# Problem3.2.2f - x ê L L „ x = (4) b n = 1 ÅÅÅÅ L ‡...

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Problem 3.2.2(f) Let us first sketch the function , by taking L=0.5 Figure 1 -0.4 -0.2 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 The periodic extension of this function is then Figure 2 -2 -1 1 2 0.2 0.4 0.6 0.8 1 Note that the periodic extension is neither even nor odd. Thus the Fourier series expansion for f H x L is then (1) f H x L = a 0 + n = 1 a n Cos H n p x ê L L + n = 1 b n Sin H n p x ê L L Calculating the coefficients gives (2) a 0 = 1 ÅÅÅÅÅÅÅÅ 2 L 0 L 1 x = 1 ÅÅÅÅ 2 (3) a n = 1 ÅÅÅÅ L 0 L Cos H n p

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Unformatted text preview: x ê L L „ x = (4) b n = 1 ÅÅÅÅ L ‡ L Sin H n p x ê L L „ x = -1 ÅÅÅÅÅÅÅ n p Cos H n p x ê L L x = L + 1 ÅÅÅÅÅÅÅ n p Cos H n p x ê L L x = = J 1-Cos H n p L ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅ n p N Thus (5) b n = 2 ÅÅÅÅÅÅÅÅ n p for n odd b n = for n even 2 Problem3.2.2f.nb...
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## This note was uploaded on 02/18/2010 for the course ENGINEEING 32145 taught by Professor Stroeve during the Fall '09 term at Universidad de Carabobo.

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Problem3.2.2f - x ê L L „ x = (4) b n = 1 ÅÅÅÅ L ‡...

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