A difficulty with mounting a woofer and tweeter side by side or one above the other is that the path that the sound has to travel from each of the loudspeakers to a listener will be different in different parts of the listening room. Hence, in the vicinity of the crossover frequencies, cancellation of the sound will result at some parts of the room, and addition will occur at others.
To avoid this effect, the loudspeakers are sometimes mounted
concentrically, i.e., the tweeter is placed behind and on the axis of the woofer (see
Fig. 7.52). In this arrangement, the diaphragm of the woofer acts as a horn and the tweeter usually has a phase plug in front of it. Horn loudspeakers will be discussed in greater detail in
Chapter 9.
Calculate the box volume and port dimensions needed to give a Chebyshev frequency response with 0.01 dB ripple and a maximum sound pressure of
SPL
max = 100
dB SPL at 1 m. Show that to make the port volume ten times the volume displacement needed to produce 100 dB SPL, the optimum bore diameter is around 9 cm. Also, calculate the cut-off frequency.
Hint: Obtain
QTS
from Eq. (6.10). Obtain
VAB
and
fB
from
Table 7.4. Calculate the maximum peak pressure and volume displacement required to produce that pressure using formulas from “Summary of bass-reflex design” on p. 334. Let the port volume be
VP
= 10
V
max and use
Eqs. (7.97) and
(7.96) to calculate the approximate length
t and cross-sectional area
SP
respectively so that the diameter is
dP=2SP/π−−−−−√
. Using a diameter of 9 cm, then calculate the exact length
t from
Eq. (7.98). Obtain the cut-off frequency
f3dB
from
Table 7.4.