69.152,0.0611463},{69.185,0.0613012},{69.218,0.0616191},{69.252,0.0615444},{69.285,0.0618623},{69.318,0.0620204},{69.351,0.0621785},{69.384,0.06246},{69.417,0.0625516},{69.45,0.0626448},{69.484,0.0625004},{69.517,0.0625919},{69.551,0.0625077},{69.584,0.0628557},{69.617,0.0630423},{69.65,0.0632304},{69.683,0.0634186},{69.716,0.0636083},{69.749,0.063798},{69.781,0.064219},{69.815,0.0641775},{69.848,0.0643672},{69.881,0.0647169},{69.914,0.0649034},{69.947,0.0650916},{69.981,0.0650469},{70.014,0.0652351},{70.048,0.065412},{70.08,0.0659248},{70.113,0.0662079},{70.146,0.0666526},{70.179,0.0669357},{70.211,0.0675735},{70.245,0.0677236},{70.278,0.0681065},{70.311,0.0683279},{70.345,0.0684795},{70.378,0.0689399},{70.41,0.0694242},{70.443,0.0696725},{70.476,0.0700807},{70.509,0.0703306},{70.542,0.0706581},{70.576,0.0709664},{70.609,0.0713477},{70.641,0.072117},{70.674,0.0724967},{70.707,0.0729983},{70.74,0.0733147},{70.773,0.0737894},{70.807,0.0738729},{70.84,0.0743492},{70.873,0.0748017},{70.905,0.0754776},{70.938,0.0759174},{70.971,0.0763589},{71.004,0.0768003},{71.037,0.0770851},{71.071,0.0772968},{71.104,0.077743},{71.136,0.0784236},{71.17,0.0787969},{71.202,0.0794364},{71.235,0.0798161},{71.268,0.0801958},{71.301,0.0807354},{71.335,0.0808791},{71.368,0.0813759},{71.401,0.081849},{71.434,0.0823206},{71.467,0.0827937},{71.5,0.0832652},{71.533,0.0837842},{71.566,0.084327},{71.6,0.0847952},{71.632,0.085574},{71.665,0.0862767},{71.698,0.0869382},{71.731,0.0874176},{71.764,0.0880569},{71.797,0.0885364},{71.831,0.0889381},{71.864,0.0895316},{71.897,0.0900997},{71.93,0.0906677},{71.962,0.0916349},{71.995,0.092203},{72.028,0.0929959},{72.061,0.0936669},{72.094,0.0944994},{72.128,0.0950927},{72.161,0.0959236},{72.194,0.0967117},{72.226,0.0977153},{72.259,0.0984781},{72.292,0.0992425},{72.326,0.0997677},{72.359,0.100592},{72.392,0.101612},{72.424,0.102709},{72.458,0.103492},{72.491,0.104351},{72.524,0.105329},{72.557,0.106281},{72.59,0.107233},{72.623,0.108186},{72.656,0.109137},{72.688,0.110454},{72.722,0.111212},{72.754,0.112434},{72.787,0.113425},{72.821,0.114023},{72.854,0.115609},{72.887,0.11701},{72.919,0.118166},{72.952,0.119087},{72.985,0.120647},{73.019,0.121519},{73.052,0.122638},{73.085,0.124236},{73.118,0.125355},{73.152,0.126241},{73.184,0.12771},{73.217,0.128761},{73.249,0.130048},{73.282,0.1311},{73.315,0.132311},{73.348,0.133561},{73.382,0.134444},{73.415,0.136361},{73.448,0.138117},{73.481,0.139392},{73.514,0.14127},{73.547,0.142484},{73.58,0.143859},{73.613,0.145075},{73.647,0.146214},{73.679,0.147801},{73.712,0.149138},{73.745,0.150314},{73.778,0.151652},{73.811,0.153147},{73.844,0.154569},{73.878,0.155968},{73.911,0.157598},{73.944,0.160028},{73.977,0.161818},{74.01,0.163267},{74.042,0.165097},{74.075,0.166852},{74.109,0.168215},{74.142,0.16981},{74.176,0.171334},{74.209,0.17341},{74.241,0.175559},{74.274,0.178596},{74.307,0.180671},{74.34,0.183791},{74.373,0.185997},{74.407,0.188137},{74.44,0.191785},{74.473,0.194151},{74.506,0.19644},{74.539,0.198997},{74.572,0.201552},{74.605,0.204268},{74.638,0.208585},{74.671,0.21142},{74.705,0.214154},{74.738,0.217443},{74.771,0.220733},{74.804,0.225942},{74.837,0.229501},{74.87,0.233596},{74.904,0.239381},{74.937,0.243794},{74.97,0.250129},{75.003,0.254638},{75.036,0.259594},{75.069,0.266791},{75.102,0.272708},{75.135,0.280225},{75.168,0.286304},{75.201,0.294463},{75.233,0.301094},{75.266,0.309735},{75.3,0.316702},{75.333,0.324539},{75.366,0.332695},{75.4,0.3427},{75.433,0.352938},{75.466,0.363334},{75.499,0.373573},{75.531,0.383886},{75.564,0.392686},{75.597,0.400206},{75.63,0.405805},{75.663,0.414404},{75.697,0.423062},{75.73,0.432275},{75.762,0.441723},{75.795,0.450937},{75.828,0.460329},{75.861,0.469574},{75.894,0.478018},{75.927,0.485343},{75.961,0.491474},{75.994,0.495995},{76.027,0.497268},{76.06,0.490378},{76.093,0.48685},{76.125,0.474117},{76.159,0.465758},{76.192,0.454619},{76.225,0.432919},{76.258,0.414738},{76.291,0.391278},{76.324,0.368156},{76.357,0.344888},{76.39,0.323379},{76.423,0.303791},{76.456,0.286443},{76.489,0.271034},{76.522,0.250037},{76.556,0.229126},{76.589,0.209889},{76.622,0.191613},{76.655,0.175856},{76.689,0.166238},{76.722,0.153977},{76.755,0.146995},{76.788,0.141294},{76.821,0.132634},{76.853,0.128412},{76.886,0.12476},{76.919,0.121907},{76.952,0.118894},{76.985,0.11618},{77.018,0.113294},{77.0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which looks like this plotted:

How can I fit both peaks at the same time using NonLinearFit? And also how can I find the area under the curve for both peaks?

EDIT: I would say that the answer provided by @MarcoB is great and the only thing remaining to ask is if somone knows what equation would be most ideal to better fit both peaks?. I appreciate it in advanced.

EDIT2: I tried using the software origin to try to find what is the best peak for both peaks and it seems that the function BWF is the best for it as written below:

Can someone help me implement this equation with NonLinearFit?

• Not everybody here speaks physics. You might want to spell out BWF.
– JimB
May 27 '20 at 3:39
• @JimB BWF stands for "Breit Wigner Fano" Function. This seems to be a good function for both peaks. I appreciate your help as well. I must add here that y0 is just the baseline. And it has 4 parameters: H,Xc,qw and w
– John
May 27 '20 at 3:43
• Can you at least tell us the physical origin of these signals? May 27 '20 at 5:16
• Please put that information in your comment into the question. May 27 '20 at 7:24
• @MarcoB the physical original are from a differential scanning calorimetry signal. The x axis is temperature and the yaxis is heat capacity. The peaks represent the melting of two different species in the sample.
– John
May 27 '20 at 13:43

## 2 Answers

Using pretty much the same code from my answer to your last question:

fit = NonlinearModelFit[
data,
height1 Exp[-(x - peakposition1)^2/peakwidth1^2] +
height2 Exp[-(x - peakposition2)^2/peakwidth2^2] +
baseline,
{
{height1, 0.5}, {peakposition1, 76}, {peakwidth1, 2},
{height2, 1.3}, {peakposition2, 92}, {peakwidth2, 1},
{baseline, 0}
}, x
];

fit["BestFitParameters"]

(* Out: {height1 -> 0.364615, peakposition1 -> 75.7626, peakwidth1 -> 0.967389,
height2 -> 1.19066,  peakposition2 -> 91.4855, peakwidth2 -> 0.306273,
baseline -> 0.0786113}*)

Show[
ListPlot[data, PlotStyle -> Black, PlotRange -> All],
Plot[
fit[x], Evaluate@Flatten@{x, MinMax[data[[All, 1]]]},
PlotStyle -> Red, PlotRange -> All
]
]


It has to be said, though, that these peaks are obviously non-gaussian. Their fitting will not do you much good apart from perhaps finding their max position (which you could also achieve with FindPeaks). For instance, you really should not try to obtain the areas of these peaks from these fits, as they will be quite wrong. You should spend some time trying to figure out which analytical shape your peaks should conform to, from the theory behind your experiment.

Here is the same idea, using the Breit-Wigner-Fano line shape you suggested:

ClearAll[bwf]
bwf[x_, y0_, h_, xc_, q_, w_] := y0 + h (1 + (x - xc)/(q w))^2 / (1 + ((x - xc)/w)^2)

fitbwf =
NonlinearModelFit[
data,
bwf[x, y0, h1, xc1, q1, w1] +
bwf[x, y0, h2, xc2, q2, w2],
{y0,
{h1, 0.4}, {xc1, 76}, q1, {w1, 1},
{h2, 1.2}, {xc2, 91}, q2, {w2, 0.3}
}, x,
MaxIterations -> 1000
]

Plot[
fitbwf[x],
Evaluate@ Flatten@ {x, MinMax[ data[[All,1]] ]},
PlotRange -> All, PlotStyle -> Red,
Prolog -> {PointSize[0.01], Black, Point[data]}
]


I can’t post a picture of the resulting plot above (I’m on mobile), but it’s not that much better than Gaussians. You may have better luck if you manually provide better starting values for the asymmetry parameters q1 and q2.

• MarcoB. Thank you! Yes, I tried the same but the fittings do not look very well and that's the reason I also asked this question. How can I make the fitting better using this code?. Also part of the question is that I am not quite sure what would be the best equation that describe them both to properly do the fitting
– John
May 27 '20 at 1:51
• @John As I mentioned at the end, these are not Gaussian-shape peaks (they are asymmetric, they have a left tail, etc). You need to figure out what shape they should be, and then you can use the same approach I use here, but with a different fitting model. May 27 '20 at 1:53
• MarcoB thanks again. Is there any way in Mathematica it can be known what is the best equation that describe both peaks or to find that out?. I was thinking that perhaps there was a way in Mathematica but I do not know.
– John
May 27 '20 at 1:55
• Also, could you tell me how to find the area individually for each peak even the gaussian equation you used? Thanks
– John
May 27 '20 at 1:58
• @John No, there is no way for Mathematica (or for any other general-purpose fitting software) to make that decision for you. In fact, many peak shapes will probably fit; it is in the eye of the experimenter, with the help of your knowledge of the associated physics, that you can make that decision. For the area, you can use the same techniques from the previous answer, but I strongly caution you against doing so, as the answers would be practically meaningless from this fit. May 27 '20 at 2:10

I am not a physicist but the need (craving? compulsion?) for the area under a parametric curve when there's such a poor fit makes no sense.

Your data is pretty dense (lots of observations uniformly spaced) so why not pick a reasonable baseline (another concept I don't understand as the left and right side of the peaks seem to have different levels) and then just find the mean of the response variable, subtract the chosen baseline, and finally multiply by the width of the peak? If you had a good fit, that's essentially what you'd get.

For the left peak:

left = Select[data, #[[1]] < 82 &];
baseline = Min[left]
(* 0.0429378 *)
width = Max[left[[All, 1]]] - Min[left[[All, 1]]]
(* 13.946 *)
area = (Mean[left[[All, 2]]] - baseline)*width
(* 1.07861 *)

• that is a very good input and you are right! This is a simpler way to find the areas. Thank you
– John
May 27 '20 at 4:44