Explanation Given data: The given diagram is shown in Figure 1. The given waveform is shown in Figure 2. Calculation: The expression for the function f ( t ) calculated from the figure is given by, f ( t ) = { 10 0 ≤ t ≤ π / 2 0 2 ≤ t ≤ π The time period of the given function is, T = π s Calculate the Fourier coefficient for the given function. The expression for a 0 is given by, a 0 = 1 T ∫ 0 T f ( t ) d t Substitute π for T in the above expression. a 0 = 1 π ∫ 0 π f ( t ) d t Break the limits for the defined range of the given function. a 0 = 1 π [ ∫ 0 π 2 10 d t + ∫ π 2 π 0 d t ] = 1 π ∫ 0 π 2 10 d t = 10 π [ t ] 0 π 2 = 10 π [ π 2 − 0 ] Further solve the above expression for a 0 . a 0 = 5 The expression for a n is given by, a n = 2 T ∫ 0 T f ( t ) cos n ω 0 t d t (1) Here, ω 0 is the angular frequency. The angular frequency is given by, ω 0 = 2 π T Substitute π for T in the above expression. ω 0 = 2 π π = 2 rad / s Substitute π for T and 2 for ω 0 in equation (1). a n = 2 π ∫ 0 π f ( t ) cos n ω 0 t d t Break the limits for the defined range of the given function. a n = 2 π [ ∫ 0 π 2 10 cos n 2 t d t + ∫ π 2 π 0 cos n 2 t d t ] = 2 π ∫ 0 π 2 10 cos n 2 t d t = 20 π ∫ 0 π 2 cos n 2 t d t = 20 π [ sin n 2 t 2 n ] 0 π 2 Further solve the above expression as, a n = 20 π [ sin ( n 2 ( π 2 ) ) 2 n − sin 0 ] = 20 π [ sin ( n π ) 2 n − 0 ] = 20 π [ sin ( n π ) 2 n ] Substitute 0 for sin ( n π ) in the above expression. a n = 20 π [ 0 2 n ] = 0 The expression for b n is given by, b n = 2 T ∫ 0 T f ( t ) sin n ω 0 t d t Substitute π for T and 2 for ω 0 in above expression. b n = 10 π ∫ 0 π 10 f ( t ) sin n 10 t d t Break the limits for the defined range of the given function. b n = 2 π [ ∫ 0 π 2 10 sin n 2 t d t + ∫ π 2 π 0 sin n 2 t d t ] = 2 π ∫ 0 π 2 10 sin n 2 t d t = 20 π ∫ 0 π 2 sin n 2 t d t = − 20 π [ cos n 2 t 2 n ] 0 π 2 Further solve the above expression. b n = − 20 π [ cos ( n 2 ( π 2 ) ) 2 n − cos 0 2 n ] = − 20 π [ cos ( n π ) − 1 2 n ] Substitute ( − 1 ) n for cos ( n π ) in the above expression. b n = − 20 π [ ( − 1 ) n − 1 2 n ] For odd values of n , ( − 1 ) n = − 1 b n = − 20 π [ − 1 − 1 2 n ] = 20 π n For even values of n , ( − 1 ) n = 1 b n = − 20 π [ 1 − 1 2 n ] = 0 The expression for current is given by, i s ( t ) = a 0 + ∑ n = 1 ∞ a n cos n ω 0 t + b n sin n ω 0 t Substitute 5 for a 0 , 0 for a n , 10 for ω 0 and 20 π n for b n in the above expression. i s ( t ) = 5 + ∑ n = 1 ∞ 0 cos n 2 t + 20 π n sin n 2 t = 5 + ∑ n = 1 ∞ 20 π n sin 2 n t = 5 + 20 π ∑ n = 1 ∞ sin 2 n t n For n t h harmonic, the dc component of the function is zero. I s n = 20 π n sin 2 n t = 20 π n ∠ − 90 ο = − j 20 π n The expression for the capacitive reactance is given by, X C = − j 1 n ω 0 C Substitute 2 for ω 0 and 2 F for C in the above expression. X C = − j 1 2 n ( 2 ) = − j 1 4 n The expression of current for forced response at the n t h harmonic frequency using current divider rule is given by, I f n = I s n ( R R + X C ) Substitute − j 20 n π for I s n , − j 1 4 n for X C and 1 for R in the above expression

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ISBN: 9780073545516

Author: Hayt, William H. (william Hart), Jr, Kemmerly, Jack E. (jack Ellsworth), Durbin, Steven M.

Publisher: Mcgraw-hill Education,

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Chapter 17 Solutions

Engineering Circuit Analysis

Ch. 17.1 - Let a third-harmonic voltage be added to the...Ch. 17.1 - A periodic waveform f(t) is described as follows:...Ch. 17.2 - Prob. 3P Ch. 17.2 - Prob. 4P Ch. 17.3 - Prob. 5P Ch. 17.3 - Prob. 6P Ch. 17.4 - Prob. 7P Ch. 17.5 - Prob. 8P Ch. 17.5 - Prob. 9P Ch. 17.6 - Prob. 10P

Ch. 17.6 - Prob. 11P Ch. 17.7 - Prob. 12P Ch. 17.7 - Prob. 13P Ch. 17.8 - Find (a) F5sin23t); (b) FAsin20t); (c)...Ch. 17.9 - Prob. 15P Ch. 17.10 - Prob. 16P Ch. 17 - Determine the fundamental frequency, fundamental...Ch. 17 - Plot multiple periods of the first, third, and...Ch. 17 - Calculate a0 for the following: (a) 4 sin 4t; (b)...Ch. 17 - Prob. 4E Ch. 17 - Prob. 5E Ch. 17 - Prob. 6E Ch. 17 - Prob. 7E Ch. 17 - With respect to the periodic waveform sketched in...Ch. 17 - Prob. 9E Ch. 17 - Prob. 10E Ch. 17 - A half-sinusoidal waveform is shown in Fig. 17.31,...Ch. 17 - Plot the line spectrum (limited to the six largest...Ch. 17 - Prob. 13E Ch. 17 - Prob. 14E Ch. 17 - Prob. 15E Ch. 17 - Prob. 16E Ch. 17 - Prob. 17E Ch. 17 - Prob. 18E Ch. 17 - The nonperiodic waveform g(t) is defined in Fig....Ch. 17 - Prob. 20E Ch. 17 - Prob. 21E Ch. 17 - Prob. 22E Ch. 17 - Prob. 23E Ch. 17 - Prob. 24E Ch. 17 - Prob. 25E Ch. 17 - Prob. 26E Ch. 17 - Prob. 27E Ch. 17 - Prob. 28E Ch. 17 - Prob. 29E Ch. 17 - Prob. 30E Ch. 17 - Prob. 31E Ch. 17 - Prob. 32E Ch. 17 - Prob. 33E Ch. 17 - Prob. 34E Ch. 17 - Prob. 35E Ch. 17 - Prob. 36E Ch. 17 - Use the Fourier transform to obtain and plot the...Ch. 17 - Prob. 38E Ch. 17 - Prob. 39E Ch. 17 - Prob. 40E Ch. 17 - For g(t) = 3etu(t), calculate (a) G(j); (b) ().Ch. 17 - Prob. 42E Ch. 17 - Prob. 43E Ch. 17 - Prob. 44E Ch. 17 - Prob. 45E Ch. 17 - Prob. 46E Ch. 17 - Find F(j) if f(t) is given by (a) 2 cos 10t; (b)...Ch. 17 - Prob. 48E Ch. 17 - Prob. 49E Ch. 17 - Prob. 50E Ch. 17 - Prob. 51E Ch. 17 - Prob. 52E Ch. 17 - Prob. 53E Ch. 17 - If a system is described by transfer function h(t)...Ch. 17 - Prob. 55E Ch. 17 - (a) Design a noninverting amplifier having a gain...Ch. 17 - Prob. 57E Ch. 17 - Prob. 58E Ch. 17 - Prob. 59E Ch. 17 - Prob. 60E Ch. 17 - Prob. 61E Ch. 17 - Prob. 62E Ch. 17 - Design an audio amplifier with gain of 10, using...

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