Biomedical Engineering Reference
In-Depth Information
a
2
¼ K
s
X
YF
X
=
S
S
0
ðm
max
Dk
d
Þðm
max
k
d
ÞþK
s
X
YF
X
=
S
K
S
½ðDþ2k
d
Þðm
max
k
d
ÞDk
d
þm
max
½ðm
max
k
d
Þð1þK
s
X
C
s
ÞD
(18.78b)
S
a
1
¼ K
s
X
YF
X
=
S
S
0
K
S
½ðDþ2k
d
Þðm
max
k
d
ÞDk
d
þK
s
X
YF
X
=
S
K
k
d
ðDþ k
d
Þ
þK
S
k
d
m
max
ð1þK
s
X
C
s
ÞþDm
max
K
S
(18.78c)
2
S
a
0
¼ K
s
X
YF
X
=
S
K
k
d
S
0
ðDþk
d
Þ
(18.78d)
Let
S ¼ yþ
a
2
3a
3
(18.79)
Eqn
(18.77)
is reduced to
3
y
þ3a
5
y2a
4
¼ 0
(18.80)
where
2
2
9a
a
5
¼
a
1
3a
3
a
(18.81a)
2
3
3
2
27a
a
4
¼
a
0
2a
3
a
1
a
2
3
þ
a
(18.81b)
3
3
6a
The solution to
Eqn (18.79)
can be obtained by letting
y ¼ xþ
a
x
¼ x
a
5
(18.82)
x
The substitution by
Eqn (18.82)
can reduce
Eqn (18.80)
into a quadratic equation, which
can be solved easily to give two roots
3
x
¼ a
4
a
6
(18.83)
where
q
a
2
3
5
a
6
¼
4
þ a
(18.84)
These two roots give identical values to y in
Eqn (18.82)
. Therefore, one of the roots can be
used. The next step is to take the cubic roots from
Eqn (18.83)
, which yields three roots:
one real and one pair of conjugate complex roots (if a
4
þ
a
6
is real). Substituting back to S,
we obtain the solutions to
Eqn (18.77)
:
3a
3
þ
3
p
S
1
¼ y
1
þ
a
2
3a
3
¼ x
1
a
5
x
1
þ
a
2
3a
3
¼
a
2
þ
p
a
4
þ a
6
a
4
a
6
(18.85a)
3
p
p
3
p
a
4
þ a
6
p
a
4
a
6
S
2
¼ y
2
þ
a
2
3a
3
¼ x
2
a
5
x
2
þ
a
2
3a
3
¼
a
2
3a
3
þ
1
þ
i
þ
1
i
(18.85b)
3
3
2
2
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