Matakuliah
Tahun
Versi
: H0434/Jaringan Syaraf Tiruan
: 2005
:1
Pertemuan 14
APLIKASI BACK PROPAGATION
1
Learning Outcomes
Pada akhir pertemuan ini, diharapkan mahasiswa
akan mampu :
• Mendemonstrasikan Back Propagation
pada pengenalan pola.
2
Outline Materi
• Aplikasi Back Propagation
3
Example: Function
Approximation

g  p  = 1 + sin --- p 
4
t
-
p
e
+
1-2-1
Network
a
4
Network
p
1-2-1
Network
a
5
Initial Conditions
1
W  0  = – 0.27
– 0.41
1
b  0  = – 0.48
– 0.13
2
W  0  = 0.09 –0.17
2
b 0  = 0.48
3
Network Response
Sine Wave
2
1
0
-1
-2
6
-1
0
1
2
Forward Propagation
0
a = p = 1
 –0.27

 – 0.75 
–
0.48
a = f W a + b  = l ogsig 
 = logsi g

1 +
 –0.41
– 0.13 
 – 0.54 
1
1
1 0
1
a1
1
-------------------0.75
1
+
e
=
= 0.321
1
0.368
-------------0.54
------1+ e
2
2
2 1
2
a = f W a + b  = purelin ( 0.09 – 0.17 0.321 + 0.48 ) = 0.446
0.368


 
2
 
e = t – a =  1 + sin  --- p   – a =  1 + sin --- 1   – 0.446 = 1.261
4 
4 


7
Transfer Function Derivatives
–n
d 1 
e
1
1
1
1
fÝn  =
----------------- = ------------------------ =  1 – -----------------  ----------------- = 1 – a  a 

– n 
–n 
d n 1 + e –n 
–n 2
1 +e
1 +e
1 + e 
1
2
d
fÝn  =
n  = 1
dn
8
Backpropagation
2 2
2
2
s = – 2 FÝ(n ) t – a  = – 2 fÝ n2   1.261 = – 2 1  1.261 = –2.522
1
2 T 2
1
1
s = FÝ(n ) W  s =
1
0
1
s = 1 – 0.321 0.321
0
1
s = 0.218
0
1
 1 – a 1  a 1 
0
0.09
– 2.522
1
1
 1 – a 2  a 2  – 0.17
0
0.09
– 2.522
 1 – 0.368  0.368 – 0.17
0
– 0.227 = – 0.0495
0.233 0.429
0.0997
9
Weight Update
 = 0.1
T
W 2 1  = W 2 0  – s2  a1  = 0.09 –0.17 – 0.1 – 2.522 0.321 0.368
2
W  1  = 0.171 – 0.0772
2
2
2
b  1 = b 0  – s = 0.48 – 0.1 – 2.522 = 0.732
1
1
1 0 T
W  1  = W  0  – s a  = – 0.27 – 0.1 – 0.0495 1 = – 0.265
– 0.41
0.0997
– 0.420
1
1
1
b  1  = b 0  –  s = – 0.48 – 0.1 – 0.0495 = – 0.475
– 0.13
0.0997
– 0.140
10
Choice of Architecture
i
g  p  = 1 + sin ----- p 
4 
1-3-1 Network
3
2
3
i=1
2
1
1
0
0
-1
-2
-1
0
1
2
3
2
-1
-2
-1
0
1
2
0
1
2
3
i=4
2
1
1
0
0
-1
-2
i=2
-1
0
1
2
-1
-2
i=8
-1
11
Choice of Network
Architecture
6
g  p  = 1 + sin ------ p 
4 
3
2
3
1-2-1
2
1
1
0
0
-1
-2
-1
0
1
2
3
2
-1
-2
-1
0
1
2
1
2
3
1-4-1
2
1
1
0
0
-1
-2
1-3-1
-1
0
1
2
-1
-2
1-5-1
-1
0
12
Convergence
g p = 1 + sinp
3
3
5
2
2
1
3
1
5
3
4
2
0
1
4
2
0
0
0
1
-1
-2
-1
0
1
2
-1
-2
-1
0
1
2
13
Generalization
{p1, t 1}  { p2, t2 }    {pQ, tQ}

g  p  = 1 + sin --- p 
4
p = –2 –1.6 –1.2   1.6 2
3
3
1-2-1
1-9-1
2
2
1
1
0
0
-1
-2
-1
0
1
2
-1
-2
-1
0
1
2
14