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 个评论
采纳的回答
Matt J
2022-2-16
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
0 个评论
更多回答(2 个)
Walter Roberson
2022-2-17
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
2022-2-17
编辑:Walter Roberson
2022-2-17
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 个评论
Walter Roberson
2022-2-17
In your other Question I show that your revised code has no root (unless you count u = infinity)
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