1 INTERPOLASI. Direct Method of Interpolation 3 What is Interpolation ? Given (x 0,y 0 ), (x 1,y 1...

1

INTERPOLASI

Direct Method of Interpolation

3

What is Interpolation ?Given (x0,y0), (x1,y1), …… (xn,yn), find the value of ‘y’ at a value of ‘x’ that is not given.

Figure 1 Interpolation of discrete.

4

Interpolants

Polynomials are the most common choice of interpolants because they are easy to:

Evaluate Differentiate, and Integrate

5

Direct MethodGiven ‘n+1’ data points (x0,y0), (x1,y1),………….. (xn,yn),pass a polynomial of order ‘n’ through the data as

given below:

where a0, a1,………………. an are real constants. Set up ‘n+1’ equations to find ‘n+1’ constants. To find the value ‘y’ at a given value of ‘x’, simply

substitute the value of ‘x’ in the above polynomial.

.....................10n

n xaxaay

6

Example 1 The upward velocity of a rocket is given as a

function of time in Table 1.

Find the velocity at t=16 seconds using the direct method for linear interpolation.

0 0

10 227.04

15 362.78

20 517.35

22.5 602.97

30 901.67

Table 1 Velocity as a function of time.

Figure 2 Velocity vs. time data for the rocket example

s ,t m/s ,tv

7

Linear Interpolation taatv 10

78.3621515 10 aav

35.5172020 10 aav

Solving the above two equations gives,

93.1000 a 914.301 a

Hence .2015,914.3093.100 tttv

m/s 7.39316914.3093.10016 v

00 , yx

xf1

11, yx

x

y

Figure 3 Linear interpolation.

8



Find the velocity at t=16 seconds using the direct method for quadratic interpolation.

0 0

10 227.04

15 362.78

20 517.35

22.5 602.97

30 901.67



s ,t m/s ,tv

9

Quadratic Interpolation

2210 tataatv

04.227101010 2210 aaav

78.362151515 2210 aaav

35.517202020 2210 aaav

Solving the above three equations gives

05.120 a 733.171 a 3766.02 a

Quadratic Interpolation

00 , yx

11, yx

22 , yx

xf2

y

x

Figure 6 Quadratic interpolation.

10

Quadratic Interpolation (cont.)

10 12 14 16 18 20200

250

300

350

400

450

500

550517.35

227.04

y s

f range( )

f x desired

2010 x s range x desired

2010,3766.0733.1705.12 2 ttttv

2163766.016733.1705.1216 v

m/s 19.392

The absolute relative approximate error obtained between the results from the first and second order polynomial is

%38410.0

10019.392

70.39319.392

a

a

11



Find the velocity at t=16 seconds using the direct method for cubic interpolation.

0 0

10 227.04

15 362.78

20 517.35

22.5 602.97

30 901.67



s ,t m/s ,tv

12

Cubic Interpolation

33

2210 tatataatv

33

2210 10101004.22710 aaaav

33

2210 15151578.36215 aaaav

33

2210 20202035.51720 aaaav

33

2210 5.225.225.2297.6025.22 aaaav

2540.40 a 266.211 a 13204.02 a 0054347.03 a

y

x

xf3

33 , yx

22 , yx

11, yx

00 , yx

Figure 7 Cubic interpolation.

13

Cubic Interpolation (contd)

10 12 14 16 18 20 22 24200

300

400

500

600

700602.97

227.04

y s

f range( )

f x desired

22.510 x s range x desired

5.2210,0054347.013204.0266.212540.4 32 tttttv

m/s 06.392

160054347.01613204.016266.212540.416 32

v

The absolute percentage relative approximate error between second and third order polynomial is

a

%033269.0

10006.392

19.39206.392

a

14

Comparison Table

Order of Polynomial

1 2 3

m/s 16tv 393.7 392.19 392.06

Absolute Relative Approximate Error

---------- 0.38410 % 0.033269 %

Table 4 Comparison of different orders of the polynomial.

15

Distance from Velocity ProfileFind the distance covered by the rocket from t=11s to t=16s ?

5.2210,0054606.013064.0289.213810.4 32 tttttv

m 1605

40054347.0

313204.0

2266.212540.4

0054347.013204.0266.212540.4

1116

16

11

432

16

11

32

16

11

tttt

dtttt

dttvss

16

Acceleration from Velocity Profile

5.2210,0054347.013204.0266.212540.4 32 ttttFind the acceleration of the rocket at t=16s given that

5.2210 ,016382.026130.0289.21

0054347.013204.0266.212540.4

2

32

ttt

tttdt

d

tvdt

dta

2

2

m/s 665.29

16016304.01626408.0266.2116

a

Lagrange Method of Interpolation

18

What is Interpolation ?

Given (x0,y0), (x1,y1), …… (xn,yn), find the value of ‘y’ at a value of ‘x’ that is not given.

19

Interpolants


EvaluateDifferentiate, and Integrate.

20

Lagrangian Interpolation

Lagrangian interpolating polynomial is given by

n

iiin xfxLxf

0

)()()(

where ‘ n ’ in )(xf n stands for the thn order polynomial that approximates the function )(xfy

given at )1( n data points as nnnn yxyxyxyx ,,,,......,,,, 111100 , and

n

ijj ji

ji xx

xxxL

0

)(

)(xLi is a weighting function that includes a product of )1( n terms with terms of ij

omitted.

21

Example The upward velocity of a rocket is given as a

function of time in Table 1. Find the velocity at t=16 seconds using the Lagrangian method for linear interpolation.

Table Velocity as a function of time

Figure. Velocity vs. time data for the rocket example

(s) (m/s)

0 010 227.0415 362.7820 517.35

22.5 602.9730 901.67

t )(tv

22

Linear Interpolation

10 12 14 16 18 20 22 24350

400

450

500

550517.35

362.78

y s

f range( )

f x desired

x s1

10x s0


)()()(1

0ii

itvtLtv

)()()()( 1100 tvtLtvtL

78.362,15 00 tt

35.517,20 11 tt

23

Linear Interpolation (contd)

1

00 0

0 )(

jj j

j

tt

tttL

10

1

tt

tt

1

10 1

1 )(

jj j

j

tt

tttL

01

0

tt

tt

)()()( 101

00

10

1 tvtt

tttv

tt

tttv

)35.517(1520

15)78.362(

2015

20

tt

)35.517(1520

1516)78.362(

2015

2016)16(

v

)35.517(2.0)78.362(8.0

7.393 m/s.

24

Quadratic InterpolationFor the second order polynomial interpolation (also called quadratic interpolation), we

choose the velocity given by

2

0

)()()(i

ii tvtLtv

)()()()()()( 221100 tvtLtvtLtvtL

25


function of time in Table 1. Find the velocity at t=16 seconds using the Lagrangian method for quadratic interpolation.



(s) (m/s)

0 010 227.0415 362.7820 517.35

22.5 602.9730 901.67

t )(tv

26

Quadratic Interpolation (contd)

10 12 14 16 18 20200

250

300

350

400

450

500

550517.35

227.04

y s

f range( )

f x desired


,100 t 04.227)( 0 tv

,151 t 78.362)( 1 tv

,202 t 35.517)( 2 tv

2

00 0

0 )(

jj j

j

tt

tttL

20

2

10

1

tt

tt

tt

tt

2

10 1

1 )(

jj j

j

tt

tttL

21

2

01

0

tt

tt

tt

tt

2

20 2

2 )(

jj j

j

tt

tttL

12

1

02

0

tt

tt

tt

tt

27


m/s19.392

35.52712.078.36296.004.22708.0

35.5171520

1516

1020

101678.362

2015

2016

1015

101604.227

2010

2016

1510

151616

212

1

02

01

21

2

01

00

20

2

10

1

v

tvtt

tt

tt

tttv

tt

tt

tt

tttv

tt

tt

tt

tttv


a

%38410.0

10019.392

70.39319.392

a

28

Cubic Interpolation

10 12 14 16 18 20 22 24200

300

400

500

600

700602.97

227.04

y s

f range( )

f x desired


For the third order polynomial (also called cubic interpolation), we choose the velocity given by

3

0

)()()(i

ii tvtLtv

)()()()()()()()( 33221100 tvtLtvtLtvtLtvtL

29


function of time in Table 1. Find the velocity at t=16 seconds using the Lagrangian method for cubic interpolation.



(s) (m/s)

0 010 227.0415 362.7820 517.35

22.5 602.9730 901.67

t )(tv

30

Cubic Interpolation (contd) 04.227,10 oo tvt 78.362,15 11 tvt

35.517,20 22 tvt 97.602,5.22 33 tvt

3

00 0

0 )(

jj j

j

tt

tttL

30

3

20

2

10

1

tt

tt

tt

tt

tt

tt;

3

10 1

1 )(

jj j

j

tt

tttL

31

3

21

2

01

0

tt

tt

tt

tt

tt

tt

3

20 2

2 )(

jj j

j

tt

tttL

32

3

12

1

02

0

tt

tt

tt

tt

tt

tt;

3

30 3

3 )(

jj j

j

tt

tttL

23

2

13

1

03

0

tttt

tttt

tt

tt

10 12 14 16 18 20 22 24200

300

400

500

600

700602.97

227.04

y s

f range( )

f x desired


31

Cubic Interpolation (contd)

m/s06.392

97.6021024.035.517312.078.362832.004.2270416.0

97.602205.22

2016

155.22

1516

105.22

101635.517

5.2220

5.2216

1520

1516

1020

1016

78.3625.2215

5.2216

2015

2016

1015

101604.227

5.2210

5.2216

2010

2016

1510

151616

323

2

13

1

13

12

32

3

12

1

02

0

231

3

21

2

01

01

30

3

20

2

10

1

v

tvtt

tt

tt

tt

tt

tttv

tt

tt

tt

tt

tt

tt

tvtt

tt

tt

tt

tt

tttv

tt

tt

tt

tt

tt

tttv


a

%033269.0

10006.392

19.39206.392

a

32

Comparison Table

Order of Polynomial

1 2 3

v(t=16) m/s 393.69 392.19 392.06

Absolute Relative

Approximate Error

--------0.38410

%0.033269

%

33

Distance from Velocity Profile

Find the distance covered by the rocket from t=11s to

t=16s ?

,00544.013195.0265.21245.4)( 32 ttttv 5.2210 t

16

11

)()11()16( dttvss

16

11

32 )00544.013195.0265.21245.4( dtttt

1611

432

]4

00544.03

13195.02

265.21245.4[ttt

t

1605 m

)5727.2)(300065045()1388.4)(33755.7125.47(

)9348.1)(45008755.52()36326.0)(67505.10875.57()(2323

2323

tttttt

tttttttv

34


,

Find the acceleration of the rocket at t=16s given that

32 00544.013195.0265.21245.4 tttdt

dtv

dt

dta

201632.026390.0265.21 tt

2)16(01632.0)16(26390.0265.21)16( a

665.29 2/ sm

,00544.013195.0265.21245.4)( 32 ttttv 5.2210 t

Newton’s Divided Difference Method of

Interpolation

36

What is Interpolation ?

Given (x0,y0), (x1,y1), …… (xn,yn), find the value of ‘y’ at a value of ‘x’ that is not given.

37

Interpolants


EvaluateDifferentiate, and Integrate.

38

Newton’s Divided Difference Method

Linear interpolation: Given pass a linear interpolant through the data

where

),,( 00 yx ),,( 11 yx

)()( 0101 xxbbxf

)( 00 xfb

01

011

)()(

xx

xfxfb

39


function of time in Table 1. Find the velocity at t=16 seconds using the Newton Divided Difference method for linear interpolation.

Table. Velocity as a function of time


0 010 227.0415 362.7820 517.35

22.5 602.9730 901.67

)s( t )m/s( )(tv

40

Linear Interpolation

10 12 14 16 18 20 22 24350

400

450

500

550517.35

362.78

y s

f range( )

f x desired

x s1

10x s0


,150 t 78.362)( 0 tv

,201 t 35.517)( 1 tv

)( 00 tvb 78.362

01

011

)()(

tt

tvtvb

914.30

)()( 010 ttbbtv

41

Linear Interpolation (contd)

10 12 14 16 18 20 22 24350

400

450

500

550517.35

362.78

y s

f range( )

f x desired

x s1

10x s0


)()( 010 ttbbtv

),15(914.3078.362 t 2015 t

At 16t

)1516(914.3078.362)16( v

69.393 m/s

42

Quadratic InterpolationGiven ),,( 00 yx ),,( 11 yx and ),,( 22 yx fit a quadratic interpolant through the data.

))(()()( 1020102 xxxxbxxbbxf

)( 00 xfb

01

011

)()(

xx

xfxfb

02

01

01

12

12

2

)()()()(

xx

xx

xfxf

xx

xfxf

b

43


function of time in Table 1. Find the velocity at t=16 seconds using the Newton Divided Difference method for quadratic interpolation.



0 010 227.0415 362.7820 517.35

22.5 602.9730 901.67

)s( t )m/s( )(tv

44


10 12 14 16 18 20200

250

300

350

400

450

500

550517.35

227.04

y s

f range( )

f x desired


,100 t 04.227)( 0 tv

,151 t 78.362)( 1 tv

,202 t 35.517)( 2 tv

45


)( 00 tvb

04.227

01

011

)()(

tt

tvtvb

1015

04.22778.362

148.27

02

01

01

12

12

2

)()()()(

tt

tt

tvtv

tt

tvtv

b

1020

1015

04.22778.362

1520

78.36235.517

10

148.27914.30

37660.0

46


))(()()( 102010 ttttbttbbtv

),15)(10(37660.0)10(148.2704.227 ttt 2010 t

At ,16t

)1516)(1016(37660.0)1016(148.2704.227)16( v 19.392 m/s

The absolute relative approximate error a obtained between the results from the first

order and second order polynomial is

a 100x19.392

69.39319.392

= 0.38502 %

47

General Form

))(()()( 1020102 xxxxbxxbbxf

where

Rewriting))(](,,[)](,[][)( 1001200102 xxxxxxxfxxxxfxfxf

)(][ 000 xfxfb

01

01011

)()(],[

xx

xfxfxxfb

02

01

01

12

12

02

01120122

)()()()(

],[],[],,[

xx

xx

xfxf

xx

xfxf

xx

xxfxxfxxxfb

48

General FormGiven )1( n data points, nnnn yxyxyxyx ,,,,......,,,, 111100 as

))...()((....)()( 110010 nnn xxxxxxbxxbbxf

where

][ 00 xfb

],[ 011 xxfb

],,[ 0122 xxxfb

],....,,[ 0211 xxxfb nnn

],....,,[ 01 xxxfb nnn

49

General formThe third order polynomial, given ),,( 00 yx ),,( 11 yx ),,( 22 yx and ),,( 33 yx is

))()(](,,,[

))(](,,[)](,[][)(

2100123

1001200103

xxxxxxxxxxf

xxxxxxxfxxxxfxfxf

0b

0x )( 0xf 1b

],[ 01 xxf 2b

1x )( 1xf ],,[ 012 xxxf 3b

],[ 12 xxf ],,,[ 0123 xxxxf

2x )( 2xf ],,[ 123 xxxf

],[ 23 xxf

3x )( 3xf

50


function of time in Table 1. Find the velocity at t=16 seconds using the Newton Divided Difference method for cubic interpolation.



0 010 227.0415 362.7820 517.35

22.5 602.9730 901.67

)s( t )m/s( )(tv

51

Example

The velocity profile is chosen as))()(())(()()( 2103102010 ttttttbttttbttbbtv

we need to choose four data points that are closest to 16t

,100 t 04.227)( 0 tv

,151 t 78.362)( 1 tv ,202 t 35.517)( 2 tv ,5.223 t 97.602)( 3 tv

The values of the constants are found as: b0 = 227.04; b1 = 27.148; b2 = 0.37660; b3 = 5.4347×10−3

52

Example

b0 = 227.04; b1 = 27.148; b2 = 0.37660; b3 = 5.4347×10−3

0b

100 t 04.227 1b

148.27 2b

,151 t 78.362 37660.0 3b

914.30 3104347.5

,202 t 35.517 44453.0

248.34

,5.223 t 97.602

53

ExampleHence

))()(())(()()( 2103102010 ttttttbttttbttbbtv

)20)(15)(10(10*4347.5

)15)(10(37660.0)10(148.2704.2273

ttt

ttt

At ,16t

)2016)(1516)(1016(10*4347.5

)1516)(1016(37660.0)1016(148.2704.227)16(3

v

06.392 m/s

The absolute relative approximate error a obtained is

a 100x06.392

19.39206.392

= 0.033427 %

54

Comparison Table

Order of Polynomial

1 2 3

v(t=16) m/s

393.69 392.19 392.06

Absolute Relative Approximate Error

---------- 0.38502 %

0.033427 %

55

Distance from Velocity Profile

Find the distance covered by the rocket from t=11s tot=16s ?

)20)(15)(10(10*4347.5

)15)(10(37660.0)10(148.2704.227)(3

ttt

ttttv 5.2210 t

32 0054347.013204.0265.212541.4 ttt 5.2210 t So

16

11

1116 dttvss

dtttt )0054347.013204.0265.212541.4( 3216

11

16

11

432

40054347.0

313204.0

2265.212541.4

tttt

m 1605

56


Find the acceleration of the rocket at t=16s given that

32 0054347.013204.0265.212541.4)( ttttv

32 0054347.013204.0265.212541.4)()( tttdt

dtv

dt

dta

2016304.026408.0265.21 tt

2)16(016304.0)16(26408.0265.21)16( a

2/ 664.29 sm

1 INTERPOLASI. Direct Method of Interpolation 3 What is Interpolation ? Given (x 0,y 0 ), (x 1,y 1...

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