Unformatted text preview: compared to C .
L
b. For L = 500 nH, C = 1 pF, and R = 1 Ω the term R2 /L2 = 4(1012 ) whereas
is an excellent approximation. Hence
fp 1
LC = 2(1018 ) so ωp √1
LC 1
√
= 225.1 MHz
2π LC c. At fp the impedance is purely resistive and equal to Z 1
ω2 C 2 R = 500 kΩ. d. At 0.75fp direct calculation shows that the reactance is 1212 Ω. The eﬀective inductance is therefore
1212
Lef f = 2π(0.75)(225.1 MHz) = 1143 nH which is more than twice the lowfrequency value of 500 nH. More
generally, assuming that R2
L2 1
LC then
Lef f = X (ω )
1
L
2
ω
1 − f2
fp which means that the eﬀective inductance will increase rapidly as f approaches fp from below. 4...
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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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