SME
Bioinstrumentation · Problem type
Ultrasound echoes: depth and reflection
Use this for “An echo returns after 130 μs. How deep is the boundary, and what share does a fat–muscle boundary reflect?”
Diagram
d
=
ct
2
=
1,540
×
130 ×
10
-6
2
=
0.1
m
=
10
cm:
the
pulse
goes
down
and
back,
so
the
boundary
is
half
the
round
trip
deep.
R
=
(
Z₂ − Z₁
Z₂ + Z₁
)²
=
(
1.7 − 1.38
1.7 + 1.38
)²
=
1.08%
of
the
intensity
comes
back;
the
rest
goes
on
(dashed).
This is an example. Type your own values to start a new problem.
d (cm)
=
1540
m/s
×
t (μs)
2
R (%)
=
Z₂ (MRayl)
−
Z₁ (MRayl)
Z₂ (MRayl)
+
Z₁ (MRayl)
2
Formulas
d
=
1,540
×
t
20,000
10.01
=
1,540
×
130
20,000
R
=
100
×
(
Z₂
−
Z₁
Z₂
+
Z₁
)²
1.0794
=
100
×
(
1.7 − 1.38
1.7 + 1.38
)²
Assumptions
•
Sound travels at c = 1540 m/s in soft tissue, the speed scanners assume.
•
The pulse goes straight down to the boundary and straight back, so the depth is half of ct.
•
The boundary is flat and meets the beam square on; each tissue has one impedance Z = ρc.
Step-by-step
WE KNOW
t = 130 μs, Z₁ = 1.38 MRayl, Z₂ = 1.7 MRayl
FIND
depth of the boundary (d), share reflected (R)
Step 1 · Find depth of the boundary (d)
Use d = 1,540 × t ÷ 20,000
The
pulse
goes
down
and
back,
so
the
depth
is
half
of
c
×
t
;
the
20,000
halves
it
and
turns
m/s
times
μs
into
cm.
d
=
1,540
×
130
20,000
d
=
200,200
20,000
d
=
10.01
cm
Step 2 · Find share reflected (R)
Use R = 100 × ((Z₂ − Z₁) ÷ (Z₂ + Z₁))²
The mismatch of the impedances over their sum, squared, is the share of the intensity sent back.
R
=
100
×
(
1.7 − 1.38
1.7 + 1.38
)²
R
=
100
×
(
0.32
3.08
)
2
R
=
100
×
0.1039
2
R
=
100
×
0.010795
R
=
1.0794%
Check
Put the numbers back in. Both sides should match.
10.01
=
1,540
×
130
20,000
✓
1.0794
=
100
×
(
1.7 − 1.38
1.7 + 1.38
)²
✓
Topic check
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