Measure time difference peaks and stores it

Hi,

Can anyone help me with a code to measure the time difference from the moment the peak is 1 till it goes back to zero and then stores it. Then it repeats again for the next peak and the next till it goes over all the peaks. So for example the graph below shows the first peak occurs at around 15197 seconds and ends at 15562 seconds, hence time difference is 365s which is then stored in a matrix. Then the code continues to the next peak and measures that time then stores it. It keeps doing that until it stores all those times.

find peak doesn't work well for my data to use widths and all that. I need some type of for and if loop. Any help appreciated

Answer 1

[Star Strider]

It would help to have the data.

One approach would be to use the islocalmax function two times using the 'FlatSelection','first' with one function call and 'FlatSelection','last' with the second function call. Use the find function to get the numerical indices for each, then sort them and use reshape to create a (2xN) or (Nx2) matrix from them, if desired.

t = linspace(0, 10).';                                      % Code Assumes Column Vectors
y = sin(2*pi*0.5*t) >= 0;

p1 = islocalmax(y, 'FlatSelection','first');
p2 = islocalmax(y, 'FlatSelection','last');
numidx = sort([find(p1) find(p2)]);

time_diff = diff(t(numidx),[],2)

time_diff = 4×1

0.9091 0.9091 0.9091 0.9091

figure
plot(t, y)
ylim([0 1.1])

The times are the same here because this funciton is periodic.

The pulsewidth function may also be helpful.

.

[mpz]

@Star Strider I attached the data. Column 1 is the time (x-axis) and column is the output(y-axis). Based on my plot I only have 5 peaks but I got more time. Is there a way to only output the time diffference for peaks. Could you please take a look at my code below

clear all;clc;close all
% load('seperatedata.mat')
% data1 = sep{3,1}; % extracts for specific test

ycut = data1(:,3) > 7.5 & data1(:,3) < 9; % Determine what parts of the data is within the acceptable range
figure(5001)
plot(data1(:,1),ycut) % plots logic data
figure(5002)
plot(data1(:,1),data1(:,3))

figure(5003)
yyaxis right
plot(data1(:,1),ycut)
yyaxis left
plot(data1(:,1),data1(:,3))

% method 1 testing
% c=1;
% t=data1(:,1);
% while any(ycut>=1)
%     f1=find(ycut>=1,1);
%     ycut(1:f1-1)=[];
%     f2=find(ycut<=0,1);
%     ycut(1:f2-1)=[];
%     deltaT(c)=t(f2)-t(f1);%provides an array of the peak time periods
%     t(1:f2-1)=[];
%     c=c+1;
% end

% method 2 testing
t=data1(:,1);
p1 = islocalmax(ycut, 'FlatSelection','first');
p2 = islocalmax(ycut, 'FlatSelection','last');
numidx = sort([find(p1) find(p2)]);
time_diff = diff(t(numidx),[],2)

Here is the output I got.

time_diff =

  369.5200
  299.9560
    0.6200
  236.0960
    2.6040
  308.2640
  470.4560
    0.7440
    

Here is approximately what I expected for the time difference based on my data1.mat

369.2
299.5
236.4
307.5
471

[Star Strider]

It is difficult to know how to deal with those data, since they are extremely noisy. I got the impression from the posted plot that they were limited to be between 0 and 1 and had flat tops. So I chose a threshold value of 100 (it can be any value you want) and created the data to have exactly those characteristics.

% t = linspace(0, 10).';                                      % Code Assumes Column Vectors
% y = sin(2*pi*0.5*t) >= 0;

LD = load(websave('data1','https://www.mathworks.com/matlabcentral/answers/uploaded_files/1101295/data1.mat'))

LD = struct with fields:
    data1: [56060×3 double]

t = LD.data1(:,1);
y = LD.data1(:,2);

yi = y;
thrshld = 100;
yi(yi < thrshld) = 0;                                       % Threshold Data
yi(yi >= thrshld) = 1;


p1 = islocalmax(yi, 'FlatSelection','first');
p2 = islocalmax(yi, 'FlatSelection','last');
numidx = ([find(p1) find(p2)]);

time_diff = diff(t(numidx),[],2);
StartTime = t(p1);
EndTime = t(p2);

Results = table(StartTime,EndTime,time_diff)

Results = 323×3 table
    StartTime    EndTime    time_diff
    _________    _______    _________

      11462       11463       1.736  
      11473       11475       1.736  
      11499       11513       13.64  
      12938       16656      3718.3  
      16657       16661       3.472  
      16661       16674      12.524  
      16674       16676       1.364  
      16676       16677       0.868  
      16678       16678       0.248  
      16678       16680        1.24  
      16680       16681        0.62  
      16681       16681       0.248  
      16682       16683        0.62  
      16685       16685           0  
      16685       16687       2.356  
      16688       16688           0  

figure
% plot(t, y)
plot(t, yi)
hold on
plot(t(p1), yi(p1), '>g')
plot(t(p2), yi(p2), '<r')
hold off
xlabel('Time')
ylabel('Thresholded Amplitude')
title('All Spikes')
ylim([0 1.1])

figure
% plot(t, y)
plot(t, yi)
hold on
plot(t(p1), yi(p1), '>g')
plot(t(p2), yi(p2), '<r')
hold off
% ylim([0 1.1])
axis([[1.14 1.16]*1E+4  0 1.1])
xlabel('Time')
ylabel('Thresholded Amplitude')
title('First Three Spikes')
p1idx = find(p1);
First3 = p1idx(1:3);
text(t(First3), yi(First3)/2, compose('\\Delta t = %7.3f',time_diff(1:3)), 'Horiz','left', 'Rotation',45)

At the chosen threshold level, the code detected 323 spikes. The ‘StartTime’ and ‘EndTime’ values in the table are rounded to the nearest integer. Change the format to see them with greater precision.

.

Answer 2

[David Hill]

I assume you have an t-array and a y-array.

c=1;
while any(y>=1)
    f1=find(y>=1,1);
    y(1:f1-1)=[];
    f2=find(y<=0,1);
    y(1:f2-1)=[];
    deltaT(c)=t(f2)-t(f1);%provides an array of the peak time periods
    t(1:f2-1)=[];
    c=c+1;
end

[mpz]

@David Hill if you don't mind could you add short descriptions of what each part of the code does so I can understand and possibly modify for my needs

[mpz]

I copy and pasted your code but the time, t, gave me continous time result. Is it possible to make it only output the time difference when there is a peak of 1 only and ignores the others when the line is at zero?

Answer 3

[mpz]

Yes that data is noisy that's why I used a boolean to only pick the ones that met my threshold. Thanks

[Star Strider]

As always, my pleasure!