Explanation Definition used: Fourier series with period 2 L : The Fourier series of a function with period 2 L is obtained as f ( x ) = a 0 + a 1 cos ( π x L ) + a 2 cos ( 2 π x L ) + ⋅ ⋅ ⋅ + a n cos ( n π x L ) + ⋅ ⋅ ⋅ + b 1 sin ( π x L ) + b 2 sin ( 2 π x L ) + ⋅ ⋅ ⋅ + b n sin ( n π x L ) + ⋅ ⋅ ⋅ The n th coefficients are obtained as a 0 = 1 2 L ∫ − L L f ( x ) d x a n = 1 L ∫ − L L f ( x ) cos ( n π x L ) d x b n = 1 L ∫ − L L f ( x ) sin ( n π x L ) d x Result used: For an even function, the Fourier series contains only cosine terms and possibly constant term. For an odd function, the Fourier series contains only sine terms and no constant term. Calculation: It is given that the pulsating force function is F ( t ) = { 0 − 2 ≤ t < 0 , 1 ≤ t < 2 8 0 ≤ t < 1 . This function is neither even nor odd; hence all the coefficients are to be calculated. The period of the function is 2 L = 4 ⇒ L = 2 . Obtain the value of a 0 as follows. a 0 = 1 2 ( 2 ) ∫ − 2 0 0 d x + 1 2 ( 2 ) ∫ 0 1 8 d x + 1 2 ( 2 ) ∫ 1 2 0 d x = 2 [ x ] 0 1 = 2 [ 1 − 0 ] = 2 Obtain the value of a n as follows. a n = 1 ( 2 ) ∫ − 2 0 0 cos n π x 2 d x + 1 ( 2 ) ∫ 0 1 8 cos n π x 2 d x + 1 ( 2 ) ∫ 1 2 0 cos n π x 2 d x = 4 [ 2 n π sin n π x 2 ] 0 1 = 8 n π [ sin n π 2 − 0 ] = 8 n π sin n π 2 However, for all even values of n , the values of a n are 0. Thus, for n = 1 , the value of b 1 becomes 8 ( 1 ) π sin ( 1 ) π 2 = 8 π ; for n = 3 , the value of b 3 becomes 8 ( 3 ) π sin ( 3 ) π 2 = − 8 3 π

Mylab Math With Pearson Etext -- 18 Week Standalone Access Card -- For Basic Technical Mathematics With Calculus

11th Edition

ISBN: 9780135902912

Author: Allyn J. Washington

Publisher: PEARSON

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Chapter 30.7, Problem 27E

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Chapter 30 Solutions

Mylab Math With Pearson Etext -- 18 Week Standalone Access Card -- For Basic Technical Mathematics With Calculus

Ch. 30.1 - Prob. 1PE Ch. 30.1 - Prob. 2PE Ch. 30.1 - Prob. 1E Ch. 30.1 - Prob. 2E Ch. 30.1 - Prob. 3E Ch. 30.1 - Prob. 4E Ch. 30.1 - Prob. 5E Ch. 30.1 - Prob. 6E Ch. 30.1 - Prob. 7E Ch. 30.1 - Prob. 8E

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