Lecture11b

Lecture11b - 0306-381 Applied Programming Lecture Eleven...

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Unformatted text preview: 0306-381 Applied Programming Lecture Eleven •Curve Fitting Using MS-Excel FInterpolation Interpolation Process of estimating values using between known values • Uses information known about the environment rather than fitting measurements • Used for interface of disparate systems • Typically on-line process 3 Interpolation Example Design of digital display data accurate to 0.25 • Display: integer values from 0 to 10 • Sensor: outputs values such as 0.00, 0.25, 0.50, 0.75, 1.00, etc. • Problem: What is most appropriate integer? Interpolation Example 10 9 8 7 6 5 4 3 2 1 0 0.00 2.00 4.00 6.00 8.00 10.00 4 y a l p s i D Display2 Display3 Display1 Sensor Output Interpolation Characteristics How does interpolation compare to and contrast with curve fitting? • Data points: interpolation uses only a few points, and resulting function includes those points • Processing: interpolation is on-line • Accuracy: generally less accurate than curve fitting Given some idea of what curve should look like, use “live” data measurements to interpolate the desired value on the spot. 5 Interpolation Function Determine function that includes known points: f (x) • n known points: {(xi, yi)}, i Î [1, n] • f (xi) = yi " i Î [1, n] 6 Linear Lagrange Interpolation Interpolation function is a straight line: f (x) • Interpolates based on 2 data points • Equation of straight line through these 2 points x - x2 x - x1 f ( x) = y1 + y2 x1 - x2 x2 - x1 7 Quadratic Lagrange Interpolation Interpolation function is quadratic: f (x) • Interpolates based on 3 data points • Quadratic function passes through these 3 points ( x - x2 )( x - x3 ) ( x - x1 )( x - x3 ) ( x - x1 )( x - x2 ) f ( x) = y1 + y3 y2 + ( x1 - x2 )( x1 - x3 ) ( x2 - x1 )( x2 - x3 ) ( x3 - x1 )( x3 - x2 ) 8 Quadratic Lagrange Interpolation Example Three data points (N = 3) • (x1, y1) = (15, 10) • (x2, y2) = (20, 11) • (x3, y3) = (25, 20) ( x - 20)( x - 25) ( x - 15)( x - 25) ( x - 15)( x - 20) f ( x) = 10 + 11 + 20 (15 - 20)(15 - 25) (20 - 15)(20 - 25) (25 - 15)(25 - 20) 9 Lagrange Interpolation Lagrange interpolating function: f (x) • Interpolates based on N data points • Polynomial of degree N - 1 N Õ( x - x j ) Õ ( xi - x j ) j =1 j ¹i j =1 j ¹i N N f ( x) = å i =1 yi • Computational complexity increases quickly with N – 2 points: 4 Add/Sub., 2 Mult., 2 Div. – 3 points: 8 Add/Sub., 9 Mult., 3 Div. – n points: O(n2) Add/Sub., O(n2) Mult., O(n) Div. 10 ...
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This note was uploaded on 04/27/2010 for the course EECC 0306-381 taught by Professor Roymelton during the Spring '10 term at RIT.

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