EMLecture4 - Lecture 4 Notes, 95.657 Electromagnetic Theory...

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Unformatted text preview: Lecture 4 Notes, 95.657 Electromagnetic Theory I, Fall 2011 Dr. Christopher S. Baird, UMass Lowell 1. Orthogonal Functions and Expansions- In the interval ( a , b ) of the variable x , a set of real or complex functions U n ( x ) where n = 1, 2, ... are orthogonal if: a b U n * x U m x dx = 0, m n- When m = n , the integral is nonzero. The functions are orthonormal if normalized to one: a b U n * x U m x dx = nm- An arbitrary, integrable function f ( x ) can be expanded in a series of the orthonormal functions U n ( x ) according to: f x = n = 1 N a n U n x - To find the expansion coefficients a n , we multiply both sides by the function U * M ( x ), integrate, and use the orthonormality property: f x U m * x = n = 1 N a n U n x U m * x a b f x U m * x dx = n = 1 N a n a b U n x U m * x dx a b f x U m * x dx = n = 1 N a n nm a b f x U m * x dx = a m- Interchange the arbitrary label m for n and get the final form: f ( x )= n = 1 N a n U n ( x ) where a n = a b f ( x ) U n * ( x ) dx Finite Series Expansion- If the functions form a complete set, and all functions that are useful in physics do, then the series expansion becomes a more accurate representation of the function f ( x ) as more terms in the series are kept. The most accurate is the infinite series: f ( x )= n = 1 a n U n ( x ) where a n = a b f ( x ) U n * ( x ) dx Infinite Series Expansion- If the interval ( a , b ) is expanded to be infinite, than the orthogonal functions become a continuum of functions, the index variable n becomes a continuous variable k , and the orthogonality condition becomes normalized to the Dirac delta function: U k * x U k ' x dx = k k ' f ( x )= A ( k ) U k ( x ) dk where A ( k )= f ( x ) U n * ( x ) dx Infinite Continuous Expansion 2. Fourier Series- The most commonly used orthogonal functions are sines and cosines, constituting Fourier series.- Start with a general expansion in terms of sines and cosines over the interval (- a /2, a /2): f x = n = A n cos k n x B n sin k n x - In order for the series to be a valid representation of the function in the interval, the series must be periodic outside the interval, so that f (- a /2) = f ( a /2). Using this requirement leads to: n = A n cos k n a / 2 cos k n a / 2 B n sin k n a / 2 sin k n a / 2 =- This must be true independent of A n and B n , so that the coefficients must be zero: sin k n a / 2 sin k n a / 2 = sin k n a / 2 = k n a / 2 = n k n = 2 n a The series now becomes: f x = n = A n cos 2 n x a B n sin 2 n x a To find the coefficients, multiply both sides by...
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EMLecture4 - Lecture 4 Notes, 95.657 Electromagnetic Theory...

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