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2013
 

 

“MINTERM” and 
“MAXTERM” 

Tugas Sistem Digital 
      

ANDREAS | 2117200710  

SISTEM KOMPUTER 
 

Tugas 1 
“Minterm” 
F(ABC)=∑(0,2,4,5,6) 

 
TABEL EKSPRESI SOP

 
 
 
 
 
 
 

A 
0 
0 
0 
0 
1 
1 
1 
1 

Input 
B 
0 
0 
1 
1 
0 
0 
1 
1 

C 
0 
1 
0 
1 
0 
1 
0 
1 

output 
KONDISI FUNGSI AND 
1 
A'B'C' 
0 
‐ 
1 
A'BC' 
0 
‐ 
1 
AB'C' 
1 
AB'C 
1 
ABC' 
0 
‐ 

Persamaan Boolean : 
F = ∑ Fi 
   =F0 +F2 +F4 + F5 +F6 
   = A’B'C' + A'BC' + AB'C' + AB'C + ABC' 
Bentuk Sederhana : 
= A’B'C' +ABC+ A'BC'+ABC + AB'C' +ABC+ AB'C +ABC+ ABC'+ABC 
=A’(B⊚C)+A(B’C’+BC)+AC(B’B)+AB(C’+C) 
= A’(B⊚C)+A+AC+AB 
 
Rangkaian Logika : 
 
 
 
 
 

Tugas 2 
“Minterm” 
F(ABCD)=∑(0,1,2,4,5,6,9,12,13,14) 
 
 
 

TABEL EKSPRESI SOP 
A 

B 
0 
0 
0 
0 
0 
0 
0 
0 
1 
1 
1 
1 
1 
1 
1 

Input 
C 
0 
0 
0 
0 
1 
1 
1 
1 
0 
0 
0 
0 
1 
1 
1 

D
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1

Output 
FUNGSI AND 

KONDISI
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0

1
1
1
0
1
1
1
0
0
1
0
0
1
1
1

 A’B’C’D’ 
A'B'C'D 
A'B'CD' 
 ‐‐ 
A'BC'D’ 
A'BC’D 
A'BCD’ 
 ‐ 
 ‐ 
AB'C’D 
 ‐ 
‐ 
ABC'D' 
ABC’D 
ABCD’ 
1 

1 

1

1

0 ‐

 

Persamaan Boolean : 
F = ∑ Fi 
=F0 +F1 +F2 + F4 +F5+ F6 +F9 +F12 + F13 +F14 
= A’B’C’D’ + A'B'C'D + A'B'CD' + A'BC'D’ + A'BC’D + ABCD’ + AB'C’D + ABC'D' + ABC'D+ ABCD’ 

Bentuk Sederhana : 
= A’B’C’D’ +ABCD+ A'B'C'D + ABCD+  A'B'CD' + ABCD+ A'BC'D’ + ABCD+  A'BC’D + ABCD+  ABCD’ + 
ABCD+ AB'C’D + ABCD+  ABC'D' + ABCD+  ABC'D+ ABCD+  ABCD’ +ABCD 
= 

A’(B⊚C⊚D)+C(A’B’D’+ABD)+B(A’C’D’+ACD)+BC(A’C’+AC)+BC{A’D’+AD)+AD(B’C’+BC)+AB(C’D
’+CD)+ABD(C’+C)+ABC(D’+D) 
=A’(B’C’+BC)+A’+A(D’+B) 
= A’(B⊚C⊚D)+C+B+BC+BC+AD+AB+ABD+ABD 

Rangkaian Logika : 
 

 
 
 
 
 
 
 
 

Tugas 3 
??? 
 
 
 
 
 
 

A 
0 
0 
0 
0 
1 
1 
1 
1 
 

Input 
B 
C 
0
0
1
1
0
0
1
1

0
1
0
1
0
1
0
1

KONDISI
1
0
1
0
1
1
1
0

output 
FUNGSI OR 
A'+B'+C' 
  
A'+B+C' 
  
A+B'+C' 
A+B'+C 
A+B+C' 
  

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cara menghitung Minterm dan maxterm aljabar boolean