How to find peaks in a data vector?

Hello,

I have the following data

data = [9 12 8 11 10 10 11 12 11 11 12 8 11 9 9 12 8 11 10 12 8 11 12 8 11 10 18 20 10 18 20 24 28 30 31 32 33 33 34 33 32 34 35 32 33 31 30 37 38 39 40 39 40 38 37 40 41 38 37 36 33 32 34 35 32 33 31 30 32 34 31 30 28 26 25 23 20 18 15 12 11 10 10 11 12 11 11 9 12 8 11 10 12 8 9 11 7 9 10 7 3 4 2 1 5 4 3 7 8 9 10 5 4 6 11 12 8 11 12 12 13 14 15 12 13 15 15 16 17 18 19 17 16 15 19 20 21 20 18 17 16 19 21 18 20 24 28 30 31 32 33 30 31 35 36 38 40 42 41 41 42 46 48 47 45 42 40 41 43 40 39 39 38 36 32 34 31 30 28 26 25 23 20 18 15 12 11 10 12 8 9 11 7 9 10 12 11 10 11 10 11 12 11 11 9 12 8 11 10 10 11 12 11 11 9 12 8 11 10 14 15 18 17 16 15 18 19 14 15 17 18 19 20 21 19 18 17 16 15 14 12 15 15 15 16 17 18 18 19 20 20 20 21 19 17 16 17 18 19 22 24 26 29 30 35 37 40 42 44 44 46 48 48 49 50 50 51 52 53 51 52 55 58 57 51 56 57 52 50 49 52 52 51 49 48 46 46 42 43 45 41 42 40 41 43 40 38 37 34 32 28 25 19 15 14 13 12 8 5 3 7 8 9 11 12 10 7 9 14 13 11 10 8 5 4 3 5 7 8 9 10 12 11 6 8 10 7 6 8];

and I need to find exactly 3 peaks in these data, for example:

How can I do that?

Note 1: I always need to find 3 peaks (this value doesn't change), but the X and Y axis range may change.

Note 2: The last peak/curve will always be higher than the other two.

Note 3: I don't need to get the value itself, I need to get the index of the value.

Answer 1

[Star Strider]

If you have the Signal Processing Toolbox, use the findpeaks function. It will do what you want. You will have to experiment with its name-value pair arguments to get the ‘correct’ result.

[Kelvin Prado]

Thanks a lot Star Strider. It worked:

findpeaks(data,'MinPeakDistance',20,'NPeaks',3,'MinPeakHeight',mean(data))

But I need to manually set the 'MinPeakDistance' argument based on a previous knowledge of my data.

Do you know if there's some another way that I can find these peaks without using this argument?

[Star Strider]

My pleasure.

There are two possible ways to do what you want, depending on how involved you want your code to get. The main peaks look periodic, so taking the fft (link), finding the dominant peak, determining the period from it (inverse of the frequency), and using that as 'MinPeakDistance' is one way I can imagine to automate it. If you subtract the mean of your data from the rest of your data before calculating the fft, you will eliminate the d-c (constant) offset ‘0 Hz’ peak, making it easier to find the lowest-frequency dominant peak.

A slightly more involved option (and the one I would use) is to take the fft to identify the dominant frequency and create a FIR bandpass filter or IIR (Chebyshev Type II) bandpass filter to eliminate the d-c offset and high-frequency noise. If all your data have the same sampling frequency and the same essential spectrum, the bandpass filter approach could (no promises) eliminate the need for any name-value pair arguments.

[Image Analyst]

Not sure I understand why you don't want to use the 'MinPeakDistance' argument. What's wrong with using that to pass in some argument value based on previous knowledge? Why do you need some other way?

[Adam]

Have you tried experimenting with the 'MinPeakProminence' attribute? These 3 peaks should all be quite identifiable by that and maybe a more simple measure to work out a prominence. based on the mean of your data or some percentile value.

[Kelvin Prado]

• Thanks again Star Strider, I will try to use the fft to find the dominant peak and determine the 'MinPeakDistance'. It seems to be relatively easy (in Matlab) and I think it solves my problem.
• Image Analyst, cause I need to create an algorithm that automatically finds the peaks. The user of my algorithm doesn't need to have a previous knowledge about the data it self, just need to know what time (X axis) each peak occurs.
• Thanks Adam, I will take a look at the 'MinPeakProminence' attribute. I didn't know about that.

[Star Strider]

Our pleasure!