Variable vs constant frequency analysis

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Ethan G
Ethan G on 28 Apr 2022
Answered: Mathieu NOE on 28 Apr 2022
I'm currently writing a function for signal analysis at a very intro-level. As one of its functions, this program is supposed to differenciate between a square voltage vs time wave with constant frequency and one with significantly varying frequency (an "abnormal" signal).
This abnormal signal in the case that im trying to solve typically either gains or loses frequency and stays that was for a long time.
How would I reasonably differenciate between the two in a way where I could, for instance, assign:
is_normal = 1; if the frequency is constant,
is_normal = 0; if the frequency is varying over time.
Thanks to any that answer! I can give a little more info if requested, and I'm not working with much knowledge of ffts/power series, so if those are used in your answer, I'd really appreciate a breakdown of what's actually happening. Cheers!

Accepted Answer

Mathieu NOE
Mathieu NOE on 28 Apr 2022
see my little demo below - no need for fft , I simply compute the cycle by cycle period (by crossing detection) on two signals , one with constant frequency , the second is a sweeping sine.
Remains to implement a criteria for "normal" frequency range and fluctuation (the steady frequency signal will have a very low period flectuation so this should be fairly easy to compare to the non stationnary signal)
hope it helps
% full width half max demo
% dummy data
x= 10*(0:n-1)/n;
y1 = sin(6*x -0.5);
y2 = sin(x.^2 -0.5);
threshold = max(y1)/2; % half amplitude
[t0_pos1,s0_pos1,t0_neg1,s0_neg1]= crossing_V7(y1,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"
[t0_pos2,s0_pos2,t0_neg2,s0_neg2]= crossing_V7(y2,x,threshold,'linear'); % positive (pos) and negative (neg) slope crossing points
% periods
d1 = diff(t0_pos1)
period1 = [d1(1) d1];
d2 = diff(t0_pos2)
period2 = [d2(1) d2];
subplot(3,1,1),plot(x,y1,'b',t0_pos1,s0_pos1,'dr',t0_neg1,s0_neg1,'dg','linewidth',2,'markersize',12);grid on
legend('signal','signal positive slope crossing points','signal negative slope crossing points');
subplot(3,1,2),plot(x,y2,'b',t0_pos2,s0_pos2,'dr',t0_neg2,s0_neg2,'dg','linewidth',2,'markersize',12);grid on
legend('signal','signal positive slope crossing points','signal negative slope crossing points');
subplot(3,1,3),plot(t0_pos1,period1,t0_pos2,period2,'linewidth',2,'markersize',12);grid on
legend('signal 1 period','signal 2 period');
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 = [];

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