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Unformatted text preview: C 3 . (d) Solve for the worst case lower cuto f frequency in Hz using the equation f L = q p 2 pole 2 f 2 zero (e) Which capacitor dominates in setting f L ? r ib r x r 1 ( ) R te . r ib 1.301 10 4 = r in R p2 r ib R p2 R 1 R 2 , , r in 1.05 10 4 = 1 R S r in C 1 . 1 2.026 10 3 = f 1 1 2 . 1 . f 1 78.567 = 2 R C R L C 2 . 2 0.044 = f 2 1 2 . 2 . f 2 3.599 = 3p R p2 R E r' e , R 3 C 3 . 3p 0.019 = f 3p 1 2 . 3p . f 3p 8.27 = 3z R E R 3 C 3 . 3z 0.83 = f 3z 1 2 . 3z . f 3z 0.192 = 1 f L f 1 2 f 2 2 f 3p 2 2 f 3z 2 . f L 79.082 = 2...
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This note was uploaded on 05/26/2011 for the course ECE 3050 taught by Professor Hollis during the Summer '08 term at Georgia Institute of Technology.
 Summer '08
 HOLLIS

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