title: Active-Filter Cookbook
author:  Donald E. Lancaster 
description: This Don Lancaster classic is by far the best-selling active filter book of all time. It gives you everything you need to know to build active lowpass, bandpass, and highpass filters.<br /><br /><br /> An active filter needs no inductors. Instead, op amps, resistors, and capacitors are used for better results. Advantages include lower cost, easy tuning, simple design, and modularity. Lancaster's Active Filter Cookbook includes practical elements such as working circuits, ready-to-use design tables, tuning, and real-world applications, making it easy to use and apply. You'll find both instant design and the mathematics behind coverage. Microcomputer pioneer Don Lancaster is the author of 34 books, 2 videos and countless articles and columns in technical magazines. He also heads Synergetics, an Arizona-based design and consulting firm.<br />
zlib_text: 15
  Active Filter Circuits
  Assessment Problems
  AP 15.1 H(s) = −(R2/R1)s
  s + (1/R1C)
  1
  R1C = 1 rad/s;
  R1 = 1 Ω,
  ·. .
  C = 1 F
  R2
  R1
  = 1,
  ·. .
  R2 = R1 = 1 Ω
  ·. .
  Hprototype(s) =
  −s
  s + 1
  AP 15.2 H(s) =
  −(1/R1C)
  s + (1/R2C) = −20,000
  s + 5000
  1
  R1C = 20,000;
  C = 5 µF
  ·. .
  R1 =
  1
  (20,000)(5 × 10−6) = 10 Ω
  1
  R2C = 5000
  ·. .
  R2 =
  1
  (5000)(5 × 10−6) = 40 Ω
  15–1
  
  15–2
  CHAPTER 15. Active Filter Circuits
  AP 15.3 ωc = 2πfc = 2π × 104 = 20,000π rad/s
  ·. .
  kf = 20,000π = 62,831.85
  C′ =
  C
  kfkm
  ·. .
  0.5 × 10−6 =
  1
  kfkm
  ·. .
  km =
  1
  (0.5 × 10−6)(62,831.85) = 31.83
  AP 15.4 For a 2nd order prototype Butterworth high pass ﬁlter
  H(s) =
  s2
  s2 +
  √
  2s + 1
  For the circuit in Fig. 15.25
  H(s) =
  s2
  s2 +
  �
  2
  R2C
  �
  s +
  �
  1
  R1R2C2
  �
  Equate the transfer functions. For C = 1F,
  2
  R2C =
  √
  2,
  ·. .
  R2 =
  √
  2 = 1.414 Ω
  1
  R1R2C2 = 1,
  ·. .
  R1 = 1
  √
  2 = 0.707 Ω
  AP 15.5 Q = 8, K = 5, ωo = 1000 rad/s, C = 1 µF
  For the circuit in Fig 15.26
  H(s) =
  −
  �
  1
  R1C
  �
  s
  s2 +
  �
  2
  R3C
  �
  s +
  � R1 + R2
  R1R2R3C2
  �
  =
  Kβs
  s2 + βs + ω2
  o
  β =
  2
  R3C ,