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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 = 1 Ω , · . . C = 1 F R 2 R 1 = 1 , · . . R 2 = R 1 = 1 Ω · . . H prototype ( s ) = s s + 1 AP 15.2 H ( s ) = (1 /R 1 C ) s + (1 /R 2 C ) = 20 , 000 s + 5000 1 R 1 C = 20 , 000; C = 5 µ F · . . R 1 = 1 (20 , 000)(5 × 10 6 ) = 10 Ω 1 R 2 C = 5000 · . . R 2 = 1 (5000)(5 × 10 6 ) = 40 Ω 15–1

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15–2 CHAPTER 15. Active Filter Circuits AP 15.3 ω c = 2 πf c = 2 π × 10 4 = 20 , 000 π rad/s · . . k f = 20 , 000 π = 62 , 831 . 85 C = C k f k m · . . 0 . 5 × 10 6 = 1 k f k m · . . k m = 1 (0 . 5 × 10 6 )(62 , 831 . 85) = 31 . 83 AP 15.4 For a 2nd order prototype Butterworth high pass filter 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 s + 1 R 1 R 2 C 2 Equate the transfer functions. For C = 1 F, 2 R 2 C = 2 , · . . R 2 = 2 = 1 . 414 Ω 1 R 1 R 2 C 2 = 1 , · . . R 1 = 1 2 = 0 . 707 Ω AP 15.5 Q = 8 , K = 5 , ω o = 1000 rad/s , C = 1 µ F For the circuit in Fig 15.26 H ( s ) = 1 R 1 C s s 2 + 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 · . . R 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 = 16 , 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 = 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 = 0 . 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 + (1 /R 2 C 2 ) Likewise Z i = (1 /C 1 ) s + (1 /R 1 C 1 ) · . . H ( s ) = (1 /C 2 )[ s + (1 /R 1 C 1 )] [ s + (1 /R 2 C 2 )](1 /C 1 ) = C 1 C 2 [ s + (1 /R 1 C 1 )] [ s + (1 /R 2 C 2 )] [b] H ( ) = C 1 C 2 + (1 /R 1 C 1 ) + (1 /R 2 C 2 ) H ( j 0) = C 1 C 2 R 2 C 2 R 1 C 1 = 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 amplifier 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 = 0 because of the ideal op amp. At the same time the gain of the ideal op amp is infinite so we have the indeterminate form 0 · ∞ . Although ω = is indeterminate we can reason that for finite large values of ω H ( ) will approach C 1 /C 2 in value. In other words, the circuit approaches a purely capacitive inverting amplifier with a gain of ( 1 /jωC 2 ) / (1 /jωC 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 + (1 /R 1 C 1 )] H ( s ) = (1 /C 2 ) [ s + (1 /R 2 C 2 )] · s R 1 [ s + (1 /R 1 C 1 )] = 1 R 1 C 2 s [ s + (1 /R 1 C 1 )][ s + (1 /R 2 C 2 )] [b] H ( ) = 1 R 1 C 2 + 1 R 1 C 1 + 1 R 2 C 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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