Digital Logic Design

Week 4

Boolean algebra

Boolean algebra

• Boolean algebra• Laws and rules• De Morgan’s theorem• Analysis of logic circuits• Standard forms• Project 1 preparation

Boolean algebra

George BooleBritish mathematician and philosopher.An Investigation of the Laws of Thought on Which are Founded the Mathematical Theories of Logic and Probabilities (1854)

Claude ShannonAmerican electrical engineer.First to apply Boole’s work to design of logic circuits (1937)

Boolean algebra is the mathematics of digital systems deals with values 0 and 1, and just three operations: addition,

multiplication, and complement

Boolean operations

Boolean addition differs from binary addition! There is no carry in Boolean addition

0 + 0 = 00 + 1 = 11 + 0 = 11 + 1 = 1

0 · 0 = 00 · 1 = 01 · 0 = 01 · 1 = 1

0’ = 11’ = 0

Booleanaddition

Booleanmultiplication Complement

Boolean operations

0 + 0 = 00 + 1 = 11 + 0 = 11 + 1 = 1

0 · 0 = 00 · 1 = 01 · 0 = 01 · 1 = 1

0’ = 11’ = 0

Booleanaddition

Booleanmultiplication Complement

ANDOR NOT

Sums and products

Sum terms are generated by OR gates

Examples of sum terms:A+B A+B’ A+B+C’

Variables are symbols representing Boolean values 0,1 A, B and C are variables

Literals are variables or complement of variables A, B’ and C’ are literals

Sums and products

• Boolean addition is the same as the OR function, so…– sum term is 1 if one or more of the literals are 1– sum term is 0 only if each literal is 0

Example: Determine the values of the variablesA,B,C,D that make the sum term A+B’+C+D’ = 0Literals A, B’, C and D’ must all equal zero if sum=0A = B’ = C = D’ = 0 A=0, B=1, C=0, D=1

Sums and products

Product terms are generated by AND gates

Examples of product terms:AB AB’ ABC’

Example: Determine the values of the variablesA,B,C,D that make the product term AB’CD’ = 1Literals A, B’, C and D’ must all equal 1 if product=1A = B’ = C = D’ = 1 A=1, B=0, C=1, D=0

A

BX

Remember: A·B often written simply as AB

Boolean algebra

• Boolean algebra• Laws and rules• De Morgan’s theorem• Analysis of logic circuits• Standard forms• Project 1 preparation

Commutative laws

For addition: A + B = B + AThe order in which variables are ORed makes no difference

For multiplication: A · B = B · AThe order in which variables are ANDed makes no difference

Associative laws

For addition: A + (B + C) = (A + B) + C

For multiplication: A(BC) = (AB)C

Distributive law

Expresses the process of factoring

A(B + C) = AB + AC

Rules of Boolean algebra Rules 1–9 from definitions of AND, OR and NOT Rules 10–12 derived from simpler rules and three laws

12. (A + B)(A + C) = A + BC

1. A + 0 = A2. A + 1 = 1

3. A · 0 = 0

4. A · 1 = A

5. A + A = A

7. A · A = A

6. A + A = 1

8. A · A = 0

9. A = A=

10. A + AB = A

11. A + AB = A + B

A couple of the simpler rulesA + 0 = A– A variable ORed with 0 is always equal to the variable

A · 1 = A A variable ANDed with 1 is always equal to the variable

Proving rules 10–12• Rule 10: A + AB = A

A + AB = A·1 + AB Rule 4: A·1 = A= A(1 + B) Factoring (Distributive

Law)= A·1 Rule 2: 1+B = 1= A Rule 4: A·1 = A

Complete the following truth table, and hence prove A+AB=A

A B AB A+AB-------------------------------------0 00 11 01 1

Proving rules 10–12• Rule 11: A + A’B = A+B

A + A’B = (A + AB) + A’B Rule 10: A = A+AB= (AA + AB) + A’B Rule 7: A = AA= AA + AB + AA’ + A’B Rule 8: adding

AA’=0= (A+A’)(A+B) Factoring= 1·(A+B) Rule 6: A+A’=1= A+B Rule 4: drop the 1

Proving rules 10–12• Rule 12: (A+B)(A+C) = A + BC

Complete the following truth table & prove (A+B)(A+C) = A+BC

A B C A+B A+C (A+B)(A+C) BCA+BC

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

Duality Principle

• Any Boolean equality remains valid if we exchange:

1 ↔ 0AND ↔ OR

– This explains why rules 1–8 appear in pairs

Example: Apply the duality principle to Rule 1: A+0 = A

A + 0 = A → A · 1 = A (Rule 4)

Boolean algebra

• Boolean algebra• Laws and rules• De Morgan’s theorem• Analysis of logic circuits• Standard forms• Project 1 preparation

De Morgan’s Theorem

A · B = A + B

A + BA

BAB

A

B

NAND Negative-OR

A + B = A · B

ABA

BA + B

A

B

NOR Negative-AND

Extends to more than two variables Example:

Each variable can also represent a combination of other variables Example:

DCBAABCD

BCACABBCACAB

Applying De Morgan’s Theorem

BCACABBCACAB

BCACAB

CBACBA

CABACBCA

BACBCA

Applying De Morgan’s Theorem

Use De Morgan’s theorem to simplify the expression

FEDCBA

Boolean algebra

• Boolean algebra• Laws and rules• De Morgan’s theorem• Analysis of logic circuits• Standard forms• Project 1 preparation

Boolean expression for a logic circuit

• Begin at the circuit inputs and work towards final output

• Write expression for each gate output

Constructing a truth table for a logic circuit

00

0

0

0

0

0

Truth tableA B C D A(B+CD)---------------------------------------------0 0 0 0 00 0 0 1 ......1 1 1 1 ...

I do NOT recommend using this method to

construct the truth table!It’s far too slow

Constructing a truth table for a logic circuit

A·(B+C·D)=1 Equals 1 only if A=1 and B+C·D=1 So expression equals 0 for top half of truth table,

where A=0 B+C·D=1

Equals 1 when B=1 or C·D=1 or both B=1 and C·D=0 → B C D=100,101,110 B=0 and C·D=1 → B C D=011 B=1 and C·D=1 → B C D=111

We can now fill in the truth table quickly

Use of “standard forms” will simplify

the creation of a truth table from logic expression

A B C D A·(B+C·D)-------------------------------------------------------------0 0 0 0 00 0 0 1 00 0 1 0 00 0 1 1 00 1 0 0 00 1 0 1 00 1 1 0 00 1 1 1 01 0 0 0 01 0 0 1 01 0 1 0 01 0 1 1 11 1 0 0 11 1 0 1 11 1 1 0 11 1 1 1 1

A=0

B=0 and C·D=1

B=1 and C·D=0

B=1 and C·D=1

Boolean algebra

• Boolean algebra• Laws and rules• De Morgan’s theorem• Analysis of logic circuits• Standard forms• Project 1 preparation

Standard forms

All Boolean expressions can be converted into either of two standard forms:

Sum-of-products (SOP) form Product-of-sums (POS) form

Makes the following much easier and systematic: Evaluation Simplification Implementation

Truth table from Boolean expression

Fewer gates for same function

Logic gates to implement expression

SOP this week; POS in week 5

Sum-of-products (SOP)Sum-of-products (SOP) expressions are product terms summed by Boolean addition

Examples:

The following expressions are not in SOP form:

ABCAB DCBCDEABC

ACCBABA

ABCAB CDBA

Remember: “summed” by Boolean addition means OR-ed together

AND/OR implementation of SOPSOP expressions simply require ORing outputs of AND gates

Domain of a Boolean expression is the set of variables in the expressionExample: domain of the expression for X in the

figure above is A, B, C, D

Conversion to SOP form

Convert the following expression to SOP form:

CBA

Any logic expression can be changed to SOP form simply by applying Boolean algebra

CBCACBACBA

Standard SOP formStandard SOP expressionAll variables in the domain appear in each product term in the expression

Example:

The following expression is not in standard SOP form:

Why not? Missing or in first term, and missing or in second term

DCABDCBACDBA

DCBADBACBA D DC C

Product terms involving all

variables in the domain are called

minterms

Conversion to standard SOP formProduct terms missing variables can always be expanded to standard form using Boolean algebra rules and

Example:ABCBAX

1XX X1X

First term is missing the variable C

ABCCCBA ABCCBACBA

Truth tables from standard SOP expressions Truth table: list of all

possible input combinations and corresponding output (0 or 1)

Place a 1 in output column for each input combination that makes the standard SOP expression a 1 0 in remaining

rows

A B C X product term

----------------------------------0 0 0 10 0 1 10 1 0 00 1 1 01 0 0 01 0 1 01 1 0 01 1 1 1

CBACBA

ABC

ABCCBACBAX

Boolean algebra

• Boolean algebra• Laws and rules• De Morgan’s theorem• Analysis of logic circuits• Standard forms• Project 1 preparation

Project 1 preparation• On Sunday (16 Nov) , the class will be given a

Boolean expression involving 4 variables (A,B,C,D) • Project 1

Marked by demonstration to me in week 5 theory section 5 parts, each of equal value:a) Expand given expression to sum-of-products (SOP) formb) Create truth table for expressionc) Expand SOP into standard SOP formd) Implement original expression in Logisim using gatese) Enter SOP expression into Logisim, and confirm that the

resulting truth table produced by Logisim matches your truth table in part b)

Project 1— sample question

a) Expand to SOP:b) Truth table: Z=1 only for ABCD=0101,0111,1101c) Standard SOP: d) Logisim circuit:

e) In Logisim: Project > Analyze Circuitenter expression at Expression tabview truth table at Table tab

BDABCZ DCBBDAZ

Week04_projectdemo.circ

DCBADCABDCBABCDAZ