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258190169(金币+20): 多谢大侠 你的QQ是多少可以和你联系一下嘛 2011-12-16 22:12:03
cenwanglai(金币+5, 计算强帖+1): 谢谢给予帮助~ 2011-12-20 09:07:46
引用回帖:
2楼: Originally posted by 258190169 at 2011-12-16 16:49:37:
本人编写的程序如下但是无法运行,由于是新手,希望高手帮忙调试一下:
function PenicilliumEst
clear all;
t=[0,10,30,50,70,90,110,130,150,160];
y=[0,0.23211,0.45906,0.68601,0.92328,1.21213,1.32561, ...

CODE:
function PenicilliumEst
clear all;
t=[0,10,30,50,70,90,110,130,150,160];
y=[0,0.23211,0.45906,0.68601,0.92328,1.21213,1.32561,1.34624,1.39782,1.398];
y0=0;

% Nonlinear least square estimate using lsqnonlin()
beta0=[0.001 0.001];
lb=[0 0];ub=[inf inf];
[beta,resnorm,residual,exitflag,output,lambda,jacobian] = ...
    lsqnonlin(@Func,beta0,lb,ub,[],t,y,y0);         
ci = nlparci(beta,residual,jacobian);
beta
% result
fprintf('\n Estimated Parameters by Lsqnonlin():\n')
fprintf('\t k1 = %.4f ± %.4f\n',beta(1),ci(1,2)-beta(1))
fprintf('\t k2 = %.4f ± %.4f\n',beta(2),ci(2,2)-beta(2))
fprintf('  The sum of the residual squares is: %.1e\n\n',sum(residual.^2))

% plot of fit results
tspan = [0  max(t)];
[tt yc] = ode45(@ModelEqs,tspan,y0,[],beta);
tc=linspace(0,max(t),200);
yca = spline(tt,yc,tc);
plot(t,y,'ro',tc,yca,'r-');
hold on
xlabel('Time');
ylabel('Concentration');
hold off
% =======================================
function f1 = Func(beta,t,y,y0)        % Define objective function
tspan =t;
[tt yy] = ode45(@ModelEqs,tspan,y0,[],beta);
yc= spline(tt,yy,t);
f1=y-yc;
% ==================================
function dydt = ModelEqs(t,y,beta)          % Model equations
dydt = (4.41/96485-(4.41*beta(1)+beta(2)*4.41/96485)*y)/(1+4.41*beta(1)*t);

改的可以运行了,但是初值不合适
The more you learn, the more you know, the more you know, and the more you forget. The more you forget, the less you know. So why bother to learn.
3楼2011-12-16 17:31:51
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