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UK EE 221 - Lecture 10 - Filter Network

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Lecture 10: Filter Networks (continued)Vijay Singh∗February 11, 2003AbstractContinuation of the discussion of filters — we will see an example.1 ExampleA telephone transmission system suffers from 60 Hz interference caused by a nearby powerutility lines. We want to filter out the 60 Hz interference using a notch filter.VinReqZ1VO+_+_Figure 1: Telephone system.Reqrepresents the equivalent resistance of the telephone system.Let us say that the telephone wires are carrying an audio signal at f = 1000 Hz,amplitude 0.2 V.vin(t) = 0.2 sin(2π1000t) + 1 sin(2π60t) (1)Let Req= 50 Ω for example.We wantVOVin−→ 0 at f = 60 HzVOVin−→ 1 at f = 1000 Hz∗Professor and Chairman, Department of Electrical & Computer Engineering, University of Kentucky,Lexington, KY, USA. E-mail: [email protected] Lecture presented on February 12, 2003. Typesetin LATEX1But,VOVin=ReqReq+ Z1So, we want:Req¿ Z1at f = 60 HzReqÀ Z1at f = 1000 HzZ1can be chosen to be a L − C circuit with a resonance frequency of 60 Hz.LCZ11Figure 2:1√LC= 60 Hz, at f =60 Hz, Z11−→ ∞.LCZ12Figure 3:1√LC=60 Hz, at f = 60Hz, Z12−→ 0.Since we want Z1to be large at 60 Hz, we pick the parallel resonance circuit of Figure 1(that is, Figure 4).LCVin+_VOPAB+_ReqFigure 4: .Picking design values for L and CThe impedance between P and A is given byZ11=jωL ×1jωCjωL +1jωC=jωL1 − ω2LC(2)2We want to pick L and C values in such a way that1 = ω2LC (3)at f = 60 Hz. Then Z11−→ ∞ andVOVin−→ 0.Letting ω = 2πf = 2π × 60 in Equation 3, we getLC =1(2π × 60)2Select C = 100µF, then L =1(2π × 60)2× 100 µF= 70.3 mHSelect C = 5µF, then L ≈ 1400 mHSelect C = 1µF, then L ≈ 7 HC L jωL @ f = 1 kHz1jωC@ f = 1 kHz100 µF 70.3 mH j420 −j1.65 µF 1.4 H j8400 −j321 µF 7.3 H j42, 000 −j1601 mF 7 mH j42 −j0.16At 1 kHz,jLω = j70.3 × 10−3× 2π × 103≈ j420 Ω1jωC=1j100 × 10−6× 103× 2π=5jπ≈1.6jΩAt f = 60 Hz,jLω = j25.2 Ω1jωC=25.2j= −j25.2


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UK EE 221 - Lecture 10 - Filter Network

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