Penyerdehanakan gerbang Kombinasi ( combinational Gate )
Sederhanakan Soal Latihan Berikut :
1. Y = A’ C’ + A B C’
2. Y = A’ B’ C D’ + A’ B’C’ D’
3. Y = A’ D + A B D
4. Y = ( A’ + B ) ( A + B )
Jawab :
Y=A' C' + A B C'
Y=(A' A) C' + B C'
Y=A' C' + B C'
TABEL
A
|
B
|
A’
|
C’
|
B C’
|
A’ C’
|
A’ C’ + B C
|
0
|
0
|
1
|
1
|
1
|
0
|
1
|
0
|
0
|
1
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
1
|
1
|
1
|
0
|
1
|
1
|
0
|
0
|
0
|
0
|
1
|
0
|
0
|
1
|
0
|
0
|
0
|
1
|
0
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
0
|
1
|
0
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
0
|
0
|
1. Y = A’ C’ + A B C’
Y = ( A’ A ) C’ + B C’
Y = A’ C’ + B C’
Tabel :
A
|
B
|
A’
|
C’
|
B C’
|
A’ C’
|
A’ C’ + B C
|
0
|
0
|
1
|
1
|
1
|
0
|
1
|
0
|
0
|
1
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
1
|
1
|
1
|
0
|
1
|
1
|
0
|
0
|
0
|
0
|
1
|
0
|
0
|
1
|
0
|
0
|
0
|
1
|
0
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
0
|
1
|
0
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
0
|
0
|
2. Y = A’ B’ C D’ + A’ B’ C’ D’
Y = ( A’ + A ) ( B’ + B’ ) ( C + C’ ) ( D’ + D’ )
Y = A’ B’ 0 D’
Y = A’ B’ D’
Tabel :
C
|
A’
|
B’
|
C’
|
D’
|
A’ B’ D’
|
0
|
1
|
1
|
1
|
1
|
1
|
0
|
1
|
1
|
1
|
0
|
0
|
1
|
1
|
1
|
0
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
0
|
1
|
0
|
1
|
1
|
0
|
0
|
1
|
0
|
1
|
0
|
0
|
1
|
1
|
0
|
0
|
1
|
0
|
1
|
1
|
0
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
0
|
0
|
0
|
1
|
1
|
0
|
0
|
1
|
0
|
1
|
0
|
1
|
0
|
1
|
0
|
1
|
0
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
0
|
0
|
0
|
0
|
1
|
0
|
0
|
1
|
0
|
0
|
0
|
1
|
0
|
1
|
0
|
0
|
0
|
0
|
0
|
3. Y = A’ D + A B D
Y = ( A’ A ) D + B D
Y = A’ D + B D
Tabel :
A
|
B
|
D
|
A’
|
A’ D
|
A B D
|
A’ D + A B D
|
0
|
0
|
0
|
1
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
0
|
1
|
0
|
1
|
0
|
1
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
0
|
0
|
0
|
1
|
0
|
1
|
0
|
0
|
0
|
0
|
1
|
1
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
0
|
0
|
1
|
1
|
4. Y = ( A’ + B ) ( A + B )
Y = ( A’ A + A’ B ) ( B A + B B )
Y = ( A’ + A B ) ( B A + B )
Y = ( A’ + B ) B
Tabel :
A
|
B
|
A’
|
A’ + B
|
( A’ + B ) B
|
1
|
1
|
0
|
1
|
1
|
1
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
1
|
0
|
0
|
1
|
1
|
0
|
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