With we obtain from 14141 exercise if the order of

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that satisfies the requirement is 3. With , we obtain from (14.141): Exercise If the order of the filter must not exceed 2, how much attenuation can be obtained from to ? Given filter specifications and hence a template, we can readily choose and in (14.138) to arrive at an acceptable Butterworth approximation. But how do we translate (14.138) to a trans- fer function and hence a circuit realization? Equation (14.138) suggests that the corresponding transfer function contains no zeros. To obtain the poles, we make a reverse substitution, , and set the denominator to zero: (14.145) That is, (14.146) This polynomial has roots given by (14.147) but only the roots having a negative real part are acceptable (why?): (14.148)

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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 761 (1) Sec. 14.5 Approximation of Filter Response 761 How are these poles located in the complex plane? As an example, suppose . Then, (14.149) (14.150) As shown in Fig. 14.49(a), the poles are located at , i.e., their real and imaginary parts are ω j ω σ +135 135 ω j ω σ 0 π n 0 (a) (b) Figure 14.49 Locations of poles for (a) second-order, and (b) -th order Butterworth filter. equal in magnitude. For larger values of , each pole falls on a circle of radius and bears an angle of with respect to the next pole [Fig. 14.49(b)]. Having obtained the poles, we now express the transfer function as (14.151) where the factor in the numerator is included to yield . Example 14.25 Using a Sallen and Key topology as the core, design a Butterworth filter for the response derived in Example 14.24. Solution With and , the poles appear as shown in Fig. 14.50(a). The complex conjugate poles and can be created by a second-order SK filter, and the real pole by a simple RC section. Since (14.152) (14.153) the SK transfer function can be written as (14.154)
BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 762 (1) 762 Chap. 14 Analog Filters j ω σ (a) (b) p 2 p 1 p 3 π 3 π 3 out V in V 228 pF 4.52 pF 109.8 pF 1 k 1 k 1 k Figure 14.50 (14.155) That is, (14.156) (14.157) In Eq. (14.61), we choose and to obtain . From Eq. (14.62), to obtain , we have some freedom, e.g., k and pF. The reader is encouraged to verify that this design achieves low sensitivities. The real pole, , is readily created by an RC section: (14.158) For example, k and pF. Figure 14.50(b) shows the resulting design. Exercise If the 228-pF capacitor incurs an error of 10%, determine the error in the value of . The Butterworth response is employed only in rare cases where no ripple in the passband can be tolerated. We typically allow a small ripple (e.g., 0.5 dB) so as to exploit responses that provide a sharper transition slope and hence a greater stopband attenuation. The Chebyshev response is one such example.

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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 763 (1) Sec. 14.5 Approximation of Filter Response 763 14.5.2 Chebyshev Response The Chebyshev response provides an “equiripple” passband behavior, i.e., with equal local max- ima (and equal local minima). This type of response specifies the magnitude of the transfer function as: (14.159) where sets the amount of ripple and denotes the “Chebyshev polynomial” of -th or- der. We consider as the “bandwidth” of the filter. These polynomials are expressed recursively
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