Transcript Spline Interpolation Method Power Point

Spline Interpolation Method
Major: All Engineering Majors
Authors: Autar Kaw, Jai Paul
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Transforming Numerical Methods Education for STEM
Undergraduates
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1
Spline Method of
Interpolation
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What is Interpolation ?
Given (x0,y0), (x1,y1), …… (xn,yn), find the
value of ‘y’ at a value of ‘x’ that is not given.
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Interpolants
Polynomials are the most common
choice of interpolants because they
are easy to:
Evaluate
Differentiate, and
Integrate.
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Why Splines ?
f ( x) 
1
1  25 x
2
Ta ble : S ix e q uid is tant ly s pace d poin ts in [-1, 1]
x
1
1  25 x
- 1.0
0.038461
- 0.6
0.1
- 0.2
0.5
0.2
0.5
0.6
0.1
1.0
5
y 
0.038461
2
Fig u re : 5
th
orde r po ly no mia l vs . e xact fu nct io n
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Why Splines ?
1.2
0.8
y
0.4
0
-1
-0.5
0
0.5
1
-0.4
-0.8
x
19th Order Polynomial
f (x)
5th Order Polynomial
Figure : Higher order polynomial interpolation is a bad idea
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Linear Interpolation
G ive n
 x 0 , y 0 ,  x 1 , y 1 ,......,  x n  1 , y n 1  x n , y n  , fit
linea r sp lines to the da ta. T his s im p ly invo lves
fo r m ing the co ns ec utive data thro ugh s tra ight lines. S o if the abo ve data is give n in a n a sce nd ing
order, the linea r sp lines a re give n b y
 yi
 f ( xi ) 
Fig u re : L ine ar s p line s
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Linear Interpolation (contd)
f ( x )  f ( x0 ) 
 f ( x1 ) 
f ( x1 )  f ( x 0 )
x1  x 0
f ( x 2 )  f ( x1 )
x2  x1
( x  x 0 ),
x 0  x  x1
( x  x 1 ),
x1  x  x 2
.
.
.
 f ( x n 1 ) 
f ( x n )  f ( x n 1 )
x n  x n 1
( x  x n  1 ), x n  1  x  x n
N o te th e t erm s o f
f ( xi )  f ( x i1 )
xi  x i1
in the ab o ve func t io n are s im p ly s lo p es b etween x i 1 and x i .
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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 linear splines.
Table Velocity as a
function of time
t (s)
0
10
15
20
22.5
30
9
(m/s)
0
227.04
362.78
517.35
602.97
901.67
v (t )
Figure. Velocity vs. time data
for the rocket example
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Linear Interpolation
t 0  15 ,
v ( t 0 )  362 . 78
t1  20,
v ( t1 )  517 . 35
v (t )  v (t 0 ) 
v (t 1 )  v (t 0 )
t1  t 0
550
517.35
500
(t  t 0 )
y s
f ( range )
 362 . 78 
517 . 35  362 . 78
20  15
( t  15 )

f x desired
362.78
v (16 )  362 . 78  30 . 913 (16  15 )
 3 9 3. 7
10

400
v ( t )  362 . 78  30 . 913( t  15 )
At t  16 ,
450
350
10
12
x s  10
0
14
16
18
x s  range  x desired
20
22
24
x s  10
m /s
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1
Quadratic Interpolation
G ive n
 x 0 , y 0 ,  x 1 , y 1 ,......,  x n  1 , y n 1 ,  x n , y n  , fit q uad ratic sp lines
thro ugh the data. T he sp lines
are give n b y
f ( x )  a 1 x  b1 x  c 1 ,
2
 a 2 x  b2 x  c2 ,
2
x 0  x  x1
x1  x  x 2
.
.
.
 a n x  bn x  cn ,
2
x n 1  x  x n
F ind a i , bi , c i , i  1, 2, … , n
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Quadratic Interpolation (contd)
Eac h q uadr atic sp line goes thro ugh tw o co nsec utive data po ints
2
a 1 x 0  b1 x 0  c 1  f ( x 0 )
a 1 x 1  b 1 x 1  c 1  f ( x1 )
2
.
.
.
2
a i x i 1  b i x i 1  c i  f ( x i 1 )
2
a i xi  bi xi  c i  f ( xi )
.
.
.
2
a n x n  1  b n x n  1  c n  f ( x n 1 )
2
a n x n  bn xn  cn  f ( x n )
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T his co nd itio n give s 2 n eq uatio ns
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Quadratic Splines (contd)
T he firs t der ivatives o f tw o q ua d ratic sp lines a re co ntinuo us a t the inte r ior po ints.
F or exa m p le, the der ivative o f the firs t sp line
a 1 x  b1 x  c 1
2
is
2 a1 x  b1
T he der ivative o f the seco nd sp line
a 2 x  b 2 x  c 2 is
2
2 a2 x  b2
and the tw o are eq ua l at x  x 1 giving
2 a 1 x 1  b 1  2 a 2 x1  b 2
2 a1 x 1  b1  2 a 2 x1  b 2  0
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Quadratic Splines (contd)
S im ilarly at the other interior points,
2 a 2 x 2  b 2  2 a 3 x 2  b3  0
.
.
.
2 a i x i  b i  2 a i 1 x i  b i 1  0
.
.
.
2 a n 1 x n 1  b n 1  2 a n x n 1  b n  0
W e have (n -1) su ch equ ations. T he total num ber of equations is ( 2 n )  ( n  1)  ( 3 n  1) .
W e can assum e that the first spline is linear, that is
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a1  0
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Quadratic Splines (contd)
This gives us ‘3n’ equations and ‘3n’ unknow ns. O nce w e find the ‘3n’ constants,
w e can find the function at an y value of ‘x’ using the splines,
f ( x )  a 1 x  b1 x  c 1 ,
2
 a 2 x  b2 x  c 2 ,
2
x 0  x  x1
x1  x  x 2
.
.
.
 a n x  bn x  c n ,
2
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x n 1  x  x n
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Quadratic Spline Example
The upward velocity of a rocket is given as a function of time.
Using quadratic splines
a) Find the velocity at t=16 seconds
b) Find the acceleration at t=16 seconds
c) Find the distance covered between t=11 and t=16 seconds
Table Velocity as a
function of time
t (s)
0
10
15
20
22.5
30
16
(m/s)
0
227.04
362.78
517.35
602.97
901.67
v (t )
Figure. Velocity vs. time data
for the rocket example
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Solution
v ( t )  a 1 t  b1 t  c1 ,
2
0  t  10
 a 2 t  b2 t  c 2 ,
10  t  15
 a 3 t  b3 t  c 3 ,
15  t  20
 a 4 t  b4 t  c 4 ,
20  t  22 . 5
 a 5 t  b5 t  c 5 ,
22 . 5  t  30
2
2
2
2
Let us set up the equations
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Each Spline Goes Through
Two Consecutive Data Points
v ( t )  a 1 t  b1 t  c1 , 0  t  10
2
a 1 ( 0 )  b1 ( 0 )  c 1  0
2
a 1 (10 )  b1 (10 )  c 1  227 . 04
2
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Each Spline Goes Through
Two Consecutive Data Points
a 2 (10 )  b 2 (10 )  c 2  227 . 04
2
t
s
0
v(t)
m/s
0
a 2 (15 )  b 2 (15 )  c 2  362 . 78
10
15
20
22.5
227.04
362.78
517.35
602.97
a 3 ( 20 )  b 3 ( 20 )  c 3  517 . 35
2
a 3 (15 )  b3 (15 )  c 3  362 . 78
2
2
a 4 ( 20 )  b 4 ( 20 )  c 4  517 . 35
2
a 4 ( 22 . 5 )  b 4 ( 22 . 5 )  c 4  602 . 97
2
a 5 ( 22 . 5 )  b5 ( 22 . 5 )  c 5  602 . 97
2
30
901.67
a 5 ( 30 )  b 5 ( 30 )  c 5  901 . 67
2
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Derivatives are Continuous at
Interior Data Points
2
v ( t )  a 1 t  b1 t  c1 , 0  t  10
 a 2 t  b 2 t  c 2 ,10  t  15
2
d
dt
a t
1
2
 b1t  c1


t  10
 2 a1t  b1  t 10
d
dt
a
t  b2 t  c 2
2
2

t  10
  2 a 2 t  b 2  t 10
2 a 1 10   b1  2 a 2 10   b 2
20 a 1  b1  20 a 2  b 2  0
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Derivatives are continuous at
Interior Data Points
At t=10
2 a 1 (10 )  b1  2 a 2 (10 )  b 2  0
At t=15
2 a 2 (15 )  b 2  2 a 3 (15 )  b 3  0
At t=20
2 a 3 ( 20 )  b 3  2 a 4 ( 20 )  b 4  0
At t=22.5
2 a 4 ( 22 . 5 )  b 4  2 a 5 ( 22 . 5 )  b 5  0
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Last Equation
a1  0
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Final Set of Equations
 0
100

0

 0
 0
 0

 0
 0
 0

 0
 20
 0

0

 0
 1
23
0
1
0
0
0
0
0
0
0
0
0
0
10
1
0
0
0
0
0
0
0
0
0
0
0
0
100
10
1
0
0
0
0
0
0
0
0
0
225
15
1
0
0
0
0
0
0
0
0
0
0
0
0
225
15
1
0
0
0
0
0
0
0
0
0
400
20
1
0
0
0
0
0
0
0
0
0
0
0
0
400
20
1
0
0
0
0
0
0
0
0
0
506 . 25
22 . 5
1
0
0
0
0
0
0
0
0
0
0
0
0
506 . 25
0
0
0
0
0
0
0
0
0
0
0
900
1
0
 20
1
0
0
0
0
0
0
0
0
0
0
30
1
0
 30
1
0
0
0
0
0
0
0
0
0
0
40
1
0
 40
1
0
0
0
0
0
0
0
0
0
0
45
1
0
 45
0
0
0
0
0
0
0
0
0
0
0
0
0   a1 
 0 
0
0   b1   227 . 04 
  

0
0 c1
227 . 04
  

0
0   a 2   362 . 78 
0
0   b 2   362 . 78 
    517 . 35 
0
0 c2
  

0
0   a 3   517 . 35 
0
0   b 3    602 . 97 
22 . 5 1   c 3   602 . 97 
  

30
1   a 4   901 . 67 
0
0   b4   0 
0
0  c4   0 
  

0
0 a5
0
  

1
0   b5   0 
0
0   c 5   0 
0
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Coefficients of Spline
24
i
ai
bi
ci
1
0
22.704
0
2
0.8888
4.928
88.88
3
−0.1356
35.66
−141.61
4
1.6048
5
0.20889
−33.956 554.55
28.86
−152.13
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Quadratic Spline Interpolation
Part 2 of 2
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Final Solution
0  t  10
v ( t )  22 . 704 t ,
 0 . 8888 t  4 . 928 t  88 . 88 ,
10  t  15
  0 . 1356 t  35 . 66 t  141 . 61 ,
15  t  20
 1 . 6048 t  33 . 956 t  554 . 55 ,
20  t  22 . 5
 0 . 20889 t  28 . 86 t  152 . 13 ,
22 . 5  t  30
2
2
2
2
26
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Velocity at a Particular Point
a) Velocity at t=16
0  t  10
v ( t )  22 . 704 t ,
 0 . 8888 t  4 . 928 t  88 . 88 ,
10  t  15
  0 . 1356 t  35 . 66 t  141 . 61 ,
15  t  20
 1 . 6048 t  33 . 956 t  554 . 55 ,
20  t  22 . 5
 0 . 20889 t  28 . 86 t  152 . 13 ,
22 . 5  t  30
2
2
2
2
v 16    0 . 1356 16   35 . 66 16   141 . 61
2
 394 . 24 m/s
27
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Acceleration from Velocity Profile
b) The quadratic spline valid at t=16 is
given by
a (16 ) 
d
dt
v (t )
t  16
2
v t    0 . 1356 t  35 . 66 t  141 . 61 , 15  t  20
a (t ) 
d
(  0 . 1356 t  35 . 66 t  141 . 61 )
2
dt
  0 . 2712 t  35 . 66 , 15  t  20
a (16 )   0 . 2712 (16 )  35 . 66  31 . 321 m/s
28
2
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Distance from Velocity Profile
c) Find the distance covered by the rocket from t=11s to
t=16s.
S 16   S 11  
16
 v ( t ) dt
11
v t   0 . 8888 t  4 . 928 t  88 . 88 , 10  t  15
2
  0 . 1356 t  35 . 66 t  141 . 61 , 15  t  20
2
S 16   S 11  
16
15
16
11
11
15
 v t dt   v t dt   v t dt
15

 0 . 8888 t
11
2

 4 . 928 t  88 . 88 dt 
16
  0 . 1356 t
2

 35 . 66 t  141 . 61 dt
15
 1595 . 9 m
29
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Additional Resources
For all resources on this topic such as digital audiovisual
lectures, primers, textbook chapters, multiple-choice
tests, worksheets in MATLAB, MATHEMATICA, MathCad
and MAPLE, blogs, related physical problems, please
visit
http://numericalmethods.eng.usf.edu/topics/spline_met
hod.html
THE END
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