I am trying to solve fsolve (multi-variable) but getting an error.

2 views (last 30 days)
Dhawal Beohar
Dhawal Beohar on 16 Feb 2022
Commented: Walter Roberson on 17 Feb 2022
function fval = func4uo(u)
d1=1;
n=1;
m=1;
a=1;
T=1;
PsByN_0=1;
fval = ((-1/u)*log((d1^m)/(a*n*PsByN_0*T*u)+d1^m)*a*T)/(1-a)*T;
xsol = fsolve (@(u) func4uo(u), 0)
ERROR: Not enough input arguments.
  14 Comments
Dhawal Beohar
Dhawal Beohar on 17 Feb 2022
Thanks ! but some other errors,
function main
u0 = 1;
u = fzero(@func4uo,u0)
end
function fval = func4uo(u)
d1=10;
n=10^-11.4;
m=2.7;
a=0.5;
T=1;
PsByN_0dB=20;
PsByN_0=10.^(PsByN_0dB/10);
fval = ((-1./u)*log((d1^m)./(a*n*PsByN_0*T*u)+d1^m)*a*T)./(1-a)*T - (1./u)*log(expint(-PsByN_0*u))*exp(-PsByN_0*u);
end
Error using fzero (line 328)
Function value at starting guess must be finite and real.
Error in main (line 39)
u = fzero(@func4uo,u0)

Sign in to comment.

Accepted Answer

Matt J
Matt J on 16 Feb 2022
By choosing a=1, you are dividing by 1-a=0 for any input value, u.
f(0), f(1), f(2)
ans = -Inf
ans = -Inf
ans = -Inf
function fval = f(u)
d1=1;
n=1;
m=1;
a=1;
T=1;
PsByN_0=1;
fval = ((-1/u)*log((d1^m)/(a*n*PsByN_0*T*u)+d1^m)*a*T)/(1-a)*T;
end

More Answers (2)

Walter Roberson
Walter Roberson on 17 Feb 2022
There is no zero for that function.
If you use negative u, then the imaginary component of the function approaches negative infinity as u gets close to zero, and only reaches zero again as u gets to -infinity.
If you use positive u and floating point values, then the expint() overflows to infinity when you reach about 8, and the exp() term numerically goes to 0 in floating point, and inf*0 is nan.
If you use positive u with the symbolic toolbox, you can show that the real part of the function is negative until infinity is reached.
Or perhaps I should say that the root is u = +inf as in the limit the function does become 0.
format long g
U = linspace(5,8);
Z = func4uo(U);
figure(); plot(U, real(Z), 'k'); title('real'); xlim([0 10])
figure(); plot(U, imag(Z), 'r'); title('imaginary'); xlim([0 10])
func4uo(10)
ans =
NaN + NaNi
func4uo(sym(10))
ans = 
vpa(ans)
ans = 
syms u
Z = func4uo(u)
Z = 
limit(Z, u, inf)
ans = 
0
vpa(ans)
ans = 
0.0
function fval = func4uo(u)
d1=10;
n=10^-11.4;
m=2.7;
a=0.5;
T=1;
PsByN_0dB=20;
PsByN_0=10.^(PsByN_0dB/10);
fval = ((-1./u).*log((d1.^m)./(a.*n.*PsByN_0.*T.*u)+d1.^m).*a.*T)./(1-a).*T - (1./u).*log(expint(-PsByN_0.*u)).*exp(-PsByN_0.*u);
end
  1 Comment
Dhawal Beohar
Dhawal Beohar on 17 Feb 2022
Thanks Walter for the help and explanation. I might need to check equation again.

Sign in to comment.


Walter Roberson
Walter Roberson on 17 Feb 2022
Edited: Walter Roberson on 17 Feb 2022
Z = @(PS) arrayfun(@(ps) fzero(@(u)func4uo(u,ps), [0.6775499178144678 1e3]), PS)
Z = function_handle with value:
@(PS)arrayfun(@(ps)fzero(@(u)func4uo(u,ps),[0.6775499178144678,1e3]),PS)
P = linspace(-5, 1);
syms u
F = func4uo(u, P(1))
F = 
string(F)
ans = "- log(692455071077987426423013376/(275018307117627*u) + 2204244764264291/4398046511104)/u - (exp(5*u)*log(expint(5*u)))/u"
%vpasolve(F)
%{
U = Z(P);
plot(P, real(U), 'k', P, imag(U), 'r');
xlabel('Ps'); ylabel('u')
%}
function fval = func4uo(u,Ps)
d1=10;
n=10^-11.4;
m=2.7;
a=0.5;
T=1;
PsByN_0dB=20;
PsByN_0=10.^(PsByN_0dB/10);
fval = ((-1./u).*log((d1^m)./(a.*n.*PsByN_0.*T.*u)+d1.^m).*a.*T)./(1-a).*T - (1./u).*log(expint(-Ps.*u)).*exp(-Ps.*u);
end
  5 Comments
Walter Roberson
Walter Roberson on 17 Feb 2022
In your other Question I show that your revised code has no root (unless you count u = infinity)

Sign in to comment.

Products


Release

R2020a

Community Treasure Hunt

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

Start Hunting!

Translated by