[MATLAB基础] 各位大侠请帮助小弟:怎样用一个周期的正弦函数去拟合40多个点

[slgu]

见下例：
x=[-.3000, 0., .3000, .6000, .9000, 1.200, 1.500, 1.800, 2.100, 2.400, 2.700, 3., 3.300, 3.600, 3.900, 4.200, 4.500, 4.800, 5.100, 5.400, 5.700, 6., 6.300, 6.600, 6.900, 7.200, 7.500, 7.800, 8.100, 8.400, 8.700, 9., 9.300, 9.600, 9.900, 10.20, 10.50, 10.80, 11.10, 11.40, 11.70, 12., 12.30]
y=[-.1853, .2964, .3068, .8302, 1.057, 1.171, 1.027, 1.052, .9638, .8794, .4683, .3575, -.1257, -.2464, -.5395, -.6379, -.7630, -.7250, -.6585, -.6725, -.3411, -.2201, .2598e-1, .5348, .7284, .9376, 1.209, 1.182, 1.155, 1.112, .9046, .5851, .1793, -.1023, -.1916, -.6913, -.7327, -.9306, -.7009, -.7055, -.6118, -.3952, -.2453]
>> fx=@(b,x)b(1)+b(2)*sin(b(3)*x+b(4))
fx =
@(b,x)b(1)+b(2)*sin(b(3)*x+b(4))
>> b=rand(1,4)
b =
0.4168 0.6569 0.6280 0.2920
>> for l=1:3
b=lsqcurvefit(fx,b,x,y)
b=nlinfit(x,y,fx,b)
end
b =
0.1581 -0.9982 0.9957 -3.1115
b =
0.1581 -0.9982 0.9957 -3.1114
b =
0.1581 -0.9982 0.9957 -3.1114
b =
0.1581 -0.9982 0.9957 -3.1114
b =
0.1581 -0.9982 0.9957 -3.1114
b =
0.1581 -0.9982 0.9957 -3.1114
>> plot(x,y,'o')
>> hold on
>> y1=b(1)+b(2)*sin(b(3)*x+b(4));%or y1=fx(b,x);
>> plot(x,y1)
>> axis tight