8/10/2019 6-01-interpolasi langsung-sjk.pdf

1/15

Direct Method of

Interpolation

1

8/10/2019 6-01-interpolasi langsung-sjk.pdf

2/15

What is Interpolation ? , , , , n, n ,

value of x that is not given.

2Figure 1 Interpolation of discrete.

8/10/2019 6-01-interpolasi langsung-sjk.pdf

3/15

Interpolants

Polynomials are the most commonchoice of inter olants because theare easy to:

Eva uate

Differentiate, and

Integrate

3

8/10/2019 6-01-interpolasi langsung-sjk.pdf

4/15

Direct Method

Given n+1 data points (x0,y0), (x1,y1),.. (xn,yn),pass a polynomial of order n through the data as given

w:

..................... nxaxaa +++=where a0, a1,. an are real constants.

et up n+1 equations to in n+1 constants. To find the value y at a given value of x, simply

substitute the value of x in the above ol nomial.

4

8/10/2019 6-01-interpolasi langsung-sjk.pdf

5/15

The upward velocity of a rocket is given as afunction of time in Table 1.

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

a e e oc y as a unc onof time.

s,t m/s,tv

0 0

10 227.04

15 362.78

20 517.35

22.5 602.97

5

30 901.67 Figure 2Velocity vs. time data for the

rocket example

8/10/2019 6-01-interpolasi langsung-sjk.pdf

6/15

Linear Interpolation

( ) taatv 10+=78.3621515 =+= aav ( )11,yx

y

( ) ( ) 35.5172020 10 =+= aav( )00

,yx

( )xf1

Solving the above two equations gives,

93.1000 =a 914.301=ax

Figure 3 Linear interpolation.

Hence

( ) .2015,914.3093.100 += tttv

6

( ) ( ) m/s7.39316914.3093.10016 =+=v

8/10/2019 6-01-interpolasi langsung-sjk.pdf

7/15

The upward velocity of a rocket is given as afunction of time in Table 2.

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

a e e oc y as a unc onof time.

s,t m/s,tv

0 0

10 227.04

15 362.78

20 517.35

22.5 602.97

7

30 901.67 Figure 5Velocity vs. time data for the

rocket example

8/10/2019 6-01-interpolasi langsung-sjk.pdf

8/15

Quadratic InterpolationQuadratic Interpolation

( ) 2210 tataatv ++=2

( )11

,yx

( )22 ,yx

y

.210 ==

( ) ( ) ( ) 78.362151515

2

210 =++= aaav

)xf235.517202020 210 =++= aaav ( )00 ,yx

x

Figure 6Quadratic interpolation.

Solving the above three equations gives

8

.0 .1 .2

8/10/2019 6-01-interpolasi langsung-sjk.pdf

9/15

Quadratic Interpolation (cont.)

450

500

550517.35

( ) 2010,3766.0733.1705.12 2 ++= ttttv

300

350

400s

f range( )

f x desired( )( ) ( ) ( )2163766.016733.1705.1216 ++=v

m/s19.392=

10 12 14 16 18 20200

250

227.04

2010 x s range, x desired,

the results from the first and second order polynomial is

70.39319.392

=

a

9

%38410.0

19.392

=

a

8/10/2019 6-01-interpolasi langsung-sjk.pdf

10/15

The upward velocity of a rocket is given as afunction of time in Table 3.

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

a e e oc y as a unc onof time.

s,t m/s,tv

0 0

10 227.04

15 362.7820 517.35

22.5 602.97

10

30 901.67 Figure 6Velocity vs. time data for the

rocket example

8/10/2019 6-01-interpolasi langsung-sjk.pdf

11/15

Cubic Interpolation

( ) 332

210 tatataatv +++= ( )33 ,yx

( )11,yx

( ) ( ) ( ) ( )332

210 10101004.22710 aaaav +++==

( ) ( ) ( ) ( )332210 15151578.36215 aaaav +++==( )xf3

( )22 ,yx( )00 ,yx

( ) ( ) ( ) ( )332

210 20202035.51720 aaaav +++==

32

x

Figure 7 Cubic interpolation.

3210 ..... ==

= = 13204.0=a 0054347.0=a

11

8/10/2019 6-01-interpolasi langsung-sjk.pdf

12/15

Cubic Interpolation (contd).,.... +++= tttttv

( ) ( ) ( ) ( )160054347.01613204.016266.212540.416 32 +++=v

600

700602.97

.

The absolute percentage relative

400

500y s

f range( )

f x desired( )

second and third order polynomial is

a

19.39206.392

10 12 14 16 18 20 22 24200

300

227.04

22.510 x s range, x desired,%033269.0

06.392

==a

12

8/10/2019 6-01-interpolasi langsung-sjk.pdf

13/15

Comparison Table

Table 4 Comparison of different orders of the polynomial.

Order of

Polynomial1 2 3

m/s16=tv 393.7 392.19 392.06

Absolute Relative

Approximate Error---------- 0.38410 % 0.033269 %

13

8/10/2019 6-01-interpolasi langsung-sjk.pdf

14/15

Distance from Velocity Profilen t e stance covere y t e roc et rom t= s to t= s

( ) 5.2210,0054347.013204.0266.212540.4 32 +++= tttt

( ) ( ) ( )1116

16

16

11

= dttvss

)0054347.013204.0266.212540.4

16432

11

32 +++= dtttt

m1605

40054347.0313204.02266.212540.411

=

+++= t

14

8/10/2019 6-01-interpolasi langsung-sjk.pdf

15/15

Acceleration from Velocity Profile

( ) 5.2210,0054347.013204.0266.212540.4 32 +++= ttttFin t e acce eration o t e roc et at t=16s given t at

( ) ( )=

d

tvdt

ta

5.2210,016382.026130.0289.21

....

2 ++=

+++=

ttt

tttdt

( ) ( ) ( )2

2

m/s665.29

16016304.01626408.0266.2116

=

++=a

15