Transcript PPT

Spline Interpolation Method
Industrial Engineering Majors
Authors: Autar Kaw, Jai Paul
http://numericalmethods.eng.usf.edu
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.
4
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Why Splines ?
1
f ( x) 
1  25 x 2
Table : Six equidistantly spaced points in [-1, 1]
x
1
1  25 x 2
-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
Figure : 5th order polynomial vs. exact function
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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
Given  x0 , y0 , x1 , y1 ,......, x n 1 , y n1  x n , y n  , fit linear splines to the data. This simply involves
forming the consecutive data through straight lines. So if the above data is given in an ascending
order, the linear splines are given by  yi  f ( xi ) 
Figure : Linear splines
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Linear Interpolation (contd)
f ( x )  f ( x0 ) 
f ( x1 )  f ( x 0 )
( x  x 0 ),
x1  x 0
x 0  x  x1
 f ( x1 ) 
f ( x 2 )  f ( x1 )
( x  x1 ),
x2  x1
x1  x  x 2
.
.
.
 f ( x n 1 ) 
f ( x n )  f ( x n 1 )
( x  x n 1 ), x n 1  x  x n
x n  x n 1
Note the terms of
f ( xi )  f ( x i 1 )
xi  x i 1
in the above function are simply slopes between xi 1 and x i .
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Example
A curve needs to be fit through the given points to fabricate
the cam. If the cam follows a straight line profile between
x = 1.28 to x = 0.66, what is the value of y at x=1.1 ? Find
using linear spline interpolation.
9
x (in.) y (in.)
2.20
0.00
1.28
0.88
0.66
1.14
0.00
1.20
–0.60 1.04
–1.04 0.60
–1.20 0.00
Cam Profile
4
3
5
2
1.4
1.2
1
6
0.8
y
Point
1
2
3
4
5
6
7
Y
7
0.6
1
0.4
X
0.2
0
-2
-1
0
1
2
x
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Linear Interpolation
x0  1.28,
y( x0 )  0.88
x1  0.66,
y( x1 )  1.14
y( x1 )  y( x0 )
y ( x)  y ( x0 ) 
( x  x0 )
x1  x0
1.14
1.1
1.05
ys
f ( range)

f x desired

1.14  0.88
 0.88 
( x  1.28)
0.66  1.28
y( x)  0.88  0.41935( x  1.28)
0.66  x  1.28
1
0.95
0.9
0.88
0.85
5
x s  10
0
0
5
x s  range  x desired
At x  1.10,
y (1.10)  0.88  0.41935(1.10  1.28)
 0.95548 in.
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x s  10
1
Linear Interpolation
Find the cam profile using linear splines.
The first linear spline connects x  1.20 and x  1.04 :
y ( x)  0.00 
0.60  0.00
( x  1.20)
 1.04  1.20
y ( x)  3.75( x  1.20)
 1.20  x  1.04
Similarly,
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y ( x)  x  1.64,
 1.04  x  0.60
y ( x)  1.04  0.26667( x  0.60),
 0.60  x  0.00
y ( x)  1.20  0.090909 x,
0.00  x  0.66
y ( x)  1.14  0.41935( x  0.66),
0.66  x  1.28
y ( x)  0.88  0.95652( x  1.28),
1.28  x  2.20
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Linear Interpolation
y ( x )  3.75( x  1.20)
 1.20  x  1.04
y ( x )  x  1.64,
 1.04  x  0.60
y ( x )  1.04  0.26667( x  0.60),
 0.60  x  0.00
y ( x )  1.20  0.090909 x,
0.00  x  0.66
y ( x )  1.14  0.41935( x  0.66),
0.66  x  1.28
y ( x )  0.88  0.95652( x  1.28),
1.28  x  2.20
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Quadratic Interpolation
Given  x0 , y0 ,  x1 , y1 ,......, x n 1 , y n 1 ,  x n , y n  , fit quadratic splines through the data. The splines
are given by
f ( x )  a1 x 2  b1 x  c1 ,
 a 2 x 2  b2 x  c2 ,
x 0  x  x1
x1  x  x 2
.
.
.
 a n x 2  bn x  cn ,
x n 1  x  x n
Find a i , bi , ci , i  1, 2, …, n
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Quadratic Interpolation (contd)
Each quadratic spline goes through two consecutive data points
a1 x 0  b1 x 0  c1  f ( x0 )
2
a1 x1 2  b1 x1  c1  f ( x1 )
.
.
.
a i xi 1  bi xi 1  ci  f ( xi 1 )
2
a i xi  bi xi  c i  f ( xi )
2
.
.
.
a n x n 1  bn x n 1  c n  f ( xn 1 )
2
a n x n  bn xn  cn  f ( x n )
2
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This condition gives 2n equations
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Quadratic Splines (contd)
The first derivatives of two quadratic splines are continuous at the interior points.
For example, the derivative of the first spline
a1 x 2  b1 x  c1 is
2 a1 x  b1
The derivative of the second spline
a 2 x 2  b2 x  c 2 is
2 a2 x  b2
and the two are equal at x  x1 giving
2 a1 x1  b1  2a 2 x1  b2
2 a1 x1  b1  2a 2 x1  b2  0
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Quadratic Splines (contd)
Similarly at the other interior points,
2a 2 x 2  b2  2a3 x 2  b3  0
.
.
.
2ai xi  bi  2ai 1 xi  bi 1  0
.
.
.
2a n 1 x n 1  bn 1  2a n x n 1  bn  0
We have (n-1) such equations. The total number of equations is (2n)  (n  1)  (3n  1) .
We can assume that the first spline is linear, that is a1  0
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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,
f ( x)  a1 x 2  b1 x  c1 ,
 a 2 x 2  b2 x  c 2 ,
x0  x  x1
x1  x  x 2
.
.
.
 a n x 2  bn x  c n ,
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x n 1  x  x n
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Example
A curve needs to be fit through the given points to fabricate
the cam. If the cam follows a straight line profile between
x = 1.28 to x = 0.66, what is the value of y at x=1.1 ? Find
using a quadratic spline interpolation.
18
x (in.) y (in.)
2.20
0.00
1.28
0.88
0.66
1.14
0.00
1.20
–0.60 1.04
–1.04 0.60
–1.20 0.00
Cam Profile
4
3
5
2
1.4
1.2
1
6
0.8
y
Point
1
2
3
4
5
6
7
Y
7
0.6
1
0.4
X
0.2
0
-2
-1
0
1
2
x
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Solution
Since there are seven data points,
six quadratic splines pass through them.
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y( x)  a1 x 2  b1 x  c1 ,
 1.20  x  1.04
 a 2 x 2  b2 x  c2 ,
 1.04  x  0.60
 a3 x 2  b3 x  c3 ,
 0.60  x  0.00
 a 4 x 2  b4 x  c4 ,
0.00  x  0.66
 a5 x 2  b5 x  c5 ,
0.66  x  1.28
 a 6 x 2  b6 x  c6 ,
1.28  x  2.20
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Solution (contd)
Setting up the equations
Each quadratic spline passes through two consecutive data points giving
a1 x 2  b1 x  c1 passes through x = −1.20 and x = −1.04,
a1 (1.20) 2  b1 (1.20)  c1  0.00
(1)
a1 (1.04) 2  b1 (1.04)  c1  0.60
(2)
Similarly,
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a2 (1.04) 2  b2 (1.04)  c2  0.60 (3)
a2 (0.60) 2  b2 (0.60)  c2  1.04 (4)
a3 (0.60) 2  b3 (0.60)  c3  1.04
(5)
a3 (0.00) 2  b3 (0.00)  c3  1.20
(6)
a4 (0.00) 2  b4 (0.00)  c4  1.20
(7)
a4 (0.66) 2  b4 (0.66)  c4  1.14
(8)
a5 (0.66) 2  b5 (0.66)  c5  1.14
(9)
a5 (1.28) 2  b5 (1.28)  c5  0.88
(10)
a 6 (1.28) 2  b6 (1.28)  c6  0.88
(11)
a 6 (2.20) 2  b6 (2.20)  c6  0.00
(12)
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Solution (contd)
Quadratic splines have continuous derivatives at the interior data points
At x = −1.04
2a1 (1.04)  b1  2a2 (1.04)  b2  0
(13)
2a2 (0.60)  b2  2a3 (0.60)  b3  0
(14)
2a3 (0.00)  b3  2a4 (0.00)  b4  0
(15)
2a4 (0.66)  b4  2a5 (0.66)  b5  0
(16)
2a5 (1.28)  b5  2a6 (1.28)  b6  0
(17)
At x = −0.60
At x = 0.00
At x = 0.66
At x = 1.28
Assuming the first spline a1 x 2  b1 x  c1 is linear,
a1  0
21
(18)
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Solution (contd)
1.2
 1.44

1.0816  1.04
 0
0

0
 0
 0
0

0
 0

0
 0
 0
0

0
 0
 0
0

0
 0
 0
0

  2.08
1

0
 0
 0
0

0
 0
 0
0

 1
0
22
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0 1.0816  1.04 1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0 0.36  0.6 1
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0.36
0
0
 0.6
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0 0.4356 0.66 1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0 0.4356 0.66 1
0 1.6384 1.28 1
0
0
0
0
0
2.08
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
 1.2
0
1
0
0
0
1.2
0
1
1
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1.32
0
1
0
0
0
 1.32
2.56
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0  a1  0.00
  

0
0 0  b1  0.60
0
0 0  c1  0.60
  

0
0 0 a2  1.04 
0
0 0  b2  1.04 
  

0
0 0  c2  1.20 
  

0
0 0  a3  1.20 
0
0 0  b3  1.14 
  

0
0 0  c3  1.14 

0
0 0 a4  0.88
  

1.6384 1.28 1  b4  0.88
4.84
2.2 1  c4  0.00
0
0 0  a5   0 
  

0
0 0  b5   0 
0
0 0  c5   0 
  

0
0 0 a6   0 
 2.56  1 0  b6   0 
0
0 0  c6   0 
0
0
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Solution (contd)
Solving the above 18 equations gives the 18 unknowns as
i
1
2
3
4
5
6
23
ai
0
−6.25
3.3611
−3.5973
3.2997
−2.8076
bi
ci
3.75
4.5
−9.25 −2.26
2.2833
1.2
2.2833
1.2
−6.8207 4.2043
8.8138 −5.8018
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Solution (contd)
Therefore, the splines are given by
y ( x)  3.75 x  4.5,
 1.20  x  1.04
 6.25 x 2  9.25 x  2.26,
 1.04  x  0.60
 3.3611x 2  2.2833x  1.2,
 0.60  x  0.00
 3.5973x 2  2.2833x  1.2,
0.00  x  0.66
 3.2997 x 2  6.8207 x  4.2043,
0.66  x  1.28
 2.8076 x 2  8.8138 x  5.8018, 1.28  x  2.20
24
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Comparison
1.6
1.5
y
f quadratic( range)

f quadratic x desired

1
f linear( range)

f linear x desired

0.5
f ( range)

f x desired

y 0 0.05 h
0
1
x0 0.05 w
0.5
0
0.5
1
1.5
x range  x desired  range  x desired  range  x desired
2
xl 0.05 w
Figure : Cam profile as defined by linear and quadratic splines
(dotted line represents sixth order polynomial interpolant)
25
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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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