- Identify the peak as y-values greater than the upper 95% confidence interval for the mean of all the data;
- Integrate the entire record using cumtrapz;
- Do a linear regression on all the integrated data up to the first index of the peak;
- Use that regression to detrend the integrated data;
- Use the identified indices of the peak to calculate the approximate integral of the peak.

# Area between baseline and data peak

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Hi,

I am trying to determine the area under a peak curve, see image. The baseline illustrated in image is not zero, and I would like to subtract this from the area under the peak. The data is from a picture and been summed to obtain this profile. I want to determine the intensity by summing the area under the peak.

1. Is there a way to determine the baseline (y- value) automatically by code.

2. How do I determine the area minus the baseline? I use sum to determine the area under the peak curve.

##### 0 Comments

### Accepted Answer

Star Strider
on 22 Sep 2014

Edited: Star Strider
on 22 Sep 2014

Here is how I would do it:

% Create Data

x = 0:480;

y = 1E4*(0.5+rand(size(x)));

y(400:450) = 1.3E+5*exp(-(x(400:450)-425).^2/50)+y(400:450);

% Generate Statistics & Identify Peak

ysts = [mean(y) median(y); mean(y)-1.96*std(y)/sqrt(length(y)) mean(y)+1.96*std(y)/sqrt(length(y))];

[ypk, pki] = max(y);

pkiv = find(y >= ysts(2,2)); % Find Peak Indices

Iy = cumtrapz(x, y-ysts(1,1)); % Integrate Entire Record

Iydb = [ones(pkiv(1),1) x(1:pkiv(1))']\Iy(1:pkiv(1))';

Iyd = Iy' - [ones(size(x))' x']*Iydb; % Detrend Using Baseline

Ipk = Iy(max(pkiv)) - Iy(min(pkiv)); % Find Peak Area

% Plot Data & Detrended Integral & Selected Statistics

figure(1)

plot(x, y)

hold on

plot(x, Iyd, '-g')

hold off

legend('Data', 'Detrended Integral', 'Location', 'NW')

text(100, 1E+6, sprintf('Peak Area = %23.15E', Ipk))

The idea is reasonably straightforward:

This approach may not generalise well if most of the baseline information is not at x-values less than the peak, because it depends on that information to detrend the integral.

The plot produces:

##### 7 Comments

Image Analyst
on 28 Jul 2017

When you do \Iy you're doing matrix division. Did you really want element-by-element division? If so, use dot slash instead of slash.

Also ones() is returning a column vector, so make sure x is a row vector, since you transpose x to be a column vector and want to stitch it to the right of ones.

Also be away of automatic expansion of your ly vector since you're dividing a pkiv(1)-by-2 vector by a vector array.

### More Answers (2)

Andrew Reibold
on 22 Sep 2014

Edited: Andrew Reibold
on 22 Sep 2014

The area under a curve is the definition of an integral. You will want to take the integral and subtract the baseline * baseline_width to find the area.

I'm not sure if its really scientific to determine the baseline just from raw data. Maybe this entire unique data set has some factor that shifts every point in some direction which makes you perceive the baseline somewhere that it isn't. I'm sure you could take the mean of non-peaked data or something, but normally the baseline is determined from the original function, historical data, or whatever you are stepping off from.

Image Analyst
on 22 Sep 2014

##### 2 Comments

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