Finding sequences representing a sin-function in a timeseries

Hi,

I got a problem which I am not able to solve on my own and I hope for some help here. Thank you already in advance!

The Problem:

I got a timeseries which includes way data from a test rig (further information probably won't help) and I want a script which finds intervals with different sinus-like motions on it's own. I got Intervals without any motion, with motion but not sinus-like and the said intervals with sinus-like motion. The script should return the Interval Borders of the parts with sinus-like motion. If the sinus function changes, but it still stays a sinus function, it should create a new interval as well.

I hope the problem is understandable and someone has an idea how to tackle this problem. I would really appreciate your help!

Kind regards

Answer 1

[Mathieu NOE]

hello

maybe this ?

a simple approach based on signal envelope

here I make a comparison using the matlab built in envelope function vs a Fex submission, that better fills my need in this case

Envelope1.1 - File Exchange - MATLAB Central (mathworks.com)

the diamond markers gives you the identified start and stop moments of sine signals , which are assumed to be of higher amplitude than the background noise (may need some filtering on real signals !)

each sine block can be extracted and saved if neede

t = 0:0.01:10;
x = 0.01*randn(size(t)); % some background noise
% create some sine periods
ind1 = 100:300;
f1 = 1.58;
x(ind1) = exp(-0.2*t(ind1)).*sin(2*pi*f1*t(ind1)+ t(ind1).^2);

ind1 = ind1+400;
f1 = 3.23;
x(ind1) = exp(0.03*t(ind1)).*sin(2*pi*f1*t(ind1)+ t(ind1).^2);

% Call function envelope to 
% obtain the envelope data
%--------------------------------------------
[up,down] = envelope(x,200,'analytic'); % matlab built in function
[up2,down2] = envelope2(t,abs(x),'linear'); % Fex submission (bottom section)

figure(1)
plot(t,x,t,up,'k',t,up2,'r')
legend('signal','envelope','envelope2');

% now let's find the start / stop time index for the sine blocks

threshold = 0.2*max(abs(x)); % 20% of peak amplitude

[t0_pos1,s0_pos1,t0_neg1,s0_neg1]= crossing_V7(up2,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(2)
plot(t,x,t,up2,'r',t0_pos1,s0_pos1,'dr',t0_neg1,s0_neg1,'dg','linewidth',2,'markersize',12);grid on
legend('signal','envelope2','sine start point','sine stop point');

% extract and save in cells sine portions

for ci = 1:numel(t0_pos1)
    ind = (t>=t0_pos1(ci) & t<=t0_neg1(ci));
    ti = t(ind);
    xi = x(ind);
    figure(2+ci)
    plot(ti,xi);
    out{ci} = [ti(:) xi(:)];
end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%ù

function [up,down] = envelope2(x,y,interpMethod)

%ENVELOPE gets the data of upper and down envelope of the known input (x,y).
%
%   Input parameters:
%    x               the abscissa of the given data
%    y               the ordinate of the given data
%    interpMethod    the interpolation method
%
%   Output parameters:
%    up      the upper envelope, which has the same length as x.
%    down    the down envelope, which has the same length as x.
%
%   See also DIFF INTERP1

%   Designed by: Lei Wang, <[e-mail]>, 11-Mar-2003.
%   Last Revision: 21-Mar-2003.
%   Dept. Mechanical & Aerospace Engineering, NC State University.
% $Revision: 1.1 $  $Date: 3/21/2003 10:33 AM $

if length(x) ~= length(y)
    error('Two input data should have the same length.');
end

if (nargin < 2)|(nargin > 3),
 error('Please see help for INPUT DATA.');
elseif (nargin == 2)
    interpMethod = 'linear';
end

 
% Find the extreme maxim values 
% and the corresponding indexes
%----------------------------------------------------
extrMaxIndex = find(diff(sign(diff(y)))==-2)+1;
extrMaxValue = y(extrMaxIndex);

% Find the extreme minim values 
% and the corresponding indexes
%----------------------------------------------------
extrMinIndex = find(diff(sign(diff(y)))==+2)+1;
extrMinValue = y(extrMinIndex);

up = extrMaxValue;
up_x = x(extrMaxIndex);

down = extrMinValue;
down_x = x(extrMinIndex);

% Interpolation of the upper/down envelope data
%----------------------------------------------------
up = interp1(up_x,up,x,interpMethod); 
down = interp1(down_x,down,x,interpMethod); 
end

[Mathieu NOE]

hello

Problem solved ?