ch15_ism - Active Filter Circuits 15 Assessment Problems AP...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
15 Active Filter Circuits Assessment Problems AP 15.1 H ( s )= ( R 2 /R 1 ) s s +(1 /R 1 C ) 1 R 1 C =1 rad/s ; R 1 =1Ω , · .. C F R 2 R 1 , · .. R 2 = R 1 · .. H prototype ( s s s +1 AP 15.2 H ( s (1 /R 1 C ) s /R 2 C ) = 20 , 000 s + 5000 1 R 1 C =20 , 000; C =5 µ F · 1 = 1 (20 , 000)(5 × 10 6 ) =10Ω 1 R 2 C = 5000 · 2 = 1 (5000)(5 × 10 6 ) =40Ω 15–1
Background image of page 1

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

View Full DocumentRight Arrow Icon
15–2 CHAPTER 15. Active Filter Circuits AP 15.3 ω c =2 πf c π × 10 4 =20 , 000 π rad/s · .. k f , 000 π =62 , 831 . 85 C 0 = C k f k m · .. 0 . 5 × 10 6 = 1 k f k m · m = 1 (0 . 5 × 10 6 )(62 , 831 . 85) =31 . 83 AP 15.4 For a 2nd order prototype Butterworth high pass flter H ( s )= s 2 s 2 + 2 s +1 For the circuit in Fig. 15.25 H ( s s 2 s 2 + ± 2 R 2 C ² s + ± 1 R 1 R 2 C 2 ² Equate the trans±er ±unctions. For C =1 F, 2 R 2 C = 2 , · .. R 2 = 2=1 . 414 Ω 1 R 1 R 2 C 2 , · 1 = 1 2 =0 . 707 Ω AP 15.5 Q =8 ,K =5 o = 1000 rad/s ,C µ F For the circuit in Fig 15.26 H ( s ³ 1 R 1 C ´ s s 2 + ³ 2 R 3 C ´ s + R 1 + R 2 R 1 R 2 R 3 C 2 ! = Kβs s 2 + βs + ω 2 o β = 2 R 3 C , · R 3 = 2 βC β = ω o Q = 1000 8 = 125 rad/s
Background image of page 2
Problems 15–3 · .. R 3 = 2 × 10 6 (125)(1) =16 k = 1 R 1 C · 1 = 1 KβC = 1 5(125)(1 × 10 6 ) =1 . 6 k ω 2 o = R 1 + R 2 R 1 R 2 R 3 C 2 10 6 = (1600 + R 2 ) (1600)( R 2 )(16 , 000)(10 6 ) 2 Solving for R 2 , R 2 = (1600 + R 2 )10 6 256 × 10 5 , 246 R 2 , 000 ,R 2 =65 . 04 Ω AP 15.6 ω o = 1000 rad/s ; Q =4 ; C =2 µ F H ( s )= s 2 +(1 /R 2 C 2 ) s 2 + " 4(1 σ ) RC # s + ± 1 R 2 C 2 ² = s 2 + ω 2 o s 2 + βs + ω 2 o ; ω o = 1 RC ; β = 4(1 σ ) RC R = 1 ω o C = 1 (1000)(2 × 10 6 ) = 500 Ω β = ω o Q = 1000 4 = 250 · .. 4(1 σ ) RC = 250 4(1 σ ) = 250 RC = 250(500)(2 × 10 6 )=0 . 25 1 σ = 0 . 25 4 =0 . 0625; · σ . 9375
Background image of page 3

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

View Full DocumentRight Arrow Icon
15–4 CHAPTER 15. Active Filter Circuits Problems P 15.1 Summing the currents at the inverting input node yields 0 V i Z i + 0 V o Z f =0 · .. V o Z f = V i Z i · .. H ( s )= V o V i = Z f Z i P 15.2 [a] Z f = R 2 (1 /sC 2 ) [ R 2 +(1 /sC 2 )] = R 2 R 2 C 2 s +1 = (1 /C 2 ) s /R 2 C 2 ) Likewise Z i = (1 /C 1 ) s /R 1 C 1 ) · ( s (1 /C 2 )[ s /R 1 C 1 )] [ s /R 2 C 2 )](1 /C 1 ) = C 1 C 2 [ s /R 1 C 1 )] [ s /R 2 C 2 )] [b] H ( C 1 C 2 " /R 1 C 1 ) /R 2 C 2 ) # H ( j 0) = C 1 C 2 ± R 2 C 2 R 1 C 1 ² = R 2 R 1 [c] H ( j C 1 C 2 j j ! = C 1 C 2 [d] As ω 0 the two capacitor branches become open and the circuit reduces to a resistive inverting amplifer having a gain oF R 2 /R 1 . As ω →∞ the two capacitor branches approach a short circuit and in this case we encounter an indeterminate situation; namely v n v i but v n because oF the ideal op amp. At the same time the gain oF the ideal op amp is infnite so we have the indeterminate Form 0 ·∞ . Although ω = is indeterminate we can reason that For fnite large values oF ωH ( ) will approach C 1 /C 2 in value. In other words, the circuit approaches a purely capacitive inverting amplifer with a gain oF ( 1 /jωC 2 ) / (1 1 ) or C 1 /C 2 .
Background image of page 4
Problems 15–5 P 15.3 [a] Z f = (1 /C 2 ) s +(1 /R 2 C 2 ) Z i = R 1 + 1 sC 1 = R 1 s [ s /R 1 C 1 )] H ( s )= (1 /C 2 ) [ s /R 2 C 2 )] · s R 1 [ s /R 1 C 1 )] = 1 R 1 C 2 s [ s /R 1 C 1 )][ s /R 2 C 2 )] [b] H ( 1 R 1 C 2 ± + 1 R 1 C 1 ²± + 1 R 2 C 2 ² H ( j 0)=0 [c] H ( j )=0 [d] As ω 0 the capacitor C 1 disconnects v i from the circuit. Therefore v o = v n =0 .
Background image of page 5

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

View Full DocumentRight Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/26/2008 for the course ELEN 214 taught by Professor Zou during the Spring '08 term at Texas A&M.

Page1 / 56

ch15_ism - Active Filter Circuits 15 Assessment Problems AP...

This preview shows document pages 1 - 6. Sign up to view the full document.

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