kuliah 2 sistem digital
TRANSCRIPT
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PENGKODEAN LAINNYA
Decimal 8,4,2,1 Excess3 8,4,-2,-1 Gray 0 0000 0011 0000 0000 1 0001 0100 0111 0100 2 0010 0101 0110 0101 3 0011 0110 0101 0111 4 0100 0111 0100 0110 5 0101 1000 1011 0010 6 0110 1001 1010 0011 7 0111 1010 1001 0001 8 1000 1011 1000 1001 9 1001 1100 1111 1000
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KODE DENGAN PENDETEKSI KESALAHAN
2
Desimal
BCD Dengan paritas genap
Dengan paritas gasal
0 0000 0000 0 0000 1
1 0001 0001 1 0001 0
2 0010 0010 1 0010 0
3 0011 0011 0 0011 1
4 0100 0100 1 0100 0
5 0101 0101 0 0101 1
6 0110 0110 0 0110 1
7 0111 0111 1 0111 0
8 1000 1000 1 1000 0
9 1001 1001 0 1001 1
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KODE HAMMING (DETEKSI DAN KOREKSI KESALAHAN)
Data: 0 1 1 0 (6) d3d2d1d0
Posisi : 1 2 3 4 5 6 7 p1 p2 d3 p4 d2 d1 d0
p1 p2 0 p4 1 1 0
p1 bertanggung jawab pada posisi: 1,3,5,7p2 bertanggung jawab pada posisi: 2,3,6,7p4 bertanggung jawab pada posisi: 4,5,6,7
p1 : p1 + 0 + 1 + 0 = genap p1 = 1p2 : p2 + 0 + 1 + 0 = genap p2 = 1p4 : p4 + 1 + 1 + 0 = genap p4 = 0
Kode Hamming:1 1 0 0 1 1 0
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MISAL KODE HAMMING PARITAS GENAP DARI BCD ADALAH 1 1 1 0 1 1 0 , BERAPA NILAI BCD TSB?
Posisi : 1 2 3 4 5 6 7 p1 p2 d3 p4 d2 d1 d0
1 1 1 0 1 1 0
p1 : 1 + 1 + 1 + 0 = ganjil salahp2 : 1 + 1 + 1 + 0 = ganjil salahp4 : 0 + 1 + 1 + 0 = genap benar
Yang benar: 1 1 0 0 1 1 0Data : 0110 (6)
Bit yang salah adalah posisi: 3 ????
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Gerbang Logika
IN OUT
0 1
1 0
OUTIN
INVERTER
X F
Y
X Y
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1
0 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1
1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
Ada 16 kemungkinan fungsi F
F0 – F15
0X AND Y X Y
X XOR Y
X OR Y X = YNOT Y
NOT X
X NAND YNOT(X AND Y)
1
X NOR YNOT(X OR Y)
Truth Table
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Gerbang Logika
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Exclusive OR (NOR)
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Teori Aljabar Boole (1)
Elementer 1. x + 0 = x 1d. x . 1 = x
2. x + x’ = 1 2d. x . x’ = 0
3. x + x = x 3d. x . x = x
4. x + 1 = 1 4d. x . 0 = 0
5, (x’)’ = x
Commutative 6. x + y = y + x 6d. x . y = y . x
Assocoative 7. x+(y+z)=(x+y)+z 7d. x(yz)=(xy)z
Distributive 8. x(y+z)=xy+xz 8d. x+(yz)=(x+y)(x+z)
Teori De Morgan 9. (x + y)’ = x’y’ 9d. (xy)’ = x’ + y’
Absorption 10. x + xy = x 10d. x(x+y) = x
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Teori Aljabar Boole (2)
Secara umum teori De Morgan dapat ditulis sebagai:
F’(X1,X2,…,Xn,0,1,+,◦) = F(X1’,X2’,…,Xn’,1,0, ◦,+)
Dualitas suatu pernyataan logika didapatkan dengan mengganti 1 dengan 0, 0 dengan 1, + dengan ◦, ◦ dengan +, dengan semua variabel tetap
F(X1,X2,…,Xn,0,1,+,◦) ⇔ F(X1,X2,…,Xn,1,0, ◦,+)
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Bukti teori De Morgan: (x + y)’ = x’y’
x y x + y (x+y)’ x’ y’ x’y’
0 0 0 1 1 1 1
0 1 1 0 1 0 0
1 0 1 0 0 1 0
1 1 1 0 0 0 0
Dengan tabel kebenaran
Dengan diagram Venn
x
x’y’
(x+y)’
y
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Contoh penyederhanaan
F = ABC + A’B’C + A’BC + ABC’ + A’B’C’
G = [(BC’ + A’D)(AB’ + CD’)]’ = (BC’ + A’D)’ + (AB’ + CD’)’ = (BC’)’(A’D)’ + (AB’)’(CD’)’ = (B’+C)(A+D’) + (A’+B)(C’+D) = AB’+AC+B’D’+CD’+A’C’+A’D+BC’+BD = 1 (dari mana???)
= (AB + A’B’)C + BC + (AB + A’B’)C’ = (A B)’ + BC⊕
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Bentuk kanonis Sum Of Product (SOP) & Product Of Sum (POS)
Des A B C F1
0 0 0 0 0
1 0 0 1 0
2 0 1 0 0
3 0 1 1 1
4 1 0 0 1
5 1 0 1 1
6 1 1 0 1
7 1 1 1 1
Dalam bentuk SOP:
F1=A’BC+AB’C’+AB’C+ABC’+ABC = ∑(m3,m4,m5,m6,m7)
= ∑(3,4,5,6,7)
Dalam bentuk POS:
F1=(A+B+C)(A+B+C’)(A+B’+C) = Л(M0,M1,M2)
= Л(0,1,2)
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0
0
1
1
0
0
1
1
P
111
011
101
001
110
010
100
000
CB A
Tuliskan bentuk SOP & POS
Bentuk SOP:
P = A’B’C’ + A’B’C + AB’C’ +AB’C = ∑(m0,m1,m4,m5) = ∑(0,1,4,5)
Bentuk POS:
P = (A+B’+C)(A+B’+C’)(A’+B’+C)(A’+B’+C’) = Л(M2,M3,M6,M7) = Л(2,3,6,7)
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Pemetaan antar SOP & POS
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Bentuk SOP:
P = A’B’C’ + A’B’C + AB’C’ +AB’C = A’B’ + AB’ = B’
P = (A+B’+C)(A+B’+C’)(A’+B’+C)(A’+B’+C’)
= (A+AB’+AC’+AB’+B’+B’C’+AC+B’C) *
(A’+A’B’+A’C’+A’B’+B’C’+A’C+B’C)
= (A+B’)(A’+B’) = AA’ + AB’ + A’B’ + B’
= B’
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Standard SOP & POS
Sum of Product (SOP) Product of Sum (POS)
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Bentuk Nonstandar
Bentuk Nonstandar (tidak dalam SOP maupun POS)
Bentuk SOP
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Implementasi
Implementasi tiga level vs. Implementasi dua level
Implementasi dua level lebih disukai karena alasan delay
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Penyederhanaan dengan menggunakan Peta-K (Karnaugh Map)
Peta-K dengan 2 variabel
m0 m1
m2 m3
xy
0
0 1
1
x’y’ x’y
xy’ xy
xy
0
0 1
1
1
1
xy
0
0 1
1
x’y + xy = (x’ + x)y = y
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Peta-K dengan 3 & 4 variabel
Peta-K dengan 3 variabel
F=A’B’C’+B’CD’+A’BCD’+AB’C’=
x’y’z’ x’y’z x’yz x’yz’
xy’z’ Xyz’ xyz xyz’
xyz
0
1
00 01 11 10
1
1 1 1 1
xyz
0
1
00 01 11 10
Peta-K dengan 4 variabel
F1= ∑(3,4,5,6,7) = x + yz
x yz
1 1 1
1
1 1 1A
C
B
DB’C’ B’D’
A’CD’
=B’C’+B’D’+A’CD’
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Peta-K dengan 5 & 6 variabelPeta-K dengan 5 variabel
F(A,B,C,D,E)=∑(0,2,4,6,9,11,13,15,17,21,25,27,29,31)
=BE+AD’E+A’B’E’
0 1 3 2 6 7 5 4
8 9 11 10 14 15 13 12
24 25 27 26 30 31 29 20
16 17 19 18 22 23 21 20A
DD
C
E E
B
1 1 1 1
1 1 1 1
1 1 1 1
1 1A
DD
CE E
B
AB
CDE
Untuk peta-K dengan 6 variabel, baca buku teks
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PEKERJAAN RUMAH1. Dari buku Morris Mano: soal no. 2.5; 2.6; 2.7; 2.11;
2.13; 2.142. Idem soal no. 3.4; 3.5; 3.7; 3.9