320 lines
9.4 KiB
Markdown
320 lines
9.4 KiB
Markdown
# TD1
|
||
![[CI-EEA21EA5 - Elec_Num - CH1 - Système binaire - TD.pdf]]
|
||
|
||
## Execice SB1. Conversion
|
||
1.1 Conversion base 10 vers base B
|
||
1. $(10)_{10} = (?)_5$
|
||
$10 = 2\times5^1+0\times5^0\implies(10)_{10}=(20)_5$
|
||
2. $(5)_{10} = (?)_3$
|
||
$5 = 1\times3^1+2\times3^0\implies(5)_{10}=(12)_3$
|
||
3. $(123)_{10} = (?)_{16}$
|
||
$123=7\times16^1+11\times16^0\implies(123)_{10}=(7B)_{16}$
|
||
$(123)_{10}=(?)_2$
|
||
$123=1\times2^6+1\times2^5+1\times2^4+1\times2^3+0\times2^2+1\times2^1+1\times2^0$
|
||
$\implies(123)_{10}=(1111011)_2$
|
||
4. $(568)_{10}=(?)_{2}$
|
||
$568=1\times2^9+0\times2^8+0\times2^7+0\times2^6+1\times2^5+1\times2^4+1\times2^3+0\times2^2+0\times2^10\times2^0$
|
||
$\implies(568)_{10}=(1000111000)_{2}$
|
||
|
||
# Nombres positifs et négatifs
|
||
En binaire signé, le nombre 8 $(1000)_2$ devient $(01000)_{BS}$.
|
||
$(110000)_{BS}=-8$.
|
||
|
||
Pour $n$ bits + 1 bit de signe, on peut compter de de $-2^{n-1}$ à $2^{n-1}$.
|
||
|
||
Pb: Un nombre et son négatif ne s'additionnent pas à 0.
|
||
$(9)_{10} + (-9)_{10}=0$, mais $(01001)_2+(11001)_2=(00010)_2=(2)_{10}$
|
||
|
||
Complément à 2
|
||
2 méthodes:
|
||
1. Obtenir le complément à 1 puis ajouter 1.
|
||
Ex: C1 de $(0000 0010)_2\rightarrow(1111 1101)_2$
|
||
C2 =(C1 + 1) $\rightarrow(1111 1101)_2+(0000 0001)_2=(1111 1110)_2$
|
||
2. Conserver tous les bits à partir de la droite jusqu'au premier 1 inclus et inverser tous les bits suivants.
|
||
Ex: C2 de $(0000 0010)_2\rightarrow(\boxed{1111 11}10)_2$
|
||
|
||
## Exercice SB3:
|
||
2. Exprimer -12 sur 5 bits
|
||
$(-12)_{10}=(?)_{BS}$ sur 5 bits
|
||
$(12)_{10}=(01100)_{BS}$
|
||
C2 de $(01100)_{BS}$ = $(10100)_{BS}$
|
||
$(01100)_{BS}+(10100)_{BS}=(00000)_{BS}$
|
||
|
||
3.a)
|
||
$(33)_{10}=(?)_2$
|
||
$(33)_{10}=(100001)_2$ -> Il faut 6 bits
|
||
|
||
b)
|
||
$(100001)_2=(0100001)_{BS}$ -> Il faut 7 bits
|
||
|
||
c)
|
||
$(-33)_{10}=(?)_{BS}$
|
||
C2 de $(0100001)_{BS}$ -> $(1011111)_{BS}$
|
||
|
||
## Exercice SB4
|
||
$(568)_{10}$ = ($0101$ $0110$ $1000$)$_{DCB}$
|
||
|
||
---
|
||
|
||
Soit un mot binaire codé sur $10$ bits.
|
||
$(Val_{Max})_2=0b1111111111$
|
||
$(Val_{Max})_{10}=2^{10}-1=1023$.
|
||
$(Val_{Min})_2=0b0000000000$
|
||
$(Val_{Max})_{10}=0$.
|
||
|
||
Pour un convertisseur analogique-numérique, qui encode $0$ à $5V$ sur 10 bits,
|
||
614 serait obtenu pour environ $3V$.
|
||
|
||
$4V$ donnerait $818.4$, soit $0b1100110010$.
|
||
|
||
|
||
| A | B | A Et B | A Ou B |
|
||
| :-: | :-: | :----: | :----: |
|
||
| 0 | 0 | 0 | 0 |
|
||
| 1 | 0 | 0 | 1 |
|
||
| 0 | 1 | 0 | 1 |
|
||
| 1 | 1 | 1 | 1 |
|
||
|
||
Calculer:
|
||
1) $x+xy=x(1+y)=x$.
|
||
2) $x(x+y)=x$.
|
||
3) $(x+\overline{y})y=xy$
|
||
4) $(x+y)(x+z)=x+xz+yx+yz=x+yz$
|
||
5) $(x+y)(x+\overline{y})=x+x\overline{y}+yx+y\overline{y}=x$
|
||
|
||
# TD 2.
|
||
![[Pasted image 20251013170054.png]]
|
||
|
||
$A=\overline{x}yz+x\overline{y}z+xy\overline{z}+xyz$
|
||
$=\overline{x}yz+x\overline{y}z+xy\overline{z}+xyz+xyz+xyz$
|
||
$=yz(\overline{x}+x)+xz(\overline{y}+y)+xy(\overline{z}+z)$
|
||
$=yz+xz+xy$
|
||
$B=xy+\overline{x}y\overline{z}+yz$
|
||
$=xy(z+\overline{z})+\overline{x}y\overline{z}+yz(x+\overline{x})$
|
||
$=xyz+xy\overline{z}+\overline{x}y\overline{z}+xyz+\overline{x}yz$
|
||
$=xy+\overline{x}y=y(x+\overline{x})=y$
|
||
|
||
![[Pasted image 20251013170215.png]]
|
||
|
||
$C=(x+z)(\overline{x}+y)$
|
||
$=x\overline{x}+xy+z\overline{x}+zy$
|
||
$=0+xy+z\overline{x}+zy$
|
||
$=xy(z+\overline{z})+\overline{x}z(y+\overline{y})+yz(x+\overline{x})$
|
||
$=xyz+xy\overline{z}+\overline{x}yz+\overline{x}\overline{y}z+xyz+\overline{x}yz$
|
||
$=xyz+xy\overline{z}+\overline{x}yz+\overline{x}\overline{y}z$
|
||
$=xy(z+\overline{z})+\overline{x}z(y+\overline{y})$
|
||
$=xy+\overline{x}z$
|
||
$D=(x\overline{y}+z)(x+\overline{y})z$
|
||
$=(xx\overline{y}+x\overline{y}\overline{y}+xz+\overline{y}z)z$
|
||
$=(x\overline{y}+x\overline{y}+xz+\overline{y}z)z$
|
||
$=x\overline{y}z+x\overline{y}z+xzz+\overline{y}zz$
|
||
$=x\overline{y}z+xz+\overline{y}z$
|
||
$=z(x+\overline{y}+x\overline{y})$
|
||
$=z(x+\overline{y})$
|
||
|
||
|
||
Théorème de Morgan
|
||
$\overline{a•b}=\overline{a}+\overline{b}$
|
||
$\overline{a+b}=\overline{a}•\overline{b}$
|
||
|
||
![[Pasted image 20251015091234.png]]
|
||
|
||
![[Pasted image 20251015094501.png]]
|
||
1.
|
||
a)
|
||
Table de vérité de $F$ et $G$:
|
||
|
||
| $x$ | $y$ | $F(x,y)$ | $G(x,y)$ |
|
||
| --- | --- | -------- | -------- |
|
||
| 0 | 0 | 0 | 0 |
|
||
| 0 | 1 | 1 | 1 |
|
||
| 1 | 0 | 1 | 1 |
|
||
| 1 | 1 | 1 | 1 |
|
||
b)
|
||
Par De Morgan, $F(x,y)=x+\overline{x}y$
|
||
$F(x,y)=\overline{\overline{x+\overline{x}y}}$
|
||
$F(x,y)=\overline{\overline{x}•\overline{\overline{x}y}}$
|
||
$F(x,y)=\overline{\overline{x}•(\overline{\overline{x}}+\overline{y})}$
|
||
$F(x,y)=\overline{\overline{x}•(x+\overline{y})}$
|
||
$F(x,y)=\overline{\overline{x}x•\overline{x}\overline{y}}$
|
||
$F(x,y)=\overline{\overline{x}x}+\overline{\overline{x}\overline{y}}$
|
||
$F(x,y)=\overline{\overline{x}\overline{y}}$
|
||
$F(x,y)=\overline{\overline{x}}+\overline{\overline{y}}$
|
||
$F(x,y)=x+y$
|
||
|
||
# TD3
|
||
![[Pasted image 20251015102351.png]]
|
||
![[Pasted image 20251015102357.png]]
|
||
$A=\overline{b}$
|
||
|
||
![[Pasted image 20251015102532.png]]
|
||
$B=\overline{a}$
|
||
|
||
![[Pasted image 20251015102540.png]]
|
||
$C=\overline{c}$
|
||
|
||
![[Pasted image 20251015102551.png]]
|
||
$D=\overline{a}\overline{c}+\overline{a}\overline{b}$
|
||
|
||
![[Pasted image 20251015102611.png]]
|
||
$E=c$
|
||
|
||
![[Pasted image 20251015102624.png]]
|
||
$F=1$
|
||
|
||
![[Pasted image 20251015102631.png]]
|
||
$G=d$
|
||
|
||
![[Pasted image 20251015102641.png]]
|
||
$H=\overline{c}d+ab$
|
||
|
||
![[Pasted image 20251015102647.png]]
|
||
$I=\overline{c}$ j
|
||
|
||
![[Pasted image 20251015110635.png]]
|
||
1.
|
||
|
||
| a/bc | 00 | 01 | 11 | 10 |
|
||
| :--: | --- | --- | --- | --- |
|
||
| 0 | 0 | 0 | 1 | 0 |
|
||
| 1 | 1 | 0 | 1 | 1 |
|
||
$F=a\overline{c}+bc$
|
||
|
||
---
|
||
# Circuits combinatoires
|
||
![[CI-EEA21EA5 - Elec_Num - CH4 - Circuits combinatoires - TD.pdf]]
|
||
|
||
## Exercice CC1. Comparateur binaire 1 bit
|
||
1. À partir de la table de vérité d’un comparateur binaire d’un bit, obtenir les équations logiques des sorties en fonction des entrées.
|
||
|
||
| A | B | Égalité | A>B | A<B |
|
||
| --- | --- | ------- | --- | --- |
|
||
| 0 | 0 | 1 | 0 | 0 |
|
||
| 0 | 1 | 0 | 0 | 1 |
|
||
| 1 | 0 | 0 | 1 | 0 |
|
||
| 1 | 1 | 1 | 0 | 0 |
|
||
$F_{Égalité}=AB+\overline{AB}$
|
||
$F_{A>B}=\overline{A}B$
|
||
$F_{A>B}=A\overline{B}$
|
||
|
||
2. Dessiner le circuit électronique du comparateur.
|
||
![[Drawing 2025-11-19 09.29.12.excalidraw|1000]]
|
||
|
||
## Exercice CC5. Étude de circuits à base de multiplexeurs
|
||
|
||
### 5.1 Premier cas d'étude
|
||
1.
|
||
|
||
| Entrée | A | B | $F_1$ |
|
||
| ------ | --- | --- | ----- |
|
||
| 0 | 0 | 0 | 1 |
|
||
| 1 | 0 | 1 | 0 |
|
||
| 2 | 1 | 0 | 1 |
|
||
| 3 | 1 | 1 | 0 |
|
||
$F_1=\overline{B}$
|
||
|
||
| Entrée | C | D | $F_2$ |
|
||
| ------ | --- | --- | ----- |
|
||
| 0 | 0 | 0 | 0 |
|
||
| 1 | 0 | 1 | e |
|
||
| 2 | 1 | 0 | e |
|
||
| 3 | 1 | 1 | 1 |
|
||
|
||
| e/CD | 00 | 01 | 11 | 10 |
|
||
| ---- | --- | --- | --- | --- |
|
||
| 0 | 0 | 0 | 1 | 0 |
|
||
| 1 | 0 | 1 | 1 | 1 |
|
||
$F_2=CD+e(C+D)$
|
||
|
||
|
||
2.
|
||
$S=\overline{\overline{F_1}F_2}$
|
||
$S=\overline{\overline{\overline{B}}(CD+eC+eD)}$
|
||
|
||
3.
|
||
$S=\overline{B(CD+eC+eD)}$
|
||
$S=\overline{BCD+BeC+BeD}$
|
||
$S=\overline{BCD}•\overline{BeC}•\overline{BeD}$
|
||
|
||
|
||
## Exercice CC8. Machine à café & thé
|
||
3 entrées: $C$, $T$ et $J$.
|
||
4 sorties: $L$ (Led rouge), $Sc$ (Sortie café), $St$ (Sortie thé) et $B$ (bip sonore)
|
||
|
||
| $C$ | $T$ | $J$ | - | $S_c$ | $S_t$ | $L$ | $B$ |
|
||
| --- | --- | --- | --- | ----- | ----- | --- | --- |
|
||
| 0 | 0 | 0 | - | 0 | 0 | 1 | 0 |
|
||
| 0 | 0 | 1 | - | 0 | 0 | 0 | 0 |
|
||
| 0 | 1 | 0 | - | 0 | 0 | 1 | 0 |
|
||
| 0 | 1 | 1 | - | 0 | 1 | 0 | 0 |
|
||
| 1 | 0 | 0 | - | 0 | 0 | 1 | 0 |
|
||
| 1 | 0 | 1 | - | 1 | 0 | 0 | 0 |
|
||
| 1 | 1 | 0 | - | 0 | 0 | 1 | 0 |
|
||
| 1 | 1 | 1 | - | 0 | 0 | 1 | 1 |
|
||
$B=CTJ$
|
||
$L=\overline{J}+CT$
|
||
$S_c=C\overline{T}J$
|
||
$S_t=CT\overline{J}$
|
||
|
||
3.
|
||
![[Drawing 2025-11-21 10.25.20.excalidraw]]
|
||
|
||
4.
|
||
$L=\overline{\overline{L}}=\overline{\overline{\overline{J}+CT}}=\overline{J+\overline{CT}}$
|
||
|
||
## Exercice CC11. Étude d’un circuit combinatoire
|
||
1.
|
||
|
||
| a | b | c | d | - | x | y | z |
|
||
| --- | --- | --- | --- | --- | --- | --- | --- |
|
||
| 0 | 0 | 0 | 0 | - | X | X | X |
|
||
| 0 | 0 | 0 | 1 | - | X | X | X |
|
||
| 0 | 0 | 1 | 0 | - | X | X | X |
|
||
| 0 | 0 | 1 | 1 | - | X | X | X |
|
||
| 0 | 1 | 0 | 0 | - | X | X | X |
|
||
| 0 | 1 | 0 | 1 | - | 0 | 1 | 1 |
|
||
| 0 | 1 | 1 | 0 | - | 1 | 0 | 1 |
|
||
| 0 | 1 | 1 | 1 | - | 1 | 0 | 1 |
|
||
| 1 | 0 | 0 | 0 | - | X | X | X |
|
||
| 1 | 0 | 0 | 1 | - | 1 | 0 | 0 |
|
||
| 1 | 0 | 1 | 0 | - | X | X | X |
|
||
| 1 | 0 | 1 | 1 | - | X | X | X |
|
||
| 1 | 1 | 0 | 0 | - | 0 | 1 | 0 |
|
||
| 1 | 1 | 0 | 1 | - | 0 | 1 | 1 |
|
||
| 1 | 1 | 1 | 0 | - | 1 | 0 | 1 |
|
||
| 1 | 1 | 1 | 1 | - | 1 | 0 | 1 |
|
||
|
||
$X=C+\overline{B}$
|
||
|
||
| ab/cd | 00 | 01 | 11 | 10 |
|
||
| ----- | --- | --- | --- | --- |
|
||
| 00 | X | X | X | X |
|
||
| 01 | X | 0 | 1 | 1 |
|
||
| 11 | 0 | 0 | 1 | 1 |
|
||
| 10 | X | 1 | X | X |
|
||
|
||
$Y=B\overline{C}$
|
||
|
||
| ab/cd | 00 | 01 | 11 | 10 |
|
||
| ----- | --- | --- | --- | --- |
|
||
| 00 | X | X | X | X |
|
||
| 01 | X | 1 | 0 | 0 |
|
||
| 11 | 1 | 1 | 0 | 0 |
|
||
| 10 | 1 | 0 | X | X |
|
||
|
||
$Z=C+BD$
|
||
|
||
| ab/cd | 00 | 01 | 11 | 10 |
|
||
| ----- | --- | --- | --- | --- |
|
||
| 00 | X | X | X | X |
|
||
| 01 | X | 1 | 1 | 1 |
|
||
| 11 | 0 | 1 | 1 | 1 |
|
||
| 10 | X | 0 | X | X |
|
||
|
||
2.
|
||
$Y=\overline{X}$
|
||
|
||
3.
|
||
![[Screenshot 2025-11-27 at 10.05.18.png]]
|
||
|
||
---
|