Not able to find fzero

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Dhawal Beohar
Dhawal Beohar on 25 May 2022
Commented: Dhawal Beohar on 7 Aug 2022 at 12:49
--> 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

Accepted Answer

Torsten
Torsten on 25 May 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 Comments
Dhawal Beohar
Dhawal Beohar on 6 Aug 2022 at 15:11
ok, thanks!

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More Answers (1)

Sam Chak
Sam Chak on 25 May 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 Comments
Dhawal Beohar
Dhawal Beohar on 7 Aug 2022 at 12:49
@Torsten you haven always big help. Can you please look into this problem and help me. Thank you!
https://uk.mathworks.com/matlabcentral/answers/1775175-how-to-store-the-value-of-intersection-points-from-fzero-into-a-vector

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