The series resistance of the branch consisting of 100

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Unformatted text preview: source and load resistances must be transformed down to 5 Ω by the shunt capactors. The √ series resistance of the branch consisting of 100 Ω||jXo is 100/(1 + Q2 ), so we need 2 1 + Q = 20, or Q = 19 = 4.36. Thus Xo must be −100/4.36 = −23 Ω (rounding to the nearest Ohm). To resonate the system, the net capacitive reactance must be −50 Ω, so we need −23 − 23 + XC = −50, or XC = −4 Ω. The final circuit is shown in the Figure. j 50Ω −j 4Ω 100Ω −j 23Ω Vs −j 23Ω 100Ω Since an explicit center frequency is not specified, we can scale the filter to any center frequency. If the center frequency is chosen to be 100 MHz, then the component values are as shown in the Figure below. 398 pF 100Ω 79.6 nH 69.2 pF Vs 69.2 pF 100Ω The magnitude of the voltage across the output resistor is shown below, forVs = 1 Volt, at frequencies near the center frequency. The center frequency is slightly shifted away from 100 MHz because the component values have been rounded. 0.5 0.4 0.3...
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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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