How to make a spline with end point limits?

2 次查看（过去 30 天）

Abdul Mannan 2015-9-15

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/243074-how-to-make-a-spline-with-end-point-limits

评论： Abdul Mannan 2015-9-15

I want to make a spline of some data x (input) and y (output), with constrained slope (first derivative) and zero initial and final 2nd derivative. For example

if true
  x = [ 0, 0.125, 2*0.125,  3*0.125, 4*0.125,  5*0.125, 6*0.125, 7*0.125, 8*0.125 ];
  y = [0, sin(2*pi*f/8),  1, sin(2*pi*3*f/8), 0, sin(2*pi*5*f/8),-1,sin(2*pi*7*f/8), 0];
end

I know that I can find fifth order spline with this like

if true
  k = 8;
    sp = spapi( optknt(x,k), x, y );
    xx = 0:0.0001:1;
    yy = fnval(xx,sp); % Evaluating data for xx using given spline
    dsp = fnder(sp);  
    dyy = fnval(xx,dsp); % First Derivative
    ddsp = fnder(sp,2);  % Second Derivative
    ddyy = fnval(xx,ddsp);
    plot(xx, yy, xx, dyy, xx, ddyy), shg % Plotting all data. 
  end

Although, it is good, but I want to put constrained on second derivative at end points. How can I do that?

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Abdul Mannan 2015-9-15

在 MATLAB Online 中打开

OK. This is explained in help in "spapi" MATLAB Command. You can do it as follows.

if true
x = [ 0, 0.125, 2*0.125,  3*0.125, 4*0.125,  5*0.125, 6*0.125, 7*0.125, 8*0.125 ];
y = [0, sin(2*pi*f/8),  1, sin(2*pi*3*f/8), 0, sin(2*pi*5*f/8),-1,sin(2*pi*7*f/8), 0];
k = 6;
sp = spapi(augknt(x,6,2),[x,x,min(x),max(x)],[y,[0 2*pi*cos(2*pi*f/8), 2*pi*cos(2*pi*f/4),2*pi*cos(2*pi*3*f/8),2*pi*cos(2*pi*4*f/8),2*pi*cos(2*pi*5*f/8),2*pi*cos(2*pi*6*f/8),2*pi*cos(2*pi*7*f/8),0],0,0]);
xx = 0:0.001:1;
yy = fnval(xx,sp);
dsp = fnder(sp);  
dyy = fnval(xx,dsp);
ddsp = fnder(sp,2);  
ddyy = fnval(xx,ddsp);
plot(xx, yy, xx, dyy, xx, ddyy), shg