Explanation Given data: Refer to Figure 17.71 in the textbook. In the circuit, the voltage source v s is same as the function f 2 ( t ) shown in Figure 17.65(b). The inductance L is 1 H . The capacitor C is 1 F . Formula used: Write the general expression to calculate trigonometric Fourier series of v s ( t ) . v s ( t ) = a 0 + ∑ n = 1 ∞ ( a n cos n ω 0 t + b n sin n ω 0 t ) (1) Here, a 0 is the dc component of x ( t ) , a n and b n are the Fourier coefficients, n is an integer, and ω 0 is the angular frequency. Write the expression to calculate the fundamental angular frequency. ω 0 = 2 π T (2) Here, T is the period of the function. Write the expression to calculate the dc component of the function x ( t ) . a 0 = 2 T ∫ 0 T 2 f 2 ( t ) d t (3) Write the expression to calculate Fourier coefficients. a n = 4 T ∫ 0 T 2 f 2 ( t ) cos n ω 0 t d t (4) Calculation: The Figure 17.65(b) is redrawn as shown in Figure 1. The time period of the function in Figure 1 is T = 3 and the signal is even. Therefore, b n = 0 . From Figure 1, the Fourier series function of the waveform is defined as, f 2 ( t ) = { 1 , 0 < t < 1 2 , 1 < t < 1.5 (5) Substitute 3 for T in equation (2) to find ω 0 . ω 0 = 2 π 3 Applying equation (5) in equation (3) to find a 0 . a 0 = 2 3 [ ∫ 0 1 1 d t + ∫ 1 1.5 2 d t ] = 2 3 [ ( t ) 0 1 + 2 ( t ) 1 1.5 ] = 2 3 [ ( 1 − 0 ) + 2 ( 1.5 − 1 ) ] = 2 3 [ 1 + 1 ] Simplify the above equation to find a 0 . a 0 = 4 3 (6) Applying equation (5) in equation (4) to find a n . a n = 4 3 [ ∫ 0 1 ( 1 ) cos ( 2 n π t 3 ) d t + ∫ 1 1.5 ( 2 ) cos ( 2 n π t 3 ) d t ] { ∵ ω 0 = 2 π 3 } = 4 3 [ ( sin ( 2 n π t 3 ) ( 2 n π 3 ) ) 0 1 + ( ( 2 ) sin ( 2 n π t 3 ) ( 2 n π 3 ) ) 1 1.5 ] = 4 3 [ ( ( 3 2 n π ) sin ( 2 n π t 3 ) ) 0 1 + ( ( 6 2 n π ) sin ( 2 n π t 3 ) ) 1 1.5 ] = 4 3 [ [ ( 3 2 n π ) sin ( 2 n π ( 1 ) 3 ) − ( 3 2 n π ) sin ( 2 n π ( 0 ) 3 ) ] + [ ( 6 2 n π ) sin ( 2 n π ( 1.5 ) 3 ) − ( 6 2 n π ) sin ( 2 n π ( 1 ) 3 ) ] 1 1.5 ] Simplify the above equation as follows, a n = 4 3 [ [ ( 3 2 n π ) sin ( 2 n π 3 ) − ( 0 ) ] + [ ( 6 2 n π ) sin ( 3 n π 3 ) − ( 6 2 n π ) sin ( 2 n π 3 ) ] ] = 4 3 [ ( 3 2 n π ) sin ( 2 n π 3 ) + ( 3 n π ) sin ( n π ) − ( 3 n π ) sin ( 2 n π 3 ) ] = 4 3 [ ( 3 2 n π ) sin ( 2 n π 3 ) + 0 − ( 3 n π ) sin ( 2 n π 3 ) ] { ∵ sin ( n π ) = 0 } = 4 3 ( 3 n π ) [ ( 1 2 ) sin ( 2 n π 3 ) − sin ( 2 n π 3 ) ] Simplify the above equation as follows, a n = ( 4 n π ) sin ( 2 n π 3 ) [ 1 2 − 1 ] = ( 4 n π ) sin ( 2 n π 3 ) ( 1 − 2 2 ) = ( − 1 2 ) ( 4 n π ) sin ( 2 n π 3 ) a n = − 2 n π sin ( 2 n π 3 ) (7) Substitute equation (6) and equation (7) in equation (1) to find v s ( t )

6th Edition

ISBN: 9780078028229

Author: Charles K Alexander, Matthew Sadiku

Publisher: McGraw-Hill Education

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Chapter 17, Problem 35P

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