TABEL KEBENARAN 4-bit INPUT
NO
|
INPUT
|
OUTPUT
|
|||
w
|
x
|
y
|
z
|
F
|
|
0
|
0
|
0
|
0
|
0
|
0
|
1
|
0
|
0
|
0
|
1
|
0
|
2
|
0
|
0
|
1
|
0
|
0
|
3
|
0
|
0
|
1
|
1
|
0
|
4
|
0
|
1
|
0
|
0
|
1
|
5
|
0
|
1
|
0
|
1
|
1
|
6
|
0
|
1
|
1
|
0
|
0
|
7
|
0
|
1
|
1
|
1
|
1
|
8
|
1
|
0
|
0
|
0
|
0
|
9
|
1
|
0
|
0
|
1
|
0
|
10
|
1
|
0
|
1
|
0
|
0
|
11
|
1
|
0
|
1
|
1
|
0
|
12
|
1
|
1
|
0
|
0
|
0
|
13
|
1
|
1
|
0
|
1
|
1
|
14
|
1
|
1
|
1
|
0
|
0
|
15
|
1
|
1
|
1
|
1
|
0
|
Penyederhanaan menggunakan Aljabar Boolean :
F(w,x,y,z) = Σ (4,5,7,13) = w’xy’z’ + w’xy’z + w’xyz + wxy’z
= w’xy’ (z’+z) + w’xyz + wxy’z
= w’xy’ + w’xyz + wxy’z + w’xy’z
= w’xy’ + w’xz (y+y’) + wxy’z
= w’xy’ + w’xz + wxy’z + w’xy’z
= w’xy’ + w’xz + xy’z (w+w’)
= w’xy’ + w’xz + xy’z
Sebelum disederhanakan, persamaan Boolean :
F(w,x,y,z) = Σ (4,5,7,13) = w’xy’z’ + w’xy’z + w’xyz +
wxy’z
Persamaan Boolean hasil penyederhanaan K-Map :
F(w,x,y,z) = Σ (4,5,7,13) = w’xy’ + w’xz + xy’z
Sumber :
