karnaugh map
TRANSCRIPT
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Karnaugh MAP (K-Map)
Pokok Bahasan :1. K-map 2 variabel2. K-map 3 variabel3. K-map 4 variabel4. Penyederhanaan rangkaian dengan k-map
Tujuan Instruksional Khusus :1.Mahasiswa dapat menerangkan dan memahami cara
membuat k-map 2, 3, 4 variabel.2.Mahasiswa dapat menerangkan dan memahami cara
peng-cover-an minterm dalam sebuah k-map..3.Mahasiswa dapat menyederhanakan persamaan logika
melalui metode k-map.
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Karnaugh Map (K-Map)
•Suatu peralatan grafis yang digunakan untuk menyederhanakanpersamaan logika atau mengkonversikan sebuah tabel kebenaran menjadi sebuah rangkaian logika.
•Salah satu metode yang paling mudah untuk penyederhanaan Rangkaian Logika.
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Map Value A B Y0 0 0 A’B’
1 0 1 A’B
2 1 0 AB’
3 1 1 AB
Karnaugh Map 2 Variabel : ( A dan B )
Tabel Kebenaran
A
AB3
A’B1
AB’2
A’B’0
0
1
0
1
B
Model II
AB3
AB’2
A’B1
A’B’0
0
1
0
1
AB
Model I
Map Value
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Desain Pemetaan K- Map 2 Variabel
0
1
0
1
AB
A’ A
B’
B
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Karnaugh Map 2 Variabel : dengan minterm-mintermnya
x y F0 0 10 1 11 0 01 1 0
0 1y
x0
1
1
0 0
1
0 1y
x0
1
x’y’
xy’ xy
x’y
x
y F = Σ(m0,m1) = x’y + x’y’
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0A
1 01
B 0 101
F=AB ′+A’B 0A
1 11
B 0 101
F=AB +A′B +AB ′
0A
1 01
B 0 101
F=AB ′+A’B 0A
1 11
B 0 101 F=A+B
F=AB +A′B +AB ′
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Contoh : 1
1 0
02
01
13
AB 0
Tabel Kebenaran 1
0Map Value A B Y0 0 0 11 0 1 02 1 0 03 1 1 1
1A’B’
A’B’0
01
02
AB3
0
1
AB 0
1AB
Jadi Y = A’B’ + AB
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Contoh : 2
1 0
11
02
03
AB 0
Tabel Kebenaran 1
0Map Value A B Y0 0 0 11 0 1 12 1 0 03 1 1 0
1A’B’
A’B
A’B’0
A’B1
02
03
0
1
AB 0
1
Jadi Y = A’
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• 0 kotak terlingkupi = “0” (Low)• 1 kotak terlingkupi = 2 variabel output• 2 kotak terlingkupi = 1 variabel output• 4 kotak terlingkupi = “1” (High)• Melingkupinya harus posisi “Horisontal “
atau “vertikal” , yang dilingkupi digit ”1” dan jumlah digit “1” yang dilingkupi 2n (1, 2,4,8,16, ...)
Catatan untuk K-Map 2 Variabel
1
110 1
1
0
AB
B’ A
0 1
1
0
A
1
1 AB
A’B’
Y = AB + A’B’
Y = B’ + A
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Contoh 3:Dari Tabel Kebenaran dibawah, tulis persamaan logikanya dengan menggunakan K-map :
11
1
0
1
0
1
AB
A’ A
B’
B
Map Value A B Y0 0 0 11 0 1 12 1 0 03 1 1 1 0
1
0
1
AB
A’ A
B’
B
1
1 1 Jadi Y = A’ + B
![Page 11: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/11.jpg)
Sederhanakan persamaan logika :Y = A + AB’ + A’B Menggunakan K- map :
Contoh 4 :
0
1
0
1
AB
A’ A
B’
B1
1
111
1
0
1
0
1
AB
A’ A
B’
B
Jadi Y = A + B
11
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Karnaugh Map 3 Variabel : ( A, B dan C )
11170116101500141103010210010000
YCBAMap Value
Tabel Kebenaran
A
00 01 11
AB’C5
ABC7
A’BC3
A’B’C1
AB’C’4
ABC’6
A’BC’2
A’B’C’0
Model IIAB
00 01 11
ABC’6
ABC7
AB’C5
AB’C’4
A’BC’2
A’BC 3
A’B’C1
A’B’C’0
10
0
1
10
0
Model IBC
C
Map Value
1
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Tabel Kebenaran
11170116101500141103010210010000
YCBAMap Value
Model III Model IV
ABC7
ABC’6
AB’C5
AB’C’4
A’BC3
A’BC’2
A’B’C1
A’B’C’0
0 1
00
01
10
11
A BC
ABC7
A’BC3
ABC’6
A’BC’2
AB’C5
A’B’C1
AB’C’4
A’B’C’0
0 1
00
01
10
11
B CA
Map Value
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Desain Pemetaan K- Map 3 Variabel
00 01 11
0
1
10ABC
C
C’
BB’
A
A’
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• 0 kotak terlingkupi = “0” (Low)• 1 kotak terlingkupi = 3 variabel output• 2 kotak terlingkupi = 2 variabel output• 4 kotak terlingkupi = 1 variabel output• 8 kotak terlingkupi = “1” (High)• Melingkupinya harus posisi “Horisontal “
atau “vertikal” , yang dilingkupi digit ”1” dan jumlah digit “1” yang dilingkupi 2n (1, 2, 4, 8, ... )
Catatan untuk K- Map3 Variabel
00 01 110
1 111110
00 01 110
1 1111
10BC
A
A
A
B’
00 01 11
0
1 11
110A
+ A’BC’+ A’BCY = AB’C’
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Contoh pengcoveran
00 01AB
C0
1
11 10
C
B
A
cab
00 01 11 1001
0 0 1 00 1 1 1
cout = ab + bc + ac
F(A,B,C) = Σm(0,4,5,7)
G(A,B,C) = 0 0
0 0
1 1
1 1CB
A
1 0
0 0
0 1
1 1CB
A
A
= AC + B’C’
cab
00 01 11 1001
0 0 1 10 0 1 1
f = a
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0A
1 11
00 0101
BC
01 1
111 10
F=AB’C’ +AB ′C +ABC +ABC ′ + A’B’C + A’BC’
A B C F0 0 0 00 0 1 10 1 0 10 1 1 01 0 0 11 0 1 11 1 0 11 1 1 1
+
F=A+B ′C +BC ′0
A
1 11
00 0101
BC
01 1
111 10
F=AB’C’ +AB ′C +ABC +ABC ′ + A’B’C + A’BC’
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Contoh 1 :Tabel Kebenaran Diketahui Tabel Kebenaran seperti disamping :
Cari persamaan logikanya :
1111710116110150001401103001021100110000YCBAMap
Value
00 01 11
0
1
1 1
1 1 1
10BC
A
A’B’
AB
AC
Jadi Y = AC + AB + A’B’
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Contoh 2 :
A
Diketahui Persamaan Boolean :D = A’BC + A’BC’ + ABC’ + ABC + AB’CSederhanakan dengan metode K-map
00 01 11
0
1
1 1
1 1 1
10BC A’BC
A’BC’
ABC’
ABCAB’C
00 01 11
0
1
1 1
1 1 1
10ABC
B
ACJadi D = B + AC
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AB’CD’10
ABCD’14
A’BCD’6
A’B’CD’2
AB’CD11
ABCD15
A’BCD7
A’B’CD3
AB’C’D9
ABC’D13
A’BC’D5
A’B’C’D1
AB’C’D’8
ABC’D’12
A’BC’D’4
A’B’C’D’0
ABCD 00 01 11 10
00
01
11
10
AB’CD’10
AB’CD11
AB’C’D9
AB’C’D’8
ABCD’14
ABCD15
ABC’D13
ABC’D’12
A’BCD’6
A’BCD7
A’BC’D5
A’BC’D’4
A’B’CD’2
A’B’CD3
A’B’C’D1
A’B’C’D’0
CDAB 00 01 11 10
00
01
11
10
Map Value
A B C D Y
0 0 0 0 0
1 0 0 0 12 0 0 1 03 0 0 1 14 0 1 0 05 0 1 0 16 0 1 1 07 0 1 1 18 1 0 0 09 1 0 0 110 1 0 1 011 1 0 1 112 1 1 0 013 1 1 0 114 1 1 1 015 1 1 1 1
Model 1
Model 2
Tabel KebenaranKarnaugh Map 4 Variabel :
( A, B, C dan D )
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Dengan wxyz input
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Desain Pemetaan K- Map 4 Variabel
ABCD 00 01 11 10
00
01
11
10
A
D
C
B
C’
D’
A’
B’
![Page 23: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/23.jpg)
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Catatan untuk K-Map4 Variabel
• 0 kotak terlingkupi = “0” (Low)• 1 kotak terlingkupi = 4 variabel output• 2 kotak terlingkupi = 3 variabel output• 4 kotak terlingkupi = 2 variabel output• 8 kotak terlingkupi = 1 variabel output• 16 kotak terlingkupi = “1” (High)• Melingkupinya harus posisi “Horisontal “
atau “vertikal” , yang dilingkupi digit ”1” dan jumlah digit “1” yang dilingkupi 2n ( 1,2, 4, 8, 16, ... )
111111
11111111
ABCD 00 01 11 10
00
01
11
10
11
1111
ABCD 00 01 11 10
00
01
11
10
AC’
A’
ACD’
B’C’
ABCD’A’BCD
![Page 24: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/24.jpg)
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Contoh pengcoveran :
A' B' D + A' C + B' C D
B C' D' + A C' + A B D'
LT =EQ =GT =
K-map untuk LT K-map untuk GT
0 0
1 0
0 0
0 0D
A
1 1
1 1
0 1
0 0B
C
K-map untuk EQ
1 0
0 1
0 0
0 0D
A
0 0
0 0
1 0
0 1B
C
0 1
0 0
1 1
1 1D
A
0 0
0 0
0 0
1 0B
C
A'B'C'D' + A'BC'D + ABCD + AB'CD’
![Page 25: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/25.jpg)
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Contoh pengcoveran :
F= A′BC ′+A′CD ′+ABC+AB ′C′D ′+ABC ′+AB ′C
CD
0AB
1 10
00 010001
00 1
111 10
11 0
11110
11 1
1F=BC ′+CD ′+ AC+ AD ′
![Page 26: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/26.jpg)
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C + B’D’
Kalau digambarkan dengan system coordinate
Contoh 1• F(A,B,C,D) =
Σm(0,2,3,5,6,7,8,10,11,14,15)F =
D
A
B
AB
CD
0000
1111
1000
01111 0
0 1
0 1
0 0
1 1
1 1
1 1
1 1C
+ A’BD
![Page 27: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/27.jpg)
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111111
1
AB00 01 11 10
00
01
11
10
CDA
D
C
BA’C
ABC’D’
AB’D
Contoh 2 : Diketahui Tabel Kebenaran , cari persamaan logikanya.
Jadi Y = A’C + AB’D + ABC’D’
C12 1
113 111
10
AB00 01 11 10
00
01
11
10
CDA
D
B
Map Value
A B C D Y
0 0 0 0 0 01 0 0 0 1 02 0 0 1 0 13 0 0 1 1 14 0 1 0 0 05 0 1 0 1 06 0 1 1 0 17 0 1 1 1 18 1 0 0 0 09 1 0 0 1 110 1 0 1 0 011 1 0 1 1 112 1 1 0 0 113 1 1 0 1 014 1 1 1 0 015 1 1 1 1 0
![Page 28: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/28.jpg)
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111111
11
WXYZ 00 01 11 10
00
01
11
10
Z
111111
11
WXYZ 00 01 11 10
00
01
11
10
W
Y
Z
Y
Y
X
W
W’X’Y’Z’
YZ
WXZ’
WX’Z
Contoh 3 : Lingkarilah danTulis Persamaan Logikanya.
Jadi M = W’X’Y’Z’ + WXZ’ + WXX’Z + YZ
![Page 29: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/29.jpg)
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Physical Implementasi
A B C D
EQ
° Step 1: Truth table° Step 2: K-map° Step 3: Minimized sum-of-
products° Step 4: Implementasi dengan
gates
K-map untuk EQ
1 0
0 1
0 0
0 0D
A
0 0
0 0
1 0
0 1B
C
![Page 30: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/30.jpg)
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Poin-poin penggunaan K-map
• Tulis persamaanlogika hasil peng-coveran.
• Buat persamaan kebentuk SOP (melaluitabel kebenaran).
• Minterm-mintermnyamasukkan ke k-map ( sesuaikan jumlah kotakatau variabel input).
• Lingkari (pe-ngcover-an) yang benar.
![Page 31: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/31.jpg)
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Don’t Care • Kondisi don’t care merupakan kondisi dimana ada beberapa
kombinasi variable input yang tidak selalu dapat dinyatakan nilaioutputnya.
• Keadaan dimana nilai outputnya tersebut bisa berlogic ‘1’ atauberlogic ‘0’ yang disimbulkan dengan “X” atau “d”.
• Kegunaan dari kondisi don’t care pada penyederhanaan fungsidapat dinyatakan pada fakta bahwa dapat diset dengan logic ‘1’atau logic ‘0’, berdasar kegunaannya untuk format kelompoklogic ‘1’ yang lebih besar.
![Page 32: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/32.jpg)
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Karnaugh maps: don’t cares (cont’d)• f(A,B,C,D) = Σ m(1,3,5,7,9) + d(6,12,13)
– f = A'D + B'C'D tanpa don't cares– f = A’D + C’D dengan don't cares
0 0
1 1
X 0
X 1D
A
1 1
0 X
0 0
0 0B
C
C f0 00 11 01 10 00 11 X100110011
D0101010101010101
10100XX00
A0000000011111111
+
B0000111100001111
+
![Page 33: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/33.jpg)
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Pengcoveran dengan Don’t Cares
0AB
x x1
00 01
00
01
CD
0x 1
011 10
1x 0
111
10
11 1
x
F=A′C′D+B+AC
![Page 34: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/34.jpg)
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Bentuk ilustrasi pengkoveran
0 X
1 1
1 0
1 0D
A
1 0
0 0
1 1
1 1B
C
6 prime implicants:A'B'D, BC', AC, A'C'D, AB, B'CD
minimum cover: AC + BC' + A'B'D
essential
minimum cover: 3 essential implicants
0 0
1 1
1 0
1 0D
A
0 1
0 1
1 1
0 0B
C
5 prime implicants:BD, ABC', ACD, A'BC, A'C'D
minimum cover: 4 essential implicants
essential
minimum cover: ABC’+ACD+A’BC+A’C’D
![Page 35: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/35.jpg)
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Aplikasi K-mapPada Rangkaian Full Adder
Adder
Cin
Cout
SB
A
A B Cin S Cout0 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
+
S = A’B’Cin + A’BCin’ + A’BCin + ABCin
Cout = A’BCin + A B’Cin + ABCin’ + ABCin
= A’BCin + ABCin + AB’Cin + ABCin + ABCin’ + ABCin
= BCin + ACin + AB
= (A’ + A)BCin + (B’ + B)ACin + (Cin’ + Cin)AB= 1·BCin + 1· ACin + 1· AB
Metode Aljabar Boole
![Page 36: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/36.jpg)
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Aplikasi K-mapPada Rangkaian Full Adder
Adder
Cin
Cout
SB
A
A B Cin S Cout0 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
+A
B
Cin
0
0
0
1 1 1
01Pengisiaan digit 1 ke K-map
Karnaugh Map for Cout
![Page 37: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/37.jpg)
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Aplikasi K-mapPada Rangkaian Full Adder
A
B
Cin
Adder
Cin
Cout
SB
A
A B Cin S Cout0 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
+
0
0
0 01
1 1 1Pengcoveran pertama.
Cout = ACinKarnaugh Map untuk Cout
![Page 38: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/38.jpg)
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Aplikasi K-mapPada Rangkaian Full Adder
A
B
Cin
Adder
Cin
Cout
SB
A
A B Cin S Cout0 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
+
0
0
0 01
1 1 1Pengcoveran kedua.
Cout = Acin + ABKarnaugh Map for Cout
![Page 39: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/39.jpg)
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Aplikasi K-mapPada Rangkaian Full Adder
A
B
Cin
Adder
Cin
Cout
SB
A
A B Cin S Cout0 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
+
0
0
0 01
1 1 1
Karnaugh Map untuk Cout
Pengcoveran ketiga (seluruhnya)
Cout = ACin + AB + BCin
![Page 40: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/40.jpg)
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Aplikasi K-mapPada Rangkaian Full Adder
A
B
Cin
Adder
Cin
Cout
SB
A
A B Cin S Cout0 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
+
0
1
1 10
0 1 0
S = A’BCin’Karnaugh Map untuk S
![Page 41: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/41.jpg)
41
Aplikasi K-mapPada Rangkaian Full Adder
A
B
Cin
Adder
Cin
Cout
SB
A
A B Cin S Cout0 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
+
0
1
1 10
0 1 0
S = A’BCin’ + A’B’CinKarnaugh Map untuk S
![Page 42: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/42.jpg)
42
Aplikasi K-mapPada Rangkaian Full Adder
A
B
Cin
Adder
Cin
Cout
SB
A
A B Cin S Cout0 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
+
0
1
1 10
0 1 0
S = A’BCin’ + A’B’Cin + ABCinKarnaugh Map untuk S
![Page 43: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/43.jpg)
43
Aplikasi K-mapPada Rangkaian Full Adder
Coba anda gambar rangkaian diagramnya ?
A
B
Cin
Adder
Cin
Cout
SB
A
A B Cin S Cout0 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
+
0
1
1 10
0 1 0
S = A’BCin’ + A’B’Cin + ABCin + AB’Cin’Karnaugh untuk S Tidak bisa direduksi
![Page 44: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/44.jpg)
44
Latihan Soal 1:
Gambarlah K-map untuk setiap ekspresi logika dibawahserta sederhanakan dengan pengcoveran yang benar :
1. AB + B’C + A’B’2. AC + AC’B + BC + B’C’3. XY + X’Z + Y’Z’4. XY +YZ + XZ +X’Y’
![Page 45: Karnaugh Map](https://reader034.vdocuments.mx/reader034/viewer/2022042607/5571f30e49795947648d6c5e/html5/thumbnails/45.jpg)
45
Latihan Soal 2 :
Gambarlah K-map untuk setiap ekspresi logika dibawahserta sederhanakan dengan pengcoveran yang benar :
1. A(BC’ + C) + B(A + A’C)2. (AC + AC’B). (BC + B’C’)3. Z(XY + X’Z) . Y’Z’(X+ Z)
Catatan : cari minterm-mintermnya dulu (rubah kebentuk SOP)