Explanation Given data: Refer to given circuit and the waveform in the textbook. Formula used: Write a general expression to calculate the impedance of a resistor in s -domain. Z R = R (1) Here, R is the value of resistance. Write a general expression to calculate the impedance of an inductor in s -domain. Z L = s L (2) Here, L is the value of inductance. Write a general expression of convolution integral. y ( t ) = ∫ − ∞ ∞ x ( λ ) h ( t − λ ) d λ (3) Here, x ( λ ) is the input function, and h ( t − λ ) is the shifted impulse unit function. Calculation: The given circuit is redrawn as shown in Figure 1. Substitute 5 Ω for R in equation (1) to find Z R . Z R = 5 Substitute 2.5 H for L in equation (2) to find Z L . Z L = s ( 2.5 H ) = 2.5 s Convert the Figure 1 into s -domain as shown in Figure 2. Apply voltage division rule in Figure 2 to find V o . V o = ( 5 5 + 2.5 s ) V i = ( 5 2.5 ( s + 2 ) ) V i = ( 2 s + 2 ) V i Rearrange the above equation to find H ( s ) = V o V i . H ( s ) = V o V i = 2 s + 2 Apply inverse Laplace transform for above equation to find h ( t ) . h ( t ) = 2 e − 2 t u ( t ) { ∵ L − 1 ( 1 s + a ) = e − a t u ( t ) } Substitute t − λ for t in above equation to find h ( λ ) . h ( t − λ ) = 2 e − 2 ( t − λ ) u ( t − λ ) = 2 e − 2 t + 2 λ u ( t − λ ) = 2 e − 2 t e 2 λ u ( t − λ ) The given waveform is drawn as shown in Figure 3. Write a general expression to calculate ω . ω = 2 π T (4) Here, T is the total time period. Refer to Figure 3, the half cycle of sine wave is given. Therefore, T 2 = π 2 Simplify the above equation to find T . T = ( π 2 ) ( 2 ) = π Substitute π for T in equation (4) to find ω . ω = 2 π π = 2 Write a general expression of sinusoidal waveform. v i ( t ) = A sin ω t (5) Refer to Figure 3, the amplitude of the waveform is 20 . Substitute 20 for A , and 2 for ω in equation (5). v i ( t ) = 20 sin 2 t With the unit step function, the above equation is written as follows: v i ( t ) = { 20 sin 2 t u ( t ) 0 < t < π 2 0 t > π 2 Substitute λ for t in above equation to find v i ( λ ) . v i ( λ ) = { 20 sin 2 λ u ( λ ) 0 < λ < π 2 0 λ > π 2 Substitute v o ( t ) for y ( t ) , and v i ( λ ) for x ( λ ) in equation (3) to find v o ( t )

11th Edition

ISBN: 9780134747224

Author: Riedel

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Chapter 13, Problem 75P

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Find the value of vo at time t=2.2 s.

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Chapter 13, Problem 73P

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4v+9v+8v=-3v+6v',-5v, where vi and vo are the input and output voltage, respectively.

I decided to focus on the magnitude where I do the normalization on low and high pass and have the bandpass as dB(dB(decibel), with frequency cutoff, I manage to get accurate but have trouble controlling the frequency cutoff accurately and the bandbass isn't working properly. As such I need help.My 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 = 10e-3; % Inductance in henries (10 mH) % Transfer functions H_low = 1 ./ (1 + 1i * w * R * C); % Low-pass filter H_high = (1i * w * R * C) ./ (1 + 1i * w * R * C); % High-pass filter H_band = (1i * w * R * C) ./ (1 + 1i * w * L / R + 1i * w * R * C); % Band-pass filter % Cutoff frequency for RC filters (Low-pass and High-pass) f_cutoff_RC = 1 / (2 * pi * R * C); % Band-pass filter cutoff frequencies f_lower_cutoff = 1 / (2 * pi *…

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Chapter 13 Solutions

EBK ELECTRIC CIRCUITS

Ch. 13.2 - The parallel circuit in Example 13.1 is placed in...Ch. 13.3 - Prob. 2AP Ch. 13.3 - The energy stored in the circuit shown is zero at...Ch. 13.3 - The dc current and dc voltage sources are applied...Ch. 13.3 - Prob. 6AP Ch. 13.3 - Using the results from Example 13.7 for the...Ch. 13.3 - The energy stored in the circuit shown is zero at...Ch. 13.4 - Derive the numerical expression for the transfer...Ch. 13.5 - Find (a) the unit step and (b) the unit impulse...Ch. 13.5 - The unit impulse response of a circuit is υo(t) =...

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Find the value of v o at time t = 2.2 s .

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