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1
Pertemuan 17HOPFIELD NETWORK
Matakuliah : H0434/Jaringan Syaraf Tiruan
Tahun : 2005
Versi : 1
2
Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• Menjelaskan konsep dari Jaringan Hopfield
3
Outline Materi
• Hopfield Model.
• Lyapnov Function.
4
Hopfield Model
5
Equations of Operation
Cdni t( )
dt------------- T i j aj t( )
j 1=
S
ni t( )
Ri----------– Ii+=
ni - input voltage to the ith amplifierai - output voltage of the ith amplifierC - amplifier input capacitanceIi - fixed input current to the ith amplifier
T i j1Ri j---------= 1
Ri-----
1---
1Ri j---------
j 1=
S
+= ni f1–ai = ai f ni =
6
Network Format
RiCdni t( )
dt------------- RiT i j a j t( )
j 1=
S
ni t( )– RiI i+=
RiC= w i j RiT i j= bi RiI i=
Define:
d ni t( )
dt------------- ni t( )– wi j aj t( )
j 1=
S
bi+ +=
dn t( )dt
------------ n t( )– Wa t( ) b+ +=
a t( ) f n t( ) =
Vector Form:
7
Hopfield Network
8
Lyapunov Function
V a 12---aTWa– f
1–u ud
0
ai
i 1=
S
bTa–+=
9
Individual Derivatives
tdd 1
2---aTWa–
1
2--- aTWa
Tdadt------– Wa Tda
dt------– aTWda
dt------–= = =
tdd
f1–u ud
0
ai
aidd
f1–u ud
0
ai
td
daif
1–ai
td
daini td
dai= = =
ddt----- f
1–u ud
0
ai
i 1=
S
nTdadt------=
tdd bTa– bTa
Tdadt------– bTda
dt------–= =
Third Term:
Second Term:
First Term:
10
Complete Lyapunov Derivative
tddV a –
dn t( )dt
------------Tdadt------ –
td
dni
td
dai
i 1=
S
–td
dni
td
dai
i 1=
S
= = =
–aiddf
1–ai
td
dai
2
i 1=
S
=
tddV a a
TWdadt------– n
Tdadt------ b
Tdadt------–+ a
TW– n
TbT
–+ dadt------= =
aTW– n
TbT
–+ –dn t( )dt
------------T
=
From the system equations we know:
So the derivative can be written:
tddV a 0If thenaid
df
1–ai 0
11
Invariant Sets
Z a : dV a dt 0= a in the closure of G =
tddV a –
aiddf
1–ai
td
dai
2
i 1=
S
=
This will be zero only if the neuron outputs are not changing:dadt------ 0=
Therefore, the system energy is not changing only at the equilibrium points of the circuit. Thus, all points in Z arepotential attractors:
L Z=
12
Example
a f n 2---tan
1– n2
--------- == n
2------ tan
2---a =
R1 2 R2 1 1= =
T1 2 T2 1 1= =W 0 1
1 0=
RiC 1= =
1.4=
I1 I2 0= = b 00
=
13
Example Lyapunov Function
V a 12---aTWa– f
1–u ud
0
ai
i 1=
S
bTa–+=
12---aTWa–
12--- a1 a2
0 11 0
a1
a2
– a1a2–= =
f1–u ud
0
a i
2------
2---u tan ud
0
a i
2------ log
2---u cos
2---–a i
0
4
2--------- log
2---ai cos–= = =
V a a1a2– 4
1.42------------- log
2---a1 cos
log 2---a2 cos
+–=
14
Example Network Equations
dndt------- n– Wf n + n– Wa+= =
dn1 dt a2 n1–=
dn2 dt a1 n2–=
a12---tan
1– 1.42
-----------n1 =
a22---tan
1– 1.42
-----------n2 =
15
Lyapunov Function and Trajectory
-1 -0.5 0 0.5 1-1
-0.5
0
0.5
1
-1
-0.5
0
0.5
1
-1
-0.5
0
0.5
1
0
1
2
a1
a2
a2 a1
V(a)
16
Time Response
0 2 4 6 8 10-1
-0.5
0
0.5
1
0 2 4 6 8 10
0
0.5
1
1.5
2
t t
a1
a2
V(a)
17
Convergence to a Saddle Point
-1 -0.5 0 0.5 1-1
-0.5
0
0.5
1
a1
a2
18
Hopfield Attractors
dadt------ 0=
Va1
Va2
V ...
aSV
T
0= =
The potential attractors of the Hopfield network satisfy:
How are these points related to the minima of V(a)? Theminima must satisfy:
Where the Lyapunov function is given by:
V a 12---aTWa– f
1–u ud
0
ai
i 1=
S
bTa–+=
19
Hopfield Attractors
Using previous results, we can show that:
V a W– a n b–+ –dn t( )dt
------------= =
The ith element of the gradient is therefore:
aiV a –
td
dni –tdd
f1–ai ( ) –
aidd
f1–ai
td
da i= = =
aiddf
1–ai 0
d a t( )dt
------------ 0= V a 0=
Since the transfer function and its inverse are monotonicincreasing:
All points for which will also satisfy
Therefore all attractors will be stationary points of V(a).
20
Effect of Gain
a f n 2--- tan
1– n2
--------- = =
-5 -2.5 0 2.5 5-1
-0.5
0
0.5
1
1.4=
0.14=
14=
n
a
21
Lyapunov Function
V a 12---aTWa– f
1–u ud
0
ai
i 1=
S
bTa–+= f1–u 2
------
u2
------ tan=
-1 -0.5 0 0.5 1
0
0.5
1
1.5
1.4=
0.14=
14=
a
f1–u ud
0
ai
2------
2---
ai2
-------- cos
log4
2---------ai2
-------- coslog–= =
4
2---------a2
-------- coslog–
22
High Gain Lyapunov Function
V a 12---aTWa– b
Ta–
12---aTAa d
Ta c+ += =
V a 2 A W–= = d b–= c 0=
where
V a 12---aTWa– bTa–=
As the Lyapunov function reduces to:
The high gain Lyapunov function is quadratic:
23
Example
V a 2 W– 0 1–1– 0
= = V a 2 I– – 1–1– –
21– 1+ 1– = = =
1 1–= 2 1=z11
1= z2
1
1–=
-1 -0.5 0 0.5 1-1
-0.5
0
0.5
1
-1
-0.5
0
0.5
1
-1
-0.5
0
0.5
1-1
-0.5
0
0.5
1
a1
a2
a2 a1
V(a)