Boolean Algebra Boolean Algebra & &

Logic CircuitsLogic Circuits

Dr. Ahmed El-BialyDr. Ahmed El-Bialy

Dr. Sahar FawzyDr. Sahar Fawzy

Binary LogicBinary Logic Binary logic deals with Binary logic deals with binarybinary variables that variables that

take on two discrete values and with the take on two discrete values and with the operationsoperations of mathematical logic applied to of mathematical logic applied to these variables.these variables.

RegularAlgebra

BooleanAlgebra

Values NumbersI ntegersReal NumbersComplex Numbers

Zero (0)One (1)

Operators Add, SubtractMultiply, DivideLogarithm, etc.

ANDORComplement

Logic Logic GatesGates

Truth TableTruth Table

A truth table is a table of combinations of the binary variables showing the relationship between the values that the variables take on and the values of the result of the operation

If there are n inputs, there will be 2n rows in the table A B C = A • B

0 0 00 1 01 0 01 1 1

Represented by a dot (.) or by the absence of an operator

AND Operation

The definition of logical (.) is:

0 • 0 = 0 0 • 1 = 0

1 • 0 = 0 1 • 1 = 1

A B C = A • B

0 0 00 1 01 0 01 1 1

A

BC

It looks like multiplication, but it is not. Therefore symbol is used for AND instead of a dot.

OR OperationOR Operation

Represented by a plus (+)

The definition of logical (+) is:

0 + 0 = 0 0 + 1 = 1

1 + 0 = 1 1 + 1 = 1

It looks like addition, but it is not. Therefore symbol is used for OR instead of a plus.

A B C = A + B

0 0 00 1 11 0 11 1 1

A

B

C

Not (Complement) Operation

Represented by a bar over the variable or a prime (’)

The definition of logical (’) is:

0’ = 1

1’ = 0

If X = 1 then X’ = 0

If X = 0 then X’ = 1

X X’

NAND: the complement of the AND operation

A B C =( A • B)’

0 0 10 1 11 0 11 1 0

A

B

C

NOR: the complement of the OR operation

A B C = (A + B)'

0 0 10 1 01 0 01 1 0

A

B

C

XOR: exclusive OR is represented by

Its definition logic is:

0 0 = 0 0 1 = 1

1 0 = 1 1 1 = 0

A

BC = A B =AB’+A’B

A B A B

0 0 00 1 11 0 11 1 0

XNOR: exclusive-NOR is the complement of the exclusive-OR and expressed as:

(A B)’=AB+A’B’

Exclusive-OR Theorem:

X 0 = X X 1 = X'

X X = 0 X X' = 1

X Y = Y X

( X Y) Z = X

( Y Z ) = X

Y Z

( X Y)' = X Y' = X' Y

Odd function: multiple-variable exclusive-OR operation : is one whenever the corresponding binary truth-table values have an odd number of 1’s.

Boolean Expression

Boolean expressions are made up of variables combined by Logical operations .

Examples: [ A B ( C + B’ ) + D] B • E C’

Literals each instance of a variable

This expression has 4 variables and 10 literals:

a’bd + bcd + ac’ + a’d’

Evaluation of Boolean Expression

Boolean expressions can be represented in a truth table which lists all possible combinations of the values of all variables in the expression.

F = A’ + B C

A B C A' B C F = A' + B C

0 0 0 1 0 10 0 1 1 0 10 1 0 1 0 10 1 1 1 1 11 0 0 0 0 01 0 1 0 0 01 1 0 0 0 01 1 1 0 1 1

Boolean Expression

Boolean expressions can be transformed from an algebraic expression into a circuit diagram composed of logic gates.

F = A’ + B C

Boolean Expression

F = X + Y ( X’ +Y’)

XYX'Y'X'+Y'Y(X'+Y')X+Y(X'+Y')

001110 0

011011 1

100110 1

110000 1

X

Y

X’+Y’

Y(X’+Y’)

X+Y(X'+Y')

X+Y

 0

 1

 1

 1

But

We used 5 gates Only one gate

Basic Boolean Algebra TheoremsBasic Boolean Algebra Theorems1. X+0=X2. X+1=13. X+X=X4. X+X’=1

5. X • 1=X6. X • 0=07. X • X=X8. X • X’=0

Associative Law:12. (X + Y) + Z = X + (Y + Z) = X + Y + Z13. ( X • Y ) • Z = X • ( Y • Z) = X • Y • ZDistributive Law:14. X ( Y + Z ) = X Y + X Z

15. X+ X • Y= X16. X ( X + Y) = X17. (X + Y)(X+Z) = X+ Z•Y18. X + X’ • Y = X+Y19. XY + YZ + Y’Z =XY+Z

9. (X’)’=X Commutative Law:10. X • Y = Y • X11. X + Y = Y + X

Basic Boolean Algebra TheoremsBasic Boolean Algebra Theorems

DeMorganDeMorgan’’s Law:s Law:

( X + Y )( X + Y )’’ = X = X’’ Y Y’’ ( X Y ) ( X Y )’’ = X = X’’ + Y + Y’’

Proof:Proof: X Y X' Y' X + Y (X + Y)' X'Y' X Y (XY)' X' + Y'

0 0 1 1 0 1 1 0 1 10 1 1 0 1 0 0 0 1 11 0 0 1 1 0 0 0 1 11 1 0 0 1 0 0 1 0 0

DeMorgan’s Law- one step rule: [ f ( X1, X2, … XN, 0, 1, +, •) ]’ = f ( X1 ’, X2 ’, … XN ’, 1, 0, •, +)

Replace all variables with the inverse

Replace + with • and • with +

Replace 0 with 1 and 1 with 0

Be careful of hierarchy ( )

Methods of Boolean Expression Simplification

• Combine terms

• Eliminate terms

• Eliminate literals

• Add redundant terms if needed

A’B + A’B’C’D’ + ABCD’

A’( B + B’C’D’) + ABCD’

A’( B + C’D’) + ABCD’

A’B + A’C’D’ + ABCD’

A’C’D’ + B( A’ +ACD’)

A’C’D’ + B( A’ +CD’)

A’B + BCD’ + A’C’D’

Standard FormsStandard Forms

Product term : (e.g., XYZ) (Minterms) Sum term : (e.g., X+Y+Z) (Maxterms)

They do not imply arithmetic operations in Boolean algebra; instead, they specify the logical operation AND and OR, respectively

X Y ZX Y ZMinMinMintermsMintermsMaxMaxMaxtermsMaxterms

0 0 00 0 0m0m0XX’’YY’’ZZ’’M0M0X+Y+ZX+Y+Z

0 0 10 0 1m1m1XX’’YY’’ZZM1M1X+Y+ZX+Y+Z’’

0 1 00 1 0m2m2XX’’YZYZ’’M2M2X+YX+Y’’+Z+Z

0 1 10 1 1m3m3XX’’YZYZM3M3X+YX+Y’’+Z+Z’’

1 0 01 0 0m4m4XYXY’’ZZ’’M4M4XX’’+Y+Z+Y+Z

1 0 11 0 1m5m5XYXY’’ZZM5M5XX’’+Y+Z+Y+Z’’

1 1 01 1 0m6m6XYZXYZ’’M6M6XX’’+Y+Y’’+Z+Z

1 1 11 1 1m7m7XYZXYZM7M7XX’’+Y+Y’’+Z+Z’’

A Boolean function F is equal to 1 for each of the following binary combinations of the variables X, Y, and Z: 000, 001,100, 110.

Derive its algebraic expression

Example 1: (Sum of Product SoP)

Truth table and logic sum of minterms

X Y ZX Y ZMinMinMintermsMintermsFF

0 0 00 0 0m0m0XX’’YY’’ZZ’’11

0 0 10 0 1m1m1XX’’YY’’ZZ11

0 1 00 1 0m2m2XX’’YZYZ’’00

0 1 10 1 1m3m3XX’’YZYZ00

1 0 01 0 0m4m4XYXY’’ZZ’’11

1 0 11 0 1m5m5XYXY’’ZZ00

1 1 01 1 0m6m6XYZXYZ’’11

1 1 11 1 1m7m7XYZXYZ00

Sum-of-Product Boolean expressions for F:

F = m0+m1+m4+m6

=X’Y’Z’+X’Y’Z+XY’Z’+XYZ’

Example 2: (Sum of Product SoP)

Design a gate network with the following rules:

1. It has three inputs and one output

2. The output is one when

a) All inputs are zero’s

b) The 3rd is zero & 1st , 2nd are ones

c) The 3rd is one & 1st , 2nd are zeros

d) All inputs are one’s

X Y ZX Y ZMinMinMintermsMintermsFF

0 0 00 0 0m0m0XX’’YY’’ZZ’’11

0 0 10 0 1m1m1XX’’YY’’ZZ00

0 1 00 1 0m2m2XX’’YZYZ’’00

0 1 10 1 1m3m3XX’’YZYZ11

1 0 01 0 0m4m4XYXY’’ZZ’’11

1 0 11 0 1m5m5XYXY’’ZZ00

1 1 01 1 0m6m6XYZXYZ’’00

1 1 11 1 1m7m7XYZXYZ11

Sum-of-Product Boolean expressions for F:S = m0+m3+m4+m7

=X’Y’Z’+X’YZ+XY’Z’+XYZ

Example 3:

Design a gate network with the following rules:

1. It has Two inputs and Two outputs

2. The outputs S & C are:

a) If the two inputs are zero’s the two outputs are zeros

b) If one of the inputs is one the S output is one and the C is zero

c) If the two inputs are one’s the S output is zero the C is one

Sum-of-Product Boolean expressionsSum-of-Product Boolean expressions S =X’Y+XY’ S =X’Y+XY’ XOR XOR C =XYC =XY AND AND

Half AdderHalf Adder

Half AdderHalf Adder

00001111

00110011

SS00111100

CC00000011

+++ +

Design a gate network with the following specs:

It has three inputs (A, B, Ci) and two outputs (S, C)

1. The outputs S & C are:a) If the only one input is 1 S=1 C=0

b) If two of the inputs are 1 S=0 C=1

c) If all of the three inputs are 1 S=1 C=1

Example 4

S=A’B’C + AB’C’+ABC+A’BC’S=A’B’C + AB’C’+ABC+A’BC’

C= A’BC+AB’C+ABC+ABC’C= A’BC+AB’C+ABC+ABC’

Map Simplification Karnaugh map: a diagram made up of squares, with

each square representing one minterm of the function.

be done systematically

Simpler to find the minimum solution

Two variables K-map:

A

B 0 1

0

1A=O, B=0

A=O, B=1

A=1, B=0

A=1, B=1

Map Simplification

A B F

0 0 10 1 11 0 01 1 0

F = A’B’ + A’B

F = A’

A

B 0 1

0

1

1

1

0

0

1

1

A = 0B = 0 or 1

0

0

1

1

Example of a two-variable K-map:

Map Simplification

Three variables K-map:

The order of BC variables are 00, 01, 11 and 10. They are Gray code (i.e. only one bit is changed each time).

AB C

0

1

00 11 1001

m0 m1 m3 m2

m4 m5 m7 m6

Each minterm corresponds to a location on K-map: m0

…m7

Example of a three-variable K-map:

Map Simplification

F (A,B,C) = m ( 2, 3, 5, 7 )

11 1

1

1 1

11

11

AB C

0

1

00 11 1001

0 0

0 01

11

1

A B C F

0 0 0 00 0 1 00 1 0 10 1 1 11 0 0 01 0 1 11 1 0 01 1 1 1

AB C

0

1

00 11 1001

0 0

0 01

11

1

Map Simplification

A = 0, B = 1, C= don’t care(0 or 1)

F = A’B

C = 1, B = 1, A= don’t care

A = 1, C = 1, B= don’t care

F = A’B+CB +AC

Map Simplification

ABCD

00 11 1001

m0 m1 m3 m2

m4 m5 m7 m6

00

11

10

01

m12 m13 m15 m14

m8 m9 m11 m10

Four variables K-map :

- Note that the orders of AB and CD follow Gray code.

Each minterm corresponds to a location on K-map: m0

…m15

Example of a four-variable K-map:

Map Simplification

Simplify function F = m ( 1, 2, 3, 5, 7, 10, 14, 15 )

ABCD

00 11 1001

00

11

10

01

0

0 0

0 0

0 0 0

1 1 1

1 1

1 1

1

F =A’DF =A’D+A’B’CF =A’D+A’B’C+ABCF =A’D+A’B’C+ABC+ACD’

or

B’CD’

K-Map with “Don’t Care” Values :

- “Don’t care conditions” means unspecified

minterms of a function, which is marked with

“X”

Map Simplification

F (A,B,C,D) = m ( 1, 5, 6, 10, 11, 14 ) + d ( 0, 7, 9, 15 )= AC + BC + A’C’D

ABCD

00 1101

00

11

10

01

X

0 1

0 0

0 X 1

1 0 0

1 X

X 1

1

10

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Combinational CircuitsCombinational Circuits