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

10-28 Passive, Active, and Digital FiltersCase 3 applies. Then K 5 ¼ 1 and from Equation 10.145bs 1 ¼ 1 10 8 1sinh 0:21661042 sin 18 ¼ 0:64677 10 8 (10:164)For K 5 ¼ 1, Equation 10.126 degenerates intowhere y ¼ s=10 8 , the minimum-phase solution of which is found to ber(y)r( y) ¼ e2 C 2 5 ( jy)1 þ e 2 C 2 5 ( jy) (10:165)^r(y) ¼y 5 þ 1:25 y 3 þ 0:3125 yy 5 þ 0:706461 y 4 þ 1:49954 y 3 þ 0:693477 y 2 þ 0:459349 y þ 0:0817225(10:166)A more general solution is given byr(y) ¼Finally, we compute the equalizer back-end impedance asy 0:64677 108 ^r(y) (10:167)y þ 0:64677 108 Z 22 (y)100 ¼ F(y)A(y) r(y)¼2(yþ1) 2y 1yþ1r(x)z 2 (y)0:5y1y þ 1¼ 1:63267 y5 þ 0:576709 y 4 þ 2:15208 y 3 þ 0:533447 y 2 þ 0:554503 y þ 0:05285570:734663 y 4 þ 0:259505 y 3 þ 0:639438 y 2 þ 0:123844 y þ 0:05285571¼ 2:222 y þ11:005 y þ13:651 y þ10:8204 y þ0:2441 y þ 0:052860:02736 y þ 0:05286(10:168)The final matching network together with its terminations is presented in Figure 10.13.0.5 μHV g1.16 μH0.388 μH 3.46206 pF–3.65 μH82 pF2.22 μH100 pFμH+100 Ω100 pF100 ΩZ 22 (s)FIGURE 10.13Fifth-order Chebyshev matching network.