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Unformatted text preview: Solutions to Homework 8 P14.92 1 2 1 Let and then the input impedance is Z R Z R Ls Cs = + = + ( 29 ( 29 2 1 2 2 1 2 1 1 1 2 1 L R R Ls LCs RC s Z Z Cs R Z s R Z Z LCs RCs R R Ls Cs + + + + + = = = + + + + + + 2 Now require : 2 then L RC RC L R C Z R R + = ⇒ = = P14.95 The transfer function can also be calculated form the circuit itself. The circuit can be represented in the frequency domain as We can save ourselves some work be noticing that the 10000 ohm resistor, the resistor labeled R and the op amp comprise a noninverting amplifier. Thus ( ) ( ) a c 1 10000 R V s V s æ ö ÷ ç = + ÷ ç ÷ ç è ø Now, writing node equations, ( ) ( ) ( ) ( ) ( ) ( ) c i o a o c and 1000 5000 V s V s V s V s V s CsV s Ls + = + = Solving these node equations gives ( ) 1 5000 1 1000 10000 1 5000 1000 R C L H s s s C L æ ö ÷ ç + ÷ ç ÷ ç è ø = æ öæ ö ÷ ÷ ç ç + + ÷ ÷ ç ç ÷ ÷ ç...
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This note was uploaded on 09/30/2010 for the course ECE 2025 taught by Professor Juang during the Spring '08 term at Georgia Tech.
 Spring '08
 JUANG
 Impedance

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