Ch15s - Chapter 15 Problem Solutions 15.1(a For example vo...

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Chapter 15 Problem Solutions 15.1 (a) For example: Low-Frequency: 2 2 1 1 1 10 9 = + = = o i v R R v R R Corner Frequency: 3 5 1 5 10 3.18 10 2 π = = × = × f RC RC (b) For Example: 2 2 1 1 2 1 1 1 o i R j C v R v R R j R C ω ω = = − + So, set 2 1 15 R R = For example, 1 2 10 , 150 R k R k = Ω = Ω 5 2 3 3 1 1 1.06 10 2 2 (15 10 ) dB R C f π π = = = × × Then 70.7 = C pF 15.2 (a) 2 2 3 1 1 0.447 7 1 (2) 1 = = = = − + + v v dB A A dB f f
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