6 04 Interpolasi Spline Sjk

7/21/2019 6 04 Interpolasi Spline Sjk

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Spline Method of

n erpo a on


2/25

Why Splines ?2251)( xxf +=

Table : Six e uidistantl s aced oints in [-1 1]

x 2

251

1

x

y+

=

- . .

-0.6 0.1

-

. .

0.2 0.5

0.6 0.1

Figure : 5th order polynomial vs. exact function1.0 0.038461


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Why Splines ?

0.8

1.2

0.4

y

-0.4

-1 -0.5 0 0.5 1

-0.8

x

19th Order Polynomial f (x) 5th Order Polynomial

Figure : Higher order polynomial interpolation is a bad idea


4/25

Linear Interpolation

Given ( ) ( ) ( )( )nnnn yxyxyxyx ,,,......,,,, 111100 , fit linear splines to the data. This simply involvesforming the consecutive data through straight lines. So if the above data is given in an ascending

order, the linear splines are given by )( ii xfy =

Figure : Linear splines


5/25

Linear Interpolation (contd)

),(

)()(

)()( 001

01

0 xxxx

xfxf

xfxf

+= 10 xxx

)()( 12 xfxf

,1

12

1xx 21

.

.

.

),()()(

)( 11

1

1

+= n

nn

nn

n xxxx

xfxfxf

nn xxx 1

Note the terms of

1)()( ii xfxf 1 ii

in the above function are simply slopes between 1ix and ix .


6/25

Example

he upward velocity of a rocket is given as afunction of time in Table 1. Find the velocity at.

Table Velocity as a

function of time

(s) (m/s)

0 0

10 227.04

t )(tv

15 362.7820 517.35

22.5 602.97

Figure. Velocity vs. time datafor the rocket example

30 901.67


7/25

Linear Interpolation

550517.35,150 =t 78.362)( 0 =tv

,201=t 35.517)( 1 =tv

450

500

y s

f range( )

)()()(

)()( 001

01

0 tttt

tvtvtvtv

+=

400

f x desired )15(

1520

..78.362

+= t

)15(913.3078.362)( += ttv

10 12 14 16 18 20 22 24350362.78

x s1

10+x s0

10 x s range, x desired,

At ,16=t

)1516(913.3078.362)16( +=v

.= m/s


8/25

Quadratic Interpolation

Given ( ) ( ) ( ) ( )nnnn yxyxyxyx ,,,,......,,,, 111100 , fit quadratic splines through the data. The splines

are given by

,)( 112

1 cxbxaxf ++= 10 xxx

,222

2 cxbxa ++= 21 xxx .

.

.

,2nnn

cxbxa ++= nn xxx 1

,i

,i

,i = , , ,


9/25

Quadratic Interpolation (contd)

Each quadratic spline goes through two consecutive data points

)( 01012

01 xfcxbxa =++

2

111111 .

.

.

)( 112

1 =++ iiiiii xfcxbxa

)(2

iiiiii xfcxbxa =++ .

.

.

)( 112

1 =++ nnnnnn xfcxbxa

)(2

nnnnnn xfcxbxa =++

This condition gives 2n equations


10/25

Quadratic Splines (contd)

The first derivatives of two quadratic splines are continuous at the interior points.

For example, the derivative of the first spline

11

2

1 cxbxa ++ is 112 bxa +

The derivative of the second spline

22

2

2 cxbxa ++ is 222 bxa +

and the two are equal at 1xx= giving

212111 22 bxabxa +=+

022 212111 =+ bxabxa


11/25

Quadratic Splines (contd)Similarly at the other interior points,

022 323222 =+ bxabxa

.

.

.

022 11 =+ ++ iiiiii bxabxa

.

.

.

022 1111 =+ nnnnnn bxabxa

- . = .

We can assume that the first spline is linear, that is 01=a


12/25

Quadratic Splines (contd)

This gives us 3n equations and 3n unknowns. Once we find the 3n constants,we can find the function at any value of x using the splines,

,)( 112

1 cxbxaxf ++= 10 xxx

,222

2 cxbxa ++= 21 xxx

.

.

.

,2 nnn cxbxa ++= nn xxx 1


13/25

Quadratic Spline Example

The upward velocity of a rocket is given as a function of time.Using quadratic splinesa

b) Find the acceleration at t=16 secondsc) Find the distance covered between t=11 and t=16 seconds

Table Velocity as a

function of time

(s) (m/s)

0 0

10 227.04

t )(tv

.20 517.35

22.5 602.97

gure. e oc y vs. me a afor the rocket example

.


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Solution2= ,111

,2 ctbta ++= 1510 t

,332

3 ctbta ++= 2015 t,44

2

4 ctbta ++= 5.2220 t

,555 ctbta ++= 305.22 t


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Each Spline Goes Throughwo Consecutive Data Points

,111 cav = t2 =111

04.227)10()10( 112

1 =++ cba


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Each Spline Goes Throughwo Consecutive Data Points

2

t v(t)

s m/s

.222 =

78.362)15()15( 222

2 =++ cba

0 0

10 227.04

78.362)15()15( 332

3 =++ cba

35.5172020 2 =++ cba15 362.78

20 517.3535.517)20()20( 44

2

4 =++ cba2

22.5 602.97

30 901.67

... 44497.602)5.22()5.22( 55

2

5 =++ cba

67.901)30()30( 555 =++ cba


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Interior Data Points

,111

cav = t

,222

2 ctbta ++= 1510 t

( ) ( )222

211

2

1 ++=++ ctbtad

ctbta

d

1010 == tt

( ) ( )2211 22 +=+ btabta

( ) ( ) 2211 102102 baba +=+

020202211

=+ baba


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Derivatives are continuous at

n er or a a o n s

0)10(2)10(2 =+ babaAt t=10

0152152 =+ babaAt t=15

=At t=20

At t=22.5

.. 5544


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Last Equation

1


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Final Set of Equations 0000000000000100 1a

78.362

04.227

04.227

000000000115225000

000000000110100000

000000000000110100

2

1

1

a

c

b

= 97.602

35.517

35.517

.

00015.2225.506000000000

000120400000000000

000000120400000000

3

2

2

b

a

c

0

67.901

97.602

00000000001200120

130900000000000000

15.2225.506000000000000

4

4

3

b

a

c

0

00

01450145000000000

0000140014000000000000001300130000

5

5

4

b

ac

5c


21/25

Coefficients of Spline

i i i

1 0 22.704 0

. . .

3

0.1356 35.66

141.614 1.6048 33.956 554.55


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Final Solution,704.22)( ttv = 100 t

,88.88928.48888.02

++= tt 1510 t61.14166.351356.0 2 += tt 2015 t

,55.554956.336048.1 2 += tt 5.2220 t

13.15286.2820889.0 2 += tt 305.22 t


23/25

Velocit at a Particular Pointa) Velocity at t=16

,704.22)( ttv = 100 t2= ,...

,61.14166.351356.0 2 += tt 2015 t2 = ,... .

,13.15286.2820889.0 2 += tt 305.22 t

( ) ( ) ( ) 61.1411666.35161356.016 2 +=v.


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b) The quadratic spline valid at t=16 is

16 = tvd

a

g ven y

16=tdt

61.14166.351356.0 2 += tttv 2015 t

)61.14166.351356.0()( 2 += ttd

ta

,66.352712.0 += t 2015 t2

..=a .=


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Distance from Velocity Profile

t=16s.( ) ( )=

16

)(1116 dttvSS11

( ) 1510,88.88928.48888.02

2 ++= ttttv

,... =

( ) ( ) ( ) ( ) ( )111616 15 16

+== dttvdttvdttvSS

( ) ( )61.14166.351356.088.88928.48888.016

2

15

2

11 11 15

++++= dtttdttt

m9.1595

1511

=

6 04 Interpolasi Spline Sjk

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