Kuliah 4&5 sistem digital
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Transcript of Kuliah 4&5 sistem digital
![Page 1: Kuliah 4&5 sistem digital](https://reader035.fdocuments.net/reader035/viewer/2022081508/5562f6d0d8b42a6f598b4753/html5/thumbnails/1.jpg)
Contoh Soal1. Sederhanakan F(A,B,C,D) = ∑(0,1,2,5,8,9,10)
a. Dalam bentuk SOP
b. Dalam bentuk POS
Dalam bentuk SOP
F=B’C’+B’D’+A’C’D
A
C
B
D
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Contoh Soal1. Sederhanakan F(A,B,C,D) = ∑(0,1,2,5,8,9,10)
a. Dalam bentuk SOP
b. Dalam bentuk POS
Dalam bentuk POS
F= (A’+B’)(C’+D’)(B’+D)
A
C
B
D
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Contoh Soal
2. Sederhanakan fungsi berikut dengan menggunakan Peta-K :
F(A,B,C,D) = ∑(0,2,3,5,8,10,11)
d(A,B,C,D) = ∑(1,7,9,12)
Catatan: d(..) adalah don’t care bisa dianggap 0 atau 1
A
C
B
D
1 d 1 1
1 d
d
1 d 1 1
Tanpa don’t care: F= B’C + B’D’ + A’BC’D
Dengan don’t care: F= B’ + A’D
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Implementasi fungsi digital dengan menggunakan gerbang NAND atau NOR saja
Teori De Morgan: (x + y)’ = x’y’
Teori De Morgan: (x y)’ = x’+y’
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Implementasi dengan gerbang NAND
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Contoh: Implementasi F=AB+CD
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Contoh: Implementasi F= (AB’+A’B)(C+D’)
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Implementasi dengan gerbang NOR
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Contoh: Implementasi F=(A+B)(C+D)E
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Contoh: Implementasi F=(AB’+A’B)(C+D’)
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Metode Quine-McCluskey (Tabular)
• Proses dua langkah:– Menentukan prime implicants– Menentukan minimal cover
• Semua proses dilakukan dengan menggunakan tabel
• Implicant yang berdekatan digabung, sebagai contoh:
0100 & 1100 menghasilkan -100-100 & -101 menghasilkan -10-
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Contoh: ƒ(A,B,C,D) = Σ(0,4,5,6,7,8,9,10,13,15)
Des Biner
0
4
5
6
7
8
9
10
13
15
0000
0100
0101
0110
0111
1000
1001
1010
1101
1111
Tabel1
0000
0100 1000
0101 0110 1001 1010
0111 1101
1111
Tabel2
0-00
Tabel3
010-01-0100-10-0
01-1-101011-1-01
-11111-1
-000
*
01--
-1-1
*
*
**
**
Implication Table (untuk menentukan prime implicant)
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Coverage Table (untuk mencari minimal cover)
0,4(0-00)
0,8(-000)
8,9(100-)
8,10(10-0)
9,13(1-01)
4,5,6,7(01--)
5,7,13,15(-1-1)
0X
X
4X
X
5
X
X
6
X
7
X
X
8
X
X
X
9
X
X
Atau ƒ(A,B,C,D) = A’B + BD + AB’D’ + ??? + ???
10
X
13
X
X
15
X
ƒ(A,B,C,D) = A’B + BD + AB’D’ + AC’D + B’C’D’
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Contoh: G(A,B,C,D) =
Σ(4,5,6,8,9,10,13)d(A,B,C,D = Σ d(0,7,15)
Des Biner
0
4
5
6
7
8
9
10
13
15
0000
0100
0101
0110
0111
1000
1001
1010
1101
1111
Tabel1
0000
0100 1000
0101 0110 1001 1010
0111 1101
1111
Tabel2
0-00
Tabel3
010-01-0100-10-0
01-1-101011-1-01
-11111-1
-000
*
01--
-1-1
*
*
**
**
Implication Table (untuk menentukan prime implicant)
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Coverage Table (untuk mencari minimal cover)
0,4(0-00)
0,8(-000)
8,9(100-)
8,10(10-0)
9,13(1-01)
4,5,6,7(01--)
5,7,13,15(-1-1)
4X
X
5
X
X
6
X
8
X
X
X
9
X
X
10
X
13
X
X
ƒ(A,B,C,D) = A’B + AB’D’ + AC’D
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Soal Latihan
Sederhanakan fungsi berikut dengan menggunakan metode Quin-McCluskey:
F(A,B,C,D) = ∑(0,2,3,5,8,10,11)
d(A,B,C,D) = ∑(1,7,9)
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SINTESIS (PERANCANGAN) RANGKAIAN DIGITAL
Prosedur:
1. Pahami persoalannya dengan benar
2. Identifikasi input & outputnya
3. Tuliskan tabel kebenarannya
4. Sederhanakan fungsinya
5. Gambarkan rangkaiannya
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CONTOH-CONTOH
1. Desain rangkaian Half Adder
Input : x, yOutput : S (Sum), C (Carry)
0 00 0
0 10 1
1 00 1
1 11 0
Carry Sum
Tabel Kebenaran
x y S C0 0 0 00 1 1 01 0 1 01 1 0 1 1
1
xy
0
0 1
1
Minimisasi S ??
S = xy’ + x’y = x y⊕
C = xy
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2. Desain rangkaian Full Adder
Input : x, y, CiOutput : S, Co
S = x’y’Ci + x’yCi’ + xy’Ci’ + xyCi = x’(y Ci) + x(y Ci)’ = x y Ci⊕ ⊕ ⊕ ⊕
Co S
0 1 00 1
xy
Ci
Tabel Kebenaran
x y Ci S Co0 0 0 0 00 0 1 1 00 1 0 1 00 1 1 0 11 0 0 1 01 0 1 0 11 1 0 0 11 1 1 1 1
1 1
1 1
xy Ci
0
1
00 01 11 10
1 1 1
1
xy Ci
0
1
00 01 11 10
Co = ???
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3. Desain rangkaian yang mendeteksi validitas kode BCD
4. Desain rangkaian yang mengkonversi kode BCD ke kode Excess-3
5. Desain rangkaian dekoder BCD ke seven-segment.
bg
a
cd
e
f
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RANGKAIAN KOMBINASIONAL DENGAN MSI & LSI
Binary parallel adder
+
A3 B3
S3
+
A2 B2
S2
+
A1 B1
S1
+
A0 B0
S0C1C2C3
Full Adder
Cout
Cin
A3 A2 A1 A0 B3 B2 B1 B0
S3 S2 S1 S0
CinCout
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Binary parallel adder/subtractor
A3 A2 A1 A0
B3 B2 B1 B0
S3 S2 S1 S0
CinCout
Cin = 0, Adder
= 1, Subtractor
4-bit binary adder
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Rangkaian konverter dari BCD ke Excess-3 dengan menggunakan 4 bit adder
Cout
Cin
Input BCD
1
0
Output:
Kode Excess-3
A0
A1
A2
A3
B0
B1
B2
B3
S0
S1
S2
S3
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Rangkaian fast adder
Si
Ci+1
Bi
Ai
Ci
Pi
Gi
• Penjumlahan dengan menggunakan binary adder seperti pembahasan di atas sangat lambat karena adanya perambatan/propagasi dari carryUntuk mempercepat digunakan rangkaian carry look ahead
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Rangkaian fast adder
Dari rangkaian fast adder, bila:
Pi = Ai ⊕ Bi (carry propagate)
Gi = Ai Bi (carry generate)
Maka: Si = Pi ⊕ Ci
Ci+1 = Gi + Pi Ci
Bila C0 diketahui, maka C1, C2 dst dapat dicari sbb:
C1 = G0 + P0 C0
C2 = G1 + P1 C1 = G1 + P1 (G0 + P0 C0 ) = G1 + P1G0 + P1P0 C0
C3 = G2 + P2 C2 = G2 + P2 (G1 + P1G0 + P1P0 C0 )
= G2 + P2 G1 + P2 P1G0 + P2 P1P0 C0
C4 = G3 + P3 C3 = ???
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Implementasi Carry Lookahead
Rangkaian logika yang semakin kompleks
Pi @ 1 gate delay
Ci Si @ 2 gate delays
BiAi
Gi @ 1 gate delay
C0C0
C0
C0P0P0
P0
P0
G0G0
G0
G0
C1
P1
P1
P1
P1
P1
P1 G1
G1
G1
C2P2
P2
P2
P2
P2
P2
G2
G2
C3
P3
P3
P3
P3
G3
C4