Explanation Given Data: The transfer function is given as follows: H ( s ) = 2 s 3 + s 2 + 4 s + 2 The LC ladder network is terminated in a 1 Ω resistor. Formula used: Refer to Equation 19.82 in the textbook and write the expression for transfer function of an LC ladder network as follows: H ( s ) = − y 21 / Y L 1 + y 22 / Y L (1) Calculation: As the network is terminated with 1 Ω resistor, the value of Z L is 1 Ω . Therefore, the value of Y L is obtained as follows: Y L = 1 Z L = 1 1 Ω = 1 S Substitute 1 for Y L in Equation (1) and obtain the transfer function for the given LC ladder network as follows: H ( s ) = − y 21 / 1 1 + y 22 / 1 H ( s ) = − y 21 1 + y 22 (2) From the denominator of given transfer function, it is clear that, the given network is a 3 rd order network. Therefore, draw the network with two inductors and one capacitor as shown in Figure 1. Separate the odd and even terms in the denominator of the given transfer function and rearrange the given transfer function as follows: H ( s ) = 2 ( s 3 + 4 s ) + ( s 2 + 2 ) = 2 ( s 3 + 4 s ) [ 1 + ( s 2 + 2 s 3 + 4 s ) ] H ( s ) = 2 s 3 + 4 s 1 + s 2 + 2 s 3 + 4 s (3) Compare Equation (3) with Equation (2) and obtain the terms y 21 and y 22 as follows: y 21 = − 2 s 3 + 4 s y 22 = s 2 + 2 s 3 + 4 s As y 22 is the output admittance of the network with input short-circuited, redraw the network in Figure 1 as shown in Figure 2.

6th Edition

ISBN: 8220102801448

Author: Alexander

Publisher: YUZU

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Question

Chapter 19.9, Problem 18PP

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Realize the given transfer function using an LC ladder network terminated in a 1 Ω resistor.

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Chapter 19.9, Problem 17PP

Chapter 19, Problem 1RQ

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Chapter 19 Solutions

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS

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