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hwset5solution - ECE 453 Homework Assignment 5 Solution...

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ECE 453: Homework Assignment 5 Solution 4-13 Solution At 10 MHz, the reactance of the capacitors is j 5 . The branch consisting of C 1 in parallel with R = 50 has Q = 50 / 5 = 10 , and transforms into j 5 in series with 50 / 10 2 = 0 . 5 , as shown in Figure a. jX L j 5 j 5 0 . 5 The two capacitors can be combined in series to yield a branch consisting of j 10 in series with 0 . 5 . This branch transforms into j 10 in parallel with 200 as shown in Figure b. jX L j 10 200 a. The inductor needs to have impedance + j 10 to resonate with the parallel capacitance. b. At resonance, the impedance Z = 200 . c. Q p = 200 / 10 = 20 . 4-20 Solution If a parallel LC circuit is used, then Q = Q p = 100 || 100 / ( ω o L ) . If a series LC circuit is used, then Q = Q s = ω o L/ (100 + 100) . In part a, the largest Q results when a parallel LC is used with the smallest possible inductance. Answer for part a: Parallel LC with L = 10 nH, C = 253 . 3 pF. In part b, the largest Q reults when a series LC is used with the largest possible inductance. Answer for part b: Series LC with L = 100 nH, C = 0 . 253 pF.
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