# Exercise is it possible for the pole frequency to be

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Exercise Is it possible for the pole frequency to be five times the zero frequency while the passband gain is ten times the stopband gain? The first-order filters studied above provide only a slope of dB/dec in the transition band. For a sharper attenuation, we must seek circuits of higher order. 14.3 Second-Order Filters The general transfer function of second-order filters is given by the “biquadratic” equation: (14.23) Unlike the numerator, the denominator is expressed in terms of quantities and because they signify important aspects of the response. We begin our study by calculating the pole frequencies. Since most second-order filters incorporate complex poles, we assume , obtaining (14.24) Note that as the “quality factor” of the poles, , increases, the real part decreases while the imaginary part approaches . This behavior is illustrated in Fig. 14.19. In other words, for high ’s, the poles look “very imaginary,” thereby bringing the circuit closer to instability.

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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 734 (1) 734 Chap. 14 Analog Filters j ω σ Q ω n + ω n Figure 14.19 Variation of poles as a function of Q. 14.3.1 Special Cases It is instructive to consider a few special cases of the biquadratic transfer function that prove useful in practice. First, suppose so that the circuit contains only poles and operates as a low-pass filter (why?). The magnitude of the transfer function is then obtained by making the substitution in (14.23) and expressed as (14.25) Note that provides a slope of dB/dec beyond the passband (i.e., if ). It can be shown (Problem 11) that the response is (a) free from peaking if ; and (b) reaches a peak at if (Fig. 14.20). In the latter case, the peak magnitude ( ) ω H ω Q > 2 2 Q 2 2 < ω n Q 2 2 1 1 ω n γ 2 Figure 14.20 Frequency response of second-order system for different values of Q.. normalized to the passband magnitude is equal to . Example 14.10 Suppose a second-order LPF is designed with . Estimate the magnitude and frequency of the peak in the frequency response. Since at , we say the circuit exhibits two zeros at infinity.
BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 735 (1) Sec. 14.3 Second-Order Filters 735 Solution Since , we observe that the normalized peak magnitude is and the corresponding frequency is . The behavior is plotted in Fig. 14.21. ( ) ω H ω ω n ω n γ 2 ω n γ 2 3 Figure 14.21 Exercise Repeat the above example for . How does the transfer function in Eq. (14.23) provide a high-pass response? In a manner similar to the first-order realization in Fig. 14.12(b), the zero(s) must fall below the poles. For example, with two zeros at the origin: (14.26) we note that approaches zero as and a constant value, , as , thus pro- viding a high-pass behavior (Fig. 14.22). As with the low-pass counterpart, the circuit exhibits a peak if with a normalized value of but at a frequency of .

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