Transcript PPT

Spline Interpolation Method
Major: All 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.
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Rocket Example Results
t
(s)
0
10
15
20
22.5
30
5
v
(m/s)
0
227.04
362.78
517.35
602.97
901.67
Polynomial
Order
Velocity at
t=16 in
m/s
Absolute
Relative
Approxima
te Error
Least
Number of
Significant
Digits
Correct
1
393.69
-------------
2
392.19
0.38%
2
3
392.05
0.036%
3
4
392.07
0.0051%
3
5
392.06
0.0026%
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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
6
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
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
10
t (s)
v (t ) (m/s)
0
10
15
20
22.5
30
0
227.04
362.78
517.35
602.97
901.67
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 )  v(t 0 )  1
(t  t 0 )
t1  t 0
517.35
500
ys
f ( range)
 362.78 
517.35  362.78
fx

(t  15) desired
20  15
v (t )  362.78  30.913( t  15)
At t  16,
 393.7
11
450
400
362.78
v (16)  362.78  30.913(16  15)
550
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
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 ,
16
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
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t (s)
v (t ) (m/s)
0
10
15
20
22.5
30
0
227.04
362.78
517.35
602.97
901.67
Figure. Velocity vs. time data
for the rocket example
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Solution
2
v(t )  a1t  b1t  c1 , 0  t  10
 a2 t  b2 t  c2 , 10  t  15
2
 a3t  b3t  c3 , 15  t  20
2
 a4 t  b4 t  c4 ,
20  t  22.5
 a5 t  b5 t  c5 ,
22.5  t  30
2
2
Let us set up the equations
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Each Spline Goes Through
Two Consecutive Data Points
v(t )  a1t  b1t  c1 , 0  t  10
2
a1 (0)  b1 (0)  c1  0
2
a1 (10)  b1 (10)  c1  227.04
2
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Each Spline Goes Through
Two Consecutive Data Points
t
s
0
v(t)
m/s
0
10
15
20
22.5
227.04
362.78
517.35
602.97
30
901.67
a2 (10) 2  b2 (10)  c2  227.04
a2 (15) 2  b2 (15)  c2  362.78
a3 (15)  b3 (15)  c3  362.78
2
a3 (20)  b3 (20)  c3  517.35
2
a4 (20)  b4 (20)  c4  517.35
a4 (22.5) 2  b4 (22.5)  c4  602.97
2
a5 (22.5)  b5 (22.5)  c5  602.97
2
a5 (30)  b5 (30)  c5  901.67
2
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Derivatives are Continuous at
Interior
Data
Points
2
v(t )  a1t  b1t  c1 , 0  t  10
 a2 t  b2 t  c2 ,10  t  15
2

d
2
a1t  b1t  c1
dt


d
2

a2t  b2t  c2
dt
t 10
2a1t  b1  t 10  2a2t  b2  t 10
2a1 10  b1  2a2 10  b2

t 10
20a1  b1  20a2  b2  0
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Derivatives are continuous at
Interior Data Points
At t=10
2a1 (10)  b1  2a2 (10)  b2  0
At t=15
2a2 (15)  b2  2a3 (15)  b3  0
At t=20
2a3 (20)  b3  2a4 (20)  b4  0
At t=22.5
2a4 (22.5)  b4  2a5 (22.5)  b5  0
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Last Equation
a1  0
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Final Set of Equations
0
0 0 1 0
100 10 1 0
0
 0 0 0 100 10

 0 0 0 225 15
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
 20 1 0  20  1
 0 0 0 30 1

0
0 0 0 0
0
0 0 0 0
 1 0 0 0
0
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0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
225
15
1
0
0
0
0
400
20
1
0
0
0
0
0
0
0
0
0
0
400
20 1
0 506 .25 22 .5 1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0 0
0  30  1 0
0
0
0
0
0
0
0
40
1
0
 40
1
0
0
0
0
0
45
1
0
0
0
0
0
0
0
0
0  a1 
0  b1 
 
0 c1
 0 
227 .04 
0
0
227 .04 
0
0
  

0
0
0 a 2  362 .78 
0
0
0  b2  362 .78 
  517 .35 
0
0
0 c2
  

0
0
0  a3  517 .35 
0
0
0  b3   602 .97 
506 .25 22 .5 1   c3  602 .97 
  

900
30 1  a 4  901 .67 
0
0
0  b4   0 
0
0
0  c4   0 
  

0
0
0  a5 
0



 45
 1 0  b5   0 
0
0
0  c5   0 
0
0
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Coefficients of Spline
25
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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v(t )  22.704t ,
Final Solution
 0.8888t 2  4.928t  88.88,
 0.1356t 2  35.66t  141.61,
 1.6048t 2  33.956t  554.55,
 0.20889t 2  28.86t  152.13,
27
0  t  10
10  t  15
15  t  20
20  t  22.5
22.5  t  30
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Velocity at a Particular Point
a) Velocity at t=16
0  t  10
10  t  15
v(t )  22.704t ,
 0.8888t 2  4.928t  88.88,
 0.1356t 2  35.66t  141.61,
 1.6048t 2  33.956t  554.55,
 0.20889t 2  28.86t  152.13,
15  t  20
20  t  22.5
22.5  t  30
v16  0.135616  35.6616  141.61
2
 394.24 m/s
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Acceleration from Velocity Profile
b) The quadratic spline valid at t=16 is
given by
d
a (16)  v(t ) t 16
dt
vt   0.1356t 2  35.66t  141.61, 15  t  20
d
a (t )  ( 0.1356t 2  35.66t  141.61)
dt
 0.2712t  35.66, 15  t  20
a(16)  0.2712(16)  35.66  31.321m/s 2
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Distance from Velocity Profile
c) Find the distance covered by the rocket from t=11s to
t=16s.
16
S 16  S 11   v(t )dt
11
vt   0.8888t 2  4.928t  88.88, 10  t  15
 0.1356t 2  35.66t  141.61, 15  t  20
16
15
16
11
11
15
S 16   S 11   vt dt   vt dt   vt dt
15


16


  0.8888t 2  4.928t  88.88 dt    0.1356t 2  35.66t  141.61 dt
11
15
 1595.9 m
30
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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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