I am trying to find the full width at half max value and plot the waveform with markers

I have a voltage waveforn captured using an osciloscope and saved in a csv file. I am trying to find the full width at half max value and plot the waveform with markers. Column 4 is the time (x axis) and column 5 is the voltage data (y axis) in the csv file. Can someone please help me?

Thanks

Answer 1

[Star Strider]

Explore the options of the Signal Processing Toolboox pulsewidth function.

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[BP]

I have tried using the pulsewidth function but the description/ instruction does not explain it well.

[Star Strider]

Attach your .csv file.

EDIT — (27 Aug 2021 at 3:40)

Also see: Pulsewidth command giving empty column vector

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[BP]

I have attached the .csv file.

[Star Strider]

Assuming that ‘Var4’ and ‘Var5’ are the variables-of-interest —

T1 = readtable('https://www.mathworks.com/matlabcentral/answers/uploaded_files/723219/v%20pulse1.csv', 'VariableNamingRule','preserve')

T1 = 10000×5 table
         Var1          Var2        Var3         Var4       Var5
    _______________    _____    __________    _________    ____

    {'File Length'}    10000    {'Points'}     -3.4e-05     16 
    {0×0 char     }      NaN    {0×0 char}     -3.4e-05     16 
    {0×0 char     }      NaN    {0×0 char}     -3.4e-05      8 
    {0×0 char     }      NaN    {0×0 char}    -3.39e-05      8 
    {0×0 char     }      NaN    {0×0 char}    -3.39e-05     16 
    {0×0 char     }      NaN    {0×0 char}    -3.39e-05      8 
    {0×0 char     }      NaN    {0×0 char}    -3.39e-05     16 
    {0×0 char     }      NaN    {0×0 char}    -3.39e-05     16 
    {0×0 char     }      NaN    {0×0 char}    -3.38e-05      8 
    {0×0 char     }      NaN    {0×0 char}    -3.38e-05     16 
    {0×0 char     }      NaN    {0×0 char}    -3.38e-05      8 
    {0×0 char     }      NaN    {0×0 char}    -3.38e-05      8 
    {0×0 char     }      NaN    {0×0 char}    -3.38e-05      8 
    {0×0 char     }      NaN    {0×0 char}    -3.37e-05      0 
    {0×0 char     }      NaN    {0×0 char}    -3.37e-05     16 
    {0×0 char     }      NaN    {0×0 char}    -3.37e-05     16 

t = T1.Var4;
y = T1.Var5;

figure
plot(t, y)
grid

% 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

y = sgolayfilt(y, 3, 101);                          % Filter Out Some Of The Noise

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

% y = detrend(y, 15);

[ymx,idx] = findpeaks(y, 'MinPeakProminence',100)

ymx = 923.1334

idx = 1028

hafmax = ymx*0.5;
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 = 1×2

-1.42e-05 3.1103e-05

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

wpmtx = 7×1

4.5303e-05 0 0 0 0 0 0

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(median(xm), hafmax.*ones(1,numel(idx)), compose('  \\leftarrow Width = %.3E',wp), 'Horiz','left', 'Vert','middle', 'Rotation',-90, 'FontSize',7)
xlabel('t')
ylabel('Pressure')
title('Detrended Signal With Data')

I am not certain what you want to define as the maximum. I left that determination to findpeaks here.

Experiment to get different results.

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[BP]

This has been a great help. I am new to MATLAB. Please explane what the following steps / commands in the code does. A line by line explanation would really help me.

[ymx,idx] = findpeaks(y, 'MinPeakProminence',100)

ymx = 923.1334

idx = 1028

hafmax = ymx*0.5;

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));

[Star Strider]

Sure!

The first is just a findpeaks call. I refer you to that documentationo to understand what it does. It returns the maximum peak and the associated index into that vector for reference later.

The ‘hafmax’ variable is just that — it is the value at half the maximum (although it can be value defined on the dependent variable range, depending on what you want to do).

The loop is a bit mor complicated. To use the interp1 function correctly, the independent variable (first argument to the function) has to be monotonically increasing or decreasing. Since the objective is to find the values for both the ascending and descending limbs of the curve, the code requires a separate interp1 call for each limb as the result. That explains the two ‘idxrng’ calls, one for each section of the curve, and a separate interp1 call for each section of the curve, producing two values of ‘xm’ correspondingly. The easiest way to understand how it works is to add a plot statement to the loop to display the different sections being interpolated.

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