Force a starting point on exponential graph

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Ran Kagan
Ran Kagan il 24 Set 2022
Commentato: Davide Masiello il 27 Set 2022
Hi,
I'm trying to fit an exponential curve to a process describing a radioactive decay. This is my data:
counts=[2423970,2171372,2065862,1830553,1100899,1037972,914015,752138,684123,606126];
normalized_counts=counts./counts(1);
time=[24.4,40.1,49.2,69.9,137.1,144.4,160.6,185.4,192.7,209.7];
At first I tried using the loveable cftool, which easily managed to fit a "classic" exponential function: y=a*exp(b*x). However, I'm interested in fitting a "pure" exponential [normalized_counts=exp(-a*time)]. When I tried to insert this custom function, I got this error:
Inf computed by model function, fitting cannot continue.
Try using or tightening upper and lower bounds on coefficients.
Then I tried to force an initial condition, so that the exponent will have to pass through the point (24.4, 1) using the code below, but to no avail as well:
x0=24.4;
y0=1;
g = @(p,time)y0*exp(-p*(time-x0));
f = fit(time,normalized_counts,g)
plot(f,time,normalized_counts)
[I got some matrix size errors, tried to add ' (to transpose) to the arrays but that didn't help as well].
Would apprecaite your help!

Risposte (2)

Davide Masiello
Davide Masiello il 24 Set 2022
Modificato: Davide Masiello il 25 Set 2022
See if this can help
counts = [2423970,2171372,2065862,1830553,1100899,1037972,914015,752138,684123,606126];
normalized_counts = counts./counts(1);
time = [24.4,40.1,49.2,69.9,137.1,144.4,160.6,185.4,192.7,209.7];
p0 = 1e-3;
g = @(p,x) exp(-p*(x-time(1)));
f = fit(time',normalized_counts',g,'StartPoint',p0)
f =
General model: f(x) = exp(-p*(x-time(1))) Coefficients (with 95% confidence bounds): p = 0.007151 (0.006894, 0.007408)
plot(f,time,normalized_counts)
  7 Commenti
Davide Masiello
Davide Masiello il 27 Set 2022
My pleasure, if that's the answer you were looking for don't forget to accept it!

Accedi per commentare.


Torsten
Torsten il 25 Set 2022
Modificato: Torsten il 25 Set 2022
You must normalize to time = 0, not time = 24.4.
The initial mass is the key, not the mass when already 24.4 (whatever) have passed.
I think what you should try to do is to introduce another unknown "counts_initial" and fit the curve as
counts_normalized = exp(-p(1)*time)
with
time=[24.4,40.1,49.2,69.9,137.1,144.4,160.6,185.4,192.7,209.7];
counts_normalized = counts/counts_initial
This means that the curve you have to fit is
counts/counts_initial = exp(-p(1)*time)
thus again the general exponential
counts = counts_initial*exp(-p(1)*time)
counts = [2423970,2171372,2065862,1830553,1100899,1037972,914015,752138,684123,606126];
time = [24.4,40.1,49.2,69.9,137.1,144.4,160.6,185.4,192.7,209.7];
p0 = [3e6,1e-3];
g = @(p,x) p(1)*exp(-p(2)*x);
g1 = @(p,x) g(p,x)-counts;
p = lsqnonlin(@(p)g1(p,time),p0);
Local minimum possible. lsqnonlin stopped because the size of the current step is less than the value of the step size tolerance.
hold on
plot(time,counts,'o')
plot([0,time],g(p,[0,time]))
hold off
grid

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