我试图适应这个数据集:
#Mydataset damped sine wave data
#X ---- Y
45.80 320.0
91.60 -254.0
137.4 198.0
183.2 -156.0
229.0 126.0
274.8 -100.0
320.6 80.0
366.4 -64.0
412.2 52.0
458.0 -40.0
503.8 34.0
549.6 -26.0
595.4 22.0
641.2 -18.0
所以我首先设置适合的宏
f(x) = exp(-a*x)*sin(b*x)
然后我做得很合适
fit f(x) 'data.txt' via a,b
iter chisq delta/lim lambda a b
0 2.7377200000e+05 0.00e+00 1.10e-19 1.000000e+00 1.000000e+00
Current data point
=========================
# = 1 out of 14
x = -5.12818e+20
z = 320
Current set of parameters
=========================
a = -5.12818e+20
b = -1.44204e+20
Function evaluation yields NaN ("not a number")
获得NaN
作为结果。所以我环顾了STackOverflow,我记得因为快速增长/衰减而需要你设置初始参数以便不会出现这个错误,因此我已经在过去的问题中通过拟合指数来解决问题(正如我所说的那样{ {3}})。所以我尝试将起始参数a
和b
设置为预期参数a = 9000
,b=146000
,但结果比之前的结果更令人沮丧:
fit f(x) 'data.txt' via a,b
iter chisq delta/lim lambda a b
0 2.7377200000e+05 0.00e+00 0.00e+00 9.000000e+03 1.460000e+05
Singular matrix in Givens()
我曾经想过:“这些数字太大了,让我们试试较小的数字”。
所以我输入了a
和b
的值并再次开始拟合
a = 0.01
b = 2
fit f(x) 'data.txt' via a,b
iter chisq delta/lim lambda a b
0 2.7429059500e+05 0.00e+00 1.71e+01 1.000000e-02 2.000000e+00
1 2.7346318324e+05 -3.03e+02 1.71e+00 1.813940e-02 -9.254913e-02
* 1.0680927157e+137 1.00e+05 1.71e+01 -2.493611e-01 5.321099e+00
2 2.7344431789e+05 -6.90e+00 1.71e+00 1.542835e-02 4.310193e+00
* 6.1148639318e+81 1.00e+05 1.71e+01 -1.481123e-01 -1.024914e+01
3 2.7337226343e+05 -2.64e+01 1.71e+00 1.349852e-02 -9.008087e+00
* 6.4751980241e+136 1.00e+05 1.71e+01 -2.458835e-01 -4.089511e+00
4 2.7334273482e+05 -1.08e+01 1.71e+00 1.075319e-02 -4.346296e+00
* 1.8228530731e+121 1.00e+05 1.71e+01 -2.180542e-01 -1.407646e+00
* 2.7379223634e+05 1.64e+02 1.71e+02 8.277720e-03 -1.440256e+00
* 2.7379193486e+05 1.64e+02 1.71e+03 1.072342e-02 -3.706519e+00
5 2.7326800742e+05 -2.73e+01 1.71e+02 1.075288e-02 -4.338196e+00
* 2.7344116255e+05 6.33e+01 1.71e+03 1.069793e-02 -3.915375e+00
* 2.7327905718e+05 4.04e+00 1.71e+04 1.075232e-02 -4.332930e+00
6 2.7326776014e+05 -9.05e-02 1.71e+03 1.075288e-02 -4.338144e+00
iter chisq delta/lim lambda a b
After 6 iterations the fit converged.
final sum of squares of residuals : 273268
rel. change during last iteration : -9.0493e-07
degrees of freedom (FIT_NDF) : 12
rms of residuals (FIT_STDFIT) = sqrt(WSSR/ndf) : 150.905
variance of residuals (reduced chisquare) = WSSR/ndf : 22772.3
Final set of parameters Asymptotic Standard Error
======================= ==========================
a = 0.0107529 +/- 3.114 (2.896e+04%)
b = -4.33814 +/- 3.678 (84.78%)
correlation matrix of the fit parameters:
a b
a 1.000
b 0.274 1.000
我看到它产生了一些结果,所以我认为一切都好,但我的幸福持续了几秒钟,直到我绘制输出:here 哇。一个非常好的。
我仍在这里想知道什么是错的,以及如何使用gnuplot正确拟合阻尼正弦波数据集。
希望有人知道答案:)
答案 0 :(得分:3)