Easy way of finding zero crossing of a function

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BlueBee77
BlueBee77 2016-2-8
评论： María del Rosario González 2024-9-2
I am trying to find zero-crossings of a function in Matlab and plot the points where zero-crossing occurs. However, i am not able to find an easy way. I tried http://terpconnect.umd.edu/~toh/spectrum/PeakFindingandMeasurement.htm
and Matlab fnzeros, but i can't understand the spmak and x,y used in these function. The function for which i want to find zero crossing is Euclidean distance function. I will really appreciate if someone can tell me an easy way of doing this.
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Walter Roberson
Walter Roberson 2016-2-9
I do not see the connection to the tags about image analysis that were provided?
BlueBee77
BlueBee77 2016-2-9
Its used in an image processing project. Finding the peak points in an eclosed boundary

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回答（8 个）

Star Strider
Star Strider 2016-2-8
编辑：Star Strider 2020-7-8
If your function is a vector of values, you can use this little function to approximate them:
zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0); % Returns Zero-Crossing Indices Of Argument Vector
It is also helpful if you want to use fzero or interp1 in a loop to get the exact values.
Note that it returns the indices of the zero-crossings, so you have to do the appropriate vector addressing to get the corresponding x and y coordinates of them.
EDIT — (7 Jul 2020 at 2:54)
Another way of defining ‘zci’ is:
zci = @(v) find(diff(sign(v)));
producing the same result.
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BlueBee77
BlueBee77 2016-2-9
Yes, for this function it works. But for my function it doesn't. I dont know why because the function has values, it shouldn't be empty. I am finding this zero-crossing so that i can look for how many peak points are there in the signal. Imagine you have an irregular enclosed boundary and you want to look for all the points(peaks). The paper say they found it by looking for zero-crossing. I would really appreciate if you can help me out. This really took alot of time and no results
Star Strider
Star Strider 2016-2-9
I have no idea what your signal is. It is possible that your signal has no zero crossings, and the paper is looking at the zero-crossings of something else (perhaps the derivative of the signal). If you want to find the peaks instead, and you have the Signal Processing Toolbox, see if the findpeaks function will work for you.
For example, in my illustration, if the original signal were a cosine, the zero-crossings of the sine curve (the negative derivative of the cosine signal) would be the peaks (and valleys) of the cosine signal.

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Isaac Yuki
Isaac Yuki 2021-11-25
I wrote following function based on the zci function of @Star Strider:
function [number_zeros,zero_crossings] = findzeros(array,samplerate)
%FINDZEROS finds zerocrossings
%Finds the zeros or the nearest values to zero in a function and gives back
%as result the number of zerocrossings and an array containing median of the
%array with the positions of the value that are zero or nearst to zero in
%a zero crossing area, so its the middlest value of the zero crossing area
z = find(diff(sign(array)));
a = 1;
b = 1;
for i=2:1:length(z)
if z(i) > z(i-1)+round((samplerate/10000))+1
a = 1;
if i == 2
zci(b,a) = z(i-1);
end
zci(end+1,a) = z(i);
b = b+1;
else
zci(b,a) = z(i);
a = a+1;
end
end
number_zeros = b; %output1
zci2 = [];
zb = [];
zc = [];
zero_crossings = [];
for b = 1:1:number_zeros
zci2 = zci(b,:);
for j=1:1:length(zci2)
if zci2(j) == 0 && j ~=1
break
end
zci3(b,j) = array(zci2(j));
end
zb = find(abs(zci3(b,:)) == min(abs(zci3(b,:))));
zb = zci2(zb);
if length(zb) <= 1
zero_crossings = [zero_crossings zb]; %output2
else
zero_crossings(end+1) = zb(floor(length(zb)/2)); %outpu2
end
end
end
xszm
xszm 2019-8-10
I think that you can interpolate your data. You can find my results as follow. Thanks for Star Strider, I found it for a long time.
NO interp
uninterp.png
interp
interp.png
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Walter Roberson
Walter Roberson 2020-7-15
w1 should not be 1:length(y1) . You should be using wl (lower-case-L not digit 1), and it should be the time vector corresponding to your input signal. If you know your sampling rate, Fs, then
if size(y1,1) == 1; y1 = y1.'); end %ensure columns of signal
c = size(y1,2);
wl = (0:size(y1,1)-1) / Fs;
Iwl2=360:0.001:740; % interpolating to 0.001-nm resolution
loc_frequ1=0;
for i=1:c
y = y1(:,i);
y2=interp1(wl,y,Iwl2); % interpolating to 0.001-nm resolution
zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0); % Returns Zero-Crossing Indices Of Argument Vector
zx = zci(y2); % Approximate Zero-Crossing Indices
loczeros1=round(Iwl2(zx));
locfrequ= ismember(Iwl2,loczeros1); % find same data from Iwl2
loc_frequ1=loc_frequ1+double(locfrequ);
end

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Nick Hunter
Nick Hunter 2020-4-12
编辑：Nick Hunter 2020-7-16
I have just worked out a quicker bug proof solution, I guess:
clear;
theta = [0:7:360*4,1440]; % Angle Vector (MUST BE a ROW VECTOR)
y = sind(theta); % Signal Vector (MUST BE a ROW VECTOR)
UpZCi = @(v) find(v(1:end-1) <= 0 & v(2:end) > 0); % Returns Up Zero-Crossing Indices
DownZCi = @(v) find(v(1:end-1) >= 0 & v(2:end) < 0); % Returns Down Zero-Crossing Indices
ZeroX = @(x0,y0,x1,y1) x0 - (y0.*(x0 - x1))./(y0 - y1); % Interpolated x value for Zero-Crossing
ZXi = sort([UpZCi(y),DownZCi(y)]);
ZX = ZeroX(theta(ZXi),y(ZXi),theta(ZXi+1),y(ZXi+1));
% === Checking for zero at the ignored value ===
if y(end)==0
ZX(end+1) = theta(end);
end
% ==============================================
figure(1)
plot(theta, y, '-b')
hold on;
plot(ZX,zeros(1,length(ZX)),'ro')
grid on;
legend('Signal', 'Interpolated Zero-Crossing')
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Walter Roberson
Walter Roberson 2020-7-16
Nick:
y = [1 0 1 0 1 0 -1].';
UpZCi = @(v) find(v(1:end-1) <= 0 & v(2:end) > 0); % Returns Up Zero-Crossing Indices
DownZCi = @(v) find(v(1:end-1) >= 0 & v(2:end) < 0); % Returns Down Zero-Crossing Indices
UpZCi(y),DownZCi(y)
ans =
2
4
ans =
6
However, you cannot [] together [2;4] and [6]
Your code is assuming that the result of UpZCi and DownZCi are row vectors, but that will not be the case if y is a column vector. You did not document an orientation requirement.
Nick Hunter
Nick Hunter 2020-7-16
Thank you so much, Walter. This is a good point. I added this condition in comments.

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Mitch Lautigar
Mitch Lautigar 2020-7-7
编辑：Walter Roberson 2020-7-7
There are many ways to try and skin this problem, many people have tried and find varying different levels of success. Here's a function you can use that requires a sinusoidal waveform to approximate 0 intersections.
[out_array] = signal_manip(s_in)
sign_array = []; %predeclaration
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Following for loop checks every spot to create the following numbers:
%For positive numbers, a "1" is placed in an array.
%For negative numbers, a "-1" is placed in an array.
%For a zero, a "0" is placed in an array.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
for i = 1:length(s_in)
if s_in(1,i) > 0
curr_sign = 1;
elseif s_in(1,i) < 0
curr_sign = -1;
else
curr_sign = 0;
end %end "if s_in > 0"
sign_array = [sign_array,curr_sign]; %gives an output array that shows you all negative and positive numbers
end %end for i = 1:length(s_in)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%following for loop looks for the change points and handles the spots where a 0 occurs.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
curr = sign_array(1,1); %Starting comparison Point
out_array = [] %Predeclaration of array
for i = 2:length(s_in)
if ( sign_array(1,i) ~= curr) %if number change detected, mark it.
out_array = [out_array,i];
elseif (sign_array(1,i) == 0) %if we have a zero on the graph, mark it and skip i forward by 1 so as to not have a false hit.
out_array = [out_array,i];
i = i + 1;
end
curr = sign_array(1,i);
end
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Walter Roberson
Walter Roberson 2020-7-15
The person who posted it left out the function keyword accidentally, so it is not surprising you missed it!
In order to get the error message you are seeing when you are passing in a vector of values, then you must have passed in a column vector of values -- but as I was careful to point out before, it expects a row vector of values.
Mohamed Jamal
Mohamed Jamal 2020-7-15
Yes it works fine now but this function is returnning array of zero crossings points right? because Im not getting my signal plot with zero crossings points marked in .. not getting a plot at all , just array!

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UMAIR RASOOL
UMAIR RASOOL 2020-8-13
close all
clear all
clc
n=-10:10;
a=.9;
y=zeros(size(n));
for k=-5:5
temp=find(n-(2*k)==0); %find xaxis where it is zero(zero crossing)
y(temp)=1; %palce the value at the zero crossing
x=n.*(sum(a^k)*y);
end
plot(n,x)
xlabel('n')
ylabel('x(n)')
Maximum=max(x);
Mitch Lautigar
Mitch Lautigar 2020-7-27
So I took the code previously used and modified it to what I needed (finding zero point crossing for a sin/cos wave. The code is below
function [freq_val] = peakfind(dm_allow)
%Step 1: Load file and find zero crossings
load('s_m.mat','s_m2','t')
zci = @(v) find(v(:) .* circshift(v(:), [-1 0]) <= 0);
zx = zci(s_m2)';
rp_zx = [];
%Step 2: manipulate above values so as to find unique values.
zx2 = round(zx ./ max(zx),2);
for i = 1:length(zx2)-1
if round(zx2(1,i),3) ~= round(zx2(1,i+1),3)
rp_zx = [rp_zx,zx2(1,i+1)];
end
end
end_array = [1];
%Eliminate redundant numbers
for i = 1:length(rp_zx)
zeta = find(zx2 == rp_zx(1,i));
end_array = [end_array,zeta(1,1)];
end
actual_spots = zx(end_array);
t_val = [];
%convert above numbers to to time values.
for i = 1:length(end_array)-1
low_spot = actual_spots(1,i);
high_spot = actual_spots(1,i+1);
max_spot = max(s_m2(low_spot:high_spot));
min_spot = abs(min(s_m2(low_spot:high_spot)));
if max_spot > min_spot
beta = find(s_m2(low_spot:high_spot) == max_spot);
t_val = [t_val,t(beta+low_spot)];
elseif max_spot < min_spot
%do nothing
end
end
%convert time values to frequency values.
if length(t_val) > 1
freq_val = [];
for i = 1:length(t_val)-1
freq_val = [freq_val,1/(t_val(i+1)-t_val(i))];
end
else
error('Cycle time not long enough')
end
end
If you use my code, simply change the following:
zx = zci(s_m2)' %replace s_m2 with whatever signal you are wanting to find zero crossings for.
  1 个评论
Mathieu NOE
Mathieu NOE 2020-11-5
dear all
FYI - this is the function I use since long time - have always worked perfect for all type of signals
function [ind,t0,s0] = crossing_V6(S,t,level,imeth)
% [ind,t0,s0,t0close,s0close] = crossing_V6(S,t,level,imeth)
% 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).
%
% [ind,t0,s0] = ... also returns the data vector corresponding to
% the t0 values.
%
% [ind,t0,s0,t0close,s0close] additionally returns the data points
% closest to a zero crossing in the arrays t0close and s0close.
%
% This version has been revised incorporating the good and valuable
% bugfixes given by users on Matlabcentral. Special thanks to
% Howard Fishman, Christian Rothleitner, Jonathan Kellogg, and
% Zach Lewis for their input.
% Steffen Brueckner, 2002-09-25
% Steffen Brueckner, 2007-08-27 revised version
% Copyright (c) Steffen Brueckner, 2002-2007
% brueckner@sbrs.net
% 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 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;
terme = S(ind(ii)) * slope;
t0(ii) = t0(ii) - terme;
% I'm a bad person, so I simply set the value to zero
% instead of calculating the perfect number ;)
s0(ii) = 0;
end
end
end
% % Addition:
% % Some people like to get the data points closest to the zero crossing,
% % so we return these as well
% [CC,II] = min(abs([S(ind-1) ; S(ind) ; S(ind+1)]),[],1);
% ind2 = ind + (II-2); %update indices
%
% t0close = t(ind2);
% s0close = S(ind2);

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Adam Cook
Adam Cook 2024-7-18
Please see the zerocrossrate function from Signal Processing Toolbox. It has a special indices output that you can use to identify the specific indices where zero crossings occur:
https://www.mathworks.com/help/releases/R2024a/signal/ref/zerocrossrate.html?s_tid=doc_srchtitle#mw_62e2c7a7-b2f5-4768-9b4a-9d68e9e7968d
Note that this will just give discrete time indices. Interpolation would be required to find the continuous time indices.

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