matlab用最小二乘法求一形如y=t/(at+b)（a和b为待定系数）的多项式，使之与下列数据相拟合
数据如下t=[1 2 3 4 5 6 7 8] y=[4.00 6.40 8.00 8.80 9.22 9.50 9.70 9.68]
1.使用非线性最小二乘拟合函数lsqcurvefit拟合
t=[1 2 3 4 5 6 7 8];
y=[4.00 6.40 8.00 8.80 9.22 9.50 9.70 9.68];
fun=@(b,x)x./(b(1)*x+b(2));
x0=[0.1 0.1];
b=lsqcurvefit(fun,x0,t,y)
结果为：
b =
0.0811 0.1468
即a=0.0811 b=0.1468
2.绘图
plot(t,y,'ko');
hold on
plot(t,fun(b,t),'r-');
legend('Original data','fitted curve ')
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第1个回答 2013-11-28
First, write a file to return the value of F (F has n components).function F = myfun(x,xdata)
F = x(1)*exp(x(2)*xdata);Next, invoke an optimization routine:% Assume you determined xdata and ydata experimentally
xdata = ...
[0.9 1.5 13.8 19.8 24.1 28.2 35.2 60.3 74.6 81.3];
ydata = ...
[455.2 428.6 124.1 67.3 43.2 28.1 13.1 -0.4 -1.3 -1.5];
x0 = [100; -1] % Starting guess
[x,resnorm] = lsqcurvefit(@myfun,x0,xdata,ydata);At the time that lsqcurvefit is called, xdata and ydata are
assumed to exist and are vectors of the same size. They must be the
same size because the value F returned by fun must
be the same size as ydata.
F = x(1)*exp(x(2)*xdata);Next, invoke an optimization routine:% Assume you determined xdata and ydata experimentally
xdata = ...
[0.9 1.5 13.8 19.8 24.1 28.2 35.2 60.3 74.6 81.3];
ydata = ...
[455.2 428.6 124.1 67.3 43.2 28.1 13.1 -0.4 -1.3 -1.5];
x0 = [100; -1] % Starting guess
[x,resnorm] = lsqcurvefit(@myfun,x0,xdata,ydata);At the time that lsqcurvefit is called, xdata and ydata are
assumed to exist and are vectors of the same size. They must be the
same size because the value F returned by fun must
be the same size as ydata.
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