matlab插值法实例.doc

Several Typical Interpolation in Matlab

Lagrange Interpolation
Supposing：
x
144
169
225
y
12
13
15
If x=175, while y=?
Solution：
Lagrange Interpolation in Matlab:

function y=lagrange(x0,y0,x);
n=length(x0);m=length(x);
for i=1:m
z=x(i);
s=0.0;
for k=1:n
p=1.0;
for j=1:n
if j~=k
p=p*(z-x0(j))/(x0(k)-x0(j));
end
end
s=p*y0(k)+s;
end
y(i)=s;
end

input：

x0=[144 169 225]
y0=[12 13 15]
y=lagrange(x0,y0,175)

obtain the answer：
x0 =

144 169 225
y0 =
12 13 15
y =
13.2302

Spline Interpolation
Supposing:
x
1
4
9
16

1
2
3
4
Resolve the cubic spline function s（x）
Solution：
Input x=[ 1 4 9 6]；y=[ 1 4 9 6]；x=[ 1 4 9 6]；pp=spline(x,y)
pp =
form: 'pp'
breaks: [1 4 6 9]
coefs: [3x4 double]
pieces: 3
order: 4
dim: 1
output : pp.coefs
ans =

-0.0500 0.5333 -0.8167 1.0000
-0.0500 0.0833 1.0333 2.0000
-0.0500 -0.2167 0.7667 4.0000
It shows the coefficients of cubic spline polynomial , so:
S（x）=

Newton’s Interpolation
Supposing:
x
1
4
9

1
2
3
Resolve
Solution:
Newton’s Interpolation in matlab：

function yi=newint(x,y,xi);
n=length(x);
ny=length(y);
if n~=ny
error
end
Y=zeros(n);Y(:,1)=y';
for k=1:n-1
for i=1:n-k
if abs(x(i+k)-x(i))i));
end
end
m=length(xi);yi=zeros(1,m);
for i=1:n;
z=ones(1,m);
for k=1:i-1;
z=z.*(xi-x(k));
end
yi=yi+Y(1,i)*z;
end

input:

x=[1 4 9];
y=[1 2 3];
xi=[5 6];
yi=newint(x,y,xi);

operate the program and obtain the answer：
yi=
2.2667 2.5000

The difference between several One-dimensional Interpolations like `nearest`、`linear`、`spline`、`cubic`
Supposing ，y = sin(x).*exp(-x/5),plot the figure .
Solution:
program of MATLAB：
x = 0:1:4*pi;
y = sin(x).*exp(-x/5);
xi = 0:0.1:4*pi;
y1 = interp1(x, y, xi, 'nearest');
y2 = interp1(x, y, xi, 'linear');
y3 = interp1(x, y, xi, 'spline');
y4 = interp1(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 and obtain the figure：
(It shows that the smoothness of 'nearest' is the worst and `cubic` is the best.)