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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 36 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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 Spring '08
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