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

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

Answer 1

[Mathieu NOE]

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

[Ganesh Naik]

Hi Mathieu, thanks for your answer. I need to find x and y values for the other signal (red signal in my plot) excatly the same location of the first one. I mean for the first signal (x,y) values are (447.3275, -6.1870), now I need to find (y) value for the other signal (red) in the same location (I tried to draw a vertical line and try to use intersection point but could not be able to do it successfully). Could you please help me with it. Thanks and regards

[Mathieu NOE]

hello again

ok so this is not exactly the same request in your original post you speak of intersecting the line with a horizontal line

I was thinking of drawing a horizontal line where the first signal peak is detected and then take the intersection point of the second signal.

I am a bit confused which point of the second signal you're exactly looking for ....

because first time you speak of (x,y) first line point at 413 and -6.18, now you speak of (447.3275, -6.1870)

so which one of the 3 green points is the one you're chasing ?

[Ganesh Naik]

Hi Mathieu, sorry there was typo with my previous message. I am looking for the "y" value with respect to (x,y) 413 and -6.18. In fact I am looking for the "y" value, first green dot (left side), "vertical line" just below the red dot. I should have mentioned it as "vertical line", which i have corrected (now).

Sorry for the confusion.

Regards

[Mathieu NOE]

hello again

so the red cross is the point you're looking for; has same x coordinate as the "reference" point on the other signal ; notice the search for the min value search does not give you the "corner" point ? I would expecte that thas would be your "true" reference point (might need a bit of data smoothing )

clc
clearvars

% data = importdata('data.txt');
load('Value.txt');
x = Value(:,1);
y = Value(:,2);
t = 1:length(x);


% the point you are looking for has following coordinates
[val,ind] = min(y); % 
tx = t(ind);
xx = x(ind);

figure(1)
plot(t,x,t,y,tx,val,'+g',tx,xx,'+r','linewidth',2,'markersize',12);grid on
legend('signal x','signal y');

[Ganesh Naik]

Hi Mathieu, thanks for your answer. This is what I was looking for. I do agree with your comments regarding true reference point. I will consider some data smoothing options for that. Is there any way can I display/label both (x,y) values for both the signlas?I am going to accept the answer!!

[Mathieu NOE]

here the figure code updated for the x,y coordinates display

the +35 x axis offset is to improve the clarity of the plot , up to you to change that

figure(1)
plot(t,x,t,y,tx,val,'+g',tx,xx,'+r','linewidth',2,'markersize',12);grid on
text(tx+35,val,['(' num2str(tx) ',' num2str(val) ')'])
text(tx+35,xx,['(' num2str(tx) ',' num2str(xx) ')'])

[Mathieu NOE]

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.
%
%     © 2002 - Michele Giugliano, PhD and Maura Arsiero
%     (Bern, Friday July 5th, 2002 - 21:10)
%    (http://www.giugliano.info) (bug-reports to [e-mail])
%
% 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]

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]

you're welcome

glad I could be of some help !