Download - Materi 3 - Aljabar Boolean
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Aljabar Boolean
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• Literal: sebuah variabel atau komplemennya – X, X
• Ekpresi: literals dikombinasikan dengan AND, OR, tanda kurung, komplementasi
• Macam notasi AND : dengan menggunakan titik atau bisa juga ^
• Macam notasi OR : dengan menggunakan tanda + atau bisa juga v
– X+Y– P Q R– A + B C
• Persamaan: variabel = ekspresi– P = (A B C) C
Definisi: Ekspresi Boolean
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Tabel Kebenaran
XX YY X.YX.Y
00 00 00
00 11 00
11 00 00
11 11 11
XX YY X+X+YY
00 00 00
00 11 11
11 00 11
11 11 11
XX X’X’
00 11
11 00
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2.1 Teorema Boolean
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Aksioma• Aksioma
– kumpulan definisi dasar (A1 - A5, A1’ - A5’) minimal yang diasumsikan benar dan secara menyeluruh mendefinisikan aljabar boolean
– Dapat digunakan untuk membuktikan teorema aljabar boolean lainnya.
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Aksioma
(A1(A1))
X=0, if XX=0, if X11(A1’(A1’))
X=1, if XX=1, if X00
(A2(A2))
If X=0, then If X=0, then X’=1X’=1
(A2’(A2’))
If X=1, then X’=0If X=1, then X’=0
(A3(A3))
0 0 · 0 = 0· 0 = 0 (A3’(A3’))
1 + 1 = 11 + 1 = 1
(A4(A4))
1 1 · 1 = 1· 1 = 1 (A4’(A4’))
0 + 0 = 00 + 0 = 0
(A5(A5))
0 0 · 1 = · 1 = 1 1 · 0 = 0· 0 = 0 (A5’(A5’))
1 + 0 = 0 + 1 = 11 + 0 = 0 + 1 = 1
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Teorema variabel tunggal
Postulate 2 Postulate 2 (P1)(P1)
X + 0 = X + 0 = xx
P1P1’’
X . 1 = XX . 1 = X
Postulate 5 Postulate 5 (P5)(P5)
X + X’ = X + X’ = 11
P5P5’’
X . X’ = 0X . X’ = 0
Theorem 1 Theorem 1 (T1)(T1)
X + X = XX + X = X T1T1’’
X . X = XX . X = X
Theorem 2 Theorem 2 (T2)(T2)
X + 1 = 1X + 1 = 1 T2T2’’
X . 0 = 0X . 0 = 0
Theorem 3 Theorem 3 (T3)(T3)
(X’)’ = X(X’)’ = X
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Contoh
• Dibuktikan melalui induksi sempurna – Karena sebuah variabel boolean hanya dapat mempunyai nilai
0 dan 1, kita dapat membuktikan sebuah teorema dengan melibatkan sebuah variabel tunggal X melalui peletakan sederhana:
X = 0 atau X =1
• Contoh: (P1) X + 0 = X– X=0 : 0 + 0 = 0 benar menurut aksioma A4’– X=1 : 1 + 0 = 1 benar menurut aksioma A5’
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Teorema dua dan tiga variabel
P3P3 X+Y = Y+XX+Y = Y+X P3’P3’ X . Y = Y. XX . Y = Y. X KOMUTATIFKOMUTATIF
T4T4 (X+Y)+Z = X+(X+Y)+Z = X+(Y+Z)(Y+Z)
T4’ T4’ (X.Y).Z = X.(Y.Z)(X.Y).Z = X.(Y.Z) ASOSIATIFASOSIATIF
P4P4 X.Y+X.Z=X.(Y+Z)X.Y+X.Z=X.(Y+Z) P4’P4’ (X+Y).(X+Z)=X+Y.Z(X+Y).(X+Z)=X+Y.Z DISTRIBUTIDISTRIBUTIFF
T5T5 (X+Y)’=X’.Y’(X+Y)’=X’.Y’ T5’T5’ (X.Y)’=X’+Y’(X.Y)’=X’+Y’ DE DE MORGANMORGAN
T6T6 X+X.Y=XX+X.Y=X T6’T6’ X.(X+Y)=XX.(X+Y)=X ABSORPSIABSORPSI
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Teorema P4(Distributif)
(P4) X · Y + X · Z = X · (Y + Z)(P4’) (X + Y) · (X + Z) = X + Y · Z
• P4 : penjumlahan dari perkalian (sum of products (SOP))
• P4’ : Perkalian dari penjumlahan (product of sums (POS))
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SOP dan POS
(bentuk SOP)
V · W · Y + V · W · Z + V · X · Y + V · X · Z (bentuk POS)
(V + Y) · (V + Z) · (W + Y) · (W + Z) · (X + Y)
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Logical Gate
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Contoh Teorema DeMorgan: NAND (Negated AND or NOT AND)
• (X · Y)’ = (X’ + Y’)– (X · Y)’ dirujuk umumnya sebagai gerbang NAND
pada ekspresi gerbang logika
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Contoh Teorema DeMorgan: NAND (Negated AND or NOT AND)
• (X · Y)’ = (X’ + Y’)– (X · Y)’ dirujuk umumnya sebagai gerbang NAND
pada ekspresi gerbang logika
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Contoh Teorema DeMorgan: NOR
• (X + Y)’ = (X’ · Y’)– (X + Y)’ dirujuk sebagai gerbang NOR pada
ekspresi gerbang logika
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2.2 Fungsi Boolean
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Fungsi Boolean
• Adalah sebuah ekspresi yang terbentuk dari variabel, OR, AND, NOT, tanda kurung dan tanda persamaan.
• Contoh :F = X.Y.Z’
Fungsi F akan bernilai 1 jika X=1, Y=1 dan Z’=0.
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Contoh
F1 = X.Y.Z’F2 = X + Y’ZF3 = X’.Y’.Z + X’.Y.Z + X.Y’F4 = X.Y’ + X’Z
XX YY ZZ FF11
FF22
FF33
FF44
00 00 00 00 00 00 00
00 00 11 00 11 11 11
00 11 00 00 00 00 00
00 11 11 00 00 11 11
11 00 00 00 11 11 11
11 00 11 00 11 11 11
11 11 00 11 11 00 00
11 11 11 00 11 00 00
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Komplemen Fungsi Boolean
• Pada tabel kebenaran, tukar nilai 0 dengan 1 dan sebaliknya
• Cara cepat: komplemen fungsi dapat ditemukan melalui pertukaran “+” dan “.” serta pengkomplemenan seluruh variabel
XX YY ZZ F1F1’’
F2F2’’
F3F3’’
F4F4’’
00 00 00 11 11 11 11
00 00 11 11 00 00 00
00 11 00 11 11 11 11
00 11 11 11 11 00 00
11 00 00 11 00 00 00
11 00 11 11 00 00 00
11 11 00 00 00 11 11
11 11 11 11 00 11 11
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Manipulasi ekspresi Boolean
• Bagaimana menyatakan (A · B + C) dalam bentuk lain?
• Gunakan teorema DeMorgan …– A · B + C = ( ( A · B + C )’ )’– = ( ( A · B )’ · C’ )’– = ( ( A’ + B’ ) · C’ )’( A · B + C )’ = ( A’ + B’ ) · C’
• Sederhanakan X’Y’Z+X’YZ+XY’ ?
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LATIHAN1. Tunjukkan dengan menggunakan tabel kebenaran hukum
DeMorgan’s(XYZ)’ = X’ + Y’ +Z’
2. Sederhanakan ekspresi boolean berikut :a. x’y’ + xy + x’yb. x’yz + xzc. ( xy’ + a’d )( ab’ + cd’ )
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Latihan
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Tugas 2
• Cari Materi tentang Canonical dan Standard Forms dan kerjakan soal ini:
Ubahlah ekspresi berikut ke dalam bentuk sum of minterms dan product of maxtermsa. F(A, B, C, D) = ∑ (0, 2, 6, 11, 13, 14)b. ( AB + C )( B + C’D )
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Format Pengiriman
• Pengiriman tugas ke email yang disediakan• Dokumen : PDF / Doc• SUBJECT EMAIL : TUGAS 2 SISTEM DIGITAL• Nama file : tugas 2 _ NPM.doc /pdf