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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 02/18/2010.

Page1 / 7

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online