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Plateau followed by one phase decay

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Francesco
Francesco on 26 Feb 2024
Commented: Francesco on 25 Apr 2024
Good morning, I am trying to figure out how to compute tau constants from my data
My data could be be fitted by such plateau followed by one phase decay function:
I tried to implement it in MATLAB as follows:
x = 0:0.5:20; % time in seconds
Y0 = -0.6; % signal baseline value
Plateau = -1; % singnal plateu after trigger/stimulus, maximum change from baseline
tau = 0.6; % exponenential decay constant
K = 1/tau; % rate constant in units reciprocal of the x-axis units
X0 = 5; % trigger time
y = Plateau+(Y0-Plateau)*exp(-K*(x-X0));
figure;plot(x,y,'k');
However, I get the following result:
I would have 2 questions:
1) why cant I reproduce the one phase decay function?
2) would you know how to use the matlab funciton "fit" for such data with plateau followed by one phase decay function?
Thanks community for your kind support,
Best regards.

Answers (2)

Alan Stevens
Alan Stevens on 26 Feb 2024
Like this?
x = 0:0.5:20; % time in seconds
Y0 = -0.6; % signal baseline value
Plateau = -1; % singnal plateu after trigger/stimulus, maximum change from baseline
tau = 0.6; % exponenential decay constant
K = 1/tau; % rate constant in units reciprocal of the x-axis units
X0 = 5; % trigger time
y = Y0*(x<=X0)+(Plateau+(Y0-Plateau)*exp(-K*(x-X0))).*(x>X0);
figure;plot(x,y,'k');
  5 Comments
Alan Stevens
Alan Stevens on 26 Feb 2024
Here's a quick fit of tau and Y0. I'll leave you to tidy it up and extend it to fit X0 as well.
x = 0:0.5:20;
y = [-0.137055262721364 -0.118841612584876 -0.274602636741299 -0.117324828772196 ...
-0.173528150754918 -0.280491919000118 -0.244300356226590 -0.367583069701879 ...
-0.423274105143034 -0.529129050767333 -0.774173830727337 -0.676677606159725 ...
-0.730062482232667 -0.863905715495076 -0.831675679632950 -0.987303352625066 ...
-0.949979744575626 -0.865710605996821 -0.901728879393798 -0.877082148456042 ...
-0.944693953430828 -1.07404346760035 -0.915521627715257 -0.901789963321291 ...
-0.955365771797851 -0.941530617721837 -0.945983148775748 -1.01735658137382 ...
-0.965635004813717 -1.06321643780048 -0.956807780654745 -1.09208906741553 ...
-1.04341265165344 -1.08982901817714 -1.07984413818039 -0.934740294823467 ...
-0.960591807908718 -1.03623550995537 -0.909687220130007 -1.09290177705358 ...
-1.01208835337351];
Plateau = -1;
X0 = 2;
fn = @(x,tau,Y0)Y0*(x<=X0)+(Plateau+(Y0-Plateau)*exp(-(x-X0)/tau)).*(x>X0);
tauY0 = [1, -0.1]; % Initial guess
tauY = fminsearch(@(tauY) F(tauY,x,y), tauY0);
tau = tauY(1); Y0 = tauY(2);
yfit = fn(x,tau,Y0);
plot(x,y,'.',x,yfit), grid
xlabel('x'), ylabel('y')
text(12,-0.25,['tau = ' num2str(tau)])
text(12,-0.35,['Y0 = ' num2str(Y0)])
function Z = F(tauY,x,y)
tau = tauY(1); Y0 = tauY(2);
Plateau = -1;
X0 = 2;
yvals = zeros(1,numel(x));
for i = 1:numel(x)
t = x(i) - X0;
yvals(i) = Y0*(t<=0)+(Plateau+(Y0-Plateau)*exp(-t/tau)).*(t>0);
end
Z = norm(yvals-y);
end
Francesco
Francesco on 25 Apr 2024
Hello, I tried to implement the solution above, however I did not manage to have single function that provides all relevant parameters such as R squared, and Tau (or K).
For my analysis, Y0, X0 and Plateau are not relevant as main outputs.
Community help is greatly appreciated.

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Francesco
Francesco on 27 Feb 2024
Thanks Alan for your fantastic help, this is of great help, I guess for now this is resolved :) and I will figure out if there will be need for a function to automatically find X0. Best regards.

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