lec30

# lec30 - 1 1 Summary of Lecture#30 • Analog filters •...

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Unformatted text preview: 1 1 Summary of Lecture #30 11/04/09 • Analog filters • Filter realization based on brick wall specification • Find n and ϖ c from ϖ p , ϖ s , A max and A min • Examples • General purpose second-order active RC filters Kerwin-Huelsman-Newcomb filters (KHN filters) • Summing circuits or summers • Difference circuits or differentiators • Integrators • KHN filters 2 • Brick wall specification for LP filters ϖ p : Pass band edge frequency • Minimum transmission |H(j ϖ ) | in the pass band ϖ s : Stop band edge frequency • Maximum transmission |H(j ϖ )| in the stop band |H| pb min |H| sb max Pass band Stop band Transition band max c | ) j ( H | 2 1 | ) j ( H | ϖ = ϖ Cutoff frequency: ϖ c Transmission 3 Brick wall specification for LP filters ϖ p : Pass band edge frequency Max. atten. (loss) |H(j ϖ ) | in the pass band, A max ϖ s : Stop band edge frequency Min. atten. (loss) |H(j ϖ )| in the stop band, A min Filter order n ? Cutoff frequecny ϖ c ? A max ϖ p ϖ s A min Attenuation 4 min max 0.1A 2 n s p 0.1A 10 1 ( / ) 10 1- ≤ ϖ ϖ- A max A min ϖ p ϖ s max p 0.1A n max c ( ) 10 1 ϖ = ε =- ϖ min max 0.1A 10 0.1A s 10 p 10 1 log 10 1 n log ( )-- ≥ ϖ ϖ Eq. (21.19) Eq. (21.18) 5 Ex. 10 (Text Ex. 21.3) For the brick wall specification shown, find n and ϖ c . A max A min A min (I) (II) min max 0.1A 10 0.1A 10 s 10 10 p 10 1 99 log log 10 1 1.1143 0.5849 n 1.594 log (5) .699 log ( )-- ≥ = = = ϖ ϖ max A 2 0.1 0.1 10 1 10 1 1.5849 1 0.5849 ×- =- =- = s p 1000 5 200 ϖ π = = ϖ π min 0.1 0. A 20 1 2 10 1 10 1 10 1 99 ×- =- =- = Case ( I ) ϖ p = 200 π rad/s, A max = 2 dB ϖ s = 1000 π rad/s, A min = 20 dB 6 Case ( II ) A max A min A min (I) (II) min max 0.1A 10 0.1A 10 s 10 10 p 10 1 9999 log log 10 1 2.1164 0.5849 n 2.1164 log (10) 1.00 log ( )-- ≥ = = = ϖ ϖ max A 2 0.1 0.1 10 1 10 1 1.5849 1 0.5849 ×- =- =- = s p 2000 10 200 ϖ π = = ϖ π min 0.1 0.1 A 4 4 10 1 10 1 10 1 9999 ×- =- =- = ϖ p = 200 π rad/s, A max = 2 dB ϖ s = 2000 π rad/s, A min = 40 dB 2 7 A max A min A min (I) (II) Case ( II ) n ≥ 2.1164 Case ( I ) n ≥ 1.594 Both requirements are met if we choose n = 3. But n must be an integer. max p p 0.1A 2n 6 2 0.2 max c c ( ) ( ) 10 1 10 1 0.5849 ϖ ϖ = = ε =- =- = ϖ ϖ p 1/6 c 0.5849 0.9145 ϖ = = ϖ p c 200 218.6 rad / s 0.9145 0.9145 ϖ π ϖ = = = π 1 1 2 2 1 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 + = + + + = + + Transfer function m m 1 m 2 m m 1 m 2 1 m m 1 m 2 m 1 m 2 1 a b c d a s a s a s a s a...
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lec30 - 1 1 Summary of Lecture#30 • Analog filters •...

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