# 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.

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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 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 Algebra INVERTER XX’ If X=0 then X’=1 If X=1 then X’=0 OR ABAB C=A+B If A=1 OR B=1 then C=1 otherwise C=0 ABAB C=A·B 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 CDCD C+D [A(C+D)]’ [A(C+D)]’+BE BEBE BE A A(C+D)

5 Basic Theorems: Operations with 0 and 1 X+0 = X X0X0 C=X X+1 = 1 X1X1 C=1 X0X0 C=0 X·0 = 0 X1X1 C=X X·1 = X

6 Basic Theorems: Idempotent Laws X+X = X XXXX C=X XXXX X·X = X

7 Basic Theorems: Involution Law X (X’)’=X B C=X

8 Basic Theorems: Laws of Complementarity X+X’ = 1 X X’ C=1 X X’ C=0 X·X’ = 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) XYXY Z C YZYZ X C

11 Associative Law (XY)Z = X(YZ) XYXY Z C YZYZ X C

12 First Distributive Law X(Y+Z) = XY+XZ

13 First Distributive Law X(Y+Z) = XY+XZ

14 First Distributive Law X(Y+Z) = XY+XZ

15 First Distributive Law X(Y+Z) = XY+XZ

16 First Distributive Law X(Y+Z) = XY+XZ

17 Second Distributive Law X+YZ = (X+Y)(X+Z)

18 Second Distributive Law X+YZ = (X+Y)(X+Z)

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 = XY XY + 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’ = X XY + XY’ = X(Y + Y’) = X·1 = X X + XY = X X(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) = X X(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 + ST W = (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

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