Digital Circuits Logic Design: Course Coordinators
Digital Circuits Logic Design: Course Coordinators
Digital Circuits Logic Design: Course Coordinators
DESIGN
Online Class for II B.Tech (IT) - I Semester
AY: 2020-21
Course coordinators:
Mrs.T.Vineela
Mr.G.Amar Tej
Mr.K.Ramesh Babu
Course Objectives:
Boolean Algebra: Boolean algebra and logic gates, Basic theorems and properties
of Boolean Algebra, Boolean functions, canonical and standard forms, Universal
Gates.
Algebra?
BOOLEAN ALGEBRA
Introduction:
1) A.A=A
2) A+A=A
2) A.B=B.A
6) Associative laws:
The associative laws allow grouping of variables.
There are two associative laws. Those are:
1) (A+B)+C=A+(B+C)
2) (A.B)C=A(B.C)
7) Distributive laws:
The distributive laws allow factoring or multiplying out
of expressions.
There are two distributive laws. Those are:
1) A(B+C)=AB+AC
2) A+BC=(A+B)(A+C)
Proof(2):
RHS=(A+B)(A+C)
=AA+AC+BA+BC
=A+AC+AB+BC
=A(1+C+B)+BC
=A.1+BC (1+C+B=1+B=1)
=A+BC
=LHS
8)Redundant Literal Rule(RLR):
There are two laws.Those are:
9) Absorption laws:
There are two laws,Those are:
1) A+A.B=A
Proof: LHS= A+A.B
= A(1+B)
= A.1 (1+B=1)
=A (A.1=A)
= RHS
2) A(A+B)=A
Proof: LHS = A(A+B)
= A.A+A.B
= A+AB (A.A=1)
= A(1+B)
= A.1
=A
= RHS
Theorems of Boolean algebra
Consensus Theorem(Included Factor Theorem):
Transposition theorem:
Proof:
RHS = (A+C)(A’+B)
= A.A’+A.B+C.A’+C.B
= 0+AB+A’C+BC
= A’C+AB+BC(A+A’)
= A’C+AB+BCA+BCA’
= A’C(1+B)+AB(1+C)
= A’C+AB
= LHS
De Morgan’s Theorem:
De Morgan’s theorem represents two of the most powerful laws
in Boolean algebra.
Law1:
Law2:
DUALITY
To get dual of any expression
Replace:
1) AND 2) OR 3) NOT
4) NAND 5) NOR
6) X-OR 7) X-NOR
1) AND gate:
AND is nothing but logical multiplication
functionality.
Logic Symbol(2-input):
Truth Table:
Truth table:
4) NAND gate:
NAND gate performs complement of AND gate.
Logic Symbol:
Truth table:
Logic symbol:
Truth table:
Logic symbol:
Truth table:
Truth table:
By observing this truth table , the
output is ‘High’ or ‘1’ only when the
inputs are having even number of
‘1’s.
1) Find the dual of the functions a) F=A+B’C
b) F= x’y’+yz
To get dual of any
expression
Replace:
F1=A+B’C
Literal :
A Variable or complement of variable.
Ex:
Term:
Minimize the following Boolean expressions
1) F= AB+AB’
2) F= ABC’+ABC+A’BC
3) F= (X’+Y)(X’+Z)
4) F=AB+A’C+AB’
5) F=XY’+XY+X’Y
6) F=A’+A’BC
Simplify the following Boolean expressions:
1) F= A.B.C+A’+A.B’.C
2) F= A’B’C’+A’B’C+A’C’
3) F= [A.B’.(C+B.D)+A’.B’].C
4) F= A’C’+ABC+AC’
5) F= (A’B’+C)’+C+AB+BC
6) F= (A’+C)(A’+C’)(A+B+C’D)
7) F= ABCD+A’BD+ABC’D
8) F= XY+X’YZ’+YZ
9) F= (XY’+Z)(X+Y’)Z
10)F= A’C’D’+AC’+BCD+A’CD’+A’CD’+A’BC+AB’C’
ANSWERS:
1) F= A’+C
2) F= (A+BC)’
3) F= B’.C
4) F= C’+AB
5) F= AC’+B
6) F= A’(B+C’D)
7) F= BD
8) F= Y
9) F= Z(X+Y’)
10) F= A’D’+AC’+BCD+A’BC
Boolean function representation:
We can represent the Boolean functions in 4 forms . Those are
1) SOP(Sum Of Products)
2) POS(Product of sums)
3) Canonical SOP
4) Canonical POS
1) SOP form:
SOP is a group of product terms summed together.
Ex : F(A,B,C) = AB+AB’C+BC
2)POS form:
POS is a group of sum terms multiplied together.
Ex : F(A,B,C) = (A+B)(B+C’)
INPUTS OUTPUT
A B Y
0 0 0
0 1 1
1 0 0
1 1 1
4) Canonical POS :
If each sum term contains all the variables of the function either
in complemented or uncomplemented form then it is called as
canonical form.
POS, Boolean expressions may be
Inputs Output
generated from truth tables quite easily, by
A B C Y determining which rows of the table have
0 0 0 0 an output of 0.
0 0 1 0
0 1 0 1 F=(A+B+C).(A+B+C’).(A+B’+C’)
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 1
1) For the given truth table , find the minimal POS expression.
INPUTS OUTPUT
A B Y
0 0 1
0 1 0
1 0 1
1 1 0
Minimal to Canonical form
To convert minimal to canonical form we have to follow 3 steps:
step2: Find the variables which are absent in each minterm or maxterm.
step3: Try to get the absent variables by using laws of Boolean algebra.
Y=A+B’C Y=(A+B+C’)(A’+C)
Ex: convert the given expression to canonical POS form
1) f(A,B,C) =(A+B)(B+C)(A+C)
= (A+B+CC’)(AA’+B+C)(A+BB’+C)
= (A+B+C)(A+B+C’)(A+B+C)(A’+B+C)
(A+B+C)(A+B’+C)
= (A+B+C)(A’+B+C)(A+B’+C)(A+B+C’)
2) f(A,B,C)= A(A+B+C)
Alternative gates
NAND as OR gate: