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# 1 pf the data also show that the imaginary part of

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Unformatted text preview: ; ω = 2π 19 × 106 s−1 1 + ω 2 C 2 R2 which is easily solved for C , yielding the result C ￿ 25.1 pF. The data also show that the imaginary part of the reactance is approximately Xm = 810Ω at f = 18 MHz. Hence Z = j ωL + 810 ￿ ω L − ω CR2 ; ω = 2π 18 × 106 s−1 1 + ω 2 C 2 R2 which is easily solved for L, yielding L ￿ 9.3 µH. 3 3-6 Solution a. The impedance of the inductor is ZL = or Z=R 1 ||(R + j ω L) j ωC 1 (1 − ω 2 LC )2 + ω 2 R2 C 2 + j ωL (1 − ω 2 LC ) − R2 C/L . (1 − ω 2 LC )2 + ω 2 R2 C 2 The resonant frequency is found by setting the imaginary part of the impedance to zero, and solving for ω . The result is ￿ 1 R2 ωp = − 2. LC L ￿ 2 1 1 L Clearly, ωp ￿ √LC only if R2 ￿ LC , or when R is small...
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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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