Challenge_5

# Challenge_5 - =-= = ∑ Therefore 29 29 29 29 29 29 29 29 t...

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EEL 3135 Dr. Fred J. Taylor, Professor Lesson #5 Fourier series Challenge The Fourier series representation of x(t)=cos(k ϖ 0 t) is given to be: ( 29 ( 29 ( 29 -∞ = = = n t jn n e a t k t x 0 0 cos ϖ where a n =0 for all k ≠± n, and a ± k otherwise. Express the Fourier series of ( 29 ( 29 ( 29 -∞ = = = n t jn e b t k t y k 0 0 sin in terms of the coefficients a n of x(t). Response If you remember Euler’s equation, that states x(t)=cos(k ϖ 0 t)=(1/2)e jk ϖ 0 t + (1/2)e -jk ϖ 0 t , it would be apparent that a k = ½ and a -k = ½. Your previous studies regarding the properties of a Fourier series should lead you to the conclusion that sine and cosine waves are related to each other through a differential operator. From the given data, ( 29 ( 29 ( 29 t jn k t jn k e a e a t x 0 0 - - + = . Upon differentiating y(t) and its Fourier series: ( 29 ( 29 ( 29 t k k dt t dx t y 0 0 sin ) ( - = = and ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 t jk k t jk k t jk k t jk k k n n t jn n e jk a e jk a dt e a a e a d dt e a d t y t k k dt t dx t y 0 0 0 0 0 0 0 0 0 0 0 ) ( sin ) ( - - - - = -∞ = - + + = + + =

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Unformatted text preview: =-= = ∑ Therefore: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 t jk k t jk k e jk a e jk a t k k sin---+ =-Simplifying: InvestiGATOR 1 EEL 3135 Dr. Fred J. Taylor, Professor ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 t jk k t jk k e k jk a e k jk a t k sin ϖ----+-= or ( 29 ( 29 ( 29 ( 29 ( 29 t jk k t jk k e ja e ja t k sin--+-= Since ( 29 ∑ ∞-∞ = = n t jn n e b t y ) ( It therefore follows that: b k =-ja k b-k =ja-k The same result could be obtained using analysis based on Euler’s equation. 2...
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## This note was uploaded on 04/15/2010 for the course EEL 3135 taught by Professor ? during the Spring '08 term at University of Florida.

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Challenge_5 - =-= = ∑ Therefore 29 29 29 29 29 29 29 29 t...

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