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

10-24 Passive, Active, and Digital FiltersThe details of these derivations can be found in Ref. [3]. Thus, with K n as specified in Equation 10.131, amatch is possible if and only if the series inductance L does not exceed a critical inductance L b . To showthat any RLC load can be matched, we must demonstrate that there exists a nonnegative real s 1 such thatL b can be made at least as large as the given inductance L and satisfies the constraint (Equation 10.131)with 0 K n 1. To this end, four cases are distinguished. LetR 2 Cv c sin gL b1 ¼ 3ð1 RCv c sinh a sin g 1 Þ 2 > 0 (10:137)þR 2 C 2 v 2 c cosh2 a cos 2 g 1 vc sin g 18R sin 2 gL b2 ¼1 sin g 3ðRCv c sinh a sin g 3 Þ 2 þð1 þ 4 sin 2 g 1 Þsin g 1 sin g 3 þ R 2 C 2 v 2 c sin2 g 2 vc sinh a > 0 (10:138)Observe that both L b1 and L b2 are positive.Case 1. RCv c sinh a 2 sin g 1 and L b1 L. Under this situation, s 1 ¼ 0 and the maximumattainable K n is given by K n ¼ 1 e 2 sinh 2 n sinh 1 sinh a2 sin g 1RCv c(10:139)The equalizer back-end impedance Z 22 (s) can be expanded in a continued fraction as inEquation 10.94 with L b1 replacing L a1 and realized by the LC ladders Figure 10.8 withL 1 ¼ L b1(10:140a)C 2m L 2m 1 ¼ 4 sin g 4m 1 sin g 4mþ1v 2 c f 2m(sinh a, sinh ^a) , m < 1 (n 1) (10:140b)2C 2m L 2mþ1 ¼ 4 sin g 4mþ1 sin g 4mþ3v 2 c f 2mþ1(sinh a, sinh ^a) , m < 1 (n 1) (10:140c)2 1where m ¼ 1, 2, ...,2(n 1) , n > 1. In addition, the final reactive element can also becomputed directly by the formulasC nL n1 ¼1 ¼2(sinh na þ sinh n^a) sin g 1, n odd (10:141a)Rv c (sinh a þ sinh ^a)( sinh na sinh n^a)2R(cosh na cosh n^a) sin g 1, n even (10:141b)v c (sinh a þ sinh ^a)(cosh na þ cosh n^a)The terminating resistance is determined bysinh na sinh n^aR 22 ¼ R , n odd (10:142a)sinh na þ sinh n^acosh na cosh n^aR 22 ¼ R , n even (10:142b)cosh na þ cosh n^aCase 2. RCv c sinh a 2 sin g 1 and L b1 < L. Under this situation, s 1 is nonzero and can bedetermined by the formula