find index of missing data point/ extrapolatation

I want fo find time and index when current is 90% of peak value during rise and fall.

Its easy to find rise but during fall, current drop quickly and there is no data point/index available at 90% of peak value.

How can I get index and time at 90% fall/drop value? probably by extrapolation or data fiting. please help.

Thanks

A1= load('data.txt');

T=A1(:,1); %Time

I=A1(:,2); %Current

[Imx,iImx]=max(I); %locate peak value and index in current

i90=find(I>=Imx*0.9,1,'first'); % First 90% of Peak during rise.

ii90=find(I>=Imx*0.9,1,'last'); % Last/second 90% of peak during fall

T90R=T(i90);

T90F=T(ii90);

plot(T,I,'.'); hold on

plot(T(i90),I(i90),'bs','LineWidth',2 ,'MarkerSize',6)

plot(T(i90),I(i90),'bs','LineWidth',2 ,'MarkerSize',6)

hold off

Answer 1

[Star Strider]

Another option:

A1= load('data.txt'); 
T=A1(:,1); %Time
I=A1(:,2); %Current
[Imx,iImx]=max(I); %locate peak value and index in current
i90 = find(diff(sign(I-0.7*Imx)));                                      % Finds Both Crossing Indices (Does Not Need To Be Exact)
for k = 1:numel(i90)
    idx = i90(k)+[-2 2];
    I90(k) = interp1(I(idx), T(idx), 0.9*Imx, 'linear','extrap');       % Interpolate To Find More Precise Times
end

figure
plot(T,I)
hold on
plot(I90, 0.9*Imx*[1 1],'bs')
hold off
grid
xlabel('Time')
ylabel('Current')

producing:

Note that this will not find the indices, because there are no indices corresponding exactly to the 90% values. It will find the approximate corresponding times (within the precision limits of the data) for both of those points.

EDIT — (24 Jul 2020 at 12:53)

Corrected typographical error.

[Akhtar Rind]

Thanks, It is working but I am trying to understand how i90 worked.

i90 = find(diff(sign(I-0.7*Imx)));

if you dont mind breaking up above. How it worked to get rough crossing?

[Star Strider]

My pleasure!

The ‘i90’ calculation creates approximate zero-crossing indices, first by subtracting 0.7 from the data (because the indices are not sufficiently close on the descending part of the pulse to differentiate them easily), then takes the sign of the resulting data, and then uses diff to calculate the difference in that. Where the sign changes (from -1 to +1, or +1 to -1), the diff function detects a difference, and returns a logical vector of the approximate locations of those changes. The find function then returns the indices of the changes, calculated from the positions in the logical vector.

After that, the code uses those indices to interpolate (or extrapolate) over that region to calculate the approximate times where the pulse equals 90% of the peak.

[Akhtar Rind]

Thank you very much for this explaination.

[Star Strider]

As always, my pleasure!

Answer 2

[John D'Errico]

Use the intersections code, written by Doug Schwarz, found for download from the file exchange.

I90 = 0.9*max(I);

[Tint,Iint] = intersections(T,I,[min(T),max(T)],[I90,I90])
Tint =
                 4.118e-05
      4.78765517241379e-05
Iint =
                  69.49044
                  69.49044

https://www.mathworks.com/matlabcentral/fileexchange/11837-fast-and-robust-curve-intersections

Note that intersections uses linear interpolation to locate the points.