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anjb99

木虫 (小有名气)

[求助] 非线性拟合-参数求解,matlab,origin!!!!

实验数据:
1.8        1.279
2        1.282
2.4        1.291
2.9        1.331
3.35        1.344
4.3        1.336
5.4        1.339
6.8        1.34
7.5        1.33
8        1.296
8.5        1.221
9.5        1.128
10.5        1.091
11        1.067
11.4        1.065
第一列为溶液pH值,换算为氢离子浓度为([H+]=0.1的pH值次方),第二列为公式中A值,A是[H+]的函数,求参数A1,A2,A3,pKa1和pKa2?非常感谢各位!!!


[ Last edited by dbb627 on 2012-8-1 at 10:39 ]
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anjb99

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引用回帖:
8楼: Originally posted by dbb627 at 2012-08-01 11:25:15
你提供的数据只有一个这样的拐点,这个是可以求出那个中间点...

数据只采集了这些,可以当其中一个为已知2.4,求出图中的那个拐点吗?
9楼2012-08-01 11:49:21
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dbb627

荣誉版主 (著名写手)

【答案】应助回帖

★ ★ ★
感谢参与,应助指数 +1
csgt0: 金币+3, 多谢应助 2012-08-01 13:30:30
A1=p1
A2ka1=p2
A3ka1ka2=p3
ka1=q1
ka1ka2=q2

A=[1.8        1.279
2        1.282
2.4        1.291
2.9        1.331
3.35        1.344
4.3        1.336
5.4        1.339
6.8        1.34
7.5        1.33
8        1.296
8.5        1.221
9.5        1.128
10.5        1.091
11        1.067
11.4        1.065];
x=10.^(-A(:,1));
y=A(:,2);
===========
matlab cftool拟合如下
General model Rat22:
     f(x) = (p1*x^2 + p2*x + p3) / (x^2 + q1*x + q2)
Coefficients (with 95% confidence bounds):
       p1 =  4.247e-014  (fixed at bound)
       p2 =      0.4073  (0.2814, 0.5331)
       p3 =   6.55e-010  (fixed at bound)
       q1 =      0.3049  (0.2092, 0.4007)
       q2 =  6.088e-010  (fixed at bound)

Goodness of fit:
  SSE: 0.002082
  R-square: 0.987
  Adjusted R-square: 0.9859
  RMSE: 0.01266

[ Last edited by dbb627 on 2012-8-1 at 10:40 ]
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.
2楼2012-08-01 10:29:54
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dbb627

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★
csgt0: 金币+1, 多谢应助 2012-08-01 13:30:39
拟合结果如图所示,稍作计算可得参数的值


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2012-08-01 10:31:22
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csgt0

荣誉版主 (著名写手)

彩色挂图

【答案】应助回帖

感谢参与,应助指数 +1
可拟合的参数多,比较难找到合适的吧,一般有大概初值比较好
CODE:
function anj
xdata0=[1.8,2,2.4,2.9,3.35,4.3,5.4,6.8,7.5,8,8.5,9.5,10.5,11,11.4];
xdata=0.1.^xdata0;
ydata=[1.279,1.282,1.291,1.331,1.344,1.336,1.339,1.34,1.33,1.296,1.221,1.128,1.091,1.067,1.065];
x0=ones(5,1)+rand(5,1);

opts = optimset('Algorithm','trust-region-reflective');
problem = createOptimProblem('lsqcurvefit','objective',@fun,...
    'x0',x0,'xdata',xdata,'ydata',ydata,'options',opts);
gs = MultiStart;
[x,f] = run(gs,problem,50)
end


function y=fun(x,xdata)
y=(x(1)*xdata.^2+x(2)*x(4)*xdata+x(3)*x(4)*x(5))./(xdata.^2+x(4)*xdata+x(4)*x(5));
end

showmethemoney
4楼2012-08-01 10:48:28
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