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HOMEWORK #3 prob 4 solution 2240 F 13 EX: L C + + vi 8! speaker – vo – The above circuit is part of a simple crossover network for driving a midrange speaker having an impedance of 8ΩΩ. The circuit is described at the following web site: http://www.termpro.com/articles/xover2.html. A more in-depth discussion of crossover networks may be found at http://sound.westhost.com/lr-passive.htm. The web site describing the above bandpass filter suggests using cutoff frequencies of ƒC1 = 130 Hz and ƒC2 = 4 kHz. This results in the following values of L and C. L = 330 µH C = 150 µF Plot |Vo/Vi| versus ω. SOL'N: Though not required in this problem, we first consider how to find the values of L and C. This is a standard band-pass filter. The cutoff frequencies for this filter are as follows: ⎛ R ⎞2 1 R ωC1,2 = ± + ⎜ ⎟ + ⎝ 2L ⎠ 2L LC where R = 8 ΩΩ. The following observations simplify our calculations: ⎛ ⎞ ⎛ ⎞ ⎛ R ⎞2 1 ⎟ ⎜ R ⎛ R ⎞2 1 ⎟ R ⎜ ωC 2 − ωC1 = + ⎜ ⎟ + − − + ⎜ ⎟ + ⎜ 2L ⎝ 2L ⎠ ⎝ 2L ⎠ LC ⎟ ⎜ 2L LC ⎟ ⎝ ⎠ ⎝ ⎠ or ωC 2 − ωC1 = R L or L= R ωC 2 − ωC1 Also, ⎛ ⎞⎛ ⎞ ⎛ R ⎞ 2 1 ⎟⎜ R ⎛ R ⎞2 1 ⎟ R ⎜ ωC1ωC 2 = − + ⎜ ⎟ + + ⎜ ⎟ + ⎜ 2L ⎝ 2L ⎠ ⎝ 2L ⎠ LC ⎟⎜ 2L LC ⎟ ⎝ ⎠⎝ ⎠ or ⎛ ⎞2 2 ⎛R⎞ ⎛R⎞ 1 ⎟ 1 ωC1ωC 2 = −⎜ ⎟ + ⎜ ⎜ ⎟ + = = ω 2o ⎝ 2L ⎠ ⎜ ⎝ 2L ⎠ LC ⎟ LC ⎝ ⎠ 2 or C= 1 ωC1ωC 2 L Now we compute the cutoff frequencies in r/s: ωC1 = 2πf C1 = 2π(130) r/s = 817 r/s ωC 2 = 2πf C 2 = 2π(4 k) r/s = 25.1 kr/s Using our formulas from above yields the following: L= R 8 = ≈ 330 µH ωC 2 − ωC1 25.1k − 0.817k C= 1 1 = F ≈ 150 µF ωC1ωC 2 L 0.817k ⋅ 25.1k ⋅ 330µ and Now for the solution of the problem. The circuit is a voltage divider: V R 1 H( jω) = o = = Vi R + jωL + 1/ jωC 1+ j 1 (ωL −1/ωC) R We use the following Matlab code to plot the frequency response: % ECE2260F07_HW3p3Matlab.m % % Plot of filter's frequency response curve figure(1) omega = 1:30e1:30e3; s = j * omega; FilterResp = 1./(1 + j * (1/8)*(omega*330e-6 - 1 ./ (omega*150e-6))); plot(omega,abs(FilterResp)) axis([0, max(omega), 0, 1]) xlabel('omega') ylabel('|H|')