2005. 05. 11 2003010482 mohiuddin ahmad 2005010207 sung-bong jang interpolation ii (8.4 spline...

2005. 05. 112003010482 Mohiuddin Ahmad

2005010207 SUNG-BONG JANG

Interpolation IIInterpolation II(8.4 SPLINE INTERPOLATION)(8.4 SPLINE INTERPOLATION)

(8.5 MATLAB’s INTERPOLATION Functions)(8.5 MATLAB’s INTERPOLATION Functions)

Computer Vision and Pattern Recognition Lab.

Mohiuddin Ahmad ([email protected])ID: 2003010482

Computer Vision and Pattern Recognition Lab.Dept. of Computer Science and Engineering

Korea University

8.4 : Spline Interpolation8.4 : Spline Interpolation

mailto:[email protected]


Spline Interpolation

Interpolation qualities Piecewise polynomial Want smooth curves Local control

Local data changes have local effect Continuous with respect to the data

No wiggling if data changes slightly Low computational effort

Items to be discussed Piecewise linear interpolation Piecewise quadratic interpolation Cubic spline interpolation


Spline Interpolation

Piecewise Linear Interpolation Simplest form of piecewise polynomial interpolation

Interpolate the data with piecewise linear function

432144332211 with ),(),,(),,(),,( :points data ofSet xxxxyxyxyxyx

],[ ],,[ ],,[ ,lssubinterva threeDefine 433322211 xxIxxIxxI

43434

33

43

4

32323

22

32

3

21212

11

21

2

,

,

,

)(

xxxyxx

xxy

xx

xx

xxxyxx

xxy

xx

xx

xxxyxx

xxy

xx

xx

xP

),( 11 yx

),( 22 yx

),( 33 yx

),( 44 yx

x

y


Piecewise Linear Interpolation

Example 8.14 Piecewise Linear Interpolation

Function is continuous, but not smooth

]3 4 1 0[ ],3 2 1 0[ Using, yx

32 ,6

21 ,23

10 ,

)(

xx

xx

xx

xP

0 1 2 30

1

2

3

4

x

y

Figure 8.18 Piecewise linear interpolation


Piecewise Quadratic Interpolation

“Knots” Where the intervals meet to be the midpoints between the data

points where the function values are given

Processing 4 data points Define node points

Spacing between consecutive data points

Relationships:

432144332211 with ),(),,(),,(),,( xxxxyxyxyxyx

4543432321211 ,2/)( ,2/)( ,2/)( , xzxxzxxzxxzxz

343232121 , , xxhxxhxxh

),( 11 yx121 xxh

),( 33 yx

),( 44 yx

x

y

232 xxh 343 xxh

1z 2z 3z 4z 5z

2/

,2/

,2/

334

223

112

hxz

hxz

hxz

2/

,2/

,2/

344

233

122

hxz

hxz

hxz



We define

Since,

Equations at the interior nodes [impose continuity condition

on the polynomials]

;)()()( :],[on

;)()()( :],[on

;)()()( :],[on

;)()()( :],[on

4442

444544

3332

333433

2222

222322

1112

111211

cxxbxxaxPzzP

cxxbxxaxPzzP

cxxbxxaxPzzP

cxxbxxaxPzzP

);(422 :)( )(

);(422 :)( )(

);(422 :)( )(

3443334233

234443

2332223222

223332

1221112211

212221

yybhbhahahzPzP

yybhbhahahzPzP

yybhbhahahzPzP

4 ,3 ,2 ,1for ,)( ,re therefo,)( kycyxPcxP kkkkkkkk



Equations at the interior nodes [impose continuity condition

on the first derivative of the polynomials]

We have 6 equations & 8 unknown coefficients (a1, a2, a3, a4, b1, b2, b3, b4)

;0 :)( )(

;0 :)( )(

;0 :)( )(

4343334443

3232223332

2121112221

bbahahzPzP

bbahahzPzP

bbahahzPzP

0 and 0

0)(2 gives 0)(

0)(2 gives 0)(

Setting,

and determinecan We

)(2)( Since

41

444444

111111

41

bb

bxxaxP

bxxaxP

bb

bxxaxP kkkk

.0

,0

,0

),(42

),(422

),(42

33433

322322

21211

1233234

233

122322223

222

1212212

211

bhaha

bbhaha

bhaha

yyhbhaha

yyhbhbhaha

yyhbhaha

Equations for the coefficients at the zero slope

conditions at the end points of the interval


Piecewise Quadratic Interpolation – Example

Example : 8: 15 )3,3( ),4,2( ),1,1( ),0,0( :points Data

101100

110110

010011

201100

220110

020011

A

.0004124 r

:,,,,, tscoefficien :Solution 324321 bbaaaa

]9143.05143.24571.23714.37714.17429.0[x

on elliminatiGaussian Using

3.0] [2.5,on 3)3(4571.2)(

2.5] [1.5,on 4)2(9143.0)2(3714.3)(

1.5] [0.5,on 1)1(5143.2)1(7714.1)(

0.5] [0.0,on )0(7429.0)(

24

23

22

21

xxP

xxxP

xxxP

xxP

Piecewise interpolating polynomial

0 1 2 3 0

1

2

3

4

x

Piecewise quadratic interpolation of four data pointsy


Piecewise Cubic Interpolation

Cubic splines → piecewise cubic polynomial

Calculation of the coefficients of cubic polynomial

Cubic spline function, S(x)

iii

nnii

xxh

yxyxyxyx

nn

1

2211

valuesbetween x Spacing

),........(),........,( ......,..........),,(),,(

intervals sub 1pointsgiven

2.order toup derivative continuous with cubic piecewiseA

),(

, ),(

, , )(

)(

:form theoffunction Spline

11

1

211

nnn

iii

xxxxP

xxxxP

xxxxP

xS),( 11 yx

),( 22 yx

),( 33 yx

),( 44 yx

x

y),( nn yx

1h 2h 3h 1nh

),( 11 nn yx



Cubic spline function S(x), n-cubic polynomials Pi(x) with coefficients pi,0, pi,1, pi,2, and pi,3 satisfy the following properties:

1. S(x) consists of piecewise cubic (I)2. Piecewise cubic interpolate the given set of data points (II)3. Piecewise cubics represent a smooth continuous function (III & IV)4. Second derivative of the resulting function is continuous (V)

S(x) is piecewise cubic, S//(x) is linear on [x0, xn-1].

.

.2,....,1,0for )( V.

.2,....,1,0for )( IV.

.2,....,1,0for )( III.

.1,....,1,0for )( II.

.1,....,1,0for

)()()()()( I.

111

111

111

33,

22,1,0,

ni)(xPxP

ni)(xPxP

ni)(xPxP

niyxS

ni

xxpxxpxxppxPxS

iiii

iiii

iiii

ii

iiiiiiii

(x)P(x)SxSxx

xxxS

xx

xxxP ii

ii

ii

ii

ii

for ion interpolat LagrangeLinear ),()()( 111

1



Use S//(xi)=ai-1 , S//(xi+1)=ai and hi=xi+1 – xi, we get,

Integrating twice will introduce two constant of integration, and the cubic function is

From, definition of hi ,

Integrating P//i(x) twice and using Pi(xi) =yi & Pi (xi+1 ) =yi+1 :

Determine the value of bi and ci

.

)()(6

)(

6

)()(

write we,1,.......,1 For,

1

331

1 iiiii

ii

i

iii xxcxxb

h

xxa

h

xxaxP

ni

iiiiii axPaxP )( and )( 111

6 and ,

611 ii

i

ii

ii

i

ii

ha

h

yc

ha

h

yb

i

ii

i

iii h

xxa

h

xxaxP

1

1)(



Construction of cubic spline From property IV, 1st derivative of Pi and Pi+1 must agree at xi+1

Natural cubic spline assigns, 2nd derivative is zero at the endpoints

Error in cubic spline interpolation is |S(x) – g(x) | S(x) : spline interpolation function g(x) : function that generated the data Let, h : maximum spacing between node points, G=max|g(4)(x)|. Condition: |S(x) – g(x) | < kh4G =O(h4)

)2,...,1(636

1

1

121

111

ni

h

yy

h

yya

ha

hha

h

i

ii

i

iii

ii

iii

i

),,.........( unknowns for equations 2 : 10 naann

0 and 0 10 naa


Choice of Two Additional Conditions

End point constraints for a cubic spline

Description of the strategy Equation involving a0 and an

(i) Clamped cubic spline: specify

S/(x0), S/(xn)

(ii) Natural cubic spline

(iii) Extrapolated S//(x) to the end

points

(iv) S//(x) is constant near the end

points

(v) Specify S//(x) at each endpoint

00 a 0na

1

120 )(10 h

aahaa

2

2111

)(

n

nnnnn h

aahaa

110 , nn aaaa

)( ),( 00 nn xSaxSa

2))(( 1

003

00

axSda

h

2))(( 1

13

1

nnnhn

adxSa

n

i

ii

hyy

id 1


Example 8.16 Natural Cubic Spline Interpolation

Example 8:16

);()( 6

1

3

2

6

1 :3i

);()( 6

1

3

2

6

1 :2i

);()( 6

1

3

2

6

1 :1i

:follows as ,,for equation The spline. cubic natural afor 0 and

intervals,each for 1).8,2( and (1,1) ),2,0( ),1,1( ),4,2( :points dataConsider

3445432

2334321

1223210

32040

yyyyaaa

yyyyaaa

yyyyaaa

aaaaa

hi

4286.15 ,7143.13 ,4286.15

system nal tridiagoa gives

484

244

484

gsimplifyin and )5,..1( and ,,for values thengSubstituti :1

321

32

321

21

40

aaa

aa

aaa

aa

iyaaStep i

5714.16/

,2857.46/

,5714.36/

,46/

gives for Solving :2

344

233

122

011

ayb

ayb

ayb

ayb

bStep i

.86/

,5714.16/

,2857.46/

,5714.36/

gives cfor Solving :3

454

343

232

121

ayc

ayc

ayc

ayc

Step i


Example 8.16 Natural Cubic Spline Interpolation

Example 8:16 – cont’d

21 ),1(8)2(57.1)2(57.2

,10 ,57.1)1(29.457.2)1(29.2

,01 ),1(29.457.3)1(29.2 57.2

,12 ),2(57.3)1(4 )2(57.2

:

: tosimplifies spline cubic The

3

33

33

3

xxxx

xxxxx

xxxxx

xxxx

S(x) -2 -1 0 1 2-2

0

2

4

6

8

x

y

Figure 8.20 Cubic spline interpolant


Matlab function for Natural Cubic Spline

Matlab code


Matlab function for Natural Cubic Spline


Example 8.17 Runge Function

)251/(1)( 2xxf ];0385.01379.00.11379.00385.0[

];0.15.00.05.01[

y

x

i

ii

i

iii

ii

iii

i

h

yy

h

yya

ha

hha

h

,......, ni

1

1

121

111 636

11for

5.0 ,Here 4321 hhhh

3048.184

3808.414

3048.18 4

spline, cubic natural considered As

32

321

21

aa

aaa

aa

Using tridiagonal solution

a1 = 8.1857, a2 = -14.4381, a3 = 8.1857

Solving for bi & ci gives

b1 = 0.0770, b2 = -0.4043, b3 = 3.2032, b4 = -0.4043

c1 = -0.4043, c2 = 3.2032, c3 = -0.4043, c4 = 0.0770-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0.10.5 ),5.0(077.0)1(4043.0)1(7286.2

5.0.00 ,4043.0)5.0(2032.3)(7286.2)5.0(8127.4

0.00.5 ),5.0(2032.3)(4043.0)5.0(8127.4)(7286.2

5.01.0 ),1(4043.0)5.0(077.0)1(7286.2

)(

tosimplifies Spline Cubic

3

33

33

3

xxxx

xxxxx

xxxxx

xxxx

xS



],0385.00664.0138.03902.00.13902.0138.00664.00385.0[

],0.175.050.025.000.025.050.075.00.1[

y

x

)2,...,1(636

1

1

121

111

ni

h

yy

h

yya

ha

hha

h

i

ii

i

iii

ii

iii

i

25.0 ,Here 87654321 hhhhhhhh

3048.184

3048.184

3048.184

3048.184

3048.184

3808.414

3048.18 4


76

765

654

543

432

321

21

aa

aaa

aaa

aaa

aaa

aaa

aa

Using tridiagonal solution

a1= 1.2221; a2 = -0.6067; a3= 18.3695; a4=-38.4551; a5=18.3695; a6=-0.6067; a7=1.2221;

Consider natural cubic spline, a0 = 0; a8 = 0;


b1 = 0.1540, b2 = 0.2147, b3 = 0.5809, b4 = 0.7954, b5 = 5.6023, b6 = 0.7954, b7 = 0.5809, b8 = 0.2147

c1 = 0.2147, c2 = 0.5809, c3 = 0.7954, c4 = 5.6023, c5 = 0.7954, c6 = 0.5809, c7 = 0.2147, c8 = 0.1540



Cubic spline simplifiesp1 = 1.2221 *( x + 1.0 ).^3 / 1.5 + 0.1540*( -0.75 - x) + 0.2147 *(x + 1.0 ) ; -1 ≤x ≤ -0.75 ;

p2 = 1.2221*(-0.5 - x).^3 / 1.5 - 0.6067 * (x + 0.75 ).^3 / 1.5 + 0.2147*( -0.50 - x) + 0.5809 *(x + 0.75 ); -0.75 ≤x ≤ -0.50 ;

p3 = -0.6067*(-0.25 - x).^3 / 1.5 + 18.3695 * (x + 0.50 ).^3 / 1.5 + 0.5809*(-0.25 - x) + 0.7954 * (x + 0.5 ) ; -0.50 ≤x ≤ -0.25 ;

p4 = 18.3695*(-0.0 - x).^3 / 1.5 - 38.4551 * (x + 0.25 ).^3 / 1.5 + 0.7954 *(-0.0 - x) + 5.6023 * (x + 0.25 ); -0.25 ≤x ≤ 0.0 ;

p5 = -38.4551*(0.25 - x).^3 / 1.5 + 18.3695 * (x + 0.00 ).^3 / 1.5 + 5.6023 *(0.25 - x) + 0.7954 * (x + 0.00 ) ; 0.0 ≤x ≤ 0.25 ;

p6 = 18.3695*(0.5 - x).^3 / 1.5 - 0.6067 * (x - 0.25 ).^3 / 1.5 + 0.7954 *(0.50 - x) + 0.5809 * (x - 0.25 ) ; 0.25 ≤x ≤ 0.50 ;

p7 = -0.6067*(0.75 - x).^3 / 1.5 + 1.2221 * (x - 0.50 ).^3 / 1.5 + 0.5809 *(0.75 - x) + 0.2147 * (x - 0.50 ); 0.5 ≤x ≤ 0.75 ;

p8 = 1.2221*(1.00 - x).^3 / 1.5 + 0.0000 * (x - 0.75 ).^3 / 1.5 + 0.2147 *(1.00 - x) + 0.1540 * (x - 0.75 ) ; 0.75 ≤x ≤ 1.00 ;

-1 -0.5 0 0.5 10

0.2

0.4

0.6

0.8

1

spline function

actual Runge func.end points

x

y

Figure 8.21 Cubic spline interpolant of Runge function


Example 8.18, Chemical Reaction Product Data

Chemical reaction of data

Conclusion: Curve is smoother than higher

degree polynomial [ Fig. 8.6] Additional data points do not

really improve the appearance of original curve [Fig. 8.5]

]31.031.030.029.029.027.026.025.021.019.017.006.000.0[

]00.290.160.150.140.110.100.190.060.050.040.010.000.0[

x

x

0 0.5 1 1.5 20

0.05

0.1

0.15

0.2

0.25

0.3

0.35

x

y

Figure 8.22 Cubic spline interpolant of chemical reaction data


Example: 8.19 Difficult data

Data points:

Tridiagonal system of equations:

]0 0 0 0.87 1 0.87 0 0 0[

]2 1.5 1 0.5 0 0.5 1 1.5 2[

y

x

Using tridiagonal system

a0 = 0

a1= -1.8408

a2= 7.3633

a3= -6.7324

a4=1.8062

a5= -6.7324

a6= 7.3633

a7=-1.8408

a8 = 0

)2,...,1(636

1

1

121

111

ni

h

yy

h

yya

ha

hha

h

i

ii

i

iii

ii

iii

i

5.0 ,Here 87654321 hhhhhhhh

0.04

88.204

76.174

24.64

76.174

88.204

0.0 4


76

765

654

543

432

321

21

aa

aaa

aaa

aaa

aaa

aaa

aa


b1 = 0,

b2 = 0.1534,

b3 = -0.6136,

b4 = 2.3010,

b5 = 1.8495,

b6 = 2.3010,

b7 = -0.6136,

b8 = 0.1534

c1 = 0.1534,

c2 = -0.6136,

c3 = 2.3010,

c4 = 1.8495

c5 = 2.3010,

c6 =-0.6136,

c7 = 0.1534,

c8 = 0


Example: 8.19 Difficult data – cont’d

Resulting piecewise function made the following polynomial

0.25.1 ),5.1(0.0)2(15.03/)2(84.1

5.10.1 ),1(15.0)5.1(61.03/)1(84.13/)5.1(36.7

0.15.0 ),5.0(61.0)1(30.23/)5.0(36.7)1(24.2

5.00.0 ,30.2)5.0(85.1)(24.2)5.0(60.0

0.05.0 ),5.0(85.1)(30.2)5.0(60.0)(24.2

5.00.1 ),1(30.2)5.0(61.0)1(24.2)5.0(45.2

0.15.1 ),5.1(61.0)0.1(15.0)5.1(45.2)1(61.0

5.10.2 ),2(15.0)2(61.0

338

337

336

335

334

333

332

31

xxxxP

xxxxxP

xxxxxP

xxxxxP

xxxxxP

xxxxxP

xxxxxP

xxxP

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-0.2

0

0.2

0.4

0.6

0.8

1

1.2

x

y

Figure 8.23 Cubic spline interpolant of humped and flat data

2005. 05. 11

2005010207 SUNG-BONG JANG

Interpolation IIInterpolation II(8.5,MATLAB’s INTERPOLATION Functions)(8.5,MATLAB’s INTERPOLATION Functions)

26 / 7

■ 8.5.1 Interpolation in One-Dimensional

8.5.1.1 MATLAB function : Interp1

8.5.1.2 yy = interp1(x,y,xx,method) : interpolates using alternative methods:

‘linear‘ : Linear interpolation (default)

'nearest‘ : Nearest neighbor interpolation

‘cubic’ : Piecewise cubic Hermite interpolation

'spline‘ : Cubic spline interpolation

8.5.1.3 method : linear

x = 0:10;

y = sin(x);

xi = 0:.25:10;

yi = interp1(x,y,xi);

plot(x,y,'o',xi,yi)

0 1 2 3 4 5 6 7 8 9 10-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

27 / 7


8.5.1.4 method : nearest

x = 0:10;

y = sin(x);

xi = 0:.25:10;

yi = interp1(x,y,xi,’nearest’);

plot(x,y,'o',xi,yi)

8.5.1.5 method : cubic

x = 0:10;

y = sin(x);

xi = 0:.25:10;

yi = interp1(x,y,xi,’cubic’);

plot(x,y,'o',xi,yi)

0 1 2 3 4 5 6 7 8 9 10-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7 8 9 10-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

28 / 7


8.5.1.6 method : spline

x = 0:10;

y = sin(x);

xi = 0:.25:10;

yi = interp1(x,y,xi,’spline’);

plot(x,y,'o',xi,yi)

8.5.1.7 Faster methods

- the data are given at equally spaced intervals(i.e, x is equally spaced and monotonic): *linear,*cubic,*spline

> What is Monotonic ? : A function which is either entirely nonincreasing or nondecreasing.

A function is monotonic if its first derivative (which need not be continuous) does not

change sign.

- the data that are not uniformly spaced : interp1q.

- MATLAB ‘plot’ function : linear interpolation

0 1 2 3 4 5 6 7 8 9 10-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

29 / 7

■ 8.5.2 Interpolation in Two Dimensionals

8.5.2.1 Overview

- much more difficult than for a single variable.

- Defined on a rectangular grid : z(i,j) = f(x(i),y(j));

- Interp2 : three ways in which the interp2 can be used.

> ZI = interp2(Z,ntimes) expands Z by interleaving interpolates

between every element, working recursively for ntimes. interp2(Z) is the same as interp2(Z,1).

> ZI = interp2(Z,XI,YI) assumes that X = 1:n and Y = 1:m, where [m,n] = size(Z).

> ZI = interp2(X,Y,Z,XI,YI) returns matrix ZI containing elements corresponding to the elements of

XI and YI and determined by interpolation within the two-dimensional function specified by matrices X, Y, and Z.

30 / 7

8.5.2.2 ZI = interp2(Z,ntimes)

- expands the data matrix z by interleaving interpolanes between every element.

xx = [x1, (x2 + x2)/2, x2,….,(xn-1 + xn-1)/2, xn ]

yy = [y1, (y2 + y2)/2, y2,….,(yn-1 + yn-1)/2, yn ]

- zz = interp2(z); zz = interp2(z,k)

- Example)

x=[2,4], y =[3,5,7], z= x’y, >> interp2(z)

ans =

6 8 10 12 14

9 12 15 18 21

12 16 20 24 28

>> interp2(z,2)

ans =

6.0000 7.0000 8.0000 9.0000 10.0000 11.0000 12.0000 13.0000 14.0000

7.5000 8.7500 10.0000 11.2500 12.5000 13.7500 15.0000 16.2500 17.5000

9.0000 10.5000 12.0000 13.5000 15.0000 16.5000 18.0000 19.5000 21.0000

10.5000 12.2500 14.0000 15.7500 17.5000 19.2500 21.0000 22.7500 24.5000

12.0000 14.0000 16.0000 18.0000 20.0000 22.0000 24.0000 26.0000 28.0000

- mesh(x,y,z’)

* mesh(X,Y,Z) : draws a wireframe mesh with color determined by Z so color is proportional to surface height. If X and Y are vectors, length(X) = n and length(Y) = m, where [m,n] = size(Z).


6 10 14

12 20 28z

22.5

33.5

4

3

4

5

6

75

10

15

20

25

30

x axis

interp2 example 1

y axis

z ax

is

31 / 7


8.5.2.3 ZI = interp2(Z,XI,YI)

- the data values beiing interpolated are defined on a mesh given by the vectors x=1:n and y=1:m, where [m,n] = size(z); y gives the rows of the grid and x gives the column

- zz = interp2(z’,xx,yy’);

- Example)

x=[1 2 3]

y=[1 2 3 4 5]

z = x’*y z=

1 2 3 4 5

2 4 6 8 10

3 6 9 12 15

Xx= [1, 1.2,2,2.2,3]

Yy=[1,1.8,2,4,4.8,5]

Zz = interp2(z’,xx,yy’) =

1.00 1.20 2.00 3.00

1.80 2.16 3.96 5.40

2.00 2.40 4.40 6.00

4.00 4.80 8.80 12.00

4.80 5.76 10.56 14.40

5.00 6.00 11.00 15.00

- mesh(xx,yy,zz)

11.5

22.5

3

1

2

3

4

50

5

10

15

x axis

intep2 example 2

y axis

z ax

is

32 / 7


8.5.2.4 ZI = interp2(X,Y,Z,XI,YI),

- Interpolation on a Rectangular Grid

- Example)

x=[2,4], y =[3,5,7], z= x’y,

xx=[2, 2.4,2.6,2.8,3,3.5,4]

yy = [3,3.5,5,7]

zz= interp2(x,y’,z’,xx,yy’)

>> zz=interp2(x,y,z',xx,yy')

zz =

6.0000 7.2000 7.8000 8.4000 9.0000 10.5000 12.0000

7.0000 8.4000 9.1000 9.8000 10.5000 12.2500 14.0000

10.0000 12.0000 13.0000 14.0000 15.0000 17.5000 20.0000

14.0000 16.8000 18.2000 19.6000 21.0000 24.5000 28.0000

- mesh(xx,yy,zz)

6 10 14

12 20 28z

22.5

33.5

4

3

4

5

6

75

10

15

20

25

30

x axis

interp2 example 3

y axis

z ax

is

33 / 7


8.5.2.5 bilinear interpolation

- The value of an interpolated point is a combination of the values of the four closest points.

- the form of each rectangular subregion

z= a + bx + cy + dxy,

Region Rij, four corners : data values: (xi,yj),(xi,yi+1),(xi+1,yJ),(xi+1,yi+1), four unknowns : aij,bij,cij,dij

z(i,j) = a ij + bij xi + cijyj+dijxiyj

z(i,j+1) = a ij + bij xi + cijyj+1+dijxiyj+1

z(i+1,j) = a ij + bij xi+1 + cijyj+dijxi+1yj

z(i+1,j+1) = a ij + bij xi+1 + cijyj+1+dijxi+1yj+1

- the form of the interpolation methods

: ZI = interp2(X,Y,Z,XI,YI,method), method=> ‘cubic’,nearest’,’linear’

- interp3 : VI = interp3(X,Y,Z,V,XI,YI,ZI) interpolates to find VI, the values of the underlying three-dimensional function V at the points in arrays XI,YI and ZI. 3-dimensional

- interpn : VI = interpn(X1,X2,X3,...,V,Y1,Y2,Y3,...) interpolates to find VI, the values of the underlying multidimensional function V at the points in the arrays Y1, Y2, Y3, etc. n-dimensional

2005. 05. 11 2003010482 mohiuddin ahmad 2005010207 sung-bong jang interpolation ii (8.4 spline...

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