# full width, half max

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##### 1 Comment

Voss
on 15 Apr 2022

Edited: Voss
on 15 Apr 2022

From the error message it is clear that x has only one element. From the code it is clear that you expect x to have at least as many elements as y1 has (which is evidently more than 1, given the error).

The first step to solving this problem would be for you to investigate why x doesn't have as many elements as you thought.

### Accepted Answer

Mathieu NOE
on 19 Apr 2022

hello

the attached code can work but is limted in terms of accuracy (will depend how fine is sampled the data)

I would suggest the code below that uses linear interpolation and therefore will give more accurate results

% full width half max demo

% dummy data

n=1000;

x= 2*(0:n-1)/n;

y1 = max(0,3.28e-3*sin(6*x -0.5));

threshold = (max(y1) - min(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"

figure(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');

% x difference between t0_neg and t0_pos = FWHM

FWHM = t0_neg1 - t0_pos1

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

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

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