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

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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

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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
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

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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 .
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 .

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

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