find multiples of real number in sampled array

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Tommaso Seresini
Tommaso Seresini on 7 Apr 2022
Commented: Mathieu NOE on 12 Apr 2022
I have an array t with a serie of values with equal spacing dt. I would like to find the values of t which are multiples of a given real number b, for different values of a parameter a.
Since i am dealing with real numbers, I don't expect to find the exact multiples. So I came up with the following solution:
I find that this solution works well for some combination of numbers but for others, such as those in the example, I have some rounding errors and i miss some points. Am I doing something wrong?
The ultimate goal would be to have a function to find the nodes and antinodes of a sine with an arbitrary frequency and phase. this method works well as long as the number of periods in t is small (2,3) but fails and is inconsistend as the frequency grows.
Thank you so much for your help!

Answers (1)

Mathieu NOE
Mathieu NOE on 8 Apr 2022
I had some code laying around and found it could be used for your purpose (the sine wave problem , and btw i could not really make the connection with the posted code).
so demo with an exponential decaying sine wave - you can try with your own signal
  • nodes = points of zero crossing (here shown both positive slope and negative slope zero crossing points)
  • antinodes = local maxima (here I just picked the positive ones)
y = sin(3*x-1).*exp(-x/10);
% antinodes = find positive local maxima
[tf, P] = islocalmax(y,'MinProminence',max(y)/3);
x_antinodes = x(tf);
y_antinodes = y(tf);
% nodes = find points crossing the zero y value
threshold = 0; % your value here
[t0_pos,s0_pos,t0_neg,s0_neg]= crossing_V7(y,x,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"
plot(x,y,x_antinodes,y_antinodes,'dk',x,threshold*ones(size(x)),'k--',t0_pos,s0_pos,'dr',t0_neg,s0_neg,'dg','linewidth',2,'markersize',12);grid on
legend('signal','antinodes','nodes level threshold','positive slope nodes','negative slope nodes');
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 = [];
Mathieu NOE
Mathieu NOE on 12 Apr 2022
glad to see you have a solution that works !
have a good day

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