Plateau followed by one phase decay
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:
Here is an example:
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’);
Given example raw data below, could you please support me with fitting of the function above and paramters extraction?
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];
Best regardsGood 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:
Here is an example:
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’);
Given example raw data below, could you please support me with fitting of the function above and paramters extraction?
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];
Best regards 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:
Here is an example:
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’);
Given example raw data below, could you please support me with fitting of the function above and paramters extraction?
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];
Best regards curve fitting MATLAB Answers — New Questions