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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. 421 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 424 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.
 Spring '08
 Staff

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