How to get time for 50%, 90% and 100% of a graph?

Hi guys,

I have another question again. If you run the script which I have attached, you will get a graph and few calculations.

What I want to do is: I need to know after how many seconds the graph reaches 50% and 90% of its maximum.

I have come this far: My idea was it, to take the piece of the curve where the gradient is zero to calculate the average for 100%. From this value I would calculate the 50% and 90%. Because this values are calculated I won't find exactly them, so I'm searching for the next smaller and next bigger value. Now I would need the time, how long it took to get to 50% and to 90%.

I need to do this for 12 graphs to get more or less good average values. Can you please tell me if there is a better/easier way to get the result, or is my way ok? And how would you solve it for several graphs.

Thank you very much!

% Load cell array
load('c_test.mat');

% Take timetable from cell array
P1_100 = c{1}(100:end,1:1);

% Convert timetable to matrix
P1_100_T = timetable2table(P1_100);
P1_100_T = P1_100_T{:,2:2};

% Calculate the average for 100%
P1_100_av = mean(P1_100_T(:,:));

% Calculate 50% and 90%
P1_50 = P1_100_av*0.5;
P1_90 = P1_100_av*0.9;

% Load data for full curve
P1 = c{1}(:,1:1);

% Interpolate timetable and set the starttime at zero
dt = milliseconds(100);
IP1 = retime(P1,'regular','linear','TimeStep', dt);
StartTime = datetime('05.07.2019 00:00.000', 'InputFormat', 'dd.MM.yyyy mm:ss.SSS');
IP1.Properties.StartTime = StartTime;

scatter(IP1.Time, IP1.Volt);

% Convert interpolated timetable to matrix
IPM1 = timetable2table(IP1);
IPM1 = IPM1{:,2:2};

% Search for next smaller and next bigger value of 50%
P1_50_uG = IPM1(find(IPM1 < P1_50, 1, 'last'));
P1_50_oG = IPM1(find(IPM1 > P1_50, 1, 'first'));

% Search for next smaller and next bigger value of 90%
P1_90_uG = IPM1(find(IPM1 < P1_90, 1, 'last'));
P1_90_oG = IPM1(find(IPM1 > P1_90, 1, 'first'));

[S.M]

I just noticed, that my code works well for 50% of the maximum value. The measurements in the range of 90% are jumping a bit, so that it shows me the upper value earlier than lower value and this is wrong.

Is there any other way to get the time how long did it take to reach 50% and 90% of the maximum?

Answer 1

[Star Strider]

Only some of your records follow the model you describe.

For those that do, this works (using the code you posted to get the ‘IP1.Time’ and ‘IP1.Volt’ vectors):

Tsec = second(IP1.Time);                                                                % Convert To Seconds
Volts = IP1.Volt;                                                                       % Retrieve From Table
Voltsd = Volts - Volts(1);                                                              % Subtract Baseline Offset
VoltsL = Voltsd > 0.01;
Ix0 = find(Voltsd>0.01, 1, 'first');

fcn = @(b,x) b(1).*(1 - exp(b(2).*(x + b(3))));                                         % Model Equation
B = fminsearch(@(b) norm(Voltsd(VoltsL) - fcn(b,Tsec(VoltsL))), [1; -0.5; Tsec(Ix0);]); % Estimate Parameters
T50  = fzero(@(t) fcn(B,t) - 0.5*B(1), 1)                                               % Time At 50% Of Max
T90  = fzero(@(t) fcn(B,t) - 0.9*B(1), 1)                                               % Time At 90% Of Max

figure
plot(Tsec(VoltsL), Voltsd(VoltsL), '-')
hold on
plot(Tsec(VoltsL), fcn(B,Tsec(VoltsL)), '-r')
xl = min(xlim);
yl = min(ylim);
plot([xl T50], [1 1]*0.5*B(1),'--g', [1 1]*T50, [yl 0.5*B(1)],'--g')
plot([xl T90], [1 1]*0.9*B(1),'--c', [1 1]*T90, [yl 0.9*B(1)],'--c')
hold off
grid
xlabel('Time (s)')
ylabel('Volts')
text(T50, 0.5*B(1), sprintf('\\leftarrow 50%%: %4.1fV, %4.1fs', 0.5*B(1), T50), 'HorizontalAlignment','left', 'VerticalAlignment','middle')
text(T90, 0.9*B(1), sprintf('\\leftarrow 90%%: %4.1fV, %4.1fs', 0.9*B(1), T90), 'HorizontalAlignment','left', 'VerticalAlignment','middle')

Append my code to the code you posted to use it. I had to extract the times as seconds in order for the regression to work. It appears to g1ve good results, and estimates the time delay to onset (as ‘B(3)’) as well.

Make appropriate changes to get the results you want.

[S.M]

Hi, thanks a lot for your answer.

I just checked your code and it worked perfectly fine. Exactly something like this I was searching for.

Could you please also explain me why your plot starts from zero? Is there a reason for that? I'm sorry to bother you with this question, but my MATLAB knowledge is very elementary and I don't really understand your code fully.

[Star Strider]

My pleasure.

It begins at zero because I could not originally get my model (with an additional parameter) to reliably estimate the y-offset (essentially ‘Volts(1)’), as well as the time offset.

(I originally posted an image of the plot. I have no idea why the file name for it is showing up instead. Must be a MathWorks server problem or something.)

I updated the model and the associated code:

Tsec = second(IP1.Time);                                                                        % Convert To Seconds
Volts = IP1.Volt;                                                                               % Retrieve From Table
Voltsd = Volts - Volts(1);                                                                      % Subtract Baseline Offset
VoltsL = Voltsd > 0.01;
Ix0 = find(Voltsd>0.01, 1, 'first');

fcn = @(b,x) b(1).*(1 - exp(b(2).*(x + b(3)))) + b(4);                                          % Model Equation
B = fminsearch(@(b) norm(Volts(VoltsL) - fcn(b,Tsec(VoltsL))), [1; -0.5; Tsec(Ix0); Volts(1)]); % Estimate Parameters                                                      % Time At 99% Of Max
V50 = 0.5*(B(1)+B(4)-Volts(1))+Volts(1);
V90 = 0.9*(B(1)+B(4)-Volts(1))+Volts(1);
TC = interp1(fcn(B,Tsec(VoltsL)), Tsec(VoltsL), [V50; V90]);
T50 = TC(1);                                                                                    % Time At 50% Of Max
T90 = TC(2);                                                                                    % Time At 90% Of Max

figure
plot(Tsec, Volts, '-')
hold on
plot(Tsec(VoltsL), fcn(B,Tsec(VoltsL)), '-r')
xl = min(xlim);
yl = min(ylim);
plot([xl T50], [1 1]*V50,'--g', [1 1]*T50, [yl V50],'--g')
plot([xl T90], [1 1]*V90,'--c', [1 1]*T90, [yl V90],'--c')
hold off
grid
xlabel('Time (s)')
ylabel('Volts')
text(T50, V50, sprintf('\\leftarrow 50%%: %5.2fV, %4.1fs', V50, T50), 'HorizontalAlignment','left', 'VerticalAlignment','middle')
text(T90, V90, sprintf('\\leftarrow 90%%: %5.2fV, %4.1fs', V90, T90), 'HorizontalAlignment','left', 'VerticalAlignment','middle')

and got these results:

The times to reach the 50% and 90% values are:

T50 =
   6.009267024018119

T90 =
  12.356403870541623

in the event that the plot image again disappears.

Note that the time to reach 100% is +Inf, since the maximum value (in my code, ‘(B(1)+B(4))’) is as asymptote. You can calculate and display other limits by adding fzero and text calls similar to the ones I have already coded.

[S.M]

Hi Star Strider, thank you so much, it works! You helped me a lot. Thank you!!!

[Star Strider]

As always, my pleasure!

[S.M]

Hello again,

Star Strider I have another question again.

I used the code to extract the time for T50 and T90, which worked well for all curves. Now I got data with similar shape of graphs where I need to do the same like before. Later I will substract T50 and T90 of this new graphs from the other graphs just to check how fast the reaction time of the sensor is.

Problem with this new data is, that they have similar shape, but the slope of the graph is much bigger, so that it rises much faster.

I thought your code will work, but I always get the error message. This error I have seen before with previous data too. There the problem was, that one or two curves were longer than 60 seconds, but I could solve it.

I guess here the problem is, that the function doesn't fit to this graphs anymore or? I tried to play around with few parameters, but nothing changed. Could you please help me with the solution again?

I will add the cell array with the new data and the code, maybe you know the solution. The data is in every second row of the cell array.

Thank you very much.

Here the error msg.:

Error using griddedInterpolant
The grid vectors must contain unique points.

Error in interp1 (line 151)
        F = griddedInterpolant(X,V,method);

Error in Test_Average_3 (line 53)
TC = interp1(fcn(B,Tsec(VoltsL)), Tsec(VoltsL), [V50; V90]);

[Star Strider]

The rise is much faster with record ‘N=2’ in those data, so interp1 was crashing into the aysmptote, creating non-unique values of the interpolating variable. The solution is to truncate the data vectors to avoid the asymptote, so the calculation of ‘T50’ and ‘T90’ can proceed.

Update my code to add:

V9999 = 0.9999*(B(1)+B(4)-Volts(1))+Volts(1);
Tmax = fcn(B,Tsec(VoltsL)) <= V9999; 

so the updated section becomes:

fcn = @(b,x) b(1).*(1 - exp(b(2).*(x + b(3)))) + b(4);                                          % Model Equation
B = fminsearch(@(b) norm(Volts(VoltsL) - fcn(b,Tsec(VoltsL))), [1; -0.5; Tsec(Ix0); Volts(1)]); % Estimate Parameters
V9999 = 0.9999*(B(1)+B(4)-Volts(1))+Volts(1);                                                   % Calculate Limit To Avoid Asymptote
Tmax = fcn(B,Tsec(VoltsL)) <= V9999;                                                            % Limit Vectors To Avoid Asymptote
V50 = 0.5*(B(1)+B(4)-Volts(1))+Volts(1);
V90 = 0.9*(B(1)+B(4)-Volts(1))+Volts(1);
TauV = (1 - exp(-1))*(B(1)+B(4)-Volts(1))+Volts(1);                                             % Time Constant V
TC = interp1(fcn(B,Tsec(VoltsL(Tmax))), Tsec(VoltsL(Tmax)), [V50; V90; TauV]);
T50 = TC(1)                                                                                    % Time At 50% Of Max
T90 = TC(2)                                                                                    % Time At 90% Of Max
Tau = TC(3)                                                                                    % Time Constant   

With those revisions, my code should now work for every record, the new ones as well as the earlier ones.

I added the τ calculation in the event that it could be relevant. It you want to include it on the plot, add this plot call:

plot([xl Tau], [1 1]*TauV,'--m', [1 1]*Tau, [yl TauV],'--m')

and this text call:

text(Tau, TauV, sprintf('\\leftarrow \\tau: %5.2fV, %4.1fs', TauV, Tau), 'HorizontalAlignment','left', 'VerticalAlignment','middle')

[S.M]

Hi Star Strider,

I apologize for the delay in replying. I didn't have time to try it out.

Star Strider, I don't know what to say: You are a genius, magician or both at the same time, your code works. Thank you so much, you helped me alot. I don't know how to thank you...

Best wishes

S.M.

[Star Strider]

No worries!

Thank you for your compliments! Accepting my Answer and voting for it in addition are all the appreciation it is possible to express here, and I thank you for those! Also, yours is an interesting problem, and was fun to solve.

[S.M]

If it would be possible, I would give you more than one voting :)

The problem is interesting, yes, but my knowledge was not enough to solve it.

Once again, thank You so much. I wish You all the best!

[Star Strider]

Thank you again!

As always, my pleasure!

[S.M]

Hi Star Strider (again) :)

can you please explain how and why you have chosen exactly this model equation?

Was it because of your experience or did you try few?

Sorry for bother you again :)

[Star Strider]

That was six months ago! I had to go back and look up the plot image (that I kept).

I chose that model because it is the general model for the sort of asymptotic curve that the data describe:

I added the parameters to generalise it.

It is possilbe that it could also be derived from the differential equation that describes the circuit that created those data.

[S.M]

I'm sorry, I know that it is long back.

Oh I see, now I understand this model a bit better. Thank you so much!

[Star Strider]

As always, my pleasure!