ECE2260F08_HW4p1soln

# It differs from a standard band pass only in having a

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Unformatted text preview: nter frequency at ω0 and a bandwidth of β. It differs from a standard band-pass only in having a maximum magnitude response of 0.75 instead of 1.0. We can achieve this reduction by placing a 3.6 kΩ resistor from top to bottom just to the right of R1: R1 = 1.2 kΩ vi(t) + – + vo (t) R2 = 3.6 kΩ – Taking the Thevenin equivalent of the input source and the two R 's, we have the following equivalent circuit: RThev = 1.2 kΩ||3.6 kΩ = 900 Ω + 3.6k 3 vi ( t) = vi ( t) 3.6k + 1.2k 4 + – vo (t) – € Now we use a band-pass configuration with L and C across the terminals from top to bottom: + RThev = 900 Ω 3 v i (t) 4 € + – L C vo (t) – The center frequency of the filter is given by the usual formula (that follows from assuming the impedances of the L and C are equal but opposite, causing the parallel impedance to become infinite, or opencircuit): 1 LC ω0 = € The bandwidth of the filter is given by the standard formula with RThev substituted for R: β= 1 RThevC Solving for C and using RThev = 900 Ω yields the following value: € € C= RThevβ = 1 1 = µF 900 ⋅ 1M / 450 2 Using this value for C , and rearranging the formula for the center frequency gives the following value for L: L= € 1 1 ω 2C 0 = 1 2 1M 1 µ 1600 2 H= 1.6 2 (1k) 2 ⋅ 2 = 5.12 H 1µ...
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## This document was uploaded on 03/17/2014 for the course ECE 2260 at Utah.

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