How to find the fall time of a voltage pulse.

I have a waveforn captured using an osciloscope and saved in a csv file. I am trying to find the 20% to 80% fall time value and plot the waveform with markers. Column 4 is the time (x axis) and column 5 is the voltage data (y axis) .

Answer 1

[BP]

The csv file is attached.

[Star Strider]

See my Comment to my Answer.

.

[BP]

I am trying to find the 20% to 80% fall time as shown in the graph in the attached spreadsheet.

[BP]

When I run this section of code below, I get the error :

Error using +

Arrays have incompatible sizes for this operation.

Error in VPlot_9_2_r1 (line 104)

xm(k1) = interp1(y(idxrng2+(-3:3)), x(idxrng2+(-3:3)), hafmax(k1));

Line 104 is the last line of the above code.

Any suggestions?

for k1 = 1:numel(hafmax)

% idxrng1 = find(y(1:idx)<hafmax(:,k1), 1, 'last')

idxrng2 = find(y(idx:numel(x))<hafmax(:,k1),1,'first')+idx;

% xm(k1) = interp1(y(idxrng1+(-3:3)), x(idxrng1+(-3:3)), hafmax(k1));

xm(k1) = interp1(y(idxrng2+(-3:3)), x(idxrng2+(-3:3)), hafmax(k1));

end

[Star Strider]

My code runs without error, so I have no idea what the problem could be.

Check to see what ‘idxrng2’ returns in your code. It should have only one (scalar) value. If it has more than one value or is empty, either of those conditions could throw that error. (I assume you are using a different data set from the one posted that I am using.)

The data you are using could be too noisy. In that situation, increase the window size (third argument) to the sgolayfillt call:

y = sgolayfilt(y, 3, 121);                          % Filter Out Some Of The Noise

It might be necessary to experiment with it until there is no long a problem with what the find call returns.

.

[BP]

It showes the following

idxrng2 =

1×0 empty double row vector

What is this telling me?

[Star Strider]

It is telling you generally that no values in the data meet that criterion, or more specifically, that on the descending limb of the curve (governed by ‘idxrng2’) there are no values less than ‘hafmax’.

Be certain that you defined ‘hafmax’ in the context of the data.

If you are using the same data I use in my Answer, there should be no problems. If you are using different data, post it and I willl see what I can do so that my code will work with it.

.

[BP]

I am using the same data and the same code you used. and it's giving the following error.

Error using +

Arrays have incompatible sizes for this operation.

Error in Diff_Mode_Voltage_1kVPlot_BFP_9_2_r1 (line 43)

xm(k1) = interp1(y(idxrng2+(-3:3)), x(idxrng2+(-3:3)), hafmax(k1));

Below is the code i am using

T1 = readtable('v pulse1 with chart.xlsx');

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, 121); % 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);

vals = [0.8 0.2];

hafmax = vals*ymx

hafmax = [1 2]

for k1 = 1:numel(hafmax)

% idxrng1 = find(y(1:idx)<hafmax(:,k1), 1, 'last')

idxrng2 = find(y(idx:numel(x))<hafmax(:,k1),1,'first')+idx;

% xm(k1) = interp1(y(idxrng1+(-3:3)), x(idxrng1+(-3:3)), hafmax(k1));

xm(k1) = interp1(y(idxrng2+(-3:3)), x(idxrng2+(-3:3)), hafmax(k1));

end

format short g

xm = xm

xm = [1 2]

% wp = diff(xm,[],2);

% wpmtx = buffer(wp,7) % Show All Values

format short

figure

plot(x, y)

hold on

for k = 1:numel(hafmax)

plot([min(xlim) xm(k)], [1 1]*hafmax(k), '--r', 'LineWidth',1.5)

plot([1 1]*xm(k), [min(ylim) hafmax(k)], '--r', 'LineWidth',1.5)

end

hold off

grid

text(xm, hafmax, compose(' \\leftarrow %2.0f%% = %.5f \\mus',[80 20; xm*1E+6].'), 'Horiz','left', 'Vert','middle', 'FontSize',8)

xlabel('t')

ylabel('Pressure')

title('Detrended Signal With Data')

[Star Strider]

Eliminate the two extra assigments that were not in my original posted code, specifically:

hafmax = [1 2]

and

xm = [1 2]

and it works. Those lines were never in any code I posted here. I have no idea where they came from.

.

[BP]

That worked

[Star Strider]

If it worked, please Accept my answer.

I made no changes in my code, and simply corrected your changed copy of it.

.

Answer 2

[BP]

It prints the correct data in the following form.

hafmax =

722.4474 180.6118

xm =

-2.2256e-06 9.72e-05

xm =

1.0e-04 *

-0.0223 0.9720

I think hefmax is the 20% and 80% points =180.61 volts and 722.45 volts.

I think 1.0e-4* is the time between the rising edge and fallowing edge which occures at t=-0.0223 and t=0.9720.

Is this correct?

--------------------

With the code as writen, is there any way to make it print out the values as shown below?

20% point= 180.61 volts

80% point =722.45 volts

FWHM time= 1.0e-04

X t1=-0.0223

X t2=0.9720

------------------------

So far, this is all i've been ably to get the data to displat which is not what I need it to look like. Please give me help and suggestions.

hafmax =

722.4474 180.6118

xm =

-2.2256e-06 9.72e-05

xm =

1.0e-04 *

-0.0223 0.9720

Full Width Half riseing edge/ falling edge) = -0.000002 u sec.

Full Width Half riseing edge/ falling edge) = 0.000097 u sec.

Hafmax points high/ low) = 722.447381 Volts.

Hafmax points high/ low) = 180.611845 Volts.