{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Ch15p - Chapter 15 Exercise Solutions EX15.1 For the...

This preview shows pages 1–4. Sign up to view the full content.

Chapter 15 Exercise Solutions EX15.1 For the circuit shown in Figure 15.7 3 1 2 dB f RC π = or ( ) 6 3 3 1 1 3.979 10 2 2 40 10 dB RC f π π = = = × × For 75 K R = Then 11 5.31 10 53.1 pF C = × = 3 4 We have 1.414C 75.1 pF 0.707C 37.5 pF C C = = = = EX15.2 1 C eq f CR = or ( )( ) 5 6 1 1 10 20 10 0.5 pF c eq C f R C = = × = EX15.3 Low-frequency gain: 1 2 30 6 5 C T C = − = − = − ( )( ) ( ) 3 12 2 3 3 12 100 10 5 10 6.63 kHz 2 2 12 10 C dB dB F f C f f C π π × × = = = × EX15.4 0 6 3 0 1 2 3 1 1 6.13 10 2 3 2 (15 10 ) 3 f RC RC f π π π = = = = × × Let 0.001 μ F 1 nF C = = Then 6.13 k R = Ω so 2 8 49 k R R = = Ω EX15.5 ( ) ( ) ( ) 0 0 4 2 1 2 1 1 2 2 1 0.02 F 2 800 10 2 2 10 20 k f C RC f R C C R R R π π μ π = = = = = = Ω EX15.6

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document