Biomedical Engineering Reference
In-Depth Information
Before we dive into detailed solutions of those quite complex equations, let us try
to gain some in
tuit
ive understandin
g o
f t
hes
e equations. Let us write
A
(
w
)
as
aw
+
b
,
r
in
A
+
τ
+
+
τ
Hr
in
2
C
ii
−
(
r
in
2
GHr
in
2
where
a
=
G
(
A
−
−
A
+
)
+
m
)
,and
b
=(
i
C
ij
w
j
−
j
=
τ
+
τ
m
r
in
2
G
for
different values of a and b. In this figure,
w
is expressed as a fraction of
g
max
. Under
the conditions considered in this chapter, a is always positive, the fix point where
the line crosses the x axis is thus unstable and the only stable fix points are located
at the boundaries.
(
A
−
−
A
+
)
j
=
i
w
j
+
θ
r
in
(
A
−
−
A
+
)
. In
Figure 11.4A,
we have plotted
A
(
w
)
For
D
=
0
.
05, the corresponding
U
(
w
)
and
P
(
w
)
are given in
Figure 11.4B
and C.
U
shows
corresponding concentration of synaptic weights in the histogram. Under the con-
ditions considered in this chapter, solutions of
P
(
w
)
has local minima at both boundaries and
P
(
w
)
(
w
)
are unsaturated and
P
(
w
)
has
significant weights at both boundaries.
Therefore
U
(
0
)
≈
U
(
g
max
)
, which means
g
max
(
j
w
j
will be adjusted until this condition
is approximately satisfied. A slight deviation of this quantity from 0 will shift the
relative proportion of weights near lower and upper bounds.
a
determines the width
of two lobes of the distribution near lower and upper bound. We plot the expected
synaptic strength versus this quantity in
Figure 11.4D
to demonstrate how changing
this quantity will shift the relative proportion of weights in the two lobes and deter-
mine the mean synaptic strength. The curve is fairly steep, therefore a slight change
in the term involving
C
ij
will have big effects on the mean synaptic strength. Notice
that the central portion of the curve is almost linear. Therefore we could write, for
2
b
2
b
+
ag
max
)
/
D
≈
0. The total weight
/
D
+
ag
max
/
D
≈
0,
E
(
P
(
w
i
)) =
Sg
max
(
2
b
+
ag
max
)
/
D
+
0
.
5
g
max
r
in
A
+
τ
+
+
τ
Hr
in
2
C
ii
−
(
r
in
2
w
i
=
2
Sg
max
(
−
G
(
A
−
−
A
+
)
+
m
)
τ
+
τ
(11.43)
m
GH
r
in
2
r
in
2
−
2
(
j
=
i
C
ij
w
j
−
G
(
A
−
−
A
+
)
j
=
i
w
j
+
θ
r
in
(
A
−
−
A
+
)
w
i
)
Gr
in
2
A
2
A
2
/
(
i
w
j
−
θ
r
in
)(
−
/
τ
−
+
+
/
τ
−
)+
0
.
5
g
max
,
j
=
where S is a scaling factor around 0.25.
11.4.5
Three common scenarios and comparison to simulations
To make the mathematical expressions derived in the previous sections more intu-
itively apparent, we will consider three common scenarios and compare the results
derived from the analytical calculations to those from simulations performed with an
integrate and fire neuron receiving 1000 Poisson inputs [79].
11.4.5.1
Constant Poisson inputs
The first situation we will consider is that of Poisson inputs with the same mean rates
and no correlations between them. From Section 11.1,
r
in
2
Q
ij
(
t
)=
+
δ
(
t
)
r
in
δ
,
(11.44)
ij
and
1
e
−
t
/
τ
K
(
t
)=
.
(11.45)
m
m
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