Intergrating from negative infinity to infinity

14 vues (au cours des 30 derniers jours)
Supreeth D K
Supreeth D K le 9 Mai 2022
Hello,
I am trying to solve an equation where limits of intergration are from negative infinity to infinity.
function damp = pmb(zeta)
syms zeta
mu=4*pi*10.^-7;i=1;v=10;w=0.001;%t=0.05; r=0.032;
sigma= 5.8*10.^7; a=mu*sigma; h=0.05;p=0.01; lp=0.1;
k=sqrt(((zeta).^2)+(i*zeta*a*v));
H=(((zeta).^2)./(((zeta).^2)+(k.^2)+(2*abs(zeta)*k*coth(k*w))));
x1=exp(-(abs(zeta))*p); x2=exp(-(abs(zeta))*h);
n=3;
for I=1:n
e1=exp(i*zeta*(1+lp)); e2=exp(i*zeta*1);
jr=(1./(4*pi*((zeta).^2))).*(e1-e2)*(-i).^(n-1);
end
A=jr*mu*(1-x1)*x2;
z=(abs(zeta*A)).^2;
S= z*real(H);
end
i get the following error, while using quad function damp=quad('pmb',-inf,inf)
"The integrand function must return an output vector of the same length as the input vector."

Réponse acceptée

Chunru
Chunru le 9 Mai 2022
The following correct the syntax error in your original program. You need to check/debug the function definition to ensure it is integratable.
y=integral(@pmb,-inf,inf)
Warning: Inf or NaN value encountered.
y = Inf
function S = pmb(zeta)
%syms zeta
mu=4*pi*10.^-7; i=1;v=10; w=0.001;%t=0.05; r=0.032;
sigma= 5.8*10.^7; a=mu*sigma; h=0.05; p=0.01; lp=0.1;
k=sqrt(((zeta).^2)+(i*zeta*a*v));
H=(((zeta).^2)./(((zeta).^2)+(k.^2)+(2*abs(zeta).*k.*coth(k*w))));
x1=exp(-(abs(zeta))*p); x2=exp(-(abs(zeta))*h);
n=3;
for I=1:n
e1=exp(i*zeta*(1+lp)); e2=exp(i*zeta*1);
%whos
jr=(1./(4*pi*((zeta).^2))).*(e1-e2)*(-i).^(n-1);
end
A=mu*jr.*(1-x1).*x2;
z=(abs(zeta.*A)).^2;
S= z.*real(H);
%whos
%pause
end

Plus de réponses (0)

Catégories

En savoir plus sur MATLAB 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!

Translated by