ideas to find the amplitude from this graph?

Hi,

Im tying to find the amplitude from that graph. I have the information for 300 points with the X and Y coordinates from that graph. One idea that i have is to find the local max and local minimum for every 2 zeros. And then, the amplitude would be the sum of local max and local min for every 2 zeros.

I think i can use the function "findpeaks" to find the locals but I'm having hard time to tell matlab to find the locals every 2 zeros.

As you see, the amplitude is changing so at the end i will have to take the average.

Any ideas would be really appreciate.

Thanks!

[Image Analyst]

I'm not clear on what you want. Do you want the "envelope" of the high frequency waveform? Do you want the running average, like you'd get from conv(), from all the data, or from the upper positive and lower negative peaks , or from the absolute value of the signal? It looks like averaging all the data would just average out to zero since it's like a modulated sine wave and there are negative points that are equal in number and amplitude to the positive points.

Answer 1

[Guillaume]

Would this work to find the locals every two zeros (I haven't got the SP toolbox)?

[peaks, locations] = findpeaks(signal); %all peaks and locations
allzeros = find(~signal);
%now divide the locations into bins whose upper bounds are allzeros(2:2:end) (every two zero)
%and find out which bin they land into:  
[~, locbin] = histc(location, allzeros(2:2:end));
%use the location bins to distribute the peaks into each bin and find the max and min per bin:
maxpeaks = accumarray(locbin, peaks, [], @max);
minpeaks = accumarray(locbin, peaks, [], @min);

You may have to do some filtering if the curve is too noisy near zero.

[Max]

maybe I'm not sure how to use why you wrote, maybe something like this?

 clc
 clear all

 %-------------------------OPTIMIZATION OF TMD----------------------------%

 %Load files%

 for u = 20:20:220;

    C= strcat('TMD_',num2str(u),'.csv');
    load(C);
end

 %---C=20 N*s/m---%

 t1=TMD_20(:,2); %seconds%
 vy=TMD_20(:,3); %m/s%

 [peaks, locations] = findpeaks(vy); %all peaks and locations
allzeros = find(vy);
%now divide the locations into bins whose upper bounds are allzeros(2:2:end) (every two zero)
%and find out which bin they land into:  
[vy, locbin] = histc(locations, allzeros(2:2:end));
%use the location bins to distribute the peaks into each bin and find the max and min per bin:
maxpeaks = accumarray(locbin, peaks, vy, @max);
minpeaks = accumarray(locbin, peaks, vy, @min);*_

but as you expected I'm getting some error:

 Error using ==> accumarray
Second input VAL must be a vector with one element for each row in SUBS, or a scalar.

 Error in ==> Solidworks_Results at 25
maxpeaks = accumarray(locbin, vy, peaks, @max);*_

[Guillaume]

You made an error and some unnecessary changes when you copied my code. First,

allzeros = find(~vy);

The ~ is required to find the 0s. If you find it confusing you could also write it as:

allzeros = find(vy == 0);

After that instruction, allzeros contains the index of all the zeros of your function (maybe I should have called the variable allzeroindices), and therefore allzeros(2:2:end) is the indices of every 2nd zero. Note that I made the assumption that the first lot of peaks were between two zeros, which is probably wrong. So it may work better with allzeros(1:2:end).

Secondly, note that I discarded the first output of histc as you don't need it. You only care about find between which pair of zero the location belongs to and that's the 2nd output. You certainly don't want to assign the first output of histc to vy, your original variable. It still should be:

[~, locbin] = histc(locations, allzeros(2:2:end)); %or allzeros(1:2:end)

The third output to accumarray is the size of the output, it's certainly not vy, it's numel(locbin) which accumarray calculates automatically if you give it [], thus leave the lines as:

maxpeaks = accumarray(locbin, peaks, [], @max);
minpeaks = accumarray(locbin, peaks, [], @min);

Finally, according to the error you get you swapped some arguments of accumarray.

Note, the idea of the algorithm is to get the location of the peak and the location of every 2nd zero. histc is then used to bin the location between each 2nd zero. You only care to know in which been each location lands, not the population of the bin, hence why you only want the 2nd output. Finally, accumarray distributes the peaks in each bin and only keep the max or min per bin.

[Max]

Thanks,

Well i fixed those errors that you told me but now when I'm running the code, i have seen that it only finds the positive peaks of my data, not the negatives, therefore my maxpeaks and minpeaks are the same.

Answer 2

[Star Strider]

First, use a bandpass filter to eliminate the high-frequency noise and any baseline drift in all the channels. I would use a Butterworth design, using buttord, butter, convert to second-order-section (sos) representation, then use filtfilt.

Finding the amplitude and peaks of the signal itself is then easy with findpeaks as you’ve already discovered.

[Star Strider]

Filtering your data to remove the high-frequency noise will produce the ‘modulation envelope’ of your signal. That seems to be what you want.

The modulation envelope appears to be what you actually want to get in attempting to find the extrema of the noisy signal.

Take the fft of your signal to find the frequencies (frequency band) of your actual data and the frequency (or frequency band) of your noise so you can determine the cutoff frequency for your filter, so you can then filter out the noise.

[Image Analyst]

And it would have been easier for us to try some things if only Max had remembered to attach his csv file(s) instead of a screenshot. We can't get data from a screenshot. OK, we can using this File Exchange submission, but we're not going to . If Max doesn't make it easy for me to help him, then I'm not going to do tons of extra work for him.

[Star Strider]

A .fig file would work also, giving us the plot to discuss and the data to easily extract from it. (I also have no desire to extract data from a .jpg.)

With .fig file data, it probably would take me at most a half hour to extract the data, design a filter (Max has the Signal Processing Toolbox because he has findpeaks), produce a clean output, and post a solution.

[Max]

I'm sorry i didn't answer, i don't know i didn't receive any email from your answer.

i attached my cdv file.

[Star Strider]

Max, what is the sampling frequency Fs (Hz) or sampling interval Ts (seconds)? Is it constant?

This seems to be one channel, but with Fs or Ts it should be enough to design the filter.

[Max]

If i understand correctly, the frequency of the force that is producing this oscillations is 1 Hz. But, if you see the data, not all the periods would have the same amount of data.

[Star Strider]

What I am interested in is the frequency at which the signals were sampled. How often were the samples collected (in samples per second)?

Alternatively, what was the time interval (in milliseconds) between any two of the individual samples you collected?

[Max]

the total sample length is 30 seconds and it has steps of 0.1 seconds, 300 points in total.

But not every period has the same amount of data.

[Star Strider]

Great!

Ts=0.1; Fs=10;