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how to get Y Value from a plot having 3 curves or more as shown in figure?

조회 수: 2(최근 30일)
naveen kumar s
naveen kumar s 2021년 8월 31일
댓글: Mathieu NOE 2021년 9월 9일
Hi all,
Please give me method to solve this problem either in script or App Designer any method is lot of helpful.
Question Described : I have a plot with multiple curve (Image attached).
Input i have X axis and Curve Data, for this i need to get Y value.
Plot looks like this. see Attachment >> Now For X value (@ point Q between P1 & P2) I should find Y.
Here If i give 2 inputs
Input1 >> data of P1,P2 Or P3 Or inbetween P1 & P2
input2>> X value
Output Requried Y Value >> for above inputs. .........................................................................Thanks for Help


Mathieu NOE
Mathieu NOE 2021년 8월 31일
you can try this code
% dummy data
y1 = 3*sin(1.5*x)+ 0.01*randn(1,n);
y2 = 1+2*sin(x -0.5)+ 0.01*randn(1,n);
y_threshold = 1; % your Y target value here
[t0_pos1,s0_pos1,t0_neg1,s0_neg1]= crossing_V7(y1,x,y_threshold,'linear'); % positive (pos) and negative (neg) slope crossing points
[t0_pos2,s0_pos2,t0_neg2,s0_neg2]= crossing_V7(y2,x,y_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"
% Q point coordinated = assumed half way between the 2 curves (y1 and y2)
x_Q_point = (t0_pos1 + t0_pos2)/2;
y_Q_point = y_threshold;
plot(x,y1,'b',t0_pos1,s0_pos1,'dr',x,y2,'g',t0_pos2,s0_pos2,'*r',x_Q_point,y_Q_point,'db','linewidth',2,'markersize',12);grid on
legend('signal 1','signal 1 positive slope crossing points','signal 2','signal 2 positive slope crossing points' ,'Q point' );
function [t0_pos,s0_pos,t0_neg,s0_neg] = crossing_V7(S,t,level,imeth)
% [ind,t0,s0,t0close,s0close] = crossing_V6(S,t,level,imeth,slope_sign) % older format
% CROSSING find the crossings of a given level of a signal
% ind = CROSSING(S) returns an index vector ind, the signal
% S crosses zero at ind or at between ind and ind+1
% [ind,t0] = CROSSING(S,t) additionally returns a time
% vector t0 of the zero crossings of the signal S. The crossing
% times are linearly interpolated between the given times t
% [ind,t0] = CROSSING(S,t,level) returns the crossings of the
% given level instead of the zero crossings
% ind = CROSSING(S,[],level) as above but without time interpolation
% [ind,t0] = CROSSING(S,t,level,par) allows additional parameters
% par = {'none'|'linear'}.
% With interpolation turned off (par = 'none') this function always
% returns the value left of the zero (the data point thats nearest
% to the zero AND smaller than the zero crossing).
% check the number of input arguments
% check the time vector input for consistency
if nargin < 2 | isempty(t)
% if no time vector is given, use the index vector as time
t = 1:length(S);
elseif length(t) ~= length(S)
% if S and t are not of the same length, throw an error
error('t and S must be of identical length!');
% check the level input
if nargin < 3
% set standard value 0, if level is not given
level = 0;
% check interpolation method input
if nargin < 4
imeth = 'linear';
% make row vectors
t = t(:)';
S = S(:)';
% always search for zeros. So if we want the crossing of
% any other threshold value "level", we subtract it from
% the values and search for zeros.
S = S - level;
% first look for exact zeros
ind0 = find( S == 0 );
% then look for zero crossings between data points
S1 = S(1:end-1) .* S(2:end);
ind1 = find( S1 < 0 );
% bring exact zeros and "in-between" zeros together
ind = sort([ind0 ind1]);
% and pick the associated time values
t0 = t(ind);
s0 = S(ind);
if ~isempty(ind)
if strcmp(imeth,'linear')
% linear interpolation of crossing
for ii=1:length(t0)
%if abs(S(ind(ii))) > eps(S(ind(ii))) % MATLAB V7 et +
if abs(S(ind(ii))) > eps*abs(S(ind(ii))) % MATLAB V6 et - EPS * ABS(X)
% interpolate only when data point is not already zero
NUM = (t(ind(ii)+1) - t(ind(ii)));
DEN = (S(ind(ii)+1) - S(ind(ii)));
slope = NUM / DEN;
slope_sign(ii) = sign(slope);
t0(ii) = t0(ii) - S(ind(ii)) * slope;
s0(ii) = level;
% extract the positive slope crossing points
ind_pos = find(sign(slope_sign)>0);
t0_pos = t0(ind_pos);
s0_pos = s0(ind_pos);
% extract the negative slope crossing points
ind_neg = find(sign(slope_sign)<0);
t0_neg = t0(ind_neg);
s0_neg = s0(ind_neg);
% empty output
ind_pos = [];
t0_pos = [];
s0_pos = [];
% extract the negative slope crossing points
ind_neg = [];
t0_neg = [];
s0_neg = [];
plot :
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