I am trying to solve fsolve (multi-variable) but getting an error.
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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
Accepted Answer
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)
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
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More Answers (2)
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)
func4uo(sym(10))
vpa(ans)
syms u
Z = func4uo(u)
limit(Z, u, inf)
vpa(ans)
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
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)
P = linspace(-5, 1);
syms u
F = func4uo(u, P(1))
string(F)
%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
on 17 Feb 2022
In your other Question I show that your revised code has no root (unless you count u = infinity)
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