Answer for part b series lc with l 100 nh c 0253 pf 4

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Unformatted text preview: a series LC is used with the largest possible inductance. Answer for part b: Series LC with L = 100 nH, C = 0.253 pF. 4-21 Solution a, For the series RLC case with RS = RL = R, the loaded Q of the system will be Qo = Xo /(2R). For the parallel RLC topology, the loaded Q will be R/(2Xo ). Thus, if Xo > R, the smallest possible bandwidth will be obtained using the series LC filter. ￿ b. With the lossy inductor in place, the loaded Q of the system becomes Qo = bandwidth, BW = ￿ fo Q￿ o = fo (Q−1 o + Q−1 ) L = fo Qo (1 + Qo QL ) = BWo (1 + Qo /QL ). c. BW ￿ = BWo (1 + 1 ), so bandwidth is increased by 25%. 4 ￿ d. PL /PL = (1 + Qo −2 . QL ) e. Loss = 20 log10 (1 + 0.25) = 1.94 dB 1 Xo 2R+Xo /QL = 1 . Q−1 +Q−1 o L The 4-24 Solution The target fractional bandwidth is 0.2 (20%), so the system should be designed to have loaded Qs = 1/0.2 = 5. Since the inductive reactance is fixed at 50 Ω, the net series resistance from the source and load must be 10 Ω. This means that the 100 ...
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This homework help was uploaded on 03/13/2014 for the course ECE 453 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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