Solutions prototypes Rattrapage CRI |
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1.
N= (73.36)10
On convertit d'abord la partie entière (73)10 par les divisions successives :
73 | |2 | ||||||
-- | |||||||
1 | 36 | |2 | |||||
-- | |||||||
0 | 18 | |2 | |||||
-- | |||||||
0 | 9 | |2 | |||||
-- | |||||||
1 | 4 | |2 | |||||
-- | |||||||
0 | 2 | |2 | |||||
-- | |||||||
0 | 1 | |2 | |||||
-- | |||||||
1 | 0 |
On convertit ensuite la partie décimale (0.36)10 par les multiplications successives (3 chiffres après la virgule) :
.36 | ||
x 2 | ||
0 | = .72 | |
x 2 | ||
1 | = .44 | |
x 2 | ||
0 | = .88 |
d'où N= (73.36)10 = (1001001.010)2
En octal, on fait des regroupement par 3 bits :
d'où N= (73.36)10 = (1 001 001.010)2 = (111.2)8
En hexadécimal, on fait des regroupement par 4 bits :
d'où N= (73.36)10 = (0100 1001.0100)2 = (49.4)16
2.
On a (-73 - 35)10 = (-73)10 + (-35)10
(73)10 = (01001001)2
donc (-73)10 = (10110111) 2C2
(35)10 = (00100011)2
donc (-35)10 = (11011101) 2C2
(-73 - 35)10 =
(1 0 1 1 0 1 1 1) 2C2
+ (1 1 0 1 1 1 0 1) 2C2
= _____________
(1 0 0 1 0 1 0 0) 2C2 et on ignore la retenue.
(10010100) 2C2 est négatif. On cherche sa valeur positive en binaire et en décimale :
(10010100) 2C2 = - (01101100) 2C2 = - (4+8+32+64)10 = - (108)10
3.
On a G = (100011101)gray
On cherche d'abord la valeur binaire de G. Il y a plusieurs méthodes et on peut utiliser la suivante :
Si (gngn-1...g 1 g0)gray = (bnbn-1...b 1 b0)2 alors
bn = gn
bn-1 = bn+ gn-1
bn-2 = bn-1+ gn-2
----
b1 = b2+ g1
b0 = b1+ g0
(sachant que 0+0 =0, 1+0= 1, 1+1=0)
Donc (100011101)gray = (111101001)2 = (1+8+32+64+128+256)10 = (489)10 = (0100 1000 1001)BCD
4.
N1=-0.66
On convertit N1 en décimal par la méthode des multiplications successives. On arrête la multiplication au bout d'environ 23 opérations (taille de la mantisse) ou lorsqu'on tombe sur une séquence répétitive : :
N° | .66 | ||
x 2 | |||
1 | 1 | = .32 | |
x 2 | |||
2 | 0 | = .64 | |
x 2 | |||
3 | 1 | = .28 | |
x 2 | |||
4 | 0 | = .56 | |
x 2 | |||
5 | 1 | = .12 | |
x 2 | |||
6 | 0 | = .24 | |
x 2 | |||
7 | 0 | = .48 | |
x 2 | |||
8 | 0 | = .96 | |
x 2 | |||
9 | 1 | = .92 | |
x 2 | |||
10 | 1 | = .84 | |
x 2 | |||
11 | 1 | = .68 | |
x 2 | |||
12 | 1 | = .36 | |
x 2 | |||
13 | 0 | = .72 | |
x 2 | |||
14 | 1 | = .44 | |
x 2 | |||
15 | 0 | = .88 | |
x 2 | |||
16 | 0 | = .76 | |
x 2 | |||
17 | 1 | = .52 | |
x 2 | |||
18 | 1 | = .04 | |
x 2 | |||
19 | 0 | = .08 | |
x 2 | |||
20 | 0 | = .16 | |
x 2 | |||
21 | 0 | = .32 | |
x 2 | |||
22 | 0 | = .64 | |
x 2 | |||
23 | 1 | = .28 | |
x 2 | |||
24 | 0 | =.56 |
d'où N1= -(0.101010001111010111000010)2 = -(1.01010001111010111000010)2 x 2-1
D'où :
S1 = 1
M1=
(010 1000 1111 0101 1100 0010)2
E1= (-1 + 127)10 = (126)10 = (0111 1110)2
N1 = (1 011 1111 0 010 1000 1111 0101 1100 0010)2VF = (B F 2 8 F 5 C 2)16VF
N2= 0.47
N° | 0.47 | ||
x 2 | |||
1 | 0 | = 0.94 | |
x 2 | |||
2 | 1 | = 0.88 | |
x 2 | |||
3 | 1 | = 0.76 | |
x 2 | |||
4 | 1 | = 0.52 | |
x 2 | |||
5 | 1 | = 0.04 | |
x 2 | |||
6 | 0 | = 0.08 | |
x 2 | |||
7 | 0 | = 0.16 | |
x 2 | |||
8 | 0 | = 0.32 | |
x 2 | |||
9 | 0 | = 0.64 | |
x 2 | |||
10 | 1 | = 0.28 | |
x 2 | |||
11 | 0 | = 0.56 | |
x 2 | |||
12 | 1 | = 0.12 | |
x 2 | |||
13 | 0 | = 0.24 | |
x 2 | |||
14 | 0 | = 0.48 | |
x 2 | |||
15 | 0 | = 0.96 | |
x 2 | |||
16 | 1 | = 0.92 | |
x 2 | |||
17 | 1 | = 0.84 | |
x 2 | |||
18 | 1 | = 0.68 | |
x 2 | |||
19 | 1 | = 0.36 | |
x 2 | |||
20 | 0 | = 0.72 | |
x 2 | |||
21 | 1 | = 0.44 | |
x 2 | |||
22 | 0 | = 0.88 | |
x 2 | |||
23 | 1 | = 0.76 | |
x 2 | |||
24 | 1 | = 0.52 | |
x 2 | |||
25 | 1 | = 0.04 |
N2 = (0.0111100001010001111010111)2 = (1.11100001010001111010111)2 x 2-2
D'où :
S2 = 0
M2=
(111 0000 1010 0011 1101 0111)2
E2= (-2 + 127)10 = (125)10 = (011 1110 1)2
N2 = (0 011 1110 1 111 0000 1010 0011 1101 0111)2VF = (3 E F 0 A 3 D 7)16VF
Il faut d'abord uniformiser les exposants :
N1= -(1.01010001111010111000010)2 x 2-1
N2 = (1.11100001010001111010111)2 x 2-2= (0.111100001010001111010111)2 x 2-1
On a N1-N2 = (-N1) + (-N2) = - (|N1| + |N2| ). On transforme la soustraction en addition et le signe du résultat sera négatif :
On additionne les 2 mantisses :
(1.0101 0001 1110 1011 1000 0100)2
+ (0.1111 0000 1010 0011 1101 0111)2
= _____________________________
(10.0100 0010 1000 1111 0101 1011)2
N1 - N2 = - (10.0100 0010 1000 1111 0101 1011)2 x 2-1
= - (1.0010 0001 0100 0111 1010 1101 1)2
= - (1 + 2-3+ ...)10 = (- 1 - 1/8)10 = -(1.13)10
S = 1
M=
001 0000 1010 0011 1101 0110
E= (0 +127)10 = (127)10 = (011 1111 1)2
N1-N2 = (1 011 1111 1 001 0000 1010 0011 1101 0110)2VF= (BF90A3D6)16VF
1.
_ _ _
= a . (b + b) + a.b
_ _
= a . 1 + a.b
_ _
= a + a.b
_ _
= (a + a). (a + b)
_
= 1 . ( a + b )
_
= a + b
_ _ _
= a + a.b + a.c.(1 + b)
_ _
= a + a.b + a.c
_ _
= a + b + a.c
_ _
= a + b + c
2.
F3 = F1 . F2
_ _ _
= (a + b ). (a + b + c)
_ _ _ _ _ _ _
= a.a + a.b + a.c + b.a + b.b + b.c
_ _ _ _ _
= a.b + a.c + b.a + b.c
Tableau de Karnaught :
\ab cd \ |
00 | 01 | 11 | 10 | |
---|---|---|---|---|---|
00 | _ _ |
0 | a.b _ a.c 1 |
_ a.c _ _ b.c 1 |
|
01 | _ _ a.b _ _ b.c 1 |
0 | a.b _ a.c 1 |
_ a.c _ _ b.c 1 |
|
11 | _ _ a.b 1 |
0 | a.b 1 |
0 | |
10 | _ _ a.b 1 |
0 | a.b 1 |
0 |
_ _ _
F3 = a.b + a.b + a.c
_ _ _
F3 = ( a + b ).(a + b + c)
3.
______________
______________
_ _ _
F3 = ( a + b ).(a + b + c)
______________
_____ ________
_ _ _
=
( a + b )+(a + b + c)
______________
______________
_ _ _
F3 = a.b + a.b + a.c
____________
___ ___ ___
_ _ _
= a.b . a.b . a.c
1.
a | b | c | d | S |
---|---|---|---|---|
0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 0 | 0 |
0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 1 |
\cd ab \ |
00 | 01 | 11 | 10 | ||||
---|---|---|---|---|---|---|---|---|
00 | 0 | 0 | 0 | 0 | ||||
01 | 0 | 0 | 1 | 0 | ||||
11 | 1 | 1 | 1 | 1 | ||||
10 | 0 | 0 | 1 | 1 |
S = a.b + b.c.d + a.c
\cd ab \ |
00 | 01 | 11 | 10 | ||
---|---|---|---|---|---|---|
00 | 0 | 0 | ||||
0 | 0 | |||||
01 | 0 | 0 | 1 | 0 | ||
11 | 1 | 1 | 1 | 1 | ||
10 | 0 | 0 | 1 | 1 |
S = ( a + b).( a + c ).( a + d ).( b + c )
_____________
_____________
S = a.b + b.c.d + a.c
_____________
__ ____ ___
S = a.b . b.c.d . a.c
_____________
______
___
__ ___ ___
S = a.b . b.c . d . a.c