matlab插值法实例

1、Several Typical Interpolation in MatlabLagrange InterpolationSuppos ing：x144169225y121315If x=175, while y=?Solution :Lagrange Interpolation in Matlab: function y=lagra nge(xO,yO,x);n=le ngth(xO);m=le ngth(x);for i=1:mz=x(i);s=0.0;for k=1: np=1.0;for j=1: nif j=k p=p*(z-xO(j)/(xO(k)-xO(j);endends=p*

2、yO(k)+s;endy(i)=s;endin put:x0=144 169 225y0=12 13 15y=lagra nge(x0,y0,175)obta in the an swer:x0 =144169225y0 =121315y =13.2302Spline InterpolationSuppos ing:x14916厂 Vx1234Resolve the cubic spline function s (x)Solution :In put x= 1 4 9 6 ; y= 1 4 9 6 ; x= 1 4 9 6 ; pp=spli ne(x,y)PP =form: ppbreak

3、s: 1 4 6 9coefs: 3x4 doublepieces: 3order: 4dim: 1output : pp.coefsans =-0.05000.5333-0.81671.0000-0.05000.08331.03332.0000-0.0500-0.21670.76674.0000It shows the coefficie nts of cubic spli ne polyno mial , so:320.0008(x -1) -0.0254(x-1)0.4024(x - 1)1,仁 x _ 4,32S (x) =0.0008(x-4) -0.0183(x-4)0.2714(

4、x -4) 2,4 乞 x _ 9,320.0008(x-9) - 0.0063(x-9)0.1484(x - 9) 3,9 乞 x 一 16,Newton s InterpolationSuppos ing:x149Vx123Resolve 5 6Soluti on:Newt on s In terpolatiorim matlab :fun ctio n yi=newin t(x,y,xi);n=len gth(x);ny=le ngth(y);if n=nyerrorendY=zeros (n );Y(:,1)=y：for k=1: n-1for i=1: n-kif abs(x(i+k

5、)_x(i)epserrorendY(i,k+1)=(Y(i+1,k)-Y(i,k)/(x(i+k)-x(i);endendm=le ngth(xi);yi=zeros(1,m);for i=1: n;z=on es(1,m);for k=1:i-1;z=z.*(xi-x(k);endyi=yi+Y(1,i)* z;end in put:x=1 4 9;y=1 2 3;xi=5 6;yi=n ewi nt(x,y,xi);operate the program and obta in the an swer： yi=2.26672.5000The difference between seve

6、ral One-dimensional Interpolations like nearest、 linear、 、 Xspline、 cubicSupposing x 0,4， y = sin(x).*exp(-x/5),plot the figure .Soluti on:program of MATLAB :x = 0:1:4*pi;y = sin( x).*exp(-x/5);xi = 0:0.1:4*pi;y1 = in terp1(x, y, xi, n earest);y2 = in terp1(x, y, xi, li near);y3 = in terp1(x, y, xi, spli ne);y4 = in terp1(x, y, xi, cubic);plot(x, y , o, xi ,y1 ,g- ,xi ,y2 ,r: ,xi ,y3 ,k-., xi, y4, b-); legend(Original, Nearest, Linear, Spline, Cubic);operate the program