Biomedical Engineering Reference
In-Depth Information
Ta b l e 5 . 2
Ratios of the
mechanical parameters
between consecutive levels.
Values are presented as mean
±
standard deviation values
for the 1-15 levels,
respectively, for the 16-24
levels
Nominal
Level 1-15
Level 16-24
0
.
81
±
0
.
32
0
.
68
±
0
.
16
λ
±
±
1
/α
0
.
56
0
.
17
0
.
56
0
.
08
χ
1
.
71
±
0
.
77
1
.
55
±
0
.
29
o
1
.
71
±
0
.
77
1
.
55
±
0
.
29
It is also worth noticing that the extra parameter
L
m
has no effect in determining
the value of the fractional order, which is then similar to what we expected from the-
oretical analysis, i.e. relation (
5.20
)[
70
]. On the other hand, the effect of this extra
term becomes significant with increasing frequencies, namely after the frequency
interval where the phase variations are observed (
ω>
100 rad/s).
5.1.2 A Viscoelastic Airway Wall
In this section, we shall use the formulas derived in Chap.
4
, namely (
4.71
), (
4.72
),
(
4.74
), and (
4.73
), with values from Table
2.1
. With these values at hand, one is able
to build an electrical network as described in the previous section. The difference
from the previous case (elastic) is that now the airway tube is modeled by a
R
-
L
-
C
-
G
element, as described in Chap.
4
. This representation allows us to consider the
viscoelastic wall properties, through the elements
C
-
G
.
For the special case of the ladder network in which
Zl
m
(s)
=
R
em
+
L
em
s
and
1
/Zt
m
(s)
1
/C
em
s
, with
m
denoting a level in the respiratory tree, one can
analyze the properties of such a network. Next to the ratios defined in (
5.4
), we add
the ratio for the conductance:
=
G
em
+
G
em
+
1
G
em
1
o
=
(5.28)
where the ratio is determined similarly as for the other parameters. The nominal
ratios are given in Table
5.2
. Notice that the ratios for
C
e
and
G
e
are similar due to
the fact that the forms in (
4.74
) and (
4.73
) are the same, except for the sin and cos
terms, whose effect are thus very small.
In a similar manner as for the elastic tube case, we can write that the total admit-
tance is given by
1
/(R
e
1
+
L
e
1
s)
Y
N
(s)
=
(5.29)
G
e
1
/
[
(G
e
1
C
e
1
s
+
1
)(R
e
1
+
L
e
1
s)
]
1
+
G
e
1
/
[
(G
e
1
C
e
1
s
+
1
)(R
e
2
+
L
e
2
s)
]
1
+
G
e
2
/
[
(G
e
2
C
e
2
s
+
1
)(R
e
2
+
L
e
2
s)
]
1
+
G
e
2
/
[
(G
e
2
C
e
2
s
+
1
)(R
e
3
+
L
e
3
s)
]
+
1
...
...
G
e(N
−
1
)
/
[
(G
e(N
−
1
)
C
e(N
−
1
)
s
+
1
)(R
eN
+
L
eN
s)
]
1
+
1
+
G
eN
/
[
(G
eN
C
eN
s
+
1
)(R
eN
+
L
eN
s)
]
