Not able to find fzero
3 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Dhawal Beohar
el 25 de Mayo de 2022
Comentada: Dhawal Beohar
el 7 de Ag. de 2022
--> difference.m
function y = difference(u,d1,n,a,m,T,PsByN_0,UmaxN_0)
d1=20;
n=10^-11.4;
m=2.7;
a=0.5;
T=1;
PsByN_0dB=5;
PsByN_0=10.^(PsByN_0dB/10);
UmaxdB = 5;
UmaxN_0=10.^(UmaxdB/10);
fun1 = (-1./u)*log(((d1^m)./(a*n*PsByN_0*T*u+d1^m)*a)./(1-a));
fun2 = (1./u)*log(((-exp(u*UmaxN_0)*(exp(-PsByN_0*u)))./(u*UmaxN_0+PsByN_0*u))*(PsByN_0*u)-(PsByN_0*u*(exp(-PsByN_0*u)))*(expint(u*UmaxN_0+PsByN_0*u))+(exp(-PsByN_0*u))+((PsByN_0*u)*(exp(-PsByN_0*u)))*(expint(PsByN_0*u))+(exp(u*UmaxN_0))./((UmaxN_0/PsByN_0)+1));
y = (fun1 - fun2);
g0=fzero(@(u) difference(u,d1,n,a,m,T,PsByN_0,UmaxN_0), 10);
end
0 comentarios
Respuesta aceptada
Torsten
el 25 de Mayo de 2022
d1=20;
n=10^-11.4;
m=2.7;
a=0.5;
T=1;
PsByN_0dB=5;
PsByN_0=10.^(PsByN_0dB/10);
UmaxdB = 5;
UmaxN_0=10.^(UmaxdB/10);
fun1 = @(u) (-1./u)*log(((d1^m)./(a*n*PsByN_0*T*u+d1^m)*a)./(1-a));
fun2 = @(u) (1./u)*log(((-exp(u*UmaxN_0)*(exp(-PsByN_0*u)))./(u*UmaxN_0+PsByN_0*u))*(PsByN_0*u)-(PsByN_0*u*(exp(-PsByN_0*u)))*(expint(u*UmaxN_0+PsByN_0*u))+(exp(-PsByN_0*u))+((PsByN_0*u)*(exp(-PsByN_0*u)))*(expint(PsByN_0*u))+(exp(u*UmaxN_0))./((UmaxN_0/PsByN_0)+1));
fun =@(u) (fun1(u) - fun2(u));
g0=fzero(fun, 10);
21 comentarios
Torsten
el 6 de Ag. de 2022
I suggest you try an interval around the zero you found, e.g.
u = fzero(fun,[0.01,0.02])
U = linspace(0.01,0.02,20);
fU = fun(U);
plot(U,fU)
Más respuestas (1)
Sam Chak
el 25 de Mayo de 2022
Guess the problem that you want to solve is a complex-valued function.
function y = difference(u)
% parameters
d1 = 20;
n = 10^-11.4;
m = 2.7;
a = 0.5;
T = 1;
PsByN_0dB = 5;
PsByN_0 = 10.^(PsByN_0dB/10);
UmaxdB = 5;
UmaxN_0 = 10.^(UmaxdB/10);
% functions
fun1 = (-1./u)*log(((d1^m)./(a*n*PsByN_0*T*u + d1^m)*a)./(1 - a));
fun2 = (1./u)*log(((- exp(u*UmaxN_0)*(exp(-PsByN_0*u)))./(u*UmaxN_0 + PsByN_0*u))*(PsByN_0*u) - (PsByN_0*u*(exp(-PsByN_0*u)))*(expint(u*UmaxN_0 + PsByN_0*u)) + (exp(-PsByN_0*u)) + ((PsByN_0*u)*(exp(-PsByN_0*u)))*(expint(PsByN_0*u)) + (exp(u*UmaxN_0))./((UmaxN_0/PsByN_0) + 1));
y = (fun1 - fun2);
end
Let's try with fzero first.
[u, fval, exitflag, output] = fzero(@(u) difference(u), 10)
The exitflag = -4 indicates that complex function value was encountered while searching for an interval containing a sign change.
Next, fsolve is used.
[u, fval, exitflag, output] = fsolve(@(u) difference(u), 10)
The exitflag = -2 means that the Equation is not solved. The exit message shows that fsolve stopped because the problem appears regular as measured by the gradient, but the vector of function values is not near zero as measured by the default value of the function tolerance.
5 comentarios
Torsten
el 25 de Mayo de 2022
Yes, as said, I plotted the difference and there is no point where the difference is 0.
Ver también
Categorías
Más información sobre Optimization en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!