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lecture 31

# lecture 31 - Brickwall specification for LP filters Summary...

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1 Summary of Lecture #31 11/05/2010 HP filter realization based on brickwall spec. HP brickwall spec. LP brickwall spec. Find n and ϖ c from ϖ p , ϖ s , A max and A min . Construct a LP filter LP filter HP filter An example General-purpose 2nd-order RC active filters Why? Kerwin-Huelsman-Newcomb filters (KHN filters) Summing circuits Difference circuits Integrator circuits 1 3 Brickwall specification for LP filters ϖ p : Pass band edge frequency A max : Max. atten. (loss) |H(j ϖ ) | in the pass band ϖ s : Stop band edge frequency A min : Min. atten. (loss) |H(j ϖ )| in the stop band Filter order n ? Cutoff frequecny ϖ c ? A max ϖ p ϖ s A min 4 min max 0.1A 2n s 0.1A p 10 1 ( ) 10 1 ϖ - - ϖ max min 2 2n 2n p 2n 2n s c,min c,max 0. n c 1A 0.1A 10 1 10 1 ϖ ϖ ϖ = ϖ - ϖ = - Eq. (19.8) Eq. (19.7) A max A min ϖ p ϖ s min max 0.1A 10 0.1A s 10 p 10 1 log 10 1 n log ( ) - - ϖ ϖ 5 Brickwall specification for HP filters ϖ p : Pass band edge frequency A max : Max. atten. (loss) |H(j ϖ ) | in the pass band ϖ s : Stop band edge frequency A min : Min. atten. (loss) |H(j ϖ )| in the stop band Filter order n ? Cutoff frequecny ϖ c ? ϖ p ϖ s A min A max ϖ s ϖ p ϖ c HP brickwall spec. LP brickwall spec. 6 HP p HP ϖ Ω = ϖ p p p HP 1 ϖ = = ϖ p s s HP ϖ Ω = ϖ p ϖ ϖ 3 dB c p s 3 33 dB cutoff point c = 1 ? LP 7 min max 0.1A 2n s 0.1A p 10 1 ( ) 10 1 - - max min 2 2n 2n p 2n 2n s c,min c,max 0. n c 1A 0.1A 10 1 10 1 = Ω - = - Eq. (19.30) A max A min p s min max 0.1A 10 0.1A s 10 p 10 1 log 10 1 n log ( ) - - Eq. (19.31) LP

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2 Take a RLC filter as an example.
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lecture 31 - Brickwall specification for LP filters Summary...

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