Calculating BPM - from PPG signal

2014 3 15
2 个回答
回答已采纳
更新时间：2021 3 19
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0 个投票

Hi everyone,
I've been able to create a script that reads heart rate - displaying the value in real time on a graph.
Now i need to calculate the number of peaks in the graph and work out the BPM
Any simple script solution?
Any help will be most appreciated! Thanks!

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Ahmet YALCIN
Ahmet YALCIN 2014-3-25
please share if you got a solution to this problem. thx
willing2learn
willing2learn 2014-3-25
Hey, use try this code out:
ydpos = diff(ydata>1000)>0; % these data points are shown in the
% attached figure below
dxmax = max(xdata(ydpos==1));
dxmin = min(xdata(ydpos==1));
dx = (dxmax - dxmin)/(sum(ydpos)-1)
dx =
written by Mischa Kim
Ahmet YALCIN
Ahmet YALCIN 2014-3-31
Thank you Mischa,
I will try your suggestion… I developed the following algorithm (similar to yours) to find the peaks, and it seems also to work:
- Smooth by using 1D Gaussian filter and Convolution g = gausswin (5); g = g / sum(g); and by convolution with 'conv' --> conv_sig = conv(sig, g, 'same');
- Calculate and plot the 1st-derivative of the convolved signal
- Find the peaks of 1st-derivative signal [pks_1stderiv, loc_1stderiv_peaks]= findpeaks(conv_sig(d), 'MINPEAKHEIGHT',1,'MINPEAKDISTANCE' ,18)
Federation
Federation 2014-5-6
Hi everyone, I need to calculate Heat Rate variability (HRV) by using Frequency domain measurement and find very low frequency(VL), low frequency(LF) , high frequency(HF) and total power. For example If X is RR interval X=[0.456, 0.567, 0.457, ..........,0.770] where length X = 600 and in second. How can I find (VL for X from 0.0-0.04), (LF for X from 0.04 - 0.15),(HF for X from 0.15 - 0.40) and total power in millisecond square.
sarang bagade
sarang bagade 2014-5-6
please tell me how you displayed the value in real time on a graph
Supriya V
Supriya V 2021-3-19
Sir/Madam, could you share source code for ppg extraction from face video that is real time and along with I want to display a graph for that.

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 采纳的回答

Mischa Kim
Mischa Kim 2014-3-15
编辑：Mischa Kim 2014-3-15

1 个投票

Check out the findpeaks command. Alternatively (if you do not have access to the Signal Processing Toolbox) you could do a Fourier Transform and look at the dominant frequency components.

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willing2learn
willing2learn 2014-3-15
Hey, thanks for the fast response!
I'm not great with matlab
Using the find peaks, how am i to calculate the frequency?
Using the fft function, i have got the attached figure - HR is less than 10Hz which is reasonable.
But i need to detect and display actual pulse count - e.g '70 beats over a 60second sample'
Thanks
Mischa Kim
Mischa Kim 2014-3-15
I think you forgot to attach. If you have frequency in Hz (beats per sec) then you just need to multiply by 60 to get beats per minute. Does that help?
willing2learn
willing2learn 2014-3-15
Hi, yes that helps
my problem is that i want to collect the data (several figures) and have a script that analyses the peaks for each figure. Then give an actual value. With my fft graph how will i do it? Or with the peakfind?
Thanks
Mischa Kim
Mischa Kim 2014-3-17
编辑：Mischa Kim 2014-3-17
Looking at the first figure, what resembles one beat? Signal values greater than 1000, 500? Does the peak lasting from x = 56 to x = 71 count as one single peak?
willing2learn
willing2learn 2014-3-17
Hi, yes signal values > 1000 count as one peak - and x=56 - x=71 is a single pulse / peak.
The square wave is due to the saturation of my current circuit. When i turn down the gain the graph should look close to an ecg curve.
Thanks
Mischa Kim
Mischa Kim 2014-3-17
编辑：Mischa Kim 2014-3-17
OK. In this case you could simply use
ydpos = diff(ydata>1000)>0; % these data points are shown in the
% attached figure below
dxmax = max(xdata(ydpos==1));
dxmin = min(xdata(ydpos==1));
dx = (dxmax - dxmin)/(sum(ydpos)-1)
dx =
37.500000000000000
xdata and ydata are the data from your HR1.fig, dx is the average duration between individual beats. I don't know what your units are on the x-axis, so you need to figure out how to convert dx into beats per minute.
Does this get you what you need and answer your question?
willing2learn
willing2learn 2014-3-17
Cool! This is almost what i want, just that i don't know how convert the x-axis into time, it is currently number of samples i think - over a 10second period. Below is my code so far:
% HR_real_time
clear all;
a = arduino('COM24')
tic
toc
time_start = tic; %start taking samples
time_stop = toc; %stop taking samples
sample_time = 60; %1k samples
xstop = 1000;
x = [0,0];
while(toc < sample_time)
b = a.analogRead(0); %storing the analog values in b
x = [x,b]; %x as a matrix?, updating x - causing the shift
figure (1)
plot(x);
axis([0,xstop,0,1500]);
grid
%t = tic; % increment t by step size
drawnow;
%pks = findpeaks(x)
end
xfft = fft(x);
figure (2)
freq=0:(1000)/length(x):(1000)/2;
plot(freq,abs(xfft(1:length(x)/2+1)))
% code
end
Thanks
willing2learn
willing2learn 2014-3-19
Hi, i'm still unable to change the x-axis to time and use the code you provided. Could you further assist please?
Thanks
willing2learn
willing2learn 2014-3-19
Hi Mischa, the time axis has been calculated and your solution added to the code. I am very close! The only issue is there seems to be a loss of counts somewhere - e.g when my heart rate is around 63BPM, Matlab returns around 55...
Could you please look at the updated code to see if i have missed something or misunderstood?
Thanks a lot
% HR_real_time
clear all;
a = arduino('COM24')
tic
toc
time_start = tic; %start taking samples
time_stop = toc; %stop taking samples
sample_time = 20; %1k samples
xstop = 20;
x = [];
time_ind=[];%zeros(600,1);
i=1;
BPM = 0;
while(toc < sample_time)
b = a.analogRead(0); %storing the analog values in b
time_ind=[time_ind, toc];%New array that stores time index
i=i+1; %everytime loop runs, increment
x = [x,b]; %x as a matrix?, updating x - causing the shift
figure (1)
plot(time_ind,x);%Plot the time index
axis([0,xstop,0,1500]);
ylabel('Magnitude');
xlabel('Time');
title('Heart Beat in Time Domain');
grid
drawnow;
end
Plots with finished loop
xfft = fft(x);
figure (2)
freq=0:(1000)/length(x):(1000)/2;
plot(freq,abs(xfft(1:length(x)/2+1)));
ylabel('Magnitude');
xlabel('Frequency');
title('Heart Beat in Frequency Domain');
figure(3)
plot(time_ind);
ylabel('Magnitude');
xlabel('Samples');
title('Heart Beat');
%%Calculating Frequency
ydpos = diff(x>1000)>0;
%at pulse instant, current value is larger than previous therfore diff>0
%at every other point, diff<0 or diff=0;
dxmax = max(time_ind(ydpos==1));% dxmax assigned max time value when ydpos>0
dxmin = min(time_ind(ydpos==1));% dxmin assigned min time value when ydpos>0
dx = (dxmax - dxmin)/(sum(ydpos)-1); % calculate frequency in 1 second???
BPM = (dx*60) % Get BPM in a minute
Mischa Kim
Mischa Kim 2014-3-20
Hard to say. In HR1 update.fig I am counting 21 beats in less than 19.5 sec
21/(19.89-0.57)*60
ans =
65.217391304347828
Is this about what you'd expect?
willing2learn
willing2learn 2014-3-20
Hi, yes that's what i expected around 63 BPM, but i ended up with around 55. I just ran another test getting 25 pulses, but the code only returns 48 BPM - should be 75 (25 beats in 20 seconds).
There must be a small mistake somewhere
willing2learn
willing2learn 2014-3-21
Hi Mischa thanks a lot for your help so far.
I have increased the sampling period to 60 seconds, ran several tests and the code is 100% accurate to within 10 counts (when the pulses are manually counted).
Most times it is accurate to with 5 beats.
Any idea how to fine tune it a bit more?
Thanks
Supriya V
Supriya V 2021-3-19
Sir/Madam, could you share source code for ppg extraction from face video that is real time and along with I want to display a graph for that.

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更多回答（1 个）

T. Thinh Nguyen
T. Thinh Nguyen 2015-10-5

0 个投票

Check out the code I write in here.
https://uk.mathworks.com/matlabcentral/fileexchange/53364-heart-rate--spo2-using-ppg
The heart rate estimation is done both by FFT or Peak detection.

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Supriya V
Supriya V 2021-3-19
Sir/Madam, could you share source code for ppg extraction from face video that is real time and along with I want to display a graph for that.

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