tei106 bab 3 aljabar boolean
DESCRIPTION
Boolean AlgebraTRANSCRIPT
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3 ALJABAR BOOLEAN
3.0 Pendahuluan
3.1 Kaidah Boolean & Gerbang Logika Dasar
3.2 Fungsi Logika Boolean & Tabel Kebenaran
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3.0 Pendahuluan • George Boole (1815 - 1864), seorang ahli matematik,
memperkenalkan konsep aljabar logika melalui publikasi tulisan, 'An Investigation of the Laws of Thought‘ (1854)..
• Aljabar Logika Boolean menggunakan
pendekatan biner, dengan variabel-variabel
dua harga ( 0 – 1, Benar – Salah , On – Off,
Ya – Tidak, … ) dan tiga operasi dasar
AND (perkalian logika), OR (penjumlahan l
ogika), NOT (pengingkaran/pembalikan logika).
• Sedemikian sederhananya konsep aljabar Boolean tsb , sehingga menuai banyak kritikan dan pengabaian dari komunitas ahli matematik pada masa itu.
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Claude Shannon (1916-2001), seorang engineer ,
mempublikasikan tesis S2nya “A Symbolic Analysis of Relay and
Switching Circuits” (1938).
Tesis tsb memanfaatkan aljabar Boolean sebagai dasar
matematis untuk analisis & perancangan
rangkaian penyaklaran.
embrio perancangan komputer digital
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Aljabar Boolean memberikan dasar matematis untuk
merepresentasikan fungsi fungsi bervariabel biner (fungsi
logika biner, fungsi logika Boolean)
logika
Boolean
x1
x2
xN-1
y1
y2
yM-1 xi dan yj adalah variabel
variabel biner
Variabel biner hanya punya
dua nilai . “0” atau “1”,
“benar” atau “salah”
“ya” atau “tidak”
“ada” atau “tidak ada”
logika biner (logika dua nilai)
Fungsi logika biner :
yi = Fi(x1 , x2 , ..., xN-1)
Nilai logika biner :
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Contoh : logika pengambilan
keputusan
x1
x2
x3
y
x1
x2
x3
0 , jelek = 1 , cakep
0 , miskin = 1 , kaya
0 , bodoh = 1 , pinter
0 , mundur = 1 , jadian
y
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
x1 x2 x3 y
Tabel Kebenaran
input output
input output
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Contoh :
Dalam sebuah ruang dg kapasitas 8 orang, Fan dan AC
dikendalikan berdasarkan jumlah orang yang ada didalam ruang,
dengan logika sbb
Logika pengendalian fan &
AC
jumlah
orang
Kendali
X < 2 Fan off, AC off
2< X < 4 Fan on, AC off
4< X < 6 Fan off, AC on
X = 7 Fan on, AC on
Pengendali x y
input 8 nilai
output 4 nilai
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jumlah
orang
sandi biner
x2 x1 x0
0 0 0 0
1 0 0 1
2 0 1 0
3 0 1 1
4 1 0 0
5 1 0 1
6 1 1 0
7 1 1 1
0 , AC di“off”kan y0 =
1 , AC di“on”kan
Penyandian input : Penyandian output :
0 , fan di“off”kan y1 =
1 , fan di“on”kan
8 nilai sandi 3 bit 4 nilai sandi 2 bit
Fan AC y1 y0
off off 0 0
off on 0 1
on off 1 0
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#orang
X
input output
x2 x1 x0 y1 y0
0 0 0 0 0 0
1 0 0 1 0 0
2 0 1 0 0 0
3 0 1 1 1 0
4 1 0 0 1 0
5 1 0 1 0 1
6 1 1 0 0 1
7 1 1 1 1 1
Tabel Kebenaran
logika pengendalian fan & AC
X < 2
2< X < 4
4< X < 6
X = 7
Pengendali
x0 y0
y1
x1 x2
x3
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Fungsi fungsi logika Boolean disusun berdasarkan
operasi operasi logika dasar (operator operator)
AND, OR dan NOT.
Operasi operasi dasar Boolean
Operasi Operator
Pembalikan logika NOT, Invert
Penjumlahan logika OR
Perkalian logika AND
3.1 KAIDAH BOOLEAN
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X F
input output
X F = X’
0 1
1 0
Fungsi logika NOT : F = X’
Tabel Kebenaran NOT :
Operasi NOT (Invert , Pembalikan , Pengingkaran)
Simbol gerbang NOT :
Operasi-operasi Logika Dasar
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Input Output
X1 X0 F
0 0 0
0 1 1
1 0 1
1 1 1
Fungsi logika OR :
Tabel Kebenaran OR :
(Contoh : OR 2-input)
Simbol gerbang logika OR :
Operasi OR (Penjumlahan logika)
F = X0 + X1 + ... + XN-1
X0
X1
XN-1
F
X1
X0 F = X0 + X1
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Input Output
X2 X1 X0 F
0 0 0 0
0 0 1 1
0 1 0 1
0 1 1 1
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 1
Tabel Kebenaran OR 3-input
Tabel Kebenaran OR 4-input ?
X1
X0 F = X0 + X1+ X2
X2
seluruh kemungkinan
kombinasi 3 var. biner
X , Y, Z
8 kombinasi
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Input Output
X1 X0 F = X1 . X2
0 0 0
0 1 0
1 0 0
1 1 1
Fungsi logika AND :
Tabel Kebenaran AND :
(Contoh : AND 2-input)
Simbol gerbang AND :
Operasi AND (Perkalian logika)
F
F = X0 . X1 . ... . XN-1
X0
X1
XN-1
F = X0 . X1
X0
X1
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input out
X Y Z F
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 0
1 1 0 0
1 1 1 1
AND 3-input : F = X . Y . Z
seluruh kemungkinan
kombinasi 3 var. biner
X , Y, Z
Jumlah kombinasi = 2n ,
n = jumlah variabel biner
8 kombinasi
X
Y F = X .Y . Z
Z
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1. X + 0 = X
3. X + 1 = 1
5. X + X = X
7. X + X ’ = 1
9. (X ’)’ = X
10. X + Y = Y + X
12. X + (Y+Z ) = (X+Y ) + Z
14. X.(Y + Z ) = XY + XZ
16. (X + Y ) = X ×Y
18. X + X’.Y = X + Y
2. X 1 = X
4. X 0 = 0
6. X X = X
8. X X ’ = 0
11. X Y = Y X
13. X (Y Z ) = (X Y ) Z
15. X + (YZ ) = (X +Y)(X + Z )
17. (X ×Y)’ = X + Y
19. X.(X + Y’) = X’.Y’
Identitas (persamaan dasar) Boolean
dual
Dual adalah pasangan identitas lts15 15
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Commutative law: Urutan penulisan variabel tidak
berpengaruh terhadap hasil.
(10) X + Y = Y + X (11) X . Y = Y . X
• Associative law: Pengelompokan variabel tidak berpengaruh
terhadap hasil.
(12) X + (Y + Z) = (X + Y) + Z (13) X.(Y.Z) = (X.Y).Z
• Distributive law: Variabel bersama dapat dikeluarkan.
(14) X.( Y + Z) = X.Y + X.Z (15) X + Y.Z = (X + Y).(X + Z)
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Dual dari suatu identitas Boolean dapat diperoleh melalui
langkah langkah sebagai berikut.
1. Pertukarkan operator AND OR dalam identitas
tersebut,
2. Pertukarkan logika 0 1 dalam identitas tersebut.
3. Hasilnya adalah identitas lain yang merupakan
dual dari identitas pertama.
A + (B . C ) = (B . C ) + D A + A’ = 1
A . (B + C ) = (B + C ) . D A . A’ = 0
Bagaimana dual untuk A . (A’ + B) = A . B ?
Contoh :
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Aturan deMorgan
Aturan deMorgan memungkinkan perubahan bentuk
perkalian logika (AND) ke bentuk penjumlahan logika (OR)
dan sebaliknya.
Identitas 16 : ( X + Y )’ = X’ . Y’ Identitas 17 : ( X . Y )’ = X’ + Y’
bentuk AND
bentuk OR
bentuk AND
bentuk OR
dasar aturan deMorgan : identitas 16 dan 17
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Perluasan aturan deMorgan untuk fungsi 3 variabel
(X + Y + Z ) = (X + (Y + Z ))
= X . (Y + Z )
= X . (Y . Z )
(X + Y + Z ) = X . Y . Z
identitas 12
identitas 16
identitas 17
identitas 13
bentuk OR bentuk AND lts15 19
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(X .Y. Z) = (X . (Y . Z ))
= X + (Y . Z )
= X + (Y + Z )
(X .Y. Z) = X + Y + Z
identitas ?
identitas ?
identitas ?
identitas ?
bentuk OR bentuk AND
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Cara penulisan lain : A’ = A
( X . Y . Z ) = ( X . ( Y . Z ) )
= X + ( Y . Z )
= X + ( Y + Z )
= X + Y + Z
(X + Y + Z ) = ( X + ( Y + Z ) )
= X . ( Y + Z )
= X . ( Y . Z )
= X . Y . Z
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Aturan de Morgan dan ekivalensi gerbang logika
( X . Y )’ = X’ + Y’
= X X
Y Y
F F
( X + Y )’ = X’ . Y’
Y
X
= F Y
X F
= F F
X
Y
X
Y
X’
Y’
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1. (A B’ + A’ B)’ = (A . B’)’ . (A’ . B)’
= (A’ + B’’) . (A’’ + B’)
= (A’ + B) . (A + B’)
2. (X’ + Y’ + Z’)’
3. ((A + B + C) D)’
4. ((A B C)’ + D + E)’
5. (A’ B (C + D’) + E)’
Soal Latihan : Dengan aturan de Morgan ubahlah bentuk ekspresi Boolean dibawah ini
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Pembuktian Identitas Boolean :
(a) Dengan induksi-lengkap
Terbukti X = X + 0
Identitas (1) : Identitas (16) : (X + Y)’ = X’ . Y’
Contoh : Buktikan
X + 0 = X
Terbukti X’ . Y’ = (X + Y)’
1 + 0
0 + 0
X + 0
1 1
0 0
X
0 1 0 0 1 1 0
0
0
1
X’ . Y’
1
1
0
( X + Y)
0 0 0 1 1
0 1 0 0 1
1 1 1 0 0
(X + Y)’ Y’ X’ Y X
sama sama
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(b) Dengan menggunakan identitas2 Boolean
(14)
Buktikan identitas X Y + Y Z + X Z = X Y + X Z
X Y + Y Z + X Z = X Y + (X + X) Y Z + X Z
= X Y + X Y Z + X Y Z + X Z
= X Y (1 + Z ) + X Z ( Y + 1 )
= X Y . 1 + X Z . 1
X Y + Y Z + X Z = X Y + X Z terbukti !
(14)
(3) (3)
(7)
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Soal Latihan : Buktikan
1) X + XY = X
X + XY = X(1 + Y)
= X . 1
X + XY = X
2) X + X’Y = X +Y
3) (X + Y)(X + Z) = X + YZ
. . . . identitas 3
. . . identitas 2
identitas 14
terbukti !
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III.2 Fungsi Logika & Tabel Kebenaran
Fungsi Logika Boolean dapat dinyatakan dalam dua bentuk
ekspresi.
1. Bentuk ekspresi Sum Of Product (SOP).
F = penjumlahan suku-suku perkalian
Contoh : F = AB + B’ C + A’ B C’
2. Bentuk ekspresi Product Of Sum (POS)
F = perkalian suku-suku penjumlahan
Contoh : F = ( B + C ) . ( A + B’ + C’ )
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Bentuk Kanonis
Ekspresi SOP atau POS yang tiap sukunya mengandung seluruh
literal (variabel) input disebut ekspresi SOP atau POS bentuk
kanonis.
Contoh : untuk fungsi 3-variabel F(A,B,C),
F(A,B,C) = A B C + A B C’ + A B’C + A’ B’C + A’ B C’
F(A,B,C) = ( A’ + B + C ) . ( A + B + C ) . ( A + B’ + C’)
Setiap suku mengandung literal (variabel) A, B, dan C.
Bentuk SOP kanonis :
Bentuk POS kanonis :
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Contoh :
f(A,B,C) = A B + A’ C + A C’
= A B (C + C’) + A’ (B + B’) C + A (B + B’) C’
= ABC + ABC’ + A’BC + A’B’C + ABC’ + AB’C’
= ABC + ABC’ + A’BC + A’B’C + AB’C’
= m7 + m6 + m3 + m1 + m4
f(A,B,C) = m(1, 3, 4, 6, 7)
Bentuk Kanonis
Bentuk Non-kanonis
mi : mintermi atau suku-perkaliani.
SOP bentuk kanonis dpt diekspresikan sebagai penjumlahan minterm.
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Contoh :
f(A,B,C) = (A + B + C).(A + B’ + C).(A’ + B + C’)
= M0 . M2 . M5
f(A,B,C) = M(0, 2, 5)
Bentuk Kanonis
Mi : Maxtermi atau suku penjumlahani.
POS bentuk kanonis dpt diekspresikan sebagai perkalian Maxterm.
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Var.inp minterm
A B mi suku
0 0 m0 A’ B’
0 1 m1 A’ B
1 0 m2 A B’
1 1 m3 A B
Minterm untuk fungsi dengan 2 var.input :
Contoh :
(1) F1(A,B) = m1 + m2
= A’ B + A B’
(2) F2(A,B) = m0 + m3
= A’ B’ + A B
(3) F3(A,B) = m0 + m1 + m3
= A’ B’ + A’ B + A B
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Var.inp Maxterm
A B Mi suku
0 0 M0 (A + B)
0 1 M1 (A + B’)
1 0 M2 (A’ + B)
1 1 M3 (A’+ B’)
Maxterm untuk fungsi dengan 2 var.input :
Contoh :
(1) G1(A,B) = M1 . M2
= (A + B’).(A’+ B)
(2) G2(A,B) = M0 . M3
= (A + B).(A’ + B’)
(3) G3(A,B) = M0 . M1 . M3
= (A + B).(A + B’).(A’ + B’) lts15 32
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Var.inp minterm
A B mi suku
0 0 m0 A’ B’
0 1 m1 A’ B
1 0 m2 A B’
1 1 m3 A B
Mi = mi’
Maxterm
Mi suku
M0 (A + B)
M1 (A + B’)
M2 (A’ + B)
M3 (A’+ B’)
Mi = mi’
(A + B) = (A’ B’)’
(A + B’) = (A’ B)’
(A’ + B) = (A B’)’
(A’+ B’) = (A B)’
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Var.input minterm Maxterm
A B C mi term Mi (mi)’ term
0 0 0 m0 A’ B’ C’ M0 ( A’ B’ C’)’ (A + B + C)
0 0 1 m1 A’ B’ C M1 ( A’ B’ C)’ (A + B + C’)
0 1 0 m2 A’ B C’ M2 ( A’ B C’)’ (A + B’ + C)
0 1 1 m3 A’ B C M3 ( A’ B C)’ (A + B’ + C’)
1 0 0 m4 A B’ C’ M4 ( A B’ C’)’ (A’ + B + C)
1 0 1 m5 A B’ C M5 ( A B’ C)’ (A’ + B + C’)
1 1 0 m6 A B C’ M6 ( A B C’)’ (A’ + B’ + C)
1 1 1 m7 A B C M7 ( A B C)’ (A’ + B’ + C’)
Minterm & Maxterm untuk fungsi dengan 3 var.input :
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Ekspresi Boolean SOP/
POS ?
kanonis/non-
kanonis ?
1 F = B (A + C) (B’ + A’)
2 F = B + AB’ + B’C + A’C
3 F = (A + B) (A’ +B) (A + B)
4 F = ABC’ + A’BC + A’B’C’
5 F = AB + (B + C) (A + C)
Soal Latihan : Isi dengan jawaban yang benar !
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Ekspresi Boolean SOP/
POS
kanonis/non-
kanonis
1 F = B (A + C) (B’ + A’) POS non-kanonis
2 F = B + AB’ + B’C + A’C SOP non-kanonis
3 F = (A + B) (A’ +B) (A + B) POS kanonis
4 F = ABC’ + A’BC + A’B’C’ SOP kanonis
5 F = AB + (B + C) (A + C) mixed non-kanonis
jawaban yang benar
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Ekspresi Boolean Tabel Kebenaran
Ekspresi Boolean :
Jumlah variabel input N
Jumlah kombinasi input = 2N Tabel 2N entries
Contoh :
F(A,B,C) = A B + B’ C + A’ B C’
Jumlah variabel input N = 3
Jumlah kombinasi input = 23 = 8
A B C F
0 0 0 0 …
1 0 0 1 …
2 0 1 0 …
3 0 1 1 …
4 1 0 0 …
5 1 0 1 …
6 1 1 0 …
7 1 1 1 …
8 entries
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A B C F
0 0 0 0 . 0 + 0’ . 0 + 0’ . 0 . 0’ = 0
0 0 1 0 . 0 + 0’ . 1 + 0’ . 0 . 1’ = 1
0 1 0 0 . 1 + 1’ . 0 + 0’ . 1 . 0’ = 1
0 1 1 0 . 1 + 1’ . 0 + 1’ . 1 . 1’ = 0
1 0 0 1 . 0 + 0’ . 0 + 1’ . 0 . 0’ = 0
1 0 1 1 . 0 + 0’ . 1 + 1’ . 0 . 1’ = 1
1 1 0 1 . 1 + 1’ . 0 + 1’ . 1 . 0’ = 1
1 1 1 1 . 1 + 1’ . 1 + 1’ . 1 . 1’ = 1
Hitung harga output F untuk setiap kombinasi input :
F(A,B,C) = A B + B’ C + A’ B C’
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F = AB + B’ C + A’ B C’
1 1 0 0
1 0 1 0
0 1 1 0
0 0 0 1
1 1 0 1
1 1 1 1
1 0 1 1
0 0 0 0
F C B A
0
1
2
3
4
5
6
7
input out
Tabel Kebenaran
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Tabel Kebenaran Ekspresi Boolean
SOP kanonis Tabel Kebenaran ekspresi Boolean -- POS kanonis
Ekspresi SOP dibentuk dengan penjumlahan minterm minterm
yang membuat output F = 1 ( minterm-minterm mi(1) )
Ekspresi Booleannya : F = mi(1) S Contoh :
mi A B F
m0 0 0 1
m1 0 1 0
m2 1 0 0
m3 1 1 1
Tabel Kebenaran F = m0 + m3
= A’ B’ + A B
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Ekspresi POS dibentuk dengan :
1. Menjumlahan minterm minterm yang membuat harga F = 0
( minterm-minterm mi(0) ) untuk mendapatkan
F’ = mi(0).
2. Nyatakan F = (F’)’ = mi(0) = mi(0)’ = Mi(0)
S
’
S
mi A B F
m0 0 0 1
m1 0 1 0
m2 1 0 0
m3 1 1 1
Contoh : F’ = m1 + m2
= A’B’ + AB
F = ( F’)’ = (A’B’ + AB)’
= (A’B’)’ . (AB)’
= (A + B). (A’+B’)
De Morgan
Tabel Kebenaran
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Contoh : Dari Tabel Kebenaran dibawah ini, nyatakan ekspresi Boolean dalam bentuk SOP dan POSnya.
1 1 0 0
1 0 1 0
0 1 1 0
0 0 0 1
1 1 0 1
1 1 1 1
1 0 1 1
0 0 0 0
F C B A
0
1
2
3
4
5
6
7
Minterm minterm yang membuat
output F berharga “1” :
m1 , m2 , m5 , m6 , dan m7
F = m1 + m2 + m5 + m6 + m7
= A’B’C + A’B C’ + A B’C + A B C’ + A B C
Bentuk SOP :
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1 1 0 0
1 0 1 0
0 1 1 0
0 0 0 1
1 1 0 1
1 1 1 1
1 0 1 1
0 0 0 0
F C B A
0
1
2
3
4
5
6
7
Minterm minterm yang membuat
output F berharga “0” :
m0 , m3 dan m4
Bentuk POS :
F’ = m0 + m3 + m4
F = (F’)’ = (m0 + m3 + m4)’
= (m0)’ . (m3)’ . (m4)’
= M0 . M3 . M4
F = (A + B + C) . ( A + B’ + C’) . ( A’ + B + C ) lts15 43
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Var.input minterm Maxterm
A B C mi suku Mi (mi)’ suku
0 0 0 m0 A’ B’ C’ M0 ( A’ B’ C’)’ A + B + C
0 0 1 m1 A’ B’ C M1 ( A’ B’ C)’ A + B + C’
0 1 0 m2 A’ B C’ M2 ( A’ B C’)’ A + B’ + C
0 1 1 m3 A’ B C M3 ( A’ B C)’ A + B’ + C’
1 0 0 m4 A B’ C’ M4 ( A B’ C’)’ A’ + B + C
1 0 1 m5 A B’ C M5 ( A B’ C)’ A’ + B + C’
1 1 0 m6 A B C’ M6 ( A B C’)’ A’ + B’ + C
1 1 1 m7 A B C M7 ( A B C)’ A’ + B’ + C’
Minterm & Maxterm untuk fungsi dengan 3 var.input :
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Fungsi logika yang (inputnya) tidak terdefinisikan
lengkap
A B C F
0 0 0 1
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 0
A B C F
0 0 0 1
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 ?
kombinasi-kombinasi input yg mungkin muncul
kombinasi-kombinasi input yg tidak mungkin muncul
terdefinisikan lengkap tak terdefinisikan lengkap lts15 45
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pembangkit
kode
BCD
pembangkit
Paritas Ganjil
A
B
C
D
F
kode
BCD
diluar kode
BCD
Pembangkit kode BCD tidak akan pernah
menghasilkan kode > 1001 pada
outputnya.
Akibatnya, kombinasi input pada
pembangkit paritas ganjil tidak akan
pernah > 1001.
Contoh : A B C D
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1 lts15 46
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pembangkit
Paritas Ganjil
A
B
C
D
F
Tabel Kebenaran Pembangkit Paritas Ganjil
A B C D F
0 0 0 0 1
0 0 0 1 0
0 0 1 0 0
0 0 1 1 1
0 1 0 0 0
0 1 0 1 1
0 1 1 0 1
0 1 1 1 0
1 0 0 0 0
1 0 0 1 1
1 0 1 0 0/1
1 0 1 1 0/1
1 1 0 0 0/1
1 1 0 1 0/1
1 1 1 0 0/1
1 1 1 1 0/1
Dengan keyakinan bahwa kombinasi > 1001
tidak akan pernah muncul, maka output F
boleh diberi harga sembarang (harga don’t
care), 0 atau 1.
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Tabel Kebenaran Pembangkit Paritas Ganjil
dapat dinyatakan sbb
Dengan d = don’t care, yang boleh diberi
nilai sembarang 0 atau 1.
Dengan keyakinan bahwa kombinasi tsb
tidak akan pernah muncul, maka pemberian
harga output = d tidak akan berpengaruh
terhadap tabel kebenaran utama.
A B C D F
0 0 0 0 1
0 0 0 1 0
0 0 1 0 0
0 0 1 1 1
0 1 0 0 0
0 1 0 1 1
0 1 1 0 1
0 1 1 1 0
1 0 0 0 0
1 0 0 1 1
1 0 1 0 d
1 0 1 1 d
1 1 0 0 d
1 1 0 1 d
1 1 1 0 d
1 1 1 1 d lts15 48
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A B C D F
0 0 0 0 1
0 0 0 1 0
0 0 1 0 0
0 0 1 1 1
0 1 0 0 0
0 1 0 1 1
0 1 1 0 1
0 1 1 1 0
1 0 0 0 0
1 0 0 1 1
1 0 1 0 d
1 0 1 1 d
1 1 0 0 d
1 1 0 1 d
1 1 1 0 d
1 1 1 1 d
A B C D F
0 0 0 0 1
0 0 0 1 0
0 0 1 0 0
0 0 1 1 1
0 1 0 0 0
0 1 0 1 1
0 1 1 0 1
0 1 1 1 0
1 0 0 0 0
1 0 0 1 1
1 0 1 0 0
1 0 1 1 0
1 1 0 0 0
1 1 0 1 1
1 1 1 0 1
1 1 1 1 0
F = (m0 + m3 + m5
+ m6 + m9)
+ (m13 + m14)
(a)
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A B C D F
0 0 0 0 1
0 0 0 1 0
0 0 1 0 0
0 0 1 1 1
0 1 0 0 0
0 1 0 1 1
0 1 1 0 1
0 1 1 1 0
1 0 0 0 0
1 0 0 1 1
1 0 1 0 d
1 0 1 1 d
1 1 0 0 d
1 1 0 1 d
1 1 1 0 d
1 1 1 1 d
A B C D F
0 0 0 0 1
0 0 0 1 0
0 0 1 0 0
0 0 1 1 1
0 1 0 0 0
0 1 0 1 1
0 1 1 0 1
0 1 1 1 0
1 0 0 0 0
1 0 0 1 1
1 0 1 0 1
1 0 1 1 1
1 1 0 0 0
1 1 0 1 1
1 1 1 0 0
1 1 1 1 0
F = (m0 + m3 + m5
+ m6 + m9)
+ (m10 + m11
+ m13 )d
(b)
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(b) F = (m0 + m3 + m5 + m6 + m9) + (m10 + m11 + m13 )d
F = (m0 + m3 + m5 + m6 + m9) + (m13 + m14)d (a)
tidak pernah muncul pada input, sehingga tidak berpengaruh pada output F
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1 1 0 0
1 0 1 0
0 1 1 0
1 0 0 1
0 1 0 1
1 1 1 1
0 0 1 1
0 0 0 0
F C B A 1. Dari Tabel Kebenaran ini, turunkan fungsi
logika dalam bentuk SOP dan POS .
input out Soal Latihan :
A B C f(A,B,C)
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 1
1 0 0 1
1 0 1 0
1 1 0 1
1 1 1 1
(a) (b) lts15 52
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? X
Y F(X,Y) = ?
1
0
1
0
0
1
1
1
1
0
0
0
1
0
0
1
1
1
1
0
0
0
1
0
0
0
1
1
0
1
0
0
0
1
0
1
0
1
1
0
1 0 0 0 1
1 0 0 1 1
1 0 0 1 0
1 1 0 0 0
F0 Y X F1 F2 F3 F4 F5 F6 F8 F9 F10 F14 F7 F15
… fungsi logika yang berbeda , Fi , i = 0 , 1 , ... , 15 ,
2. Untuk sistem kombinatorial dengan N = 2 input, berapakah jumlah fungsi logika yg dpt dibentuk ?
F8 = fungsi AND F14 = fungsi OR
F6 = fungsi EXOR F7 = fungsi NAND F1 = fungsi NOR
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X Y F0
0 0 0
0 1 0
1 0 0
1 1 0
F0 = 0
X Y F1
0 0 1
0 1 0
1 0 0
1 1 0
F1 = X’. Y’
X Y F2
0 0 0
0 1 1
1 0 0
1 1 0
F2 = X’ . Y
X Y F3
0 0 1
0 1 1
1 0 0
1 1 0
F3 = X
X Y F4
0 0 0
0 1 0
1 0 1
1 1 0
F4 = X.Y’
X Y F5
0 0 1
0 1 0
1 0 1
1 1 0
F5 = Y’
X Y F6
0 0 0
0 1 1
1 0 1
1 1 0
F6 = X’.Y+X.Y’
X Y F7
0 0 1
0 1 1
1 0 1
1 1 0
F7 = X’ + Y’
F8 = ? F9 = ? . . . F15 = ? lts15 54
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1
0
1
0
0
1
1
1
1
0
0
0
1
0
0
1
1
1
1
0
0
0
1
0
0
0
1
1
0
1
0
0
0
1
0
1
0
1
1
0
1
0
0
0
1
1
0
0
1
1
1
0
0
1
0
1
1
0
0
0
F0
Y
X
F
1
F2
F
3
F4
F
5
F6
F
8
F9
F
10
F
14
F
7
F1
5
0 0 0 0 F0
0 0 0 1 F1
0 0 1 0 F2
0 0 1 1 F3
0 1 0 0 F4
0 1 0 1 F5
0 1 1 0 F6
0 1 1 1 F7
1 0 0 0 F8
1 0 0 1 F9
1 0 1 0 F10
1 0 1 1 F11
1 1 0 0 F12
1 1 0 1 F13
1 1 1 0 F14
1 1 1 1 F15
22 bit
2 22
kombinasi
Fungsi logika dengan N variabel biner menghasilkan
2 2 N fungsi logika yang berbeda
N=2
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input output
XN XN-
1
. . . X1 X0 y0 y1 y2 . . . y22N -1
0 0 0 . . . 0 0 0 1 0 . . . 1
1 0 0 . . . 0 1 0 0 1 . . . 1
2 0 0 . . . 1 0 0 0 0 . . . 1
3 0 0 . . . 1 1 0 0 0 . . . 1
: : : : : : : : : : :
2N-4 1 1 . . . 0 0 0 0 0 . . . 1
2N-3 1 1 0 1 0 0 0 . . . 1
2N-2 1 1 . . . 1 0 0 0 0 . . . 1
2N-1 1 1 . . . 1 1 0 0 0 . . . 1
Untuk N variabel input , ada 2N kombinasi 2
N k
om
bin
asi
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3. Buktikan identitas dibawah ini.
(a) ( X’ Y + X Z )(X + Y’) = X Z
(b) A’ C’ + A D + A C D’ = A’ C’ D’ + C’ D + A C
(c) X + Y’Z = X Y Z + X Y Z’ + X Y’ Z + X Y’ Z’ + X’ Y’ Z
(d) (A B)’ + (A C)’ + A’ B’ C’ = A’ + B’ + C’
4. Dengan aljabar Boolean sederhanakan ekspresi ini :
(a) F = B D + B (D + E) + D'(D + F)
(b) F = (B + B C)(B + B‘ C)(B + D)
(c) F = A B C (A B + C'(B C + A C))
(d) F = A + A B + A B’ C
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1. f = a’bc + a’bc’ + abc + ab’c’ + abc‘ = b + a (bc)’
2. f = ab + a’b + a’b’
3. f = ( a + a’) ( ab + abc’ )
4. f = ( ( a + b’ ) ( c’ + d) )’
5. f = ( a + ( bc )’ + cd )’ + (bc)‘
6. f = ( (a + b + c’ ) ( a’+ b + c’ ) )’
7. f = ( ab + a’c + bc’ )’
8. f = ( ab + a’c’ ( b + c‘) )’
5. Buktikan
= a’ b + c d’
= a b
= a’ + b
= a’d’ + b’ + c’
= b’ c
= a b’ + b’c’
= ( a’+ b’ )( a + c )
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1. f = a’bc + a’bc’ + abc + ab’c’ + abc‘ =/= b + a (bc)’
2. f = ab + a’b + a’b’
3. f = ( a + a’) ( ab + abc’ )
a b + a’ b + a’ b’ = a b + a’ b + a’ b + a’ b’
= b (a + a’) + a’ (b + b’) = b + a’
( a + a’) ( ab + abc’ ) = 1 ( a b ( 1 + c’) ) = a b
= a’ + b
= a b
4. f = ( ( a + b’ ) ( c’ + d) )’
( a+b’)’ + ( c’ + d )’ = a’ b + c d’
= a’ b + c d’
Jawaban soal no. 5
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= (( a + (b c)’ + c d) . (b c) )’
= ( a (b c) + (b c)’(b c) + c d (b c))’
= ( a b c + (b’ + c’).b c + b d c )’
= ( a b c + b b’ c+ b c c’ + b d c )’
= ( a b c + b d c )’ = (a b c)’ (b d c)’
= (a’ + b’ + c’) ( b’ + d’ + c’ )
= a’ b’ + a’ d’ + a’ c’ + b’ b’ + b’ d’ + b’ c’ + c’ b’ + c’ d’ + c’ c’
= a’ b’ + a’ d’ + a’ c’ + b’ + b’ d’ + b’ c’ + c’d’ + c’
= ( a’ b’ + b’ + b’ c’ + b’ d’) + ( a’ c’ + c’ + c’ d’) + a’ d’
= b’ ( a’ + 1 + c’ + d’) + c’ ( a’ + 1 + d’) + a’ d’
= b’ + c’ + a’ d’ terbukti
5. f = ( a + ( bc )’ + cd )’ + (bc)‘ = a’d’ + b’ + c’
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= ( a + b + c’ )’ + ( a’ + b + c’ )’
= ( a’ b’ c) + ( a b’ c )
= b’ c ( a’ + a ) = b’ c terbukti
= (a b)’ . (a’ c)’ . (b c’)’ = ( a’ + b’) ( a + c’ ) ( b’ + c)
= ( a’ a + a’ c’ + b’ a + b’ c’ ) ( b’ + c)
= ( a’ c’ + b’ a + b’ c’ ) ( b’ + c)
= a’ b’ c’ + a’ c’ c + a b’ + a b’ c + b’ b’ c’ + b’ c’ c
= a’ b’ c’ + a b’ + a b’ c + b’ c’
= b’ c’ ( a’ + 1) + a b’ (1 + c) = b’ c’ + a b’ terbukti
6. f = ( (a + b + c’ ) ( a’+ b + c’ ) )’ = b’ c
7. f = ( ab + a’c + bc’ )’ = a b’ + b’c’
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= (a b)’ . ( a’ c’ ( b + c’ ))’
= ( a’ + c’ ) ( a’ b c’ + a’ c’ c’ )’
= ( a’ + c’ ) ( (a’ b c’ )’ . ( a’ c’ )’ )
= ( a’ + c’ ) ( ( a + b’ + c ) ( a + c )
= ( a’ + c’ ) ( a a + a c + a b’ + b’ c + a c + c c )
= ( a’ + c’ ) ( a + a c + a b’ + b’ c + a c + c )
= ( a’ + c’ ) ( a + a c + a b’ + b’ c + c )
= ( a’ + c’ ) ( a ( 1 + c + b’ ) + c ( b’ + 1 )
= (a’ + c’) ( a + c ) terbukti
8. f = ( ab + a’c’ ( b + c‘) )’ = ( a’+ b’ )( a + c )
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