Find the second signal points (x, y location) exactly the same place of the first signal peak (x, y) matlab

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Ganesh Naik 2021-4-27

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评论： Mathieu NOE 2021-4-27

Value.txt

Hi all, I have two signals where I find the negative peak of one signal and need to find the corresponding x,y points for the other signal. I was thinking of drawing a "vertical line" where the first signal peak is detected and then take the intersection point of the second signal. I think there are better straight forward methods to achieve this. In my example (please refer to the figure) x,y values are 413 and -6.18. I need to find the "y" value for the other signal (red signal) in the same location.

For your reference, I have attached the sample plot and the signal.

Kind regards

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Answer 1

Mathieu NOE 2021-4-27

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https://ww2.mathworks.cn/matlabcentral/answers/814215-find-the-second-signal-points-x-y-location-exactly-the-same-place-of-the-first-signal-peak-x-y#answer_685840

在 MATLAB Online 中打开

crossing_V7.m

hello

hope this helps you see function attached

load('Value.txt');
x = Value(:,1);
y = Value(:,2);
t = 1:length(x);
threshold = min(y); % 
[t0_pos,s0_pos,t0_neg,s0_neg]= crossing_V7(x,t,threshold,'linear'); % positive (pos) and negative (neg) slope crossing points 
% ind => time index (samples)
% t0 => corresponding time (x) values 
% s0 => corresponding function (y) values , obviously they must be equal to "threshold"
figure(1)
plot(t,x,t,y,t0_pos,s0_pos,'+r',t0_neg,s0_neg,'+g','linewidth',2,'markersize',12);grid on
legend('signal x','signal y','positive slope crossing points','negative slope crossing points');
% the point we are looking for is the last 'positive slope crossing points'
xx = t0_pos(end)
yy = threshold

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Mathieu NOE 2021-4-27

在 MATLAB Online 中打开

below some example of smoothing

I guess the sgolay method should work fine in your case

clc
close all
Fs = 1000;
samples = 1000;
dt = 1/Fs;
t = (0:samples-1)*dt;
y = square(2*pi*3*t) + 0.1*randn(size(t));
% %%%%%%%%%%%%%%%%
figure(1)
N = 10;
ys = slidingavg(y, N);
plot(t,y,t,ys);legend('Raw','Smoothed');
title(['Data samples at Fs = ' num2str(round(Fs)) ' Hz / Smoothed with slidingavg' ]);
% %%%%%%%%%%%%%%%%
figure(2)
N = 10;
ys = medfilt1(y, N,'truncate');
plot(t,y,t,ys);legend('Raw','Smoothed');
title(['Data samples at Fs = ' num2str(round(Fs)) ' Hz / Smoothed with medfilt1' ]);
grid on
%%%%%%%%%%%%%%%%
figure(3)
N = 10;
ys = sgolayfilt(y,3,51);
plot(t,y,t,ys);legend('Raw','Smoothed');
title(['Data samples at Fs = ' num2str(round(Fs)) ' Hz / Smoothed with sgolayfilt' ]);
grid on
%%%%%%%%%%%%%%%%
NN = 4;
Wn = 0.1;
[B,A] = butter(NN,Wn);
figure(4)
ys = filtfilt(B,A,y);
plot(t,y,t,ys);legend('Raw','Smoothed');
title(['Data samples at Fs = ' num2str(round(Fs)) ' Hz / Smoothed with butterworth LP' ]);
grid on
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function out = slidingavg(in, N)
%   OUTPUT_ARRAY = SLIDINGAVG(INPUT_ARRAY, N)
%
%  The function 'slidingavg' implements a one-dimensional filtering, applying a sliding window to a sequence. Such filtering replaces the center value in
%  the window with the average value of all the points within the window. When the sliding window is exceeding the lower or upper boundaries of the input
%  vector INPUT_ARRAY, the average is computed among the available points. Indicating with nx the length of the the input sequence, we note that for values
%  of N larger or equal to 2*(nx - 1), each value of the output data array are identical and equal to mean(in).
%
%  *  The input argument INPUT_ARRAY is the numerical data array to be processed.
%  *  The input argument N  is the number of neighboring data points to average over for each point of IN.
%
%  *  The output argument OUTPUT_ARRAY is the output data array.
%
%     Â© 2002 - Michele Giugliano, PhD and Maura Arsiero
%     (Bern, Friday July 5th, 2002 - 21:10)
%    (http://www.giugliano.info) (bug-reports to michele@giugliano.info)
%
% Two simple examples with second- and third-order filters are
%  slidingavg([4 3 5 2 8 9 1],2)   
%  ans =
%   3.5000  4.0000  3.3333  5.0000  6.3333  6.0000  5.0000
%
%  slidingavg([4 3 5 2 8 9 1],3)
%  ans =
%   3.5000  4.0000  3.3333  5.0000  6.3333  6.0000  5.0000
%
if (isempty(in)) | (N<=0)                                              % If the input array is empty or N is non-positive,
 disp(sprintf('SlidingAvg: (Error) empty input data or N null.'));     % an error is reported to the standard output and the
 return;                                                               % execution of the routine is stopped.
end % if
if (N==1)                                                              % If the number of neighbouring points over which the sliding 
 out = in;                                                             % average will be performed is '1', then no average actually occur and
 return;                                                               % OUTPUT_ARRAY will be the copy of INPUT_ARRAY and the execution of the routine
end % if                                                               % is stopped.
nx   = length(in);             % The length of the input data structure is acquired to later evaluate the 'mean' over the appropriate boundaries.
if (N>=(2*(nx-1)))                                                     % If the number of neighbouring points over which the sliding 
 out = mean(in)*ones(size(in));                                        % average will be performed is large enough, then the average actually covers all the points
 return;                                                               % of INPUT_ARRAY, for each index of OUTPUT_ARRAY and some CPU time can be gained by such an approach.
end % if                                                               % The execution of the routine is stopped.
out = zeros(size(in));         % In all the other situations, the initialization of the output data structure is performed.
if rem(N,2)~=1                 % When N is even, then we proceed in taking the half of it:
 m = N/2;                      % m = N     /  2.
else                           % Otherwise (N >= 3, N odd), N-1 is even ( N-1 >= 2) and we proceed taking the half of it:
 m = (N-1)/2;                  % m = (N-1) /  2.
end % if
for i=1:nx,                                                 % For each element (i-th) contained in the input numerical array, a check must be performed:
 if ((i-m) < 1) & ((i+m) <= nx)                             % If not enough points are available on the left of the i-th element..
  out(i) = mean(in(1:i+m));                                 % then we proceed to evaluate the mean from the first element to the (i + m)-th.
 elseif ((i-m) >= 1) & ((i+m) <= nx)                        % If enough points are available on the left and on the right of the i-th element..
  out(i) = mean(in(i-m:i+m));                               % then we proceed to evaluate the mean on 2*m elements centered on the i-th position.
elseif ((i-m) >= 1) & ((i+m) > nx)                          % If not enough points are available on the rigth of the i-th element..
  out(i) = mean(in(i-m:nx));                                % then we proceed to evaluate the mean from the element (i - m)-th to the last one.
 elseif ((i-m) < 1) & ((i+m) > nx)                          % If not enough points are available on the left and on the rigth of the i-th element..
  out(i) = mean(in(1:nx));                                  % then we proceed to evaluate the mean from the first element to the last.
 end % if
end % for i
end

Ganesh Naik 2021-4-27

Hi Mathieu, thank you so much for your help with finding the corresponding values and displaying them on the plot. I appreciate your help with filtering signals, I will go through it in detail. Thanks again and kind regards!!!

Mathieu NOE 2021-4-27

you're welcome

glad I could be of some help !

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Find the second signal points (x, y location) exactly the same place of the first signal peak (x, y) matlab

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Find the second signal points (x, y location) exactly the same place of the first signal peak (x, y) matlab

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