1 mhz 2 lc c at fp the impedance is purely resistive

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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 low-frequency 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.

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