Can't integrate function using Matlab
1 vue (au cours des 30 derniers jours)
Afficher commentaires plus anciens
I am trying to do a basic integration of a formula for 4 different values. After doing a bit of reading, apparently I need to create a function handle but I am not getting any success. I have put my attempt below. I might be missing more though.
L=0:100;
L_bar=25;
sigma_d =[0.25, 0.5, 0.75, 1.0];
mu = log(L_bar) - 0.5*(sigma_d).^2;
L_tilda = exp(mu);
% eta_r = P.*exp(-r./L);
% I want to integrate eta from zero to infinity where P has changing
% variables to give 4 different plots.
r = 0:100;
% I was just testing below to see if a loop would run before attempting to
% plot anything
P = zeros(1,length(L));
fun = @(r) P.*exp(-r./L);
for ii=1:numel(sigma_d)
P= 1./(L.*sqrt(2*pi)*sigma_d(ii)).*exp(-log(L/L_tilda(ii)).^2/(2*sigma_d(ii)^2)) ;
eta_r = integral(fun,0,inf);
end
0 commentaires
Réponse acceptée
Torsten
le 9 Fév 2023
L_bar=25;
sigma_d =[0.25, 0.5, 0.75, 1.0];
mu = log(L_bar) - 0.5*(sigma_d).^2;
L_tilda = exp(mu);
r = 0:100;
for ii=1:numel(sigma_d)
P= @(L)1./(L.*sqrt(2*pi)*sigma_d(ii)).*exp(-log(L/L_tilda(ii)).^2/(2*sigma_d(ii)^2));
fun = @(L,r) P(L).*exp(-r./L);
eta_r(ii,:)=integral(@(L)fun(L,r),0,Inf,'ArrayValued',true);
end
plot(r,eta_r)
grid on
4 commentaires
Dyuman Joshi
le 9 Fév 2023
You will have to apply that individually -
L=0:100;
L_bar=25;
sigma_d =[0.25, 0.5, 0.75, 1.0];
mu = log(L_bar) - 0.5*(sigma_d).^2;
L_tilda = exp(mu);
r = 0:100;
LineTypes={'-' '--' ':' '-.'};
for ii=1:numel(sigma_d)
P= @(L)1./(L.*sqrt(2*pi)*sigma_d(ii)).*exp(-log(L/L_tilda(ii)).^2/(2*sigma_d(ii)^2));
fun = @(L) P(L).*exp(-r./L);
eta_r(ii,:)=integral(@(L)fun(L),0,Inf,'ArrayValued',true);
plot(r,eta_r(ii,:),'LineStyle', LineTypes{ii}, 'LineWidth', 1.5)
hold on
end
xlabel('r (\mu m)')
xticks([0 20 40 60 80 100])
ylabel('\eta(r)');
ylim([0 1])
title('Spatial Correlation Function LBAR=25')
set(gca, 'FontSize',10);
legend('\sigma_d =0.25', '\sigma_d= 0.5', '\sigma_d =0.75', '\sigma_d = 1')
Plus de réponses (1)
Dyuman Joshi
le 9 Fév 2023
You will have to explicitly change the function handle to change its definition
P=pi;
fun = @(x) P*x;
P=3;
%fun(3) is not equal to 3*3=9
fun(3)
L=0:100;
L_bar=25;
sigma_d =[0.25, 0.5, 0.75, 1.0];
mu = log(L_bar) - 0.5*(sigma_d).^2;
L_tilda = exp(mu);
% eta_r = P.*exp(-r./L);
% I want to integrate eta from zero to infinity where P has changing
% variables to give 4 different plots.
r = 0:100;
% I was just testing below to see if a loop would run before attempting to
% plot anything
for ii=1:numel(sigma_d)
P= 1./(L.*sqrt(2*pi)*sigma_d(ii)).*exp(-log(L/L_tilda(ii)).^2/(2*sigma_d(ii)^2));
fun = @(x) P.*exp(-x./L);
eta_r=integral(fun,0,Inf,'ArrayValued',true)
end
Since you are dividing by 0 (first element of L), you get a NaN and thus the warning from the integral() solver.
3 commentaires
Dyuman Joshi
le 9 Fév 2023
"I need to be integrating with respect to L"
You should have mentioned that before. By the definition of the function handle, I took it as you are integrating w.r.t r
Now as you want to integrate w.r.t L, is there a need to define L=0:100?
Or L is a variable in the definition of P as well?
Voir également
Catégories
En savoir plus sur Logical dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!