Explanation Given data: Refer to Figure 17.69 in the textbook. The Fourier series expansion of v s ( t ) is, v s ( t ) = 10 + 5 π ∑ n = 1 ∞ 1 n sin ( n π t ) (1) The inductance L is 2 H The resistor R is 10 Ω . The capacitor C is 1 4 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 ) (2) 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. Calculation: On comparing equation (1) and (2), the value of ω n is, ω n = n ω 0 = n π Refer to Figure 17.69 in the textbook. For DC source, the inductor acts as a short circuit and the capacitor as an open circuit. Therefore, the output voltage v o ( t ) is, v o ( t ) = 0 For the DC case, the output voltage is, v o = 0 Consider the given circuit for AC analysis. The impedance of the inductor is calculated as follows, Z L = j ω n L Substitute 2 for L in the above equation as follows. Z L = j 2 n π { ∵ ω n = n π } The impedance of the capacitor is calculated as follows, Z C = 1 j ω n C Substitute 1 4 for C in the above equation as follows. Z C = 1 j ω n ( 1 4 ) = 4 j n π { ∵ ω n = n π } The modified circuit is shown in Figure 1. Applying Kirchhoff’s current law in Figure 1. V o − V s 10 + V o j 2 n π + V o ( 4 j n π ) = 0 V o − V s 10 + V o j 2 n π + j n π V o 4 = 0 V o ( 1 10 + 1 j 2 n π + j n π 4 ) = V s 10 V o ( 1 10 − j 2 n π + j n π 4 ) = V s 10 Simplify the above equation as follows, V o ( 1 10 + j ( n π 4 − 1 2 n π ) ) = V s 10 V o ( 1 + j ( 2

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Chapter 17, Problem 33P

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Determine the response vo(t) of the circuit shown in Figure 17.69.

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Chapter 17, Problem 32P

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