Aljabar Boolean dan Gate
Gerbang
Transistor dan Gerbang
+ Vcc
Vout
Kolektor
Basis
Emiter
Vin
+ Vcc
Vout
V1
V2
+ Vcc
Vout
V1
V2
Inverter/
NOT
Nand
Nor
5 Gerbang Utama
A | X |
0 | 1 |
1 | 0 |
A | B | X |
0 | 0 | 1 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 0 |
A | B | X |
0 | 0 | 1 |
0 | 1 | 0 |
1 | 0 | 0 |
1 | 1 | 0 |
A | B | X |
0 | 0 | 0 |
0 | 1 | 0 |
1 | 0 | 0 |
1 | 1 | 1 |
A | B | X |
0 | 0 | 0 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 1 |
A
X
A
B
X
A
B
X
A
B
X
A
B
X
NOT
NAND
NOR
AND
OR
Inverter
AND Adalah NAND yang di invert
OR Adalah NOR yang di invert
NAND dan NOR butuh 2 T
AND dan OR butuh 3 T
Aljabar Boolean
Aljabar Boolean
Aljabar Boolean
Aljabar Boolean
Definisi Formal
Simplifikasi
Ekspresi | Rules |
C + ¬(BC) | Original Expression |
C + (¬B + ¬C) | DeMorgan's Law. |
(C + ¬C) + ¬B | Commutative, Associative Laws. |
T + ¬B | Compliment Law. |
T | Identity Law. |
Simplifikasi
Ekspresi | Rules |
¬(AB)(¬A + B)(¬B + B) | Original Expression |
¬(AB)(¬A + B) | Compliment law, Identity law. |
(¬A + ¬B)(¬A + B) | DeMorgan's Law |
¬A + (¬B)B | Distributive law. This step uses the fact that or distributes over and. It can look a bit strange since addition does not distribute over multiplication. |
¬A | Compliment, Identity |
Simplifikasi
Expresi | Rules |
(A + C)(AD + A(¬D)) + AC + C | Original Expression |
(A + C)A(D + ¬D) + AC + C | Distributive. |
(A + C)A + AC + C | Compliment, Identity. |
A((A + C) + C) + C | Commutative, Distributive. |
A(A + C) + C | Associative, Idempotent. |
AA + AC + C | Distributive. |
A + (A + T)C | Idempotent, Identity, Distributive. |
A + C | Identity, twice. |
Ekspresi Boolean
Notasi Boolan
AND OR NOT
Pemrograman && || ~
Operasi Boolean-1 · + -
Operasi Boolean-2 + ¬
Operasi Boolean-3 ^ v /
Aljabar Boolean
A | B | C | M |
0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 |
0 | 1 | 0 | 0 |
0 | 1 | 1 | 1 |
1 | 0 | 0 | 0 |
1 | 0 | 1 | 1 |
1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 |
A
M
B
C
-A
-B
-C
A
B
C
-A
-B
-C
1
2
3
4
5
6
7
-ABC
A-BC
AB-C
ABC
TK F.Mayoritas
Rangkaian TK F.Mayoritas
M=f(ABC)
Tabel Kebenaran
Aljabar Boolean
Aljabar Boolean
Aljabar Boolean
= m0 + m3 + m6 = S m (0, 3, 6)
(A+B+¬C).(A+¬B+C).(¬A+B+C).(¬A+B+¬C).(¬A+¬B+¬C)
= M1M2M4M5M7 = ? M(1, 2, 4, 5, 7)
Aljabar Boolean
Membaca Tabel Kebenaran (TK)
Elektronika Dasar
Integrated Circuit (IC)
Kemasan dengan 2 deret pin diluar dan IC di dalamnya secara teknis disebut sebagai Dual Inline Packages (DIP)
– Klasifikasi chip:
Contoh dari F.Mayoritas
Implementasi Fungsi Boolean
Step 1, Tabel Kebenaran�M=Fungsi(ABC)
A | B | C | M |
0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 |
0 | 1 | 0 | 0 |
0 | 1 | 1 | 1 |
1 | 0 | 0 | 0 |
1 | 0 | 1 | 1 |
1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 |
Step 2, Inverter dan komplemennya�M=F(ABC)
A
B
C
-A
-B
-C
A
B
C
-A
-B
-C
1
2
3
Komplemen Inverter A,B dan C
Step 3, Gambar gerbang AND�M=F(ABC)
A
B
C
-A
-B
-C
A
B
C
-A
-B
-C
1
2
3
4
5
6
7
Step 4, Hubungkan input yang sesuai dengan gerbang AND�M=F(ABC)
A
B
C
-A
-B
-C
A
B
C
-A
-B
-C
1
2
3
4
5
6
7
-ABC
A-BC
AB-C
ABC
Kemungkinan kombinasi dari
Hasil Inverter adalah –ABC,A-BC,
AB-C dan ABC
Step 5, Masukan Output gerbang AND ke Sebuah Gerbang OR�M=F(ABC)
A
M
B
C
-A
-B
-C
A
B
C
-A
-B
-C
1
2
3
4
5
6
7
-ABC
A-BC
AB-C
ABC
Rangkaian TK F.Mayoritas
M=f(ABC)
IMPLEMENTASI Fungsi Boolean (lanj.)
IMPLEMENTASI Fungsi Boolean (lanj.)
A
B
A
A
-A
-A
AB
A
B
AB
A
B
AB
A+B
A
B
Gerbang NAND
Gerbang NOR
1 var di NAN dan di NOR
= NOT
2 var di NAND= AND
2 var di NOR= OR
Ekuivalensi Rangkaian
A | B | C | AB | AC | AB+AC |
0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 0 | 0 |
0 | 1 | 1 | 0 | 0 | 0 |
1 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 0 | 1 | 1 |
1 | 1 | 0 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 1 | 1 |
A | B | C | A | B+C | A(B+C) |
0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 1 | 0 |
0 | 1 | 0 | 0 | 1 | 0 |
0 | 1 | 1 | 0 | 1 | 0 |
1 | 0 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 1 |
Rangkaiannya
A
B
C
AB
AC
AB+AC
A
B
C
B+C
A(B+C)
Hukum Aljabar Boolean
Nama | Bentuk AND | Bentuk OR |
Identitas | 1A=A | 0+A=A |
Pembatalan | 0A=0 | 1+A=1 |
Idempoten | AA=A | A+A=A |
Inversi | A-A=0 | A+-A=1 |
Komutatif | AB=BA | A+B=B+A |
Asosiatif | (AB)C=A(BC) | (A+B)+C=A+(B+C) |
Distributif | A+BC=(AB)+(AC) | A(B+C)=AB+AC |
Absorbsi | A(A+B)=A | A+AB=A |
De organ | -(AB)=-A + -B | -(A+B)=-A -B |
Alternatif gerbang NAND, NOR dan OR
-(AB)
=
(-A)+(-B)
-(A+B)
=
(-A)(-B)
(AB)
=
-((-A)+(-B))
(A+B)
=
-((-A)(-B))
3 Rangkaian menghitung XOR dengan NAND dan NOR
A | B | X |
0 | 0 | 0 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 0 |
-A
B
A
-B
-A
B
A
-B
-A
B
A
-B
Tabel XOR
K-Map
K Map
Contoh
Contoh
¬A¬BC + ¬ABC = ¬AC·(¬B + B) = ¬AC · 1 = ¬AC
AB + ¬BC + AC = (AB · 1) + (¬B C · 1) + (AC · 1)
= (AB · (¬C + C)) + (¬BC · (¬A + A)) + (AC ·
(¬B + B)) = (AB¬C + ABC) + (¬A¬BC + A¬BC)
+ (A¬BC + ABC) = AB¬C + ABC + ¬A¬BC
+ A¬BC = ¬A¬BC + A¬BC + AB¬C + ABC
Contoh