Explanation Definition used: Fourier series with period 2 π : The Fourier series expansion of the function f ( x ) with period 2 π is given as: f ( x ) = a 0 + a 1 cos x + a 2 cos 2 x + ⋅ ⋅ ⋅ + a n cos n x + ⋅ ⋅ ⋅ + b 1 sin x + b 2 sin 2 x + ⋅ ⋅ ⋅ + b n sin n x + ⋅ ⋅ ⋅ The n th coefficients are obtained as a 0 = 1 2 π ∫ − π π f ( x ) d x a n = 1 π ∫ − π π f ( x ) cos n x d x b n = 1 π ∫ − π π f ( x ) sin n x d x Calculation: It is given that the loudness function is L ( t ) = { 0 − π ≤ t < 0 100 t 0 ≤ t < π 2 100 ( π − t ) π 2 ≤ t < π . Obtain the value of a 0 as follows. a 0 = 1 2 π ∫ − π 0 ( 0 ) d t + 1 2 π ∫ 0 π 2 100 t d t + 1 2 π ∫ π 2 π 100 ( π − t ) d t = 50 π ∫ 0 π 2 t d t + 50 π ∫ π 2 π ( π − t ) d t = 25 π [ t 2 ] 0 π 2 + 25 π [ 2 π t − t 2 ] π 2 π = 25 π [ ( π 2 ) 2 − 0 ] + 25 π [ 2 π ( π ) − ( π ) 2 − 2 π ( π 2 ) + ( π 2 ) 2 ] Further simplified as, a 0 = 25 π [ π 2 4 ] + 25 π [ 2 π 2 − π 2 − π 2 + π 2 4 ] = 25 π [ π 2 4 ] + 25 π [ π 2 4 ] = 25 2 π Obtain the value of a 1 as follows. a 1 = 1 π ∫ − π 0 ( 0 ) cos t d t + 1 π ∫ 0 π 2 100 t cos t d t + 1 π ∫ π 2 π 100 ( π − t ) cos t d t = 100 π ∫ 0 π 2 t cos t d t + 100 π ∫ π 2 π ( π − t ) cos t d t = 100 π [ t sin t + cos t ] 0 π 2 + 100 π [ ( π − t ) sin t − cos t ] π 2 π = 100 π [ π 2 sin π 2 + cos π 2 − 0 sin 0 − cos 0 ] + 100 π [ ( π − π ) sin π − cos π − ( π − π 2 ) sin π 2 + cos π 2 ] Further simplified as, a 1 = 100 π [ π 2 − 1 ] + 100 π [ − ( − 1 ) − ( π − π 2 ) + 0 ] = 100 π [ π 2 − 1 ] − 100 π [ π 2 − 1 ] = 0 Obtain the value of a 2 as follows

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 30.6, Problem 24E

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To find: The Fourier series of the function that represents the loudness.

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Chapter 30.6, Problem 23E

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

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