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

10-20 Passive, Active, and Digital Filters0.5 μH97.5 Ω0.341 μH 1.53 μH 1.34 μH104 pF 198 pF+V g–N100 pFZ 22 (s)100 ΩFIGURE 10.9Sixth-order Butterworth matching network.and from Equation 10.93,L a2 ¼ 1:401 mH > L ¼ 0:5 mH (10:111)Case 3 applies, and we have K 5 ¼ 1 and from Equation 10.100bs 1 ¼ 0:381966 10 8 (10:112)The normalized reflection coefficient is found to be(y 0:381966)y 5r(y) ¼(y þ 0:381966) ðy 5 þ 3:23607y 4 þ 5:23607y 3 þ 5:23607y 2 þ 3:23607y þ 1Þ(10:113)where y ¼ s=10 8 . Finally, we compute the equalizer back-end impedance asZ 22 (s)100 ¼ F(y)z 2 (y)A(y) r(y)2(0:5 yþ1)¼20:5 y 10:5 y0:5 yþ1r(s)10:5 y þ 1¼ 2:573 y5 þ 4:1631 y 4 þ 5:177 y 3 þ 4:2136 y 2 þ 2:045 y þ 0:381972:8541 y 4 þ 4:618 y 3 þ 4:118 y 2 þ 2:045 y þ 0:381971¼ 0:9015 y þ11:949 y þ11:821 y þ10:8002 y þ1:005 y þ 0:38220:9944 y þ 0:3822(10:114)The final matching network together with its terminations is presented in Figure 10.10.Example 10.7LetR ¼ 100 V, C ¼ 50 pF, L ¼ 3 mF (10:115a)n ¼ 4, v c ¼ 10 8 rad=s (10:115b)