MATLABnlinfit秩亏

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收藏|2019/05/27 05:03

满意回答

2019/05/27 05:13

应该是有关数学排列的问题。

邱红

其他回答(2)
  • 主要是数学排列
    回答于 2019/05/27 06:10
  • >> t=0:30;>> x=[107.507 109.300 111.026 112.704 114.333 115.823 117.171 118.517 119.850 121.121 122.389 123.626 124.761 125.786 126.743 127.627 128.453 129.227 129.988 130.756 131.448 132.129 132.802 133.450 134.091 134.735 135.404 136.072 136.782 137.462 138.271];>> syms c d>> [c,d]=solve(c/(1+(c/107.507-1)*exp(-5*d))==115.823,c/(1+(c/107.507-1)*exp(-8*d))==119.850,c,d);>> c=double(c)c = 109.738281373644 - 1.60177395988749i>> d=double(d)d = -0.150875720473954 - 2.06096070058854i>> b0=[c,d];>> [b,r,j]=nlinfit(t,x,fun,b0);警告: 秩亏,秩 = 1,tol = 5.675835e-13。> In nlinfit>LMfit (line 587) In nlinfit (line 284)警告: 秩亏,秩 = 1,tol = 4.209312e-13。> In nlinfit>LMfit (line 587) In nlinfit (line 284)警告: 秩亏,秩 = 1,tol = 4.033439e-13。> In nlinfit>LMfit (line 587) In nlinfit (line 284)警告: 秩亏,秩 = 1,tol = 4.015428e-13。> In nlinfit>LMfit (line 587) In nlinfit (line 284)警告: Some columns of the Jacobian are effectively zero at the solution, indicating that themodel is insensitive to some of its parameters. That may be because those parameters are notpresent in the model, or otherwise do not affect the predicted values. It may also be due tonumerical underflow in the model function, which can sometimes be avoided by choosing betterinitial parameter values, or by rescaling or recentering. Parameter estimates may beunreliable.> In nlinfit (line 381)
    回答于 2019/05/27 05:43
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