How to use findpeaks to find the width of the peak at 1/e^2.

I have a data and I want to find the peak value at 1/e^2. What I use is the following code.

x=field5(:,1);
y=field5(:,2);
e = exp(1)
[pks,locs,fwhm,proms] = findpeaks(y, x);
[~,i] = sort(pks,'descend');
peak_1 = pks(i);

esquared = max(peak_1)*1/e^2;  
z = abs(y-esquared);
[p,loc] = findpeaks(-z);
plot(x,y,'b',x(loc),y(loc),'+r');
angle = diff(x(loc))/2

What I want to get is the width of the peak at the value of 1/e^2 (like FWHM but lower), but I get wrong results. The two points are not at the correct locations expecting them to be infront of each other.

[dpb]

I use findpeaks only to locate the actual peak location(s) of interest and then interp1 to locate the intersection with the desired level if want precise intersection(*).

It would be a useful enhancement if the ' WidthReference' named parameter would take an absolute level or percentage of the peak height besides the two named references. I've always wondered why TMW didn't do that from the git-go.

(*) I know I had a utility function coded in prior occupation/when still had a day job, but I didn't seem to get it off the corporate machine along with a fair amount of other things I've occasionally missed since. But, it's not difficult to implement.

Answer 1

[Mathieu NOE]

hello

this would be my suggestion

this code will find the peak value then from there you can tell at which y threshold value you want to search for crossing points

then x values difference will give you the peak width

clc
clearvars

% dummy data
n=100;
x=(0:n-1)/n;
y = 0.75*sin(5*x -0.5);

[max_val,locs] = max(y); % find the peak and its x axis location
threshold = max_val/2; % your value here

[t0_pos1,s0_pos1,t0_neg1,s0_neg1]= 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"

figure(1)
plot(x,y,'b',x(locs),max_val,'dr',t0_pos1,s0_pos1,'db',t0_neg1,s0_neg1,'dg','linewidth',2,'markersize',12);grid on
legend('signal 1','signal 1 peak value','signal 1 positive slope crossing points','signal 1 negative slope crossing points'  );

% time difference between crossing points
time_width = - t0_pos1 + t0_neg1

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

[Vahram Voskerchyan]

thanks for the answer it worked. Now I have to figure out what's the code.

[Mathieu NOE]

My pleasure

if you have questions don't hesitate

all the best