Biomedical Engineering Reference
In-Depth Information
. Substituting
I
m
ðÞ¼
K
s
e
st
0
The Laplace transform of the current pulse is
1
I
m
(
s
Þ
into the
node equation and rearranging terms yields
V
m
0ðÞ
e
st
0
K
ð
1
Þ
V
m
þ
þ
V
m
(
s
Þ¼
1
C
m
R
TH
1
C
m
R
TH
1
C
m
R
TH
s
þ
sC
m
s
þ
sC
m
R
TH
s
þ
V
m
0ðÞ¼
V
TH
, gives
Performing a partial fraction expansion, and noting that
0
@
1
A
þ
V
TH
s
1
s
1
e
st
0
V
m
(
s
Þ¼
KR
TH
1
1
C
m
R
TH
s
þ
Transforming back into the time domain yields the solution
u
u
e
t
t
o
t
R
TH
C
m
e
V
m
(
t
Þ¼
V
TH
þ
R
TH
K
1
1
t
t
0
Þ
(
t
Þ
R
TH
K
R
TH
C
m
(
The ionic current (
I
i
) and capacitive current (
I
c
) are shown in the following figure, where
I
i
¼
V
m
V
TH
R
Th
I
c
¼
C
m
dV
m
dt
20
I
i
I
c
15
10
5
0
0
0.003
0.006
0.009
0.012
0.015
−
5
−
10
−
15
−
20
Time (s)
A current pulse of 6 ms
(top) and 2 ms (bottom) using parameters for the frog skeletal muscle. The time constant is
Continued
Shown in the following figures are graphs of
V
m
in response to a 15
m
