Transcript PPT

Direct Method of
Interpolation
Computer Engineering Majors
Authors: Autar Kaw, Jai Paul
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
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1
Direct 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.
Figure 1 Interpolation of discrete.
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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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Direct Method
Given ‘n+1’ data points (x0,y0), (x1,y1),………….. (xn,yn),
pass a polynomial of order ‘n’ through the data as given
below:
y  a0  a1 x  ....................  an x .
n
where a0, a1,………………. an are real constants.
 Set up ‘n+1’ equations to find ‘n+1’ constants.
 To find the value ‘y’ at a given value of ‘x’, simply
substitute the value of ‘x’ in the above polynomial.
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Example
A robot arm with a rapid laser scanner is doing a quick quality check
on holes drilled in a rectangular plate. The hole centers in the plate that
describe the path the arm needs to take are given below.
If the laser is traversing from x = 2 to x = 4.25 in a linear path, find
the value of y at x = 4 using the direct method for linear interpolation.
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x (m)
y (m)
2
4.25
5.25
7.81
9.2
10.6
7.2
7.1
6.0
5.0
3.5
5.0
Figure 2 Location of holes on the
rectangular plate.
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Linear Interpolation
yx   a0  a1 x
7.2
y 2.00  a0  a1 2.00  7.2
y 4.25  a0  a1 4.25  7.1
7.2
7.18
7.16
ys
f ( range)

f x desired

7.14
7.12
Solving the above two equations gives,
a0  7.2889
a1  0.044444
7.1
7.1
7.08
5
x s  10
0
Hence
7
0
5
10
x s  range  x desired
y x   7.2889  0.044444 x, 2.00  x  4.25
y 4.00   7.2889  0.0444444.00   7.1111 in.
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x s  10
1
Example
A robot arm with a rapid laser scanner is doing a quick quality check
on holes drilled in a rectangular plate. The hole centers in the plate that
describe the path the arm needs to take are given below.
If the laser is traversing from x = 2 to x = 4.25 in a linear path, find
the value of y at x = 4 using the direct method for quadratic interpolation.
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x (m)
y (m)
2
4.25
5.25
7.81
9.2
10.6
7.2
7.1
6.0
5.0
3.5
5.0
Figure 2 Location of holes on the
rectangular plate.
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Quadratic Interpolation
y x   a 0  a1 x  a 2 x 2
y 2.00  a 0  a1 2.00  a 2 2.00  7.2
2
y 4.25  a0  a1 4.25  a 2 4.25  7.1
2
y 5.25  a0  a1 5.25  a 2 5.25  6.0
2
Solving the above three equations gives
a0  4.5282
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a1  1.9855
a 2  0.32479
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Quadratic Interpolation (contd)
y x   4.5282  1.9855 x  0.32479 x 2 , 2.00  x  5.25
y 4.00  4.5282  1.98554.00  0.324794.00
 7.2735 in.
7.56258
The absolute relative approximate error a obtained
between first and second order polynomial is
a 
7.2735  7.1111
 100
7.2735
2
8
7.5
ys
f ( range)

f x desired

7
6.5
 2.2327%
6
6
2
2
10
2.5
3
3.5
4
4.5
5
x s  range  x desired
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5.5
5.25
Comparison Table
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Order of
Polynomial
1
2
Location (in.)
7.1111
7.2735
Absolute Relative
Approximate Error
----------
2.2327%
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Example
A robot arm with a rapid laser scanner is doing a quick quality check
on holes drilled in a rectangular plate. The hole centers in the plate that
describe the path the arm needs to take are given below.
If the laser is traversing from x = 2 to x = 4.25 in a linear path, find
the value of y at x = 4 using the direct method using a fifth order
polynomial.
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x (m)
y (m)
2
4.25
5.25
7.81
9.2
10.6
7.2
7.1
6.0
5.0
3.5
5.0
Figure 2 Location of holes on the
rectangular plate.
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Fifth Order Interpolation
2
3
4
5


y x  a 0  a1 x  a 2 x  a3 x  a 4 x  a5 x
y 2.00   7.2  a0  a1 2.00   a 2 2.00  a3 2.00   a 4 2.00   a 5 2.00 
2
3
4
5
y 4.25  7.1  a0  a1 4.25  a 2 4.25  a3 4.25  a 4 4.25  a5 4.25
2
3
4
5
y 5.25  6.0  a 0  a1 5.25  a 2 5.25  a3 5.25  a 4 5.25  a5 5.25
2
3
4
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y 7.81  5.0  a0  a1 7.81  a 2 7.81  a3 7.81  a 4 7.81  a5 7.81
2
3
4
5
y 9.20   3.5  a 0  a1 9.20  a 2 9.20  a3 9.20   a 4 9.20   a5 9.20
2
3
4
5
y 10.60   5.0  a0  a1 10.60  a 2 10.60  a3 10.60  a 4 10.60  a5 10.60
2
13
3
4
5
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Fifth Order Interpolation (contd)
Writing the six equations in matrix form, we have
1
1

1

1
1

1
2.00
4.25
5.25
7.81
9.20
10.6
4.00
18.063
27.563
60.996
84.640
112.36
8.00
76.766
144.70
476.38
778.69
1191.0
a0  30.898 a1  41.344
a3  2.7862
16.00
32  a0  7.2
326.25 1386.6   a1   7.1
759.69 3988.4  a 2  6.0
    
3720.5 29057   a3  5.0
7163.9 65908  a 4  3.5
   
12625 133820 a5  5.0
a2  15.855
a4  0.23091 a5  0.0072923
yx   30.898  41.344 x  15.855 x 2  2.7862 x 3  0.23091x 4  0.0072923x 5 ,2  x  10.6
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Fifth Order Interpolation (contd)
yx   30.898  41.344 x  15.855 x 2  2.7862 x 3  0.23091x 4  0.0072923x 5 ,2  x  10.6
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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/direct_met
hod.html
THE END
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