TRINITY INSTITUTE OF PROFESSIONAL STUDIESSector – 9, Dwarka Institutional Area, New

Delhi-75Affiliated Institution of G.G.S.IP.U, Delhi

DIGITAL ELECTRONICSPaper Code :- BCA 106

Keywords:Boolean Algebra, logic, Laws

By :-HARI MOHAN JAIN

TRINITY INSTITUTE OF PROFESSIONAL STUDIESSector – 9, Dwarka Institutional Area, New Delhi-75

2

Boolean Algebra

TRINITY INSTITUTE OF PROFESSIONAL STUDIESSector – 9, Dwarka Institutional Area, New Delhi-75

3Example #1: Boolean AlgebraSimplify the following Boolean expression and note the Boolean theorem used at each step. Put the answer in SOP form.

D C C BA A F 1

SOP – Sum Of Product (Sum of MINTERMS)

Y = AB + BC

POS – Product Of Sum (Product of MAXTERMS)

Y = (A+B) (B+C)

TRINITY INSTITUTE OF PROFESSIONAL STUDIESSector – 9, Dwarka Institutional Area, New Delhi-75

4Example #1: Boolean AlgebraSimplify the following Boolean expression and note the Boolean theorem used at each step. Put the answer in SOP form.

D C C BA A F 1

Solution

BA BA

0 BA D 0 BA

D C C BA

D C C BA A

1

1

1

1

1

1

FFFFF

F

; Theorem #3

; Theorem #4

; Theorem #1

; Theorem #5

TRINITY INSTITUTE OF PROFESSIONAL STUDIESSector – 9, Dwarka Institutional Area, New Delhi-75

5Example #2: Boolean Algebra

Simplify the following Boolean expression and note the Boolean theorem used at each step. Put the answer in SOP form.

0 B A 1 B A C C B C B BF 2

TRINITY INSTITUTE OF PROFESSIONAL STUDIESSector – 9, Dwarka Institutional Area, New Delhi-75

6Example #2: Boolean Algebra

Simplify the following Boolean expression and note the Boolean theorem used at each step. Put the answer in SOP form. 0 B A 1 B A C C B C B BF 2

Solution

B A C BF

B A C B F

0 B A C B F

0 B A B A C B F

0 B A 1 B A C B F

0 B A 1 B A C B C B F

0 B A 1 B A C C B C B BF

2

2

2

2

2

2

2

; Theorem #3 (twice)

; Theorem #7

; Theorem #2

; Theorem #1

; Theorem #5

TRINITY INSTITUTE OF PROFESSIONAL STUDIESSector – 9, Dwarka Institutional Area, New Delhi-75

7Example #3: Boolean Algebra

Simplify the following Boolean expression and note the Boolean theorem used at each step. Put the answer in SOP form.

T) R)(S (R T RF3

TRINITY INSTITUTE OF PROFESSIONAL STUDIESSector – 9, Dwarka Institutional Area, New Delhi-75

8Example #3: Boolean Algebra

Simplify the following Boolean expression and note the Boolean theorem used at each step. Put the answer in SOP form. T) R)(S (R T RF3

Solution

R S TF

R S 1TF

R S S 1TF

R S S R RTF

T S R S T RT RF

T S R S T R 0 T RF

T S R S T R R R T RF

T RS R T RF

3

3

3

3

3

3

3

3

; Theorem #12B

; Theorem #4

; Theorem #5

; Theorem #12A

; Theorem #8

; Theorem #6

; Theorem #2

TRINITY INSTITUTE OF PROFESSIONAL STUDIESSector – 9, Dwarka Institutional Area, New Delhi-75

9Example #4: Boolean Algebra

Simplify the following Boolean expression and note the Boolean theorem used at each step. Put the answer in SOP form.

S Q P S Q P S P F4

TRINITY INSTITUTE OF PROFESSIONAL STUDIESSector – 9, Dwarka Institutional Area, New Delhi-75

10Example #4: Boolean AlgebraSimplify the following Boolean expression and note the Boolean theorem used at each step. Put the answer in SOP form. S Q P S Q P S P F4

Q P PS F

Q P 1 PS F

Q P Q 1 PS F

S Q P QP S P F

S Q P Q S P F

S Q P S Q S P F

S Q P S Q P S P F

4

4

4

4

4

4

4

; Theorem #12A

; Theorem #13C

; Theorem #12A

; Theorem #12A

; Theorem #6

; Theorem #2

Solution

TRINITY INSTITUTE OF PROFESSIONAL STUDIESSector – 9, Dwarka Institutional Area, New Delhi-75

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• Boolean identities

AND Function OR Function NOT function00=0 0+0=001=0 0+1=110=0 1+0=111=1 1+1=1A0=0 A+0=A0A=0 0+A=AA1=A A+1=11A=A 1+A=1AA=A A+A=A

0 AA 1AA

10

0 1

AA

TRINITY INSTITUTE OF PROFESSIONAL STUDIESSector – 9, Dwarka Institutional Area, New Delhi-75

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Commutative law Absorption law

Distributive law De Morgan’s law

Associative law Note also

• Boolean laws

ABBABAAB

))(()(

CABABCABCABCBA

CBACBACABBCA

)()()()(

ABAAAABA

)(

BABABABA

ABBAABABAA

)(