a. Explanation Given data: The input admittance is Y ( s ) . The admittance has a pole at s = − 3 and zero at s = − 1 . The value of Y ( ∞ ) is 0.25 S . Formula used: Write a general expression to calculate transfer function. H ( s ) = K ( s − z 1 ) ( s − z 2 ) ...... ( s − z n ) ( s − p 1 ) ( s − p 2 ) ...... ( s − p n ) . (1) Here, K is the constant, z 1 , z 2 , .... z n are the zeros of system, and p 1 , p 2 , .... p n are the poles of system. Calculation: From the given data, the admittance Y ( s ) has one pole and one zero. Therefore, equation (1) becomes, H ( s ) = K ( s − z 1 ) ( s − p 1 ) Substitute − 1 for z 1 , − 3 for p 1 , and Y ( s ) for H ( s ) in above equation to find Y ( s ) . Y ( s ) = K ( s − ( − 1 ) ) ( s − ( − 3 ) ) Y ( s ) = K ( s + 1 ) s + 3 (2) Simplify the equation (2) as follows, Y ( s ) = K s ( 1 + 1 s ) s ( 1 + 3 s ) = K ( 1 + 1 s ) ( 1 + 3 s ) Substitute ∞ for s in above equation to find Y ( ∞ ) b. To determine Find the expression of the current i ( t ) through Y ( s ) .

6th Edition

ISBN: 9780078028229

Author: Charles K Alexander, Matthew Sadiku

Publisher: McGraw-Hill Education

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Chapter 16, Problem 104CP

To determine

Find the function of Y(s).

To determine

Find the expression of the current i(t) through Y(s).

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Chapter 16, Problem 103CP

Chapter 16, Problem 105CP

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Find the function of Y ( s ) .

Solution Summary: The author explains that the input admittance Y(s) has one pole and one zero.

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Find the function of Y(s).

Find the expression of the current i(t) through Y(s).

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Chapter 16 Solutions