Value to get back to initial

Hello!!

Shown in the image, I have a exponential decay of fluid due to dialysis. I want to stop the dialysis at 75mg/Liter (shown at the intersection at the red line and blue line) and want to see how long it takes from that point to go back up to 150mg/Liter. How can i do that?

[Mark Sherstan]

To clarify you want to find the intersection point at y = 75 and y = 150 and find the time between those two points?

[Umber Sheikh]

yes

Answer 1

[Adam Danz]

This solution finds the time window that starts and ends when dialysis is at 75 and 150, respectively. I created fake data to work with. Then you can define the dialysis window. 'Duration' is the duration of the window along the x axis that starts and ends according to the dialysis window along the y axis. Lastly, a plot is produced to show the window for confirmation.

% Create fake data to work with
time = 0 : .1 : 50; 
x = linspace(10.5, -2, length(time)); 
dialysis = exp(x/2); 

% define dialysis window
window = [75, 150]; 

% Find closest index to the dialysis window
% Assumes 'dialysis' is a row vector.  Otherwise, remove transposes.
[~, minIdx] = min(abs([dialysis', dialysis'] - window)); 

% Calculate time between the dialysis window
duration = time(minIdx(1)) - time(minIdx(2)); 

% Verify by plotting results
figure
plot(time, dialysis); 
hold on
plot(time([1,1;end,end]), dialysis([minIdx; minIdx]), 'm-') %dialysis window
plot(time([minIdx; minIdx]), dialysis([minIdx', minIdx']), 'm-') %time window
title(sprintf('Fake Data; time window = %.2f', duration))