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In-Depth Information
the longitudinal resistance is r
L
. By ignoring the final points, the voltage varies
according to the following relation [
16
,
65
]
c
M
A
j
dV
j
dt
V
j
r
M
=A
j
V
j
C
1
V
j
4r
L
h=.d
j
C
1
/
C
V
j
1
V
j
4r
L
h=.d
j
C
1
/
D
C
(1.71)
Dividing by h the two last terms (coupling terms) one obtains
h
d
jC1
V
j
C
1
V
j
d
j
V
j
V
j
1
(1.72)
4r
L
h
4r
L
h
As h!0 one obtains the diffusion operator
4r
L
@x
.d
2
.x/
@
@x
/
@
(1.73)
Moreover, by dividing both members of Eq. (
1.71
) with
dx
, cable's equation comes
to the form
c
M
@
@t
D
V
r
M
1
4r
L
d.x/
@x
.d
2
.x/
@
@x
/
@
C
(1.74)
It is noted that the term
d
j
.V
j
1
V
j
/
4r
L
h
has current dimensions and at the limit h!0 it
corresponds to longitudinal current I
L
D
d
2
.x/
4r
L
@V
@x
.
If d.x/ D d is the constant diameter of the segment, then it is possible to multiply
the two segments with r
M
, and this leads to the linear cable equation
@
@t
DV C
2
@
2
V
(1.75)
@x
2
q
d
r
M
4r
L
is the space constant and D r
M
c
M
is the time constant.
Considering that as x!1 the voltage of the membrane approaches zer
o
, then
the steady-state solution of the cable's equation takes the form V.x/D Ae
where D
x
.
1.7.3
The Infinite Length Cable
The relation given in Eq. (
1.75
) is modified in case of an infinite length cable.
Including an excitation current to the previous equation
@
@t
C V.x;t/
2
@
2
V
(1.76)
D r
M
I.x;t/
@x
2
where 1 <x<C1 and V.x;0/ D V
0
.x/. The current I.x;t/ has units
A=cm
2
. For this linear differential equation of the 1st order, the solution is [
16
,
65
]
V
0
.k/ C .
r
M
/
R
0
e
.1C
2
k
2
/
.
t
s/
V.k;t/D e
.1
C
2
k
2
/t
I.;s/
ds
(1.77)
