Explanation Given data: Refer to Figure 14.7 in the textbook. Formula used: Write a general expression to calculate the impedance of resistor in s -domain. Z R = R (1) Here, R is the value of the resistor. 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 the inductor. Write a general expression to calculate the impedance of a capacitor in s -domain. Z C = 1 s C (3) Here, C is the value of the capacitor. Calculation: The given circuit is redrawn as Figure 1. Substitute 10 for R in equation (1) to find Z R 1 . Z R 1 = 10 Substitute 10 for R in equation (1) to find Z R 2 . Z R 2 = 10 Substitute 2 for L in equation (2) to find Z L . Z L = 2 s Substitute 50 m for C in equation (3) to find Z C . Z C = 1 s ( 50 m ) = 1 s ( 50 × 10 − 3 ) { ∵ 1 m = 10 − 3 } = 1 s ( 0.05 ) = 20 s The s -domain circuit of the Figure 1 is drawn as Figure 2. Refer to Figure 2, the total current I i ( s ) is the addition of the current I 1 and I 2 . That is, I i ( s ) = I 1 + I 2 (4) Refer to Figure 1, the resistor R 1 and the capacitor C is connected in series form. Therefore, the impedance Z 1 in Figure 2 is expressed as following using the respective values of the resistor and capacitor. Z 1 = 10 + 20 s Refer to Figure 1, the resistor R 2 and the inductor L is connected in series form. Therefore, the impedance Z 2 in Figure 2 is expressed as following using the respective values of the resistor and inductor. Z 2 = 10 + 2 s The Figure 2 is redrawn as Figure 3 using the impedances Z 1 and Z 2 . Refer to Figure 3, the impedances Z 1 and Z 2 are connected in parallel with the voltage source. For the parallel connection, the voltage is same. Write the expression to calculate the current I 1 for the circuit in Figure 3. I 1 = V o ( s ) Z 1 Write the expression to calculate the current I 2 for the circuit in Figure 3. I 2 = V o ( s ) Z 2 Substitute V o ( s ) Z 1 for I 1 and V o ( s ) Z 2 for I 2 in equation (4) to find I i ( s ) . I i ( s ) = V o ( s ) Z 1 + V o ( s ) Z 2 = V o ( s ) ( 1 Z 1 + 1 Z 2 ) Substitute 10 + 20 s for Z 1 and 10 + 2 s for Z 2 in above equation to find I i ( s ) . I i ( s ) = V o ( s ) ( 1 ( 10 + 20 s ) + 1 ( 10 + 2 s ) ) = V o ( s ) ( ( 10 + 2 s ) + ( 10 + 20 s ) ( 10 + 2 s ) ( 10 + 20 s ) ) = V o ( s ) ( 20 + 2 s + 20 s 100 + 200 s + 20 s + 40 ) = V o ( s ) ( 20 s + 2 s 2 + 20 s 140 + 200 s + 20 s ) Simplify the above equation to find I i ( s ) . I i ( s ) = V o ( s ) ( 20 s + 2 s 2 + 20 s 140 s + 200 + 20 s 2 s ) = V o ( s ) ( 2 ( s 2 + 10 s + 10 ) 2 ( 10 s 2 + 70 s + 100 ) ) = V o ( s ) ( s 2 + 10 s + 10 10 s 2 + 20 s + 50 s + 100 ) = V o ( s ) ( s 2 + 10 s + 10 10 s ( s + 2 ) + 50 ( s + 2 ) ) Simplify the above equation to find I i ( s )

6th Edition

ISBN: 9780078028229

Author: Charles K Alexander, Matthew Sadiku

Publisher: McGraw-Hill Education

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Question

Chapter 14.2, Problem 2PP

To determine

Find the transfer function Vo(ω)Ii(ω) for the circuit shown in Figure 14.7.

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Chapter 14.2, Problem 1PP

Chapter 14.4, Problem 3PP

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Chapter 14 Solutions

Fundamentals of Electric Circuits

Ch. 14.2 - Obtain the transfer function VoVs of the RL...Ch. 14.2 - Prob. 2PP Ch. 14.4 - Draw the Bode plots for the transfer function...Ch. 14.4 - Sketch the Bode plots for H()=50j(j+4)(j+10)2 Ch. 14.4 - Construct the Bode plots for H(s)=10s(s2+80s+400)Ch. 14.4 - Obtain the transfer function H() corresponding to...Ch. 14.5 - A series-connected circuit has R = 4 and L = 25...Ch. 14.6 - A parallel resonant circuit has R = 100 k, L = 50...Ch. 14.6 - Calculate the resonant frequency of the circuit in...Ch. 14.7 - For the circuit in Fig. 14.40, obtain the transfer...

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Find the transfer function V o ( ω ) I i ( ω ) for the circuit shown in Figure 14.7.

Solution Summary: The author explains how to calculate the transfer function V_o(omega

Find the transfer function Vo(ω)Ii(ω) for the circuit shown in Figure 14.7.

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Chapter 14 Solutions