Medical Imaging - Introduction to Biomedical Engineering - page 1049

The impedance, Z 2 , can be found from

v 2 ¼ ð

1 þ RF Þ Z

Z 2 ¼ p

2 1

ð 16 : 14 Þ

1 RF

Finally, the right-hand side of Eq. (16.14) can be solved to obtain

RF ¼ Z

Z

1
Z 2 þ Z 1

2

ð 16 : 15a Þ

A transmission factor, TF , can be determined from

TF ¼ 1 þ RF ð 16 : 15b Þ

or

2 Z 2

TF ¼ ð 16 : 15c Þ

Z þ Z

1 2

EXAMPLE PROBLEM 16.2
Find the reflection and transmission factors for the case of a free, a perfectly matched, and a
rigid boundary. What is the acoustic pressure transferred under these conditions? Assume water
as medium 1.

Solution
For a free or air-type boundary or open-circuit condition,

Z 2 ¼ 0, so from Eq. (16.15a), there will

be a 180 inversion of the incident wave, or RF ¼ -1. Here, the reflected wave cancels the incident,

so TF ¼ 0. For a matched condition, Z 2 ¼ Z 1 , RF ¼ 0 or no reflection, and TF ¼ 1 for perfect ampli-

tude transfer. For a rigid boundary, , corresponding to a short-circuit condition or a stress-

free boundary, and the incident wave will be reflected back, or

Z 2 ¼1

RF ¼þ 1, without phase inversion.

In this case, TF ¼ 2.

Oblique Waves at a Liquid-Liquid Boundary
What happens when the incident wave is no longer normal to the boundary? This situa-
tion is depicted in Figure 16.7, in which a single-frequency longitudinal wave traveling in a
liquid medium 1 is incident at an angle to a boundary with a different liquid medium 2 in
the plane . At the boundary, pressure and particle velocity are continuous. The tangen-

tial components of wavenumbers also must match, so along the boundary,

x-z

k

x ¼ k

1 sin

y I ¼ k

2 sin

y T ¼ k

1 sin

y R ð 16 : 16a Þ

1

where k 1 and k 2 are the wavenumbers for mediums 1 and 2, respectively. The reflected

angle, y R , is equal to the incident angle, y I , and an acoustic Snell's law results,

sin

y T ¼ c

y I

1
c 2

ð 16 : 16b Þ

sin