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lecture 30

# lecture 30 - Summary of Lecture#30 Examples of 1st 2nd and...

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1 Summary of Lecture #30 11/03/2010 Examples of 1st-, 2nd- and 3 rd - order active RC opt. amp. filters Low pass filters (Sallen Key filters) High pass filters Band pass filters Band rejection filters All pass filters 1 0 1 0 2 2 1 0 2 2 1 2 a s a Bilinear equation, H (s) s b a s a s a Biquadratic equaiton 1 1 , H (s) s b s b + = + + = + + First-order RC active filters v out (t) + + + - + + _ v in (t) C _ R 1 R 2 out 2 1 in 1 V (s) (R / R )s H(s) V (s) s (1/(R C)) = = - + x o HP v out (t) + + + - + + _ v in (t) _ R 1 C R 2 out in 1 2 V (s) 1 1 H(s) V (s) R C s (1/(R C)) = = - + x LP Leaky integrator circuit Second-order active RC op. amp. LP filters Sallen and Key active LP filters x x 2 1 2 1 2 2 1 2 1 1 2 2 1 2 1 in out C C R R 1 s ) C R K 1 C R 1 C R 1 ( s C C R R K ) s ( V ) s ( V ) s ( H + - + + + = = A B R R 1 K + = 4 Consider V - of the inverting terminal. 2 1 2 1 2 2 1 2 1 1 2 2 1 2 1 in out C C R R 1 s ) C R K 1 C R 1 C R 1 ( s C C R R K V V ) s ( H + - + + + = = A out A B R V V R R - = + Due to virtual short A a out A B R V V R V R V + - = = = + KCL at note V a : KCL at node V b : a b a 2 2 V V V 0 0 R 1/(C s) - - = b in b out b a 1 1 2 V V V V V V 0 R 1/(C s) R - - - + = A B R R 1 K + = From these equations 5 In Lecture #26, we chose R A → ∞ and R B = 0 , R 1 = R 2 = 1 , then K = 1 and By choosing C 1 = 2 F and C 2 = 1/ 2 F , then H(s) becomes a normalized Butterworth transfer function.

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