Menentukan Nilai SOP dan POS dengan tabel kebenaran serta K-Map :
Berikut adalah cara yang digunakan untuk menyelesaikan persamaan logika menggunakan metode SOP dan POS.
Fungsi : R(a,g,n)=∏(2,6,7)
1. POS
r(a,g,n) = ∏ (2,6,7)
r(a,g,n) = M2.M6.M7
r(a,g,n) = (a+g’+n)(a’+g’+n)(a’+g’+n’)
SOP
r’(a,g,n) = ∏(0,1,3,4,5)
r’(a,g,n) = (M0.M1.M3.M4.M5)’
(M0’+M1’+M3’+M4’+M5’)
(a+g+n)’+(a+g+n’)’+(a+g’+n’)’+(a’+g+n)’+(a’+g+n’)’
(a’g’n’)+(a’g’n)+(a’gn)+(ag’n’)+(ag’n)
(m0+m1+m3+m4+m5)
∑(0,1,3,4,5)
2.
Tabel Kebenaran POS
a
|
g
|
n
|
a’
|
g’
|
n’
|
(a+g’+n)
|
(a’+g’+n)
|
(a’+g’+n’)
|
(a+g’+n)(a’+g’+n)(a’+g’+n’)
|
0
|
0
|
0
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
1
|
1
|
1
|
0
|
1
|
1
|
1
|
1
|
0
|
1
|
0
|
1
|
0
|
1
|
0
|
1
|
1
|
0
|
0
|
1
|
1
|
1
|
0
|
0
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
0
|
1
|
0
|
1
|
0
|
1
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
1
|
1
|
0
|
1
|
0
|
1
|
1
|
1
|
0
|
0
|
0
|
1
|
1
|
0
|
0
|
Tabel Kebenaran SOP
a
|
g
|
n
|
a’
|
g’
|
n’
|
(a’g’n’)
|
(a’g’n)
|
(a’gn)
|
(ag’n’)
|
(ag’n)
|
(a’g’n’)+(a’g’n)+(a’gn)+(ag’n’)+(ag’n)
|
0
|
0
|
0
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
0
|
1
|
0
|
0
|
1
|
1
|
1
|
0
|
0
|
1
|
0
|
0
|
0
|
1
|
0
|
1
|
0
|
1
|
0
|
1
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
0
|
0
|
0
|
0
|
1
|
0
|
0
|
1
|
1
|
0
|
0
|
0
|
1
|
1
|
0
|
0
|
0
|
1
|
0
|
1
|
1
|
0
|
1
|
0
|
1
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
1
|
0
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
3.
K-MAP POS
r(a,g,n)
= (a+g’+n)(a’+g’+n)(a’+g’+n’)
gn
a
|
00
|
01
|
11
|
10
|
0
|
1
|
1
|
1
|
0
|
1
|
1
|
1
|
0
|
0
|
Penyederhanaan : (g’+n)(g’+a’) = (g’+na’)
K-MAP SOP
r(a,g,n) = (a’g’n’)+(a’g’n)+(a’gn)+(ag’n’)+(ag’n)
gn
a
|
00
|
01
|
11
|
10
|
0
|
1
|
1
|
1
|
0
|
1
|
1
|
1
|
0
|
0
|
Penyederhanaan : g’+na’
Rangkaian Logika
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