HEART RATE from heart sound

Question

Monday David 2024-2-28

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2088201-heart-rate-from-heart-sound

评论： William Rose 2024-3-4

Screenshot 2024-02-28 124958.png

Hi, Please I need help to remove errors on peak detected, here is the code I used.

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Cris LaPierre 2024-2-28

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.

Find local maxima and minima in the live editor

Monday David 2024-2-29

移动：Star Strider 2024-2-29

adultCASE1.zip

Here is my audio file.

Kindly assist on it.

Thanks

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Answer 1

William Rose 2024-2-28

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2088201-heart-rate-from-heart-sound#answer_1418471

编辑：William Rose 2024-2-29

在 MATLAB Online 中打开

heartSoundsApexNormalBell10s.zip

@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

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William Rose 2024-2-29

@Monday David,

I think my script makes a few errors when applied to adultCASE1. In 12 out of 14 beats in this recording, the script identifies the second heart sound (S2). In the other two beats, it identifies the first heart sound (S1). In this subject, S2 is generally louder than S1. The reverse is true in the recording I uploaded ("normal" from Univ. of Michigan library). If you are just trying to estimate mean rate, and the script identifies one sound or the other, but not both, then it doesn't make much difference, but if you are trying to estimate HR variability (which I do not recommend with heart sounds), you'd get incorrect results, because you would conclude there some short beats and some long beats, even if that was not really true. The image below is an annortated version of the image inthe answer above. Most of the red asterisks are on the second sound (S2). The red circles show where the script identified S1 instead of S2; the green arrows shows the S2 that comes after the S1s that were identified.

Ideally, all the red asterisks would be on S1s (and they are, in the UMich recording), or they would all be on S2s.

I noticed the absence of an S1 before the third S2, when listening to adultCASE1. That missing S1 corresponds to the flat spot in the plot below, from 2.0 to 2.4 seconds. Maybe something was blocking the sound during this time.