fittype函數是基于最小二乘法進行數據擬合的嗎？這個精度如何改進呢？我的原始數據如下：

x=[1.57079632700000

1.43965228700000

1.30438927000000

1.20635412600000

1.11818009200000

1.04588855400000

0.979758029000000

0.921411672000000

0.839101944000000

0.765449050000000

0.688846549000000

0.573462832000000

0.456141800000000

0.349921062000000

0.251816104000000

0.153777470000000

0.0522726110000000

0

-0.0522726110000000

-0.153777470000000

-0.251816104000000

-0.349921062000000

-0.456141800000000

-0.573462832000000

-0.688846549000000

-0.765449050000000

-0.839101944000000

-0.921411672000000

-0.979758029000000

-1.04588855400000

-1.11818009200000

-1.20635412600000

-1.30438927000000

-1.43965228700000

-1.57079632700000]'

y=[0

0.0466150000000000

0.148860000000000

0.257980000000000

0.376420000000000

0.502290000000000

0.626270000000000

0.729550000000000

0.821830000000000

0.880210000000000

0.927810000000000

0.974550000000000

0.993000000000000

0.992290000000000

0.974820000000000

0.950290000000000

0.922520000000000

0.914770000000000

0.922520000000000

0.950290000000000

0.974820000000000

0.992290000000000

0.993000000000000

0.974550000000000

0.927810000000000

0.880210000000000

0.821830000000000

0.729550000000000

0.626270000000000

0.502290000000000

0.376420000000000

0.257980000000000

0.148860000000000

0.0466150000000000

0]'

用fittype進行擬合，代碼如下

syms t

f=fittype('(c0+c1*t+c2*t^2+c3*t^3+c4*t^4)*cos(t)','independent','t','coefficients',{'c0','c1','c2','c3','c4'});

cfun=fit(x,y,f) %顯示擬合函數，數據必須為列向量形式

xi=-pi/2:0.1*pi:pi/2;

yi=cfun(xi);

figure

plot(xi,yi,'b-',x,y,'g*');

title('擬合函數圖形');

結果如下圖所示

cfun =

General model:

cfun(t) = (c0+c1*t+c2*t^2+c3*t^3+c4*t^4)*cos(t)

Coefficients (with 95% confidence bounds):

c0 =? ?? ?0.9474??(0.9274, 0.9674)

c1 =??-4.287e-09??(-0.04692, 0.04692)

c2 =? ?? ?0.8196??(0.7295, 0.9098)

c3 =? ?3.162e-09??(-0.05159, 0.05159)

c4 =? ???-0.6442??(-0.713, -0.5754)

但是從圖像上明顯可以看到，在中間部位凹下去的地方模擬精度很差，請問按照我自定義的函數進行擬合怎樣才能提高這個精度呢

unt.jpg

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2020-8-11 18:04 上傳

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