isqnonlin command: not enough input parameters error

조회 수: 15 (최근 30일)
Bilge Kaan Atay
Bilge Kaan Atay 2019년 3월 21일
댓글: Bilge Kaan Atay 2019년 3월 23일
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."

채택된 답변

Stephan
Stephan 2019년 3월 21일
편집: Stephan 2019년 3월 21일
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

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Interpolation에 대해 자세히 알아보기

Community Treasure Hunt

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

Start Hunting!

Translated by