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Unformatted text preview: ECE329 Fall 2008 Homework 14 — Solution December 11, 2008 1. Let us consider two capacitors C 1 = 100 pF and C 2 connected at both ends of a quarterwavelength section of a 50ohm transmission line. This circuit is designed to resonate at f o = 200 MHz . a) The quarterwavelength section is used to transform the impedance of the second capacitor, Z 2 = 1 jω o C 2 , into the impedance of an equivalent inductor Z eq = Z 2 o Z 2 = jω o C 2 Z 2 o = jω o L eq . The equivalent inductor L eq = C 2 Z 2 o together with C 1 constitute a parallel LC circuit that resonates at ω o = 1 √ L eq C 1 . Solving for C 2 , we can find that C 2 = 1 C 1 Z 2 o ω 2 o = 2 . 533 pF . b) Since v p = 2 3 c and f o = 200 MHz then λ o = v p /f o = 1 m . Thus, the length of the transformer is l = λ o 4 = 25 cm . c) Yes, the circuit built in part (a) does resonate at frequencies other than 200 MHz . For this purpose, let us first calculate the input impedance of the capacitor C 2 . Since Z 2 = 1 jωC 2 and Γ L = Z L Z o Z L + Z o = 1 jωC 2 50 1 jωC 2 + 50 = 1 jωC 2 · 50 1 + jωC 2 · 50 , we can find that Γ in = Γ L e 2 jβl = e 2 jωl/v p 1 jωC 2 · 50 1 + jωC 2 · 50 and also that Z in = Z o 1+Γ in 1 Γ in . This impedance in parallel with Z 1 = 1 jωC 1 constitute a parallel LC circuit that resonates if the following condition is satisfied 1 Z 1 + 1 Z in = jωC 1 + 1 Z in = 0 ....
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 Spring '08
 FRANKE
 Electrical impedance, Impedance matching, Standing wave, zo, ZL, Antenna tuner

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