Biomedical Engineering Reference
In-Depth Information
Note that
Z L is positive for forward traveling waves and negative for backward traveling
waves. For fresh water at 20 C,
1.48 MegaRayls (10 6 kg/m 2 sec), and
c L ¼
1,481 m/s,
Z L ¼
998 kg/m 3 . The subscript
r ¼
designates longitudinal waves, the type that is most impor-
tant in imaging. In a longitudinal wave, the changes in wave amplitude are aligned along
the direction of propagation with positive parts of sinusoidal waves, called compressional
half cycles, and negative ones, called rarefactional half cycles, as depicted in Figure 16.5b.
The instantaneous intensity is
L
I L ¼ pv * ¼ pp * = Z L ¼ vv * Z L
ð
16
:
10
Þ
To determine the amplitude of a reflected wave, a solution similar to that of Eq. (16.7) can
be applied. Consider the problem of a single frequency acoustic plane wave propagating in
an ideal fluid medium a distance
to a boundary with a different medium, as shown in
Figure 16.6. A plane wave is a sinusoidal-type wave with infinite extent in the lateral direc-
tions (
d
x
and
y
). For the example shown, the propagating medium has a wavenumber
k 1 and
an impedance
k 2 .As
shown in Figure 16.6, the waves to the left of this second medium can be described by a
combination of forward and backward traveling waves
Z 1 , whereas the second medium has impedance
Z 2 and a wavenumber
p ¼ p 0 exp
ð i ðo t k L z ÞÞ þ RFp 0 exp
ð i ðo t þ k L z ÞÞ
ð
16
:
11
Þ
which is a modified version of Eq. (16.7),
k L ¼ k 1 , and RF is a reflection factor for the ampli-
tude of the negative going wave.
To determine the factor
, it is useful to make analogies between acoustic variables and
more well-known electrical parameters: pressure to voltage and particle velocity to electri-
cal current. As shown in Figure 16.6, a source is situated at
RF
z ¼ d
, and the second medium is
represented by a real load of impedance
Z 2 located at
z ¼
0, with a semi-infinite length. By
analogy, the pressure at
z ¼
0 is like a voltage drop across
Z 2 , so from Eq. (16.11), drop out
common exp (
i o t
) terms to obtain
p
¼ p
ð
1
þ
RF
Þ
ð
16
:
12
Þ
2
0
The particle velocity there is like the sum of currents flowing in opposite directions
corresponding to the two wave components. From Eq. (16.9),
v
¼ð
1
RF Þ p
= Z
ð
16
:
13
Þ
2
0
1
P
TF x P
RF x P
Medium 1
k 1 , Z 1 , z = d
Medium 2
k 2 , Z 2 , z = 0
FIGURE 16.6 One-dimensional model of wave propagation at a boundary between media having impedances
of
Z 1 and
Z 2 .
Search WWH ::




Custom Search