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Tunable Constant Q Band-Pass Filter Design Using Q and K Values

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Tunable Constant Q Band-Pass Filter Design Using Q and K Values Tunable Constant Q Band-Pass Filter Design Using q and k Values This paper describes a method for designing a tunable, coupled-resonator band-pass filter using q and k values. The method results in a filter that has a constant Q and a consistent filter characteristic as it is tuned from low to high frequency. The q and k values are either calculated or are taken from published tables. A symmetric design based on a N=2 low-pass prototype and using lossless components is assumed for this design example. Xc R C L C L R n n n n n n Figure 1. Prototype filter. Lc or Cc RS CS LS CS LS RL Figure 2. Realized filter with inductive or capacitive resonator coupling. RN = Nodal resistance CN = Nodal capacitance LN = Nodal inductance XC = Coupling reactance, either inductive (LC) or capacitive (CC). RS = RL = source and load resistance CS = shunt capacitance LS = shunt inductance CC = coupling capacitance LC = coupling inductance Given qN & kN: 휔0 Filter Q, 푄푏푝 = (1) ∆휔−3푑퐵 푘푁 퐾푁 = (2) 푄푏푝 푄푁 = 푞푁푄푏푝 (3) 푅푁 = 휔0푄푁퐿푁 (4) 1 휔0 = (5) √퐿푁퐶푁 Coupling structure and shunt elements: 퐶푁푘푁 퐶퐶 = 퐶푁퐾푁 = (6) 퐶푆 = 퐶푁 − 퐶퐶 (7) 푄푏푝 퐿푁 퐿푁푄푏푝 1 퐿퐶 = = (8) 퐿푆 = 1 1 (9) 퐾푁 푘푁 − 퐿푁 퐿퐶 To make the filter tunable, either the nodal C or nodal L (or both) must change. From (6) it is seen that with capacitive coupling, CC varies with nodal capacitance CN. From (8), the inductive coupling element, LC varies with nodal inductance LN. Conclusion: If it is desired to hold the coupling element constant while tuning a filter by varying CN , then inductive coupling would be the type to use. Conversely, if the filter is tuned with variable inductance then capacitive coupling should be used. Notice that these are true only for a filter whose Q, QBP is constant as the filter center frequency is tuned. 푅푁 But from (3) & (4): 푄푏푝 = (10) 휔0퐿푁푞푁 ∴ 푄푏푝 ∝ 1⁄휔0 In order to have the Q remain constant as the filter is tuned, then 푅푁 the numerator of (10) needs to vary linearly with frequency as the filter is tuned from 휔퐿푂to 휔퐻퐼 . This will give a filter with a non-changing transfer function and constant 푄푏푝. 푅푁 From (4), = 푄푏푝퐿푁푞푁. With 푄푏푝 held constant as 휔0 is increased, 푅푁 must increase proportionally. 휔0 We want a matching network that converts a fixed resistance on one side to one that varies proportionally with frequency on the other. LC Z Z Matching Matching R RS Network Network L LS CS CS LS Desired R of the matching network presented to the resonant nodes: R |x| ? f f Any reactive component of the matching network will be absorbed by either the Ln or Cn. Consider this narrow band equivalency: L a 2 2 휔0퐿푎 푅 = 푅 + (see Appendix) R a L 1 R 1 1 푎 @0 푅푎 Using this network as the matching circuit changes Figure 2 as follows: L a LC L a RS = Ra RL = Ra LS CS CS LS @0 @0 LC R 1 L 1 L 1 R 1 LS CS CS LS Figure 3. Filter with matching network and narrowband equivalent. R1 will be the nodal resistance, RN and will increase somewhat non-linearly with frequency but can be made to match the desired value of R1 at the two frequency tuning extremes, 휔퐿푂 and 휔퐻퐼. Between the two frequency extremes, R1 will have a positive error relative to the ideal value of RN. A simulation of the network will show whether it has an unacceptable effect on the filter’s transfer function. It can be mitigated somewhat by reducing the frequency range for the calculation of the network. This will split the error between those frequencies above and below the calculated range, which will be negative, and the those between the calculated range, which will be positive. 2 2 2 2 휔퐿푂퐿푎 휔퐻퐼퐿푎 푅1_퐿푂 = 푅푎 + (11) 푅1_퐻퐼 = 푅푎 + (12) 푅푎 푅푎 Since 푅1 increases proportionally with frequency: 휔퐻퐼 푅1_퐻퐼 = 푚푅1_퐿푂 , where 푚 = , (13) 휔퐿푂 2 2 푚휔퐿푂퐿푎 From (11), 푚푅1_퐿푂 = 푚푅푎 + 푅푎 2 2 2 푚 휔퐿푂퐿푎 From (12) and (13), 푚푅1_퐿푂 = 푅푎 + 푅푎 Equating: 2 2 2 2 2 푦푖푒푙푑푠 푚 휔퐿푂퐿푎 푚휔퐿푂퐿푎 2 2 2 푅푎 + = 푚푅푎 + → 푅푎 = 푚휔퐿푂퐿푎 푅푎 푅푎 1 푅푎 퐿푎 = (14) √푚 휔퐿푂 푚 푅 = 푅 ( ) (15) 푎 1_퐿푂 푚+1 2 푅푎 퐿1 = 퐿푎 + 2 (16) 휔0퐿푎 1 퐿푆 = 1 1 1 (17) − − 퐿푁 퐿1 퐿퐶 L1 will vary some with frequency but it is normally such a large value that its effect on Ls is minimal. To get realizable values the value of Ra will usually be increased. This can be matched to 50 ohm using a transformer. Normally one would design the filter to meet attenuation and passband requirements at the two tuning extremes, then using a spreadsheet loaded with appropriate equations, adjust the nodal inductance and perhaps filter Q to get an Ra that can be matched by a convenient transformer ratio and realizable component values. The design method shown for non-dissipative components can be extended to dissipative components and higher numbers of sections by using the corresponding q and k values associated with those components and sections. Example Design: Tunable Preselector Frequency tuning range: 118 – 152 MHz When tuned to 137 MHz we want the attenuation at 157 MHz to be at least 35 dB. According to Zverev, a N=3 Butterworth has a 3 dB to 35 dB bandwidth ratio of about 3.8. 1372 When tuned to 137 MHz, the bandwidth at the 35 dB point is 157 − ≅ 37.5 푀퐻푧. The bandwidth at the 3 157 37.5 137 dB point is = 8.9 푀퐻푧. The filter Q is then = 13.8. 3.8 9.9 The filter: La Lc Lc La Ra Lsa Csa Lsb Csb Lsa Csa Ra ≡ @휔0 Lc Lc R1 L1 Lsa Csa Lsb Csb Lsa Csa L1 R1 Flo = 118 MHz Fhi = 152 MHz Filter, n=3 Butterworth q1 = qN = 1.0 k12 = k34 = 0.7071 LN and Q were manipulated to give realizable values and to give an RA value that was a transformable value from 50 ohms. Q = 16.0 LN = 68 nH Lsa = Lsb = 75 nH La = 540 nH Ra = 450 ohm (50 ohm with 9:1 transformer) LC = 1539 nH (filter topology will reduce this to LC/4) Csa = Csb Csa_118MHz = 27 pF Csa_137MHz = 20 pF Csa_152MHz = 16 pF This final filter and its response is shown below: Appendix - Narrow-Band Approximations 2 푅푎 퐿1 = 퐿푎 + 2 (A-1) 휔0퐿푎 2 2 휔0퐿푎 푅1 = 푅푎 + (A-2) 푅푎 L a L 1 R a 2 퐿1푅1 퐿 = (A-3) 푎 푅2+휔2퐿2 @0 1 0 1 R 1 2 2 휔0퐿1푅1 푅푎 = 2 2 2 (A-4) 푅1+휔0퐿1 푅푎(푅1−푅푎) 퐿푎 = √ 2 (A-5) 휔0 퐶푏 퐶2 = 2 2 2 (A-6) 1+휔0퐶푏 푅푏 1 C 2 푅 = 푅 + (A-7) 2 푏 휔2퐶2푅 C b 0 푏 푏 R b 1 퐶푏 = 퐶2 + 2 2 (A-8) 휔 푅 퐶2 @0 0 2 R 2 푅2 푅푏 = 2 2 2 (A-9) 1+휔0퐶2 푅2 2 퐶푏 = √휔0 푅푏(푅2 − 푅푏) (A-10) B. Baker .
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