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 ) is given by, f ( t ) = { 12 0 ≤ t ≤ π / 10 0 π / 10 ≤ t ≤ π / 5 The time period of the given function is, T = π 5 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 5 for T in the above expression. a 0 = 5 π ∫ 0 π 5 f ( t ) d t Break the limits for the defined range of the given function. a 0 = 5 π [ ∫ 0 π 10 12 d t + ∫ π 10 π 5 0 d t ] = 5 π ∫ 0 π 10 12 d t = 60 π [ t ] 0 π 10 = 60 π [ π 10 − 0 ] Further solve the above expression for a 0 . a 0 = 6 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 5 for T in the above expression. ω 0 = 2 π π / 5 = 10 rad / s Substitute π 5 for T and 10 for ω 0 in equation (1). a n = 10 π ∫ 0 π 10 f ( t ) cos n ω 0 t d t Break the limits for the defined range of the given function. a n = 10 π [ ∫ 0 π 10 12 cos n 10 t d t + ∫ π 10 π 5 0 cos n 10 t d t ] = 10 π ∫ 0 π 10 12 cos n 10 t d t = 120 π ∫ 0 π 10 cos n 10 t d t = 120 π [ sin n 10 t 10 n ] 0 π 10 Further solve the above expression as, a n = 120 π [ sin ( n 10 ( π 10 ) ) 10 n − sin 0 ] = 120 π [ sin ( n π ) 10 n − 0 ] = 120 π [ sin ( n π ) 10 n ] Substitute 0 for sin ( n π ) in the above expression. a n = 120 π [ 0 10 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 5 for T and 10 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 = 10 π [ ∫ 0 π 10 12 sin n 10 t d t + ∫ π 10 π 5 0 sin n 10 t d t ] = 10 π ∫ 0 π 10 12 sin n 10 t d t = 120 π ∫ 0 π 10 sin n 10 t d t = − 120 π [ cos n 10 t 10 n ] 0 π 10 Further solve the above expression. b n = − 120 π [ cos ( n 10 ( π 10 ) ) 10 n − cos 0 10 n ] = − 120 π [ cos ( n π ) − 1 10 n ] Substitute ( − 1 ) n for cos ( n π ) in the above expression. b n = − 120 π [ ( − 1 ) n − 1 10 n ] For odd values of n , ( − 1 ) n = − 1 b n = − 120 π [ − 1 − 1 10 n ] = 24 π n For even values of n , ( − 1 ) n = 1 b n = − 120 π [ 1 − 1 10 n ] = 0 The expression for voltage is given by, v s ( t ) = a 0 + ∑ n = 1 ∞ a n cos n ω 0 t + b n sin n ω 0 t Substitute 6 for a 0 , 0 for a n , 10 for ω 0 and 24 π n for b n in the above expression. v s ( t ) = 6 + ∑ n = 1 ∞ 0 cos n 10 t + 24 π n sin n 10 t = 6 + ∑ n = 1 ∞ 24 π n sin 10 n t For n t h harmonic, the dc component of the function is zero. V s n = ∑ n = 1 ∞ 24 π n sin n 10 t = 24 π n ∠ − 90 ο = − j 24 π n From figure 1, the resistor and inductor are in series

9th Edition

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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The MATLAB code is going well but the last part in bandpass, the legend that is supposed to tell the color of both lower and upper-frequency cutoff does not align with each other. As such I need help My Matlab code: % Define frequency range for the plot f = logspace(1, 5, 500); % Frequency range from 10 Hz to 100 kHz w = 2 * pi * f; % Angular frequency % Parameters for the filters R = 1e3; % Resistance in ohms (1 kΩ) C = 1e-6; % Capacitance in farads (1 μF) L = 0.1; % Inductance in henries (chosen for proper bandpass response) % Compute cutoff frequencies f_cutoff_RC = 1 / (2 * pi * R * C); % RC low-pass/high-pass cutoff f_resonance = 1 / (2 * pi * sqrt(L * C)); % Resonant frequency of RLC Q_factor = (1/R) * sqrt(L/C); % Quality factor of the circuit % Band-pass filter cutoff frequencies f_lower_cutoff = f_resonance / (sqrt(1 + 1/(4*Q_factor^2)) + 1/(2*Q_factor)); f_upper_cutoff = f_resonance / (sqrt(1 + 1/(4*Q_factor^2)) - 1/(2*Q_factor)); % Define Transfer Functions H_low =…

1° ⑤ Aa "Human-written solution required" 2. Using the characteristics of Fig. 6.11, determine ID for the following levels of VGs (with VDS > VP): a. VGs = 0V. b. VGs=-1 V. c. VGs -1.5 V. d. VGS -1.8 V. e. VGS = -4 V. f. VGs=-6V. 3. Using the results of problem 2 plot the transfer characteristics of ID vs. VGS- 4. a. Determine Vps for VGs = 0V and Ip = 6 mA using the characteristics of Fig. 6.11. b. Using the results of part (a), calculate the resistance of the JFET for the region Ip = 0 to 6 mA for VGs =0V. c. Determine Vps for VGS = -1 V and ID = 3 mA. d. Using the results of part (c), calculate the resistance of the JFET for the region ID = 0 to 3 mA for VGs -1 V. e. Determine Vps for VGs = -2 V and ID = 1.5 mA. f. Using the results of part (e), calculate the resistance of the JFET for the region ID = 0 to 1.5 mA for VGS-2 V. g. Defining the result of part (b) as ro, determine the resistance for VGs -1 V using Eq. (6.1) and compare with the results of part (d). h. Repeat part (g)…

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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

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