Force a starting point on exponential graph
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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!
Answers (2)
Davide Masiello
on 24 Sep 2022
Edited: Davide Masiello
on 25 Sep 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)
plot(f,time,normalized_counts)
7 Comments
Ran Kagan
on 24 Sep 2022
Yes, like below right?
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];
x0 = [1 1e-3];
g = @(a,b,x) a*exp(-b*x);
f = fit(time',normalized_counts',g,'StartPoint',x0)
plot(f,time,normalized_counts)
I didn't consider this because in your code you had specified y0 as a given value rather than an adjustable parameter.
Sam Chak
on 24 Sep 2022
Yes, that is also what I interpreted "pure" exponential function, normalized_counts = exp(-a*time).
Davide Masiello
on 25 Sep 2022
@Ran Kagan I understand a bit better now, thanks. I have modified my answer accordingly, see if it helps.
Ran Kagan
on 27 Sep 2022
Davide Masiello
on 27 Sep 2022
My pleasure, if that's the answer you were looking for don't forget to accept it!
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);
hold on
plot(time,counts,'o')
plot([0,time],g(p,[0,time]))
hold off
grid
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