Explanation Given data: Refer to Figure 16.110 in the textbook. The expressions of admittances are, Y 1 = s C 1 Y 2 = 1 R 1 Y 3 = s C 2 The value of resistance ( R 1 ) is 1 k Ω . The transfer function is, V o ( s ) V s ( s ) = − ( s s + 10 ) (1) Calculation: The given circuit is drawn as shown in Figure 1. According to the virtual ground concept, the voltage at the node of non-inverting terminal is equal to ground voltage (i.e., 0 V). Apply Kirchhoff’s current law at the non-inverting terminal of the op amp. ( V s − 0 ) Y 3 + ( V o − 0 ) Y 1 + ( V o − 0 ) Y 2 = 0 V s Y 3 + V o Y 1 + V o Y 2 = 0 V s Y 3 = − V o Y 1 − V o Y 2 V s Y 3 = − V o ( Y 1 + Y 2 ) Simplify the above equation to find V o V s . V o V s = − Y 3 ( Y 1 + Y 2 ) Substitute s C 1 for Y 1 , 1 R 1 for Y 2 , and s C 2 for Y 3 in above equation to find V o V s . V o V s = − ( s C 2 ) ( s C 1 + 1 R 1 ) = − ( s C 2 ) C 1 ( s + 1 R 1 C 1 ) V o V s = − s ( C 2 C 1 ) ( s + 1 R 1 C 1 ) (2) Compare equation (1) and (2). C 2 C 1 = 1 Simplify the above equation to find C 1

EE 98: Fundamentals of Electrical Circuits - With Connect Access

6th Edition

ISBN: 9781259981807

Author: Alexander

Publisher: MCG

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Chapter 16, Problem 101P

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Chapter 16, Problem 100P

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Chapter 16 Solutions

EE 98: Fundamentals of Electrical Circuits - With Connect Access

Ch. 16.2 - Determine vo(t) in the circuit of Fig. 16.6,...Ch. 16.2 - Prob. 2PP Ch. 16.2 - Prob. 3PP Ch. 16.3 - For the circuit shown in Fig. 16.12 with the same...Ch. 16.3 - Prob. 5PP Ch. 16.3 - The initial energy in the circuit of Fig. 16.17 is...Ch. 16.4 - Prob. 7PP Ch. 16.4 - Prob. 8PP Ch. 16.4 - Prob. 9PP Ch. 16.5 - Obtain the state variable model for the circuit...

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Find the value of capacitance C 1 and C 2 using the given transfer function.

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Chapter 16 Solutions