Get y value in plot(t-s diagram)

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주희 박 on 10 Jun 2022
Answered: Star Strider on 10 Jun 2022
Hi, This is t-s diagram and I want to get y value that meet with black x line. There are 2 points in the pic.
Can I get them with simple method?

Mathieu NOE on 10 Jun 2022
hello
see my demo based on dummy data - trying to reproduce your case
code :
clc
clearvars
% dummy data
n=1000;
y= (1:800);
x = 34.1+0.35*sin(y/100 -pi/2).*exp(-y/200);
threshold = 34.11; % your value here
[t0_pos,s0_pos,t0_neg,s0_neg]= crossing_V7(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"
figure(1)
plot(x,y,'b',s0_pos,t0_pos,'*r',s0_neg,t0_neg,'dk','linewidth',2,'markersize',12);grid on
set(gca,'YDir','reverse')
xlim([33.7 34.4]);
legend('signal','signal positive slope crossing points','signal negative slope crossing points');
% y value for first point
t0_pos
% y value for second point
t0_neg
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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
error(nargchk(1,4,nargin));
% 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!');
end
% check the level input
if nargin < 3
% set standard value 0, if level is not given
level = 0;
end
% check interpolation method input
if nargin < 4
imeth = 'linear';
end
% 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;
end
end
end
% 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);
else
% empty output
ind_pos = [];
t0_pos = [];
s0_pos = [];
% extract the negative slope crossing points
ind_neg = [];
t0_neg = [];
s0_neg = [];
end
end

Star Strider on 10 Jun 2022
Try something like this —
y = linspace(-10, 800);
x = 34 + 4.5*exp(-0.005*y) .* sin(2*pi*y/700);
v = 34;
idx = find(diff(sign(x - v)));
for k = 1:numel(idx)
ixrng = idx(k)-1:idx(k)+1;
yv(k) = interp1(x(ixrng), y(ixrng), v);
end
figure
plot(x, y)
hold on
plot(v*ones(size(yv)), yv, 'rs')
hold off
set(gca, 'YDir','reverse')
xline(v)
text(v*ones(size(yv)), yv, compose(' \\leftarrow %8.3f',yv), 'Horiz','left', 'Vert','middle')
I am using created data, so it will be necessary for you to adapt your data to my code.
.