Fundamentals-of-Microelectronics-Behzad-Razavi.pdf

Exercise explain why the impedance has a zero at the

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Exercise Explain why the impedance has a zero at the origin. Now let us turn our attention to the parallel RLC tank depicted in Fig. 14.26(a). We can obtain C L 1 Z 1 R 1 2 Figure 14.26 Lossy tank. by replacing in parallel with in Eq. (14.32): (14.34) (14.35)
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 739 (1) Sec. 14.3 Second-Order Filters 739 The impedance still contains a zero at the origin due to the inductor. To compute the poles, we can factor from the denominator, thus obtaining a form similar to the denominator of (14.23): (14.36) (14.37) where and . It follows from (14.24) that (14.38) (14.39) These results hold for complex poles, i.e., if (14.40) or (14.41) On the other hand, if decreases and , we obtain real poles: (14.42) (14.43) So long as the excitation of the circuit does not alter its topology, the poles are given by (14.39) or (14.43), a point that proves useful in the choice of filter structures. Before studying different RLC filters, it is instructive to make several observations. Consider the voltage divider shown in Fig. 14.27, where a series impedance and a parallel impedance out V in V Z S Z P Figure 14.27 Voltage divider using general impedances. yield (14.44) The “topology” of a circuit is obtained by setting all independent sources to zero.
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 740 (1) 740 Chap. 14 Analog Filters We note that (a) if, at high frequencies, goes to zero and/or goes to infinity, then the circuit operates as a low-pass filter; (b) if, at low frequencies, goes to zero and/or goes to infinity, then the circuit serves as a high-pass filter; (c) if remains constant but falls to zero at both low and high frequencies then the topology yields a band-pass response. These cases are conceptually illustrated in Fig. 14.28. f ( ) ω H f ( ) ω H f ( ) ω H Z S Z P = 0 or = Z S Z P = 0 or = Z P = 0 Z p = 0 (c) (a) (b) Figure 14.28 (a) Low-pass, (b) high-pass, and (c) bandpass responses obtained from the voltage divider of Fig. 14.27. Low-Pass Filter Following the observation depicted in Fig. 14.28(a), we construct the circuit shown in Fig. 14.29, where C out V in V R L 1 1 1 Z S Z P Figure 14.29 Low-pass filter obtained from Fig. 14.27. (14.45) (14.46) This arrangement provides a low-pass response having the same poles as those given by (14.39) or (14.43) because for , it reduces to the topology of Fig. 14.26. Furthermore, the transi- tion beyond the passband exhibits a second-order roll-off because both and . The reader can show that (14.47) Example 14.14 Explain how the transfer function of (14.47) can provide a voltage gain greater than unity. Solution If the of the network is sufficiently high, the frequency response exhibits “peaking,” i.e., a gain of greater than unity in a certain frequency range. With a constant numerator, the transfer We assume and do not go to zero or infinity simultaneously.
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 741 (1) Sec. 14.3 Second-Order Filters 741 function provides this effect if the denominator falls to a local minimum. Writing the squared magnitude of the denominator as (14.48) and taking its derivative with respect to , we have (14.49) The derivative goes to zero at (14.50)
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