Finding several local maximum values in a given range and corresponding indices

If we have a dataset "y" which consists of a sum of 5 gaussian peaks as function of time t, there will be 5 local maximum values in the whole y values. Basically, I would like to obtain the y axis maximum values and their corresponding t axis values. For example peak 1, has a maximum value 5 and it corresponds to t value of 19.

One can individually find the maximum values by giving a range say

[a ,i]=max(y(1:20));% locating maxima in a given range of the first peak

value_1= t(i); % Corresponding value of time for index i1.

One can repeat this 5 times by specifying the ranges for all peaks. Is there a better way to achieve the same result as an output in a single vector, and the corresponding time for those maxima in another vector . Thanks.

[madhan ravi]

If you mean all the indices of the max value:

Indices = find(y==max(y(:))

[FW]

I have edited the question to clarify it. Basically, I would like to obtain the y axis maximum values and their corresponding t axis values. For example peak 1, has a maximum value 5 and it corresponds to t value of 19. I want to have all the 5 values of maximum and their corresponding time values?

Answer 1

[Star Strider]

Use the Signal Processing Toolbox findpeaks function, or the islocalmax (R2017b and later) function.

[FW]

Thanks, will try that. The findpeaks function will find the maximum y vaues. Is it possible to locate the corresponding values for the time as well?

[Star Strider]

As always, my pleasure.

The findpeaks function has several possible outputs, of which you need only the first two, the ‘pks’ and ‘locs’ variables. Use the ‘locs’ value (the index of the corresponding peak values, unless you provide a vector of the independent variable values) to identify the ‘time’ values.

Example —

[pks,locs] = findpeaks(y);
timev = time(locs);

That should do what you want.

[FW]

Yes this works. Thank you.

[Star Strider]

As always, my pleasure!

[FW]

I was exploring the Signal Processing Toolbox, they don't seem to have peak area measurements. Is there any function which allows us to measure peak area once we specify the integration limits.

[Star Strider]

It doesn’t calculate peak area. The closest it’s possible to get is the full-width-half-maximum calculation, the ‘w’ (peak widths) output in the documentation.

Some of your peaks overlap, so they may represent different (I assume Gaussian) functions. It might be best to use the information that findpeaks returns to fit multiple Gaussians to your data, and then calculate those areas.

Another option is to negate your signal and use findpeaks to find the valleys (peaks of the negated signal), and then use those to determine the dividing lines between the peaks. Then use the cumtrapz function and the ‘locs’ output of using findpeaks on the negated signal to determine the areas between the valley indices.

It all depends on what you want to do and how you want to define your peaks.

[MW]

That sounds good, will try that. I hope Matlab includes peak area determination just like OriginPro where one can pan the area of interest and determine its area.

[Star Strider]

Thank you.

‘I hope Matlab includes peak area determination just like OriginPro where one can pan the area of interest and determine its area.’

I doubt that’s an option, or if it is, I’ve not heard of it.

Fitting Gaussians is not difficult. See: Area under each peak for an illustration, including findpeaks calls. You can probably use that code with a few tweaks to use your own data. (I will help as necesary, since it’s my code. That code uses trapz, however it would probably not be very difficult to tweak it to use integral to calculate the areas, once the parameters of the Gaussians are known.)

Answer 2

[Image Analyst]

It looks like you already have an acceptable answer, but if you want code to fit some specified number of Gaussians to a signal, let me know - I have that, though not in a general purpose demo right now (I'd have to create that). Attach your signal if you need this.