Explanation Given data: Refer to Figure 17.72(a) and 17.72(b) in the textbook. The inductor L is 1 H . The capacitor C is 100 mF . The resistor R is 5 Ω . Formula used: Write the general expression for Fourier series expansion. f ( t ) = a 0 + ∑ n = 1 ∞ a n cos n ω 0 t + ∑ n = 1 ∞ b n sin n ω 0 t (1) Write the expression to calculate the fundamental angular frequency. ω 0 = 2 π T (2) Here, T is the period of the function. Calculation: Refer to Figure 17.72(a) in the textbook. The source voltage v s ( t ) over the period is, f ( t ) = 10 ( 1 − t ) , 0 < t < 1 (3) The time period of Figure 17.72(a) is, T = 1 Substitute 1 for T in equation (2) to find the angular frequency ω 0 . ω 0 = 2 π 1 = 2 π Finding the Fourier coefficient a 0 . a 0 = 1 T ∫ 0 T v s ( t ) d t (4) Applying equation (3) in equation (4) as follows, a 0 = 1 1 [ ∫ 0 1 10 ( 1 − t ) d t ] { ∵ T = 1 } = 10 [ t − t 2 2 ] 0 1 = 10 [ ( 1 − 1 2 ) − ( 0 − 0 2 ) ] = 10 [ ( 2 − 1 2 ) − ( 0 ) ] Simplify the equation as follows, a 0 = 10 2 = 5 Finding the Fourier coefficient a n . a n = 2 T ∫ 0 T v s ( t ) cos n ω 0 t d t (5) Applying equation (3) in equation (5) as follows, a n = 2 1 [ ∫ 0 1 10 ( 1 − t ) cos 2 n π t d t ] { ∵ T = 1 , ω 0 = 2 π } a n = 20 [ ∫ 0 1 ( 1 − t ) cos 2 n π t d t ] (6) Assume the following to reduce the equation (6). x = ∫ 0 1 ( 1 − t ) cos ( 2 π n ) t d t (7) Substitute the equation (7) in equation (6) to find a n . a n = 20 x (8) Consider the following integration formula. ∫ a b u d v = [ u v ] a b − ∫ a b v d u (9) Compare the equations (7) and (9) to simplify the equation (7). u = ( 1 − t ) d v = cos ( 2 π n t ) d t d u = − d t v = sin ( 2 π n t ) 2 π n Using the equation (9), the equation (7) can be reduced as, x = ∫ 0 1 t cos ( 2 π n ) t d t = [ ( 1 − t ) sin ( 2 π n t ) 2 π n ] 0 1 − ∫ 0 1 sin ( 2 π n t ) 2 π n ( − d t ) = [ ( 1 − 1 ) sin ( 2 π n ( 1 ) ) 2 π n − ( 1 − 0 ) sin ( 2 π n ( 0 ) ) 2 π n ] + ( 1 2 π n ) [ − cos ( 2 π n t ) 2 π n ] 0 1 = [ 0 − 0 ] − 1 ( 2 π n ) 2 [ cos ( 2 π n ( 1 ) ) − cos ( 2 π n ( 0 ) ) ] Simplify the above equation to find x . x = − 1 ( 2 π n ) 2 [ cos ( 2 π n ( 1 ) ) − cos ( 0 ) ] = − 1 ( 2 π n ) 2 [ 1 − 1 ] { ∵ sin ( 2 π ) = 0 cos ( 2 π ) = 1 cos ( 0 ) = 1 } = − 1 ( 2 π n ) 2 [ 0 ] = 0 Substitute 0 for x in equation (8) to find a n . a n = 20 ( 0 ) = 0 Finding the Fourier coefficient b n . b n = 2 T ∫ 0 T v s ( t ) sin n ω 0 t d t (10) Applying equation (3) in equation (5) as follows, b n = 2 1 ∫ 0 1 10 ( 1 − t ) sin 2 π n t d t { ∵ T = 1 , ω 0 = 2 π } b n = 20 ∫ 0 1 ( 1 − t ) sin 2 π n t d t (11) Assume the following to reduce the equation (11). y = ∫ 0 1 ( 1 − t ) sin 2 π n t d t (12) Substitute equation (12) in equation (11) to find b n . b n = 20 y (13) Compare the equations (12) and (9) to simplify the equation (12)

6th Edition

ISBN: 9780078028229

Author: Charles K Alexander, Matthew Sadiku

Publisher: McGraw-Hill Education

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Question

Chapter 17, Problem 36P

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Find the response io(t) of the circuit shown in Figure 17.72(a).

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Chapter 17, Problem 35P

Chapter 17, Problem 37P

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