Not able to find fzero
1 回表示 (過去 30 日間)
古いコメントを表示
--> 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 件のコメント
採用された回答
Torsten
2022 年 5 月 25 日
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 件のコメント
Torsten
2022 年 8 月 6 日
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)
その他の回答 (1 件)
Sam Chak
2022 年 5 月 25 日
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 件のコメント
Torsten
2022 年 5 月 25 日
Yes, as said, I plotted the difference and there is no point where the difference is 0.
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