Passive, active, and digital filters (3ed., CRC, 2009) - tiera.ru

Design of Resistively Terminated Networks 9-9those given in Equations 9.35 through 9.39 except that the roles of C’s and L’s are interchanged andR 1 and R 2 are replaced by their reciprocals.C 1 ¼2 sin p=2nR 1 (1 a)v c(9:45a)C 2m4 sin g1 L 2m ¼4m 3 sin g 4m 1v 2 cð 1 2a cos g 4m 2 þ a 2 Þ(9:45b)C 2mþ1 L 2m ¼ 4 sin g 4m 1 sin g 4mþ1v 2 cð 1 2a cos g 4m þ a 2 Þ(9:45c)for m ¼ 1, 2, ..., dn=2e. The values of the final elements can also be calculated directly byC n ¼2 sin p=2nR 2 (1 þ a)v c, n odd (9:46a)L n ¼ 2R 2 sin p=2n, n even (9:46b)(1 þ a)v c9.3 Double-Terminated Chebyshev NetworksNow, we consider the problem of synthesizing an LC ladder which when connected between a resistivesource of internal resistance R 1 and a resistive load of resistance R 2 will yield a preassigned Chebyshevtransducer power-gain characteristicG v 2 ¼ jr 21 ( jv) j 2 ¼K n1 þ e 2 Cn 2 ð v=v cÞ(9:47)with K n bounded between 0 and 1. Following Equation 9.24, the squared magnitude of the inputreflection coefficient can be written asjr 11 ( jv) j 2 ¼ 1 G(v 2 ) ¼ 1 jr 21 ( jv) j 2 ¼ 1 K n þ e 2 Cn 2 ð v=v cÞ1 þ e 2 Cn 2 ð v=v cÞ(9:48)Appealing to analytic continuation, we obtainr 11 (s)r 11 ( s) ¼ ð1K nÞ 1 þ ^e2 C 2 n ( jy)1 þ e 2 C 2 n ( jy) (9:49)wheree^e ¼ p ffiffiffiffiffiffiffiffiffiffiffiffiffi(9:50)1 K nAs in the Butterworth case, we assign LHS poles to r 11 (s) and the minimum-phase solution of Equation9.49 becomesr 11 (s) ¼^p(y)p(y)(9:51)