Issue with locating end of poorly defined peak

1 Ansicht (letzte 30 Tage)
Teshan Rezel
Teshan Rezel am 19 Jun. 2022
Kommentiert: Mathieu NOE am 22 Jun. 2022
Hi folks,
I have several graphs that look similar to the one attached. I am trying to formalise a general approach to finding the location of the start and end of each peak. The start is somewhat easy (using find, and the peak position can be found using max), since the graph starts from 0 (I am interested only in the second entry onwards).
However, the end is really tricky as it never quite gets to zero. I have tried using findpeaks, and also myEnd = find(iso(myPeakLoc:end, 1) > 1e-06); but to no avail.
Can someone advise me on an alternative route please?
  1 Kommentar
Mathieu NOE
Mathieu NOE am 20 Jun. 2022
hello
can you share your code as well (not only the data) ?
tx

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Mathieu NOE
Mathieu NOE am 20 Jun. 2022
hello again
tried this - does it help ?
load('iso.mat')
curve = 10; % which curve do you want to plot ?
data = percent_iso(2:end,curve);
% max point
[maxval,maxind] = max(data);
% start point = find first point above 0.1% of max value
[startind] = find(data>0.001*maxval,1,'first');
startval = data(startind);
% stop point= find first point below 0.1% of max value past the peak
[endind] = find(data(maxind:end)<0.001*maxval,1,'first');
endind = endind + maxind;
endval = data(endind);
plot(data,'DisplayName','percent_iso')
hold on
plot(maxind,maxval,'dr',startind,startval,'dg',endind,endval,'dk');
hold off
legend('data','max','start','end');
  2 Kommentare
Teshan Rezel
Teshan Rezel am 21 Jun. 2022
@Mathieu NOE this is perfect, thank you!
Mathieu NOE
Mathieu NOE am 22 Jun. 2022
as always, my pleasure !

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Produkte


Version

R2022a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by