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HEART RATE from heart sound

20 vues (au cours des 30 derniers jours)
Monday David
Monday David le 28 Fév 2024
Commenté : William Rose le 4 Mar 2024
Hi, Please I need help to remove errors on peak detected, here is the code I used.
  2 commentaires
Cris LaPierre
Cris LaPierre le 28 Fév 2024
Try using the Find Local Extrema task in a live script. This will allow you to interactively adjust the inputs to findpeaks, which can make it easier to identify the correct settings.
Monday David
Monday David le 29 Fév 2024
Déplacé(e) : Star Strider le 29 Fév 2024
Here is my audio file.
Kindly assist on it.
Thanks

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William Rose
William Rose le 28 Fév 2024
Modifié(e) : William Rose le 29 Fév 2024
@Monday David, If you attach the MP3 file, others can run your code. You will have to zip it first.
[Edit: The sounds which are labelled S1 in the plot below are really S2, and vice versa. This does not affect the accuracy of the estimate of heart rate.]
I would rectify the signal with abs() instead of setting negaitve values to zero. But it probably won't make a huge difference. I wonder what Fs is, because if I knew, I would undertand the plot in the image better. I cannot tell how many seconds of data I am looking at.
The plot in your image has peaks with irregular spaciong and amplitude. It certainly does not look like normal heart sounds.
Here is a 10 second recording of normal heart sounds, from the University of Michigan heart sounds library.
unzip('heartSoundsApexNormalBell10s.zip');
[x,Fs]=audioread('heartSoundsApexNormalBell10s.mp3');
t=(0:length(x)-1)'/Fs; % time vector
envDur=0.02; % envelope duration (s)
[xUpper,~]=envelope(abs(x),envDur*Fs,'rms');
[pks,locs]=findpeaks(xUpper,Fs,MinPeakDistance=0.3,MinPeakHeight=0.1);
s1pks=pks(pks>0.4); s1locs=locs(pks>0.4);
s2pks=pks(pks<=0.4); s2locs=locs(pks<=0.4);
plot(t,xUpper,'-b',s1locs,s1pks,'r*',s2locs,s2pks,'gx');
grid on; xlabel('Time (s)'); legend('amplitude','S1','S2','Location','southeast')
Estimate the HR from the S1 locations:
HR=(length(s1locs)-1)/(s1locs(end)-s1locs(1));
fprintf('Heart rate=%.1f bpm.\n',HR*60)
Heart rate=74.7 bpm.
OK
  11 commentaires
Monday David
Monday David le 4 Mar 2024
Okay. Thanks I appreciate
William Rose
William Rose le 4 Mar 2024
@Monday David, you're welcome. Good luck with your HR analysis.

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