Solution Exercice 1 / Série 3
a) Table de vérité pour n= 3 bits
Dec |
Binaire |
Gray |
|
x |
y |
z |
a |
b |
c |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
0 |
1 |
2 |
0 |
1 |
0 |
0 |
1 |
1 |
3 |
0 |
1 |
1 |
0 |
1 |
0 |
4 |
1 |
0 |
0 |
1 |
1 |
0 |
5 |
1 |
0 |
1 |
1 |
1 |
1 |
6 |
1 |
1 |
0 |
1 |
0 |
1 |
7 |
1 |
1 |
1 |
1 |
0 |
0 |
Tableaux de Karnaught :
a : |
\z
xy\ |
0 |
1 |
00 |
0 |
0 |
01 |
0 |
0 |
11 |
1 |
1 |
10 |
1 |
1 |
|
a = x |
b : |
\z
xy\ |
0 |
1 |
00 |
0 |
0 |
01 |
1 |
1 |
11 |
0 |
0 |
10 |
1 |
1 |
|
_ _
b= x.y + x.y = x ⊕ y |
c : |
\z
xy\ |
0 |
1 |
00 |
0 |
1 |
01 |
1 |
0 |
11 |
1 |
0 |
10 |
0 |
1 |
|
_ _
c= y.z + y. z = y ⊕z |
|
b) Table de vérité pour n= 3 bits
Dec |
Gray |
Binaire |
|
a |
b |
c |
x |
y |
z |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
0 |
1 |
3 |
0 |
1 |
0 |
0 |
1 |
1 |
2 |
0 |
1 |
1 |
0 |
1 |
0 |
7 |
1 |
0 |
0 |
1 |
1 |
1 |
6 |
1 |
0 |
1 |
1 |
1 |
0 |
4 |
1 |
1 |
0 |
1 |
0 |
0 |
5 |
1 |
1 |
1 |
1 |
0 |
1 |
Tableaux de Karnaught :
x : |
\c
ab\ |
0 |
1 |
00 |
0 |
0 |
01 |
0 |
0 |
11 |
1 |
1 |
10 |
1 |
1 |
|
x= a |
y : |
\c
ab\ |
0 |
1 |
00 |
0 |
0 |
01 |
1 |
1 |
11 |
0 |
0 |
10 |
1 |
1 |
|
_ _
y= a.b + a.b = a ⊕b |
z : |
\c
ab\ |
0 |
1 |
00 |
0 |
1 |
01 |
1 |
0 |
11 |
0 |
1 |
10 |
1 |
0 |
|
z= a ⊕b ⊕c |
|