Borrar filtros
Borrar filtros

Please, I would appreciate any assistance on debugging the attached code.

3 visualizaciones (últimos 30 días)
Isaac
Isaac el 1 de Abr. de 2016
Editada: Walter Roberson el 2 de Abr. de 2016
The first section of the code is okay, but I am having it difficult solving for uu and ud in the attached code. I want to continue solving for the values of uu and ud as long the conditions in the while loops are valid. Pls, see the attached file.
This is the pseudo code:
for j=1:N
k=1;
uu(k,j)=0;
Lou(k,j)=2;
p(k,j)=6000;
pQ(k,j)=asin(1+Lou(k,j));
Lh2(j)=200;
k22(k,j)=Lh2(j)./Lou(k,j);
if p(k,j)>pQ(k,j)
while Lou(k,j)<Lh2
k22f(k,j)=Lh2(j)./(Lou(k,j)+uus);
funLu(k,j)=asin(1+k22f(k,j));
testfunLu=@(x)eval(subs(funLu(k,j),uus,x))-p(k,j);
uuf(k,j)=fsolve(testfunLu,55);
uu(k+1,j)=uuf(k,j);
Lou(k+1,j)=Lou(k,j)+uu(k+1,j);
k22(k+1,j)=Lh2(j)./(Lou(k+1,j)+uu(k+1,j));
p(k+1,j)=p(k,j);
k=k+1
end
end
end
  9 comentarios
Isaac
Isaac el 1 de Abr. de 2016
Editada: per isakson el 2 de Abr. de 2016
I have tried using the fsolve this way also
testfunLu=@(x)eval(subs(funLu(k,j),uus,x))-p(k,j);
uuf(k,j)=fsolve(testfunLu,55);
Isaac
Isaac el 1 de Abr. de 2016
Editada: per isakson el 2 de Abr. de 2016
This is the pseudo code:
for j=1:N
k=1;
uu(k,j)=0;
Lou(k,j)=2;
p(k,j)=6000;
pQ(k,j)=asin(1+Lou(k,j));
Lh2(j)=200;
k22(k,j)=Lh2(j)./Lou(k,j)
if p(k,j)>pQ(k,j)
while Lou(k,j)<Lh2
k22f(k,j)=Lh2(j)./(Lou(k,j)+uus);
funLu(k,j)=asin(1+k22f(k,j));
testfunLu=@(x)eval(subs(funLu(k,j),uus,x))-p(k,j);
uuf(k,j)=fsolve(testfunLu,55);
uu(k+1,j)=uuf(k,j);
Lou(k+1,j)=Lou(k,j)+uu(k+1,j);
k22(k+1,j)=Lh2(j)./(Lou(k+1,j)+uu(k+1,j));
p(k+1,j)=p(k,j);
k=k+1
end
end
end

Iniciar sesión para comentar.

Respuestas (0)

Categorías

Más información sobre Loops and Conditional Statements en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by