Pulsewidth command giving empty column vector

Hello

I am trying to find the 66% width of different pulses of a Atrial Blood Pressure signal by using following command syntax:

pulsewidth(chunk,t,'MidPercentReferenceLevel',66)

where "chunk" is pulse and "t" is time.

when i use this command, it gives me values for some pulses width and for some pulse i get "0×1 empty double column vector". for example i am getting "0×1 empty double column vector" in the attached picture of pulse. can someone tell me which thing i am doing wrong? and can someone help me how to find 66% width without having empty column vector?

i would be thankful to you

Answer 1

[Star Strider]

It would be nice to see the waveforms that give values and those that do not.

I suspect the reason is that the ones that do not, have initial or final values that are not the same, as the one in the plot image shows. One possible way to deal with that is to get the minimum of the pressure waves for the entire recording and use that as the zero reference. For the waves that do not return values, it would then be necessary to find the maximum for that particular wave and calculatlate the 66% width.

One approach to that might be something similar to:

x = linspace(-5000, 5000, 250);
y = [7.5; 1.9; 0.9] .* exp(-(0.001*[1; 0.85; 0.6] * x).^2);
[ymx,idx] = max(y,[],2);
hafmax = ymx*0.66;
for k = 1:numel(hafmax)
    idxrng1 = find(y(k,1:idx(k))<hafmax(k), 1, 'last');
    idxrng2 = find(y(k,idx(k):numel(x))<hafmax(k),1,'first')+idx(k);
    xm(k,1) = interp1(y(k,idxrng1+(-3:3)), x(idxrng1+(-3:3)), hafmax(k));
    xm(k,2) = interp1(y(k,idxrng2+(-3:3)), x(idxrng2+(-3:3)), hafmax(k));
end

format short g
xm

xm = 3×2

-644.86 644.86 -758.58 758.58 -1074.5 1074.5

format short

figure
plot(x, y)
hold on
for k = 1:numel(hafmax)
    plot([xm(k,1) xm(k,2)], [1 1]*hafmax(k), '-k', 'LineWidth',1.5)
end
hold off
grid
xlim([-1 1]*2000)

This was designed to reply to a different problem, however it would work here as well, since it returns the independent variable values for the dependent variable equalling a specific value, so determining the width would simply mean subtracting those values (here, the columns of ‘xm’). It will be necessary to decide, likely for every waveform that does not return values, to use either the ascending or descending parts of the waveform to calculate the 66% value.

.

[Shehzaib Shafique]

Respected sir,

thank you for you reply.

Actually sir i am not expert in MATLAB so i did not get the logic of your code.

Moreover, when i try to replave "y" and "x" with my variables "chunk" and "t" respectively, i do not get results properly.

is it possible if you help me in finding the 66% width of my data pulse? i am attaching the data of pulse here.

Your help would mean a lot.

[Star Strider]

I did my best to make this a general as I could, so it would work with multiple pulses, however I cannot test that as rigorously as I would like (note that it is designed to work with row vectors):

LD = load('pulse.mat');
chunk = LD.chunk;
t = linspace(0, numel(chunk)-1, numel(chunk));

x = t(:).';                                             % Force Row Vectors
y = chunk(:).';                                         % Force Row Vectors

[ymx,idx] = findpeaks(y, 'MinPeakProminence',10);

hafmax = ymx*0.66;
for k = 1:numel(hafmax)
    idxrng1 = find(y(1:idx(k))<hafmax(k), 1, 'last');
    idxrng2 = find(y(idx(k):numel(x))<hafmax(k),1,'first')+idx(k);
    xm(k,1) = interp1(y(idxrng1+(-3:3)), x(idxrng1+(-3:3)), hafmax(k));
    xm(k,2) = interp1(y(idxrng2+(-3:3)), x(idxrng2+(-3:3)), hafmax(k));
end

format short g
xm = xm
wp = diff(xm,[],2)
format short

figure
plot(x, y)
hold on
for k = 1:numel(hafmax)
    plot([xm(k,1) xm(k,2)], [1 1]*hafmax(k), '-k', 'LineWidth',1.5)
end
hold off
grid
text(x(idx), hafmax.*ones(1,numel(idx)), compose('\\uparrow\nWidth = %.3f',wp), 'Horiz','center', 'Vert','top')

producing:

It measures the height from 0, not with respect to the baseline (since I have no idea what that is), so it may be necessary to change the code to reflect that, likely by subtracting the minimum of the entire signal from tthe entire signal, then present that corrected signal to findpeaks and the rest of the code. The resulting ‘hafmax’ and width calculations should then be correct.

Experiment with it to get the result you want.

I could have done this with the onlilne Run feature, however getting it to read .mat files takes more effort than just copying and pasting the result from running it on my computer.

.

[Shehzaib Shafique]

It worked. Thank you so much sir.

[Star Strider]

My pleasure!

If my Answer helped you solve your problem, please Accept it!

.

[Shehzaib Shafique]

Respected sir,

Thank you for your help. your code worked on all pulse accept the last one, the last pulse gives following error.

Error using matlab.internal.math.interp1

Sample points must be unique and sorted in ascending order.

Error in interp1 (line 154)

VqLite = matlab.internal.math.interp1(X,V,method,method,Xqcol);

I tried to fix it but i could not. Can you resolve this error?

Thnaks in advance

load 'filtered_abp'
fs=250;
filtered_abp = [filtered_abp zeros(1,80)]; % zero padding
time=[1:length(filtered_abp)]./fs;

[pks,loc,width,prominace]=findpeaks(filtered_abp,'MinPeakHeight',22,'annotate','extents');

invert_abp=-filtered_abp;

[inv_pks,inv_locs]=findpeaks(invert_abp,'MinPeakHeight',-20,'MinPeakDistance',50);

wind=50; % window
vec=[];

for c=2:2:length(inv_pks)
chunk=filtered_abp(inv_locs(c): inv_locs(c+1)+wind); % pulse + window

 % 66% width
t = linspace(0, numel(chunk)-1, numel(chunk));

x = t(:).';                                             % Force Row Vectors
y = chunk(:).';                                         % Force Row Vectors

[ymx,idx] = findpeaks(y, 'MinPeakProminence',10);

hafmax = ymx*0.66;
for k = 1:numel(hafmax)
    idxrng1 = find(y(1:idx(k))<hafmax(k), 1, 'last');
    idxrng2 = find(y(idx(k):numel(x))<hafmax(k),1,'first')+idx(k);
    xm(k,1) = interp1(y(idxrng1+(-3:3)), x(idxrng1+(-3:3)), hafmax(k));
    xm(k,2) = interp1(y(idxrng2+(-3:3)), x(idxrng2+(-3:3)), hafmax(k));
end

format short g
xm = xm
wp = diff(xm,[],2)
format short

% figure
% plot(x, y)
% hold on
% for k = 1:numel(hafmax)
%     plot([xm(k,1) xm(k,2)], [1 1]*hafmax(k), '-k', 'LineWidth',1.5)
% end
% hold off
% grid
% text(x(idx), hafmax.*ones(1,numel(idx)), compose('\\uparrow\nWidth = %.3f',wp), 'Horiz','center', 'Vert','top')

%vec=[vec; wp]
end

xm = 1×2

21.638 53.465

wp = 
       31.827

xm = 1×2

19.551 48.337

wp = 
       28.785

xm = 1×2

23.522 53.226

wp = 
       29.704

xm = 1×2

22.451 55.069

wp = 
       32.617

xm = 1×2

22.435 53.091

wp = 30.6561

xm = 1×2

19.4712 46.0098

wp = 26.5386

xm = 1×2

23.8133 57.1841

wp = 33.3708

xm = 1×2

22.6255 53.5973

wp = 30.9718

xm = 1×2

18.9069 44.4516

wp = 25.5446

xm = 1×2

23.7565 57.1283

wp = 33.3718

Error using matlab.internal.math.interp1
Sample points must be unique.

Error in interp1 (line 188)
VqLite = matlab.internal.math.interp1(X,V,method,method,Xqcol);

I am attaching the whole signal here

[Star Strider]

I am not certain what the problem is.

Using my original code, and detrending the data (this makes it more reliable), I get these results —

LD = load('filtered_abp[1].mat');
y = LD.filtered_abp;
t = linspace(0, numel(y)-1, numel(y));

x = t(:).';                                         % Force Row Vectors
y = y(:).';                                         % Force Row Vectors

figure
plot(t, y)
grid
xlabel('t')
ylabel('Pressure')
title('Original Signal')

y = detrend(y, 15);

[ymx,idx] = findpeaks(y, 'MinPeakProminence',10);

hafmax = ymx*0.66;
for k = 1:numel(hafmax)
    idxrng1 = find(y(1:idx(k))<hafmax(k), 1, 'last');
    idxrng2 = find(y(idx(k):numel(x))<hafmax(k),1,'first')+idx(k);
    xm(k,1) = interp1(y(idxrng1+(-3:3)), x(idxrng1+(-3:3)), hafmax(k));
    xm(k,2) = interp1(y(idxrng2+(-3:3)), x(idxrng2+(-3:3)), hafmax(k));
end

format short g
xm = xm

xm = 14×2

9.524 20.757 125.76 151.4 266.74 294.41 408.61 436.83 550.15 578.03 692.72 719.91 832.49 859.11 973.75 1002.9 1118.4 1144.9 1258.1 1283.4

wp = diff(xm,[],2);
wpmtx = buffer(wp,7)                            % Show All Values

wpmtx = 7×2

11.233 29.107 25.646 26.537 27.67 25.27 28.217 30.241 27.881 25.185 27.192 0 26.62 20.373

format short

figure
plot(x, y)
hold on
for k = 1:numel(hafmax)
    plot([xm(k,1) xm(k,2)], [1 1]*hafmax(k), '-k', 'LineWidth',1.5)
end
hold off
grid
text(x(idx), hafmax.*ones(1,numel(idx)), compose('  \\leftarrow Width = %.3f',wp), 'Horiz','left', 'Vert','middle', 'Rotation',-90, 'FontSize',7)
xlabel('t')
ylabel('Pressure')
title('Detrended Signal With Data')
xlim([-50  max(t)])

Experimnent to get different results.

.

[Shehzaib Shafique]

Now it worked. Thank you so much sir for you time and help.

[Star Strider]

As always, my pleasure!

.