Biomedical Engineering Reference
In-Depth Information
Inset:
Snapshot of the blue sample path from 323 to 326 ms shows the 0
.
2mV
discontinuities in
V
due to the synaptic inputs.
(B)
Current in top panel averaged
over 1 ms time bins. Each point represents the average current into the neuron in the
previous millisecond. For clarity of presentation, consecutive points are joined by
lines.
Right
. Histogram of currents from left panel. The smooth blue line represents
the distribution of
(
t
)
)
t
+
D
t
I
dt
,where
I
t
)
(
1
/
D
t
(
(
t
)
is the current into the cell in the
t
diffusion approximation, Equation (15.10), and D
t
1ms.
(C)
Same as
A
, but with
inputs now described by the diffusion approximation. The macroscopic structure of
the sample paths is very similar. The differences between the Poisson input and the
diffusion approximation can only be appreciated by looking at the inset.
=
of probability represents an extra probability current
S
reset
(
V
,
t
)
, that adds to the cur-
(
,
|
,
)
(
)
rent
S
V
t
V
0
t
0
associated to the sub-threshold dynamics of
V
t
. Taking this into
account, one can rewrite the Fokker-Planck equation like
∂
∂
t
∂
∂
V
[
S
reset
r
(
V
,
t
|
V
0
,
t
0
)=
−
S
(
V
,
t
|
V
0
,
t
0
)+
(
V
,
t
)]
.
(15.21)
Since this injection only results in a change of probability mass in
V
=
V
reset
, the new
current is given by
S
reset
(
V
,
t
)=
n
(
t
−
t
re f
)
Q
(
V
−
V
reset
)
.
(15.22)
To find the solution for the steady-state distribution
, we insert expression
(15.22) into the Fokker-Planck equation (15.21), and look for time independent so-
lutions by setting the left hand side of this equation to zero,
r
ss
(
V
)
s
V
2t
m
∂
2
∂
V
2
∂
V
[
(
∂
V
−
V
ss
)
t
m
r
ss
(
V
)] +
r
ss
(
V
)=
−
nd
(
V
−
V
reset
)
.
(15.23)
Solving this equation with the boundary conditions (15.19-15.20), one obtains the
following expression for the steady-state distribution [24]
exp
V
ss
s
V
V
−
V
ss
s
V
V
th
−
2
2nt
m
s
V
−
(
V
−
V
ss
)
V
r
−
V
ss
s
V
e
x
2
dx
r
ss
(
V
)=
Q
(
x
−
)
.
(15.24)
s
V
The function r
ss
(
V
)
gives the fraction of cells in a non-refractory state with depo-
larizations in
in the steady state. Taking into account also the fraction
nt
re f
of neurons in a refractory state, the steady state firing rate n can be found by
the normalization condition
(
V
,
V
+
dV
)
V
th
(
)
+
nt
re f
=
.
r
ss
V
dV
1
(15.25)
−
•
Plugging expression (15.24) into this equation and solving for n one gets
t
m
√
p
V
th
−
V
ss
s
V
Vr
−
Vss
s
V
n
e
x
2
=
t
re f
+
(
1
+
erf
(
x
))
dx
,
(15.26)
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