Boolean Logic ITI 1121 N. El Kadri. 2 What is a switching network? Switching Network X1X1 XmXm X2X2 Z1Z1 ZmZm Z2Z2 Combinatorial Network: A stateless

Boolean Logic

ITI 1121

N. El Kadri

2

What is a switching network?

SwitchingNetwork

X1

Xm

X2

Z1

Zm

Z2

Combinatorial Network: A stateless network. The output is completely determined by the values of

the input.Sequential Network: The network stores an internal

state. The output is determined by the input,and by the internal state.

3

Logic Functions: Boolean AlgebraINVERTER

X X’

X X’0 11 0

If X=0 then X’=1If X=1 then X’=0

OR

AB

C=A+B

A B C0 0 00 1 11 0 11 1 1

If A=1 OR B=1 then C=1 otherwise C=0

AB

C=A·B

A B C0 0 00 1 01 0 01 1 1

If A=1 AND B=1 then C=1 otherwise C=0

AND

4

Boolean expressions and logic circuits

Any Boolean expression can be implemented as a logic circuit.

X = [A(C+D)]’+BE

CD

C+D[A(C+D)]’ [A(C+D)]’+BE

BE

BE

AA(C+D)

5

Basic Theorems: Operations with 0 and 1

X+0 = X

X0

C=XX 0 C0 0 01 0 1

X+1 = 1

X1

C=1X 1 C0 1 11 1 1

X0

C=0

X·0 = 0

X 0 C0 0 01 0 0

X1

C=X

X·1 = X

X 1 C0 1 01 1 1

6

Basic Theorems:Idempotent Laws

X+X = X

XX

C=XX X C0 0 01 1 1

XX

C=X

X·X = X

X X C0 0 01 1 1

7

Basic Theorems: Involution Law

X

(X’)’=X

BC=X

X B C0 1 01 0 1

8

Basic Theorems:Laws of Complementarity

X+X’ = 1

XX’

C=1X X’ C 0 1 1 1 0 1

XX’

C=0

X·X’ = 0

X X’ C0 1 01 0 0

9

Expression Simplification using the Basic Theorems

X can be an arbitrarily complex expression.

Simplify the following boolean expressions as much as you can using the basic theorems.

(AB’ + D)E + 1 =(AB’ + D)(AB’ + D)’ =(AB + CD) + (CD + A) + (AB + CD)’ =

(AB’ + D)E + 1 = 1(AB’ + D)(AB’ + D)’ = 0(AB + CD) + (CD + A) + (AB + CD)’ = 1

10

Associative Law(X+Y)+Z = X+(Y+Z)

X Y Z X+Y (X+Y)+Z Y+Z X+(Y+Z)0 0 0 0 0 0 00 0 1 0 1 1 10 1 0 1 1 1 10 1 1 1 1 1 11 0 0 1 1 0 11 0 1 1 1 1 11 1 0 1 1 1 11 1 1 1 1 1 1

XY

ZC

YZ

XC

11

Associative Law

(XY)Z = X(YZ)

X Y Z XY (XY)Z YZ X(YZ)0 0 0 0 0 0 00 0 1 0 0 0 00 1 0 0 0 0 00 1 1 1 0 1 01 0 0 0 0 0 01 0 1 0 0 0 01 1 0 1 0 0 01 1 1 1 1 1 1

XY

ZC

YZ

XC

12

First Distributive LawX(Y+Z) = XY+XZ

X Y Z Y+Z X(Y+Z) XY XZ XY+XZ 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 1 1 0 1 1 1 1 0 1 1 1 0 1 1 1 1 1 1 1 1 1

13



14



15



16



17

Second Distributive LawX+YZ = (X+Y)(X+Z)

X Y Z YZ X+YZ X+Y X+Z (X+Y)(X+Z)0 0 0 0 0 0 0 00 0 1 0 0 0 1 00 1 0 0 0 1 0 00 1 1 1 1 1 1 11 0 0 0 1 1 1 11 0 1 0 1 1 1 11 1 0 0 1 1 1 11 1 1 1 1 1 1 1

18

Second Distributive LawX+YZ = (X+Y)(X+Z)

X Y Z YZ X+YZ X+Y X+Z (X+Y)(X+Z) 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 0 1 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1

19

Second Distributive Law(A different proof)

(X + Y)(X + Z) = X(X + Z) + Y(X + Z) (using the first distributive law)

= XX + XZ + YX + YZ (using the first distributive law)

= X + XZ + YX + YZ (using the idempotent law)

= X·1 + XZ + YX + YZ (using the operation with 1 law)

= X(1 + Z + Y) + YZ (using the first distributive law)

= X·1 + YZ (using the operation with 1 law)

= X + YZ (using the operation with 1 law)

20

Simplification Theorems

(X + Y’)Y = XYXY + Y’Y = XY + 0 = XY

XY’ + Y = X + Y(using the second distributive law)XY’ + Y = Y + XY’ = (Y + X)(Y + Y’) = (Y + X)·1 = X + Y

XY + XY’ = XXY + XY’ = X(Y + Y’) = X·1 = X

X + XY = XX(1 + Y) = X·1 = X

(X + Y)(X + Y’) = X(X + Y)(X + Y’) = XX + XY’ + YX + YY’ = X + X(Y’ + Y) + 0 = X + X·1 = X

X(X + Y) = XX(X + Y) = XX + XY = X·1 + XY = X(1 + Y) = X·1 = X

21

Examples

Simplify the following expressions:

W = [M + N’P + (R + ST)’][M + N’P + R + ST]

W = M + N’P

X = M + N’P Y = R + STW = (X + Y’)(X + Y)

W = XX + XY + Y’X + Y’Y

W = X·1 + XY + XY’ + 0

W = X + X(Y + Y’) = X + X·1 = X

Documents

Boolean Logic ITI 1121 N. El Kadri. 2 What is a switching network? Switching Network X1X1 XmXm X2X2 Z1Z1 ZmZm Z2Z2 Combinatorial Network: A stateless