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|')