How to chop up/segment a periodic signal?

Question

Susan 2022-9-7

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1799255-how-to-chop-up-segment-a-periodic-signal

评论： Star Strider 2022-11-29

ECG.mat

I have a normal EKG signal (a periodic signal), and I 'd like to automatically chop up the signal into individual cycles (containing the P, QRS, and T waves). Could somebody please tell me how I can do that? I don't know the length of each cycle; should I calculate these precisely? Getting an approximate cycle length using autocorrelation would be good enough? The .mat file is attached.

Many thanks in advance!

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Answer 1

Star Strider 2022-9-7

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1799255-how-to-chop-up-segment-a-periodic-signal#answer_1047010

在 MATLAB Online 中打开

This approach takes advantage of the fact that the Q-T interval in a normal EKG is less than one-half the previous R-R interval.

This will work for an EKG displaying regular sinus rhythm, however it might not work for atrial fibrillation (that would not then have an indentifable P-wave anyway).

LD = load(websave('ECG','https://www.mathworks.com/matlabcentral/answers/uploaded_files/1119640/ECG.mat'))

LD = struct with fields:

ECG: [524798×1 double]

ECG = LD.ECG;

Fs = 1024;

L = numel(ECG);

t = linspace(0, L-1, L)/Fs;

[pks,locs] = findpeaks(ECG, 'MinPeakProminence',0.5)

pks = 684×1

2.5451 2.1133 1.7806 1.5151 1.2995 1.1442 1.0095 0.9103 0.8269 0.7568

locs = 684×1

251 1019 1787 2554 3323 4091 4858 5627 6394 7162

figure

plot(t, ECG)

hold on

plot(t(locs), pks, '^r')

hold off

grid

xlim([20 22.5])

for k = 1:numel(pks)-2

RR = (locs(k+1) - locs(k));

idxrng = locs(k+1) + (-fix(RR/2) : fix(RR/2));

ECGp{k} = ECG(idxrng);

end

figure

subplot(3,1,1)

plot(ECGp{20})

grid

title('Complex 20')

subplot(3,1,2)

plot(ECGp{100})

grid

title('Complex 100')

subplot(3,1,3)

plot(ECGp{200})

grid

title('Complex 200')

You could also use these to generate an ensemble average.

.

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Star Strider 2022-9-7

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As always, my pleasure!

Sure!

LD = load(websave('ECG','https://www.mathworks.com/matlabcentral/answers/uploaded_files/1119640/ECG.mat'))

LD = struct with fields:

ECG: [524798×1 double]

ECG = LD.ECG;

Fs = 1024;

L = numel(ECG);

t = linspace(0, L-1, L)/Fs;

[pks,locs] = findpeaks(ECG, 'MinPeakProminence',0.5);

figure

plot(t, ECG)

hold on

plot(t(locs), pks, '^r')

hold off

grid

xlim([20 22.5])

for k = 1:3%numel(pks)-2

RR = (locs(k+1) - locs(k));

idxrng = locs(k+1) + (-fix(RR/2) : fix(RR/2));

ECGp{k} = ECG(idxrng);

v = zeros(1,1000);

EKGe(k,500+(-fix(RR/2) : fix(RR/2))) = ECG(idxrng);

end

EKGensavg = mean(EKGe); % Trim Zeros & End Transients

EKGensavgv = EKGensavg~=0;

EKGensavg = EKGensavg(EKGensavgv);

EKGensavg = EKGensavg(2:end-1);

EKGensstd = std(EKGe); % Trim Zeros & End Transients

EKGensstdv = EKGensstd~=0;

EKGensstd = EKGensstd(EKGensstdv);

EKGensstd = EKGensstd(2:end-1);

figure

plot(EKGensavg, '-b', 'DisplayName','EKG Ensemble Average')

hold on

plot(EKGensavg+EKGensstd, '--r', 'DisplayName','+1 Standard Deviation')

plot(EKGensavg-EKGensstd, '--r', 'DisplayName','- 1 Standard Deviation')

hold off

legend('Location','best')

The slope derives from the initial transient in the original signal that is included in the segments. Choose a subset of the record that does not include that transient to get an ensemble average without it. That edited record then becomes the ‘ECG’ vector. The rest of the code should work without requiring any changes.

.

Star Strider 2022-9-8

在 MATLAB Online 中打开

I am not certain how best to compare two different EKG cycles. From a statistical perspective, the Wilcoxon signed rank test (the signrank function) may be most appropriate, since they could be considered ‘paired’ data. Otherwise, doing a linear regression of one against the other would be an option.

Exploring those —

LD = load(websave('ECG','https://www.mathworks.com/matlabcentral/answers/uploaded_files/1119640/ECG.mat'))

LD = struct with fields:

ECG: [524798×1 double]

ECG = LD.ECG;

Fs = 1024;

L = numel(ECG);

t = linspace(0, L-1, L)/Fs;

[pks,locs] = findpeaks(ECG, 'MinPeakProminence',0.5);

figure

plot(t, ECG)

hold on

plot(t(locs), pks, '^r')

hold off

grid

xlim([20 22.5])

for k = 1:numel(pks)-2

RR = (locs(k+1) - locs(k));

idxrng = locs(k+1) + (-fix(RR/2) : fix(RR/2));

ECGp{k} = ECG(idxrng);

end

EKG020 = ECGp{20};

EKG500 = ECGp{500};

[p,h,stats] = signrank(EKG020, EKG500)

p = 2.4750e-127

h = logical

1

stats = struct with fields:

zval: 24.0048 signedrank: 2.9595e+05

mdl = fitlm(EKG020, EKG500)

mdl =

Linear regression model: y ~ 1 + x1 Estimated Coefficients: Estimate SE tStat pValue _________ __________ _______ ______ (Intercept) -0.018932 0.00019501 -97.083 0 x1 1.0032 0.0023175 432.89 0 Number of observations: 769, Error degrees of freedom: 767 Root Mean Squared Error: 0.00528 R-squared: 0.996, Adjusted R-Squared: 0.996 F-statistic vs. constant model: 1.87e+05, p-value = 0

figure

histfit(EKG020, 100)

title('Distribution Fit To ‘EKG020’')

figure

EKGdiff = EKG020-EKG500;

histfit(EKGdiff, 100)

title('Normal Distribution Fit To ‘EKG020’-‘EKG500’')

figure

EKGdiff = EKG020-EKG500;

histfit(EKGdiff(EKGdiff>0), 100, 'lognormal')

title('Lognormal Distribution Fit To ‘EKG020’-‘EKG500’')

% figure

% subplot(3,1,1)

% plot(ECGp{20})

% grid

% title('Complex 20')

% subplot(3,1,2)

% plot(ECGp{100})

% grid

% title('Complex 100')

% subplot(3,1,3)

% plot(ECGp{200})

% grid

% title('Complex 200')

So by both tests, they are not statistically significantly different.

It is important to consider how the data are distributed because statistical tests rest on certain assumptions about the data. The distrbution plots demonstrate that a single EKG segment is not normally distributed, however the difference between the two (using only positive values) comes closer to being lognormally-distributed than normally-distributed, at least by inspection. (Most physiological variables are lognormally distributed, at lelast in part because they can never be negative and in most instances are always greater than zero, the reason I chose this option.) There might be more appropriate distibutions, however the distribution needs to make sense in the context of the data, so for example fitting it to a Poisson distribution would not be appropriate, since these data are not arrival times.

I am not certain how else to compare them that would return a meaningful measure of their differences.

(My apologies for the delay — supper time and late-day news time intervened.)

.

Star Strider 2022-9-8

Thank you!

‘Do you think for the data set I have, I should choose a subset of the record that does not include that transient (ignore the transient part)? Or detrend the signal before any processing to get a signal without it? Or remove the baseline wander before any processing?’

Either would work. The detrending approach would lilely involve fitting the curve to an exponential function and then subtracting it. An alternative might be to use the detrend function with a relatively high degree polynomial. (I was not able to get a good result with a filtering approach, so ploynomial detrending might be the most appropriate.)

‘If we have two ECG cycles, does it make sense to consider one as a base and compare the second one with that? For example, the first cycle (the base one) is collected in a hospital, and the second one is in the street; I like to know how much the environment affects the ECG signal.’

That’s an interesting problem! The hospital (with few interfering distractions of any sort) could be considered the ‘baseline’ and the ambulatory EKG outside the hospital might be considered data to be compared to it. The rate and rhythm (specifically rate variation) might be the only significant variations.

However to do any specific comparison of individual segments, I would use the ensemble average as the baseline for any single segment, and compare the others to it. (If there was any length discrepancy, the longer segment would temoprarily have to be reduced to the length of the shorter segment in order to do the comparison.)

.

Star Strider 2022-9-20

编辑：Star Strider 2022-9-20

在 MATLAB Online 中打开

This is an interesting problem!

I decided to compare them by doing a Lead II ensemble average on each situation (‘Env1’, ‘Env2’, ‘Hospital’) and then use a statistical test (friedman) to see if there were any differences between them. There are differences by this test, and appear to be highly significant —

LD1 = load(websave('Env2','https://www.mathworks.com/matlabcentral/answers/uploaded_files/1129730/Env1.mat'))

LD1 = struct with fields:

Env1: [87297×8 table]

LD2 = load(websave('Env2','https://www.mathworks.com/matlabcentral/answers/uploaded_files/1129735/Env2.mat'))

LD2 = struct with fields:

Env2: [92161×8 table]

LD3 = load(websave('Hospital','https://www.mathworks.com/matlabcentral/answers/uploaded_files/1129740/Hospital.mat'))

LD3 = struct with fields:

Hospital: [30721×8 table]

Env1 = LD1.Env1;

Env2 = LD2.Env2;

Hosp = LD3.Hospital;

Fs = 1024; % Guess, Need Actual Value

L1 = size(Env1,1);

t1 = linspace(0, L1-1, L1)/Fs;

L2 = size(Env2,1);

t2 = linspace(0, L2-1, L2)/Fs;

L3 = size(Hosp,1);

t3 = linspace(0, L3-1, L3)/Fs;

ttlc = {'Env1','Env2','Hospital'};

EKGc = {Env1.II; Env2.II; Hosp.II}; % Create Array To Make Processing Easier

figure

subplot(3,1,1)

plot(t1, Env1.II)

grid

title('Env1')

xlim([0 10])

ylim([-0.2 0.6])

subplot(3,1,2)

plot(t2, Env2.II)

grid

title('Env2')

xlim([0 10])

ylim([-0.2 0.6])

subplot(3,1,3)

plot(t3, Hosp.II)

grid

title('Hospital')

xlim([0 10])

ylim([-0.2 0.6])

for k1 = 1:size(EKGc,1)

k1

ECG = EKGc{k1};

% Fs = 1024;

L = numel(ECG);

t = linspace(0, L-1, L)/Fs;

[pks,locs] = findpeaks(ECG, 'MinPeakProminence',0.5);

figure

plot(t, ECG)

hold on

plot(t(locs), pks, '^r')

hold off

grid

title(ttlc{k1})

xlim([20 22.5])

ylim([-0.2 0.6])

for k = 1:numel(pks)-2

RR = (locs(k+1) - locs(k));

idxrng = locs(k+1) + (-fix(RR/2) : fix(RR/2));

ECGp{k1}{k} = ECG(idxrng);

v = zeros(1,1000);

EKGe{k1}(k,500+(-fix(RR/2) : fix(RR/2))) = ECG(idxrng);

end

EKGensavg = mean(EKGe{k1}); % Trim Zeros & End Transients

EKGensavgv = EKGensavg~=0;

EKGensavg = EKGensavg(EKGensavgv);

EKGensavgc{k1,:} = EKGensavg(2:end-1);

end

k1 = 1

k1 = 2

k1 = 3

NrSp = size(EKGc,1);

figure

for k = 1:NrSp

subplot(NrSp,1,k)

plot(EKGensavgc{k})

grid

ylim([-0.2 0.6])

title(ttlc{k})

end

sgtitle('Ensemble AVerages')

MinSzc = min(cellfun(@(x)size(x,2),EKGensavgc)) % Minimum Length

MinSzc = 767

QF = factor(MinSzc)

QF = 1×2

13 59

EKGensavgct = cellfun(@(x)x(1:MinSzc),EKGensavgc,'Unif',0) % Trim All To Minimum Length

EKGensavgct = 3×1 cell array

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EKGensavgt = cell2mat(EKGensavgct).'; % Column Matrix For Statistics

[p,EnsTable,Stats] = friedman(EKGensavgt, QF(2))

p = 4.2669e-16

EnsTable = 5×6 cell array

{'Source' } {'SS' } {'df' } {'MS' } {'Chi-sq' } {'Prob>Chi-sq'} {'Columns' } {[1.8584e+05]} {[ 2]} {[9.2918e+04]} {[ 70.7810]} {[ 4.2669e-16]} {'Interaction'} {[5.9515e+04]} {[ 24]} {[2.4798e+03]} {0×0 double} {0×0 double } {'Error' } {[5.7618e+06]} {[2262]} {[2.5472e+03]} {0×0 double} {0×0 double } {'Total' } {[6.0071e+06]} {[2300]} {0×0 double } {0×0 double} {0×0 double }

Stats = struct with fields:

source: 'friedman' n: 13 meanranks: [96.2764 94.3859 76.3377] sigma: 51.2396

I encourage you to do other tests on them to see if you can find any differences that might be physiologically significant, as well as statistically significant. Note that the ‘meanrank’ is lower for ‘Hospital’ than for the others. That could be due to any residual noise in the first two, however the ensemble averaging should significantly reduce it. Explaining that might be worth pursuing. Also consider filtering all of them (the sgolayfilt function might be easiest and most robust, however frequency-selective filters might also work) to elimiinate as much noise as possible without compromising the EKG morphology itself, and then see if the statistical significant differences remain.

This code sets up an approach (using ensemble averages) to use for these and other records that may allow you to detect differences.

EDIT — (20 Sep 2022 at 2:46)

Minor corrections to the comments, code unchanged.

.

Star Strider 2022-11-29

在 MATLAB Online 中打开

The records do not appear to have any associated sampling frequencies or time vectors.

It would be helpful to know the sampling frequencies of each record.

Uz = unzip('https://www.mathworks.com/matlabcentral/answers/uploaded_files/1211718/EKG%20signals.zip');

for k = 1:3

EKG{k,:} = struct2table(load(Uz{k+1}));

end

EKG

EKG = 3×1 cell array

{61441×1 table} {61461×1 table} {63837×1 table}

figure

for k = 1:3

subplot(3,1,k)

plot(EKG{k}{:,1})

grid

xlim([0 3500])

end

[Sp1,idx1] = findpeaks(EKG{1}{:,1}, 'MinPeakHeight',0.65)

Sp1 = 120×1

0.7223 0.7235 0.7650 0.7234 0.7195 0.7564 0.7131 0.7115 0.7606 0.7111

idx1 = 120×1

271 782 1294 1807 2319 2831 3342 3855 4366 4878

didx1 = diff(idx1)

didx1 = 119×1

511 512 513 512 512 511 513 511 512 512

[Sp2,idx2] = findpeaks(EKG{2}{:,1}, 'MinPeakHeight',max(EKG{2}{:,1})*0.90)

Sp2 = 282×1

0.5632 0.5703 0.5764 0.5640 0.5557 0.5487 0.5526 0.5682 0.5688 0.5571

idx2 = 282×1

467 473 479 976 990 1494 1503 2013 2015 2512

didx2 = diff(idx2)

didx2 = 281×1

6 6 497 14 504 9 510 2 497 14

The records have some serious anomalies. What are the random spikes? They do not appear to be pacemaker spikes. They do not appear to have any specific characteristics, either within a specific record or between different records, such that it would be possible to eliminate them. In ‘EKG1’ they appear to have the advantage of having a greater amplitude than the R-deflections, so they can be relatively easily detected. Eliminating them then requires finding their beginning and end points and setting everything between those points to NaN and then using the fillmissing function to interpolate the missing values.

That is not the situation for ‘EKG2’ and ‘EKG3’ so I am not certain if it is possible to reliably detect those, or eliminate them. I am not aware of a specific signal processing approach that would detect them.

The one feature that distinguishes the spikes from the QRS complexes is that the QRS complexes all appear to have distinct Q and S defiections, so detecting those would be important in distinguishing the QRS deflections from the spikes. The problem arises if a spike obscures a Q or S deflection. In that situation that approach would not work for that particular QRS complex.

That is the best I can do in advising you how to solve this. Of course, once the spikes are eliminated, my previous code will work to isolate the individual QRS compelxes.

.

Star Strider 2022-11-29

编辑：Star Strider 2022-11-29

在 MATLAB Online 中打开

As always, my pleasure!

Those are actual records and not artificially corrupted?

Something is seriously wrong with the instrumentation that would allow those spikes to be recorded as part of the EKG signal (actually added to it). Be sure you are using a neutral (right leg or some other point) reference that can be subtracted from the other leads to prevent this sort of corruption.

For ‘EKG1’ the spikes are (mostly) symmetrical about the identified peak, so it is relatively straightforward to identify them and the eliminate them by setting the identified index range to NaN and then using fillmissing. Here are examples of two separate such replacements —

Uz = unzip('https://www.mathworks.com/matlabcentral/answers/uploaded_files/1211718/EKG%20signals.zip');

Fs = 1024;

for k = 1:3

EKG{k,:} = struct2table(load(Uz{k+1}));

L = numel(EKG{k});

t{k,:} = linspace(0, L-1, L).'/Fs;

end

t

t = 3×1 cell array

{61441×1 double} {61461×1 double} {63837×1 double}

EKG

EKG = 3×1 cell array

{61441×1 table} {61461×1 table} {63837×1 table}

figure

for k = 1:3

subplot(3,1,k)

plot(t{k}, EKG{k}{:,1})

grid

xlim([0 3.5])

% xlim([0 3500])

end

[Sp1,idx1] = findpeaks(EKG{1}{:,1}, 'MinPeakHeight',0.65);

t_spikes{1,:} = t{1}(idx1);

didx1 = diff(idx1);

% Lv = ischange(EKG{1}, 'Threshold',0.005);

figure

plot(t{1}, EKG{1}{:,1}, '.-')

hold on

plot(t_spikes{1}, Sp1, '^r')

idxrng{1} = idx1(1)+[-15:15]; % Identify First Spike

plot(t{1}(idxrng{1}), EKG{1}{idxrng{1},1}, '-g', 'LineWidth',2)

hold off

grid

title('Spike %3d',1)

xlim([0 0.65])

EKG{1}{idxrng{1},1} = NaN; % Assign Rnage To 'NaN'

EKG{1} = fillmissing(EKG{1}, 'makima'); % Interpolate 'NaN' Values To Eliminate First Spike

figure

plot(t{1}, EKG{1}{:,1}, '.-')

hold on

plot(t_spikes{1}, Sp1, '^r')

idxrng{1} = idx1(1)+[-15:15];

plot(t{1}(idxrng{1}), EKG{1}{idxrng{1},1}, '-g', 'LineWidth',2)

hold off

grid

title(sprintf('Spike %3d',1))

xlim([0 0.65])

figure

plot(t{1}, EKG{1}{:,1}, '.-')

hold on

plot(t_spikes{1}, Sp1, '^r')

% plot(t{1}(idxrng{10}), EKG{1}{idxrng{10},1}, '-g', 'LineWidth',2)

hold off

grid

title('Original Sample')

xlim([0 3.5])

for k = 1:numel(idx1) % Loop: Find & Eliminate All Identified Spikes

idxrng{k} = idx1(k)+[-15:15];

EKG{1}{idxrng{k},1} = NaN;

EKG{1} = fillmissing(EKG{1}, 'makima');

end

figure

plot(t{1}, EKG{1}{:,1}, '.-')

hold on

plot(t_spikes{1}, Sp1, '^r')

% plot(t{1}(idxrng{10}), EKG{1}{idxrng{10},1}, '-g', 'LineWidth',2)

hold off

grid

title('Sample With Spikes Removed')

xlim([0 3.5])

% xlim([0 0.65]+4.5)

% figure

% plot(t{1}, EKG{1}{:,1}, '.-')

% hold on

% plot(t_spikes{1}, Sp1, '^r')

% idxrng{10} = idx1(10)+[-15:15];

% % t{1}(idxrng{10})

% plot(t{1}(idxrng{10}), EKG{1}{idxrng{10},1}, '-g', 'LineWidth',2)

% hold off

% grid

% title(sprintf('Spike %3d',10))

% xlim([0 0.65]+4.5)

To figure out the index ranges for the spikes in ‘EKG1’ I simply counted the plotted points in the spike and then experimented to get these results. (I also experimented with the ischange function, however it does not give good results — detecting the beginning and ending of the spikes — here. You can experiment with findchangepts, however I doubt it wil do much better.) The only way to do this appears to manually characterise the spikes (as I did here). In ‘EKG1’ it is relatively easy to locate the spikes using findpeaks, and then using the manual indexing approach to eliminate them. That will likely not work for the other records. I have no idea what to suggest for those.

With respect to ‘EKG1’ this approach generally works (I did not examine the entire record in detail). It will be necessary for you to do something similar to the other records. Unfortunately, thare appears to be no specific way to identify the spikes in the other records, so it may be necessary for you to detect and define them (or define whatever characteristics are common to all of them) manually, however this general approach will work to eliminate them once you have solved that problem.

EDIT — Minor aesthetic changes.

.

Susan 2022-11-29

Thank you so very much again for your help! You gave me enough hints to work on other signals and figure out a way to eliminate unwanted spikes.

Yes, these records are actual records in a noisy scenario to measure an instrument's tolerance to the environment and are not artificially corrupted.

Star Strider 2022-11-29

As always, my pleasure!

I now get the impression that the intent here is to develop the instrumentation, and the EKG records are not the actual objective.

.

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