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 5Ω . The branch consisting of C 1 in parallel with R = 50 Ω has Q = 50 / 5 = 10 ,andtrans formsinto j 5 in series with 50 / 10 2 =0 . , as shown in Figure a. jX L j 5Ω j 5Ω 0 . 5Ω The two capacitors can be combined in series to yield a branch consisting of j 10 Ω in series with 0 . . This branch transforms into j 10 Ω in parallel with 200 Ω as shown in Figure b. L j 10 200Ω a. The inductor needs to have impedance + j 10 Ω to resonate with the parallel capacitance. b. At resonance, the impedance Z = 200 Ω . 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 . 253 pF. 4-21 Solution a, For the series RLC case with R S = R L = R , the loaded Q of the system will be Q o = X o / (2 R ) .F o rth e
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hwset5solution - ECE 453 Homework Assignment 5 Solution...

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