Biomedical Engineering Reference
In-Depth Information
The fixed point of this system is
m
ext
V
C Js
ss
,
s
ss
=
t
syn
f
(
+
s
V
)
.
(15.81)
To check the stability of the fixed points of this network, the standard procedure is to
consider a small perturbation of a steady state
s
=
s
ss
+
d
s
exp
(
l
t
)
,
(15.82)
where d
s
is a small perturbation that grows at a rate l. Stability of the steady state
implies
Re
0 for all possible perturbations. Inserting Equation (15.82) in Equa-
tion (15.80), we get
(
l
)
<
s
=
s
ss
.
t
syn
+
d
f
(
m
V
(
s
)
,
s
V
)
l
=
−
(15.83)
ds
Thus, the stability condition l
<
0is
s
=
s
ss
<
d
f
(
m
V
(
s
)
,
s
V
)
t
syn
.
(15.84)
ds
Equation (15.84) is a condition on the slope of the input-output function f at the
value of the input current given by
s
ss
. Since it is more intuitive to work with firing
rates, we can express it as a condition on the slope of f as a funcion of n if we note
that we only need the value of this slope at the steady state. In general
d
f
ds
ds
d
n
d
f
d
n
=
.
(15.85)
Since the output rate is an instantaneous function of
s
, we can calculate the first term
on the right hand side for all
s
(
t
)
. The second term we do not know in principle, but
on the steady state
s
ss
=
t
syn
n
ss
. Thus
s
=
s
ss
=
d
n
n
d
f
ds
t
syn
d
f
n
ss
,
(15.86)
=
so that the stability condition becomes
n
d
f
(
m
V
(
n
)
,
s
V
)
n
ss
<
1
.
(15.87)
d
n
=
Thus, for a fixed point to be stable, the slope of the output rate as a function of the
input rate, should be less than one at the fixed point. This is shown graphically in
Figure 15.4,
where f
(
m
V
(
n
)
,
s
V
)
is plotted as a function of n, both for an excitatory
network (
J
5 mV) and an inhibitory network (
J
5 mV). The external inputs
are adjusted so that there is an intersection with the diagonal at 1 Hz in both cases.
When the network is excitatory (Figure 15.4A; in this case we use m
ext
V
=
0
.
=
−
0
.
=
0mV),the
function f
(
m
V
(
n
)
,
s
V
)
raises very fast from zero rates to saturation. Thus, the slope
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