isqnonlin command: not enough input parameters error
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Bilge Kaan Atay
am 21 Mär. 2019
Kommentiert: Bilge Kaan Atay
am 23 Mär. 2019
I am trying to solve a set of equation which is consist of five equations. Here is the function file:
function [f1, f2, f3, f4, f5] = myfun(a1, a2, a3, a4, a5, x1, x2, x3, x4, x5)
f1 = x1-x2*(exp((x4*a1)/(x3*a5))-1)-((x4*a1)/x5)-a1;
f2 = x2*(exp(a2/(x3*a5))-1)+(a2/x5)-x1;
f3 = x1-x2*(exp((a3+x4*a4)/(x3*a5))-1)-((a3+x4*a4)/x5)-a4;
f4 = a3*(((x2/(x3*a5))*exp((a3+x4*a4)/(x3*a5))+(1/x5))/(1+((x4*x2)/(x3*a5))*exp((a3+x4*a4)/(x3*a5))+(x4/x5)))-a4;
f5 = (((x2/(x3*a5))*exp((x4*a1)/(x3*a5))+(1/x5))/(1+((x4*a1)/(x3*a5))*exp((x4*a1)/(x3*a5))+(x4/x5)))-(1/x5);
end
This is the main code:
% Sabit degerler
k = 1.38e-23; % Boltzman sabiti
q = 1.602e-19; % Elektronun yuku
T = 25+273; % Sicaklik (K)
Ns = 36; % Hucre Sayisi
a1 = 2.4; % Isc
a2 = 21.8; % Voc
a3 = 17.2; % Vmpp
a4 = 2.20; % Impp
a5 = (Ns*k*T)/q; % Vt
%Is0 = (a1+Ki*dt)/(exp((a2+Kv*dt)/(x3*a5))-1);
x0 = [2.4, 1.2e-7, 1.5, 0.5, 600];
lb = [2, 1e-7, 1.3, 0.3, 400];
ub = [2.6, 5e-7, 1.7, 0.7, 900];
x = lsqnonlin(@myfun, x0, lb, ub);
I saved function file as a "myfun.m" and both of the files are in the same directory. When I run the main code it gives this error expression:
"Not enough input arguments.
Error in myfun (line 2)
f1 = x1-x2*(exp((x4*a1)/(x3*a5))-1)-((x4*a1)/x5)-a1;
Error in lsqnonlin (line 206)
initVals.F = feval(funfcn{3},xCurrent,varargin{:});
Error in mohapatra (line 17)
x = lsqnonlin(@myfun, x0, lb, ub);
Caused by:
Failure in initial objective function evaluation. LSQNONLIN cannot continue."
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Akzeptierte Antwort
Stephan
am 21 Mär. 2019
Bearbeitet: Stephan
am 21 Mär. 2019
Hi,
try:
% Sabit degerler
k = 1.38e-23; % Boltzman sabiti
q = 1.602e-19; % Elektronun yuku
T = 25+273; % Sicaklik (K)
Ns = 36; % Hucre Sayisi
a1 = 2.4; % Isc
a2 = 21.8; % Voc
a3 = 17.2; % Vmpp
a4 = 2.20; % Impp
a5 = (Ns*k*T)/q; % Vt
%Is0 = (a1+Ki*dt)/(exp((a2+Kv*dt)/(x3*a5))-1);
x0 = [2.4, 1.2e-7, 1.5, 0.5, 600];
lb = [2, 1e-7, 1.3, 0.3, 400];
ub = [2.6, 5e-7, 1.7, 0.7, 900];
x = lsqnonlin(@(x)myfun(x, a1, a2, a3, a4, a5), x0, lb, ub);
function f = myfun(x, a1, a2, a3, a4, a5)
x1 = x(1);
x2 = x(2);
x3 = x(3);
x4 = x(4);
x5 = x(5);
f(1) = x1-x2*(exp((x4*a1)/(x3*a5))-1)-((x4*a1)/x5)-a1;
f(2) = x2*(exp(a2/(x3*a5))-1)+(a2/x5)-x1;
f(3) = x1-x2*(exp((a3+x4*a4)/(x3*a5))-1)-((a3+x4*a4)/x5)-a4;
f(4) = a3*(((x2/(x3*a5))*exp((a3+x4*a4)/(x3*a5))+(1/x5))/(1+((x4*x2)/(x3*a5))*exp((a3+x4*a4)/(x3*a5))+(x4/x5)))-a4;
f(5) = (((x2/(x3*a5))*exp((x4*a1)/(x3*a5))+(1/x5))/(1+((x4*a1)/(x3*a5))*exp((x4*a1)/(x3*a5))+(x4/x5)))-(1/x5);
end
Best regards
Stephan
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