Date post: | 16-Jul-2015 |
Category: |
Engineering |
Upload: | anto-jose-moyalan |
View: | 801 times |
Download: | 6 times |
Switching Theory and Logic Design
Boolean Algebra
Boolean Algebra
A set B of elements (a,b,c,….) with an equivalence
relation( =), two binary operations( + and .) and a
unary operation( complementation – denoted by ‘ ) is a
boolean algebra if and only if the following postulates
are satisfied.
Associativity
Commutativity
Distributivity
Identity
Complement
P.1 ) Associativity :-
The + and . Operators are associative
(a+b)+c = a+(b+c)
(ab)c = a(bc) = abc
P.2) Commutativity :-
The + and . Operators are commutative
a + b = b+ a
ab = ba
P.3) Distributivity :-
a + bc = (a+b)(a+c)
a( b+c) = ab+ ac
P.4) Identity Elements
There exists an identity element (denoted by 0) for the
+ operation and another (denoted by 1) for the .
Operation within B such that
a+ 0 = a
a 1=a
P.5) Complement
Each member of B has a complement within B such
that if a’ is the complement of a , then
a + a’ = 1
aa‘ = 0
Fundamental Theorms on Boolean Algebra
1) Closure of Identity Elements
For all a€B, a+1 = 1 and a0=0
2) Equality Theorem
For all a,b,c € B , if a+b = a+c and
ab= ac, then b = c
Complementary Theorem
For all a,b €B, if a+b =1 and ab=0, then a = b’ and
b=a’
DeMorgan’s Theorem
(a+b)’ = a’b’ and
(ab)’ = a’ + b’
Show That (A+B)(A+B’) = A
S T ABC + AB’C+ ABC’=A(B+C)
S T(A+B)(A+B’)(A’+C) =AC
S T AB+ A(B+C) + B( B+C) = B+ AC
S T (X + Y+XY)(X+Z) = (X+Y)(X+Z)
ST (A+ B’+A’B)C’ = C’
Boolean Constants, Variables & Functions
The Boolean algebra contains two constants 0 and 1
Any characters in lowercase or uppercase are used
as Boolean Variables
Boolean Functions
Any Boolean constants, variables, itself or combined
with any of three operations(logical sum, logical
product & complement) will give a Boolean
expression. Any boolean expression can be written
as a Boolean Function
F = A’B + (AB)’
The above mentioned Boolean algebra is also called
Switching algebra since the two constant values 0
and 1 represent off & on respectively in a circuit
Electronic Gates
An electronic circuit that has one or more inputs and
one output, and in which the electrical condition of
the output at any time is dependent on those of the
inputs at that time is called an electronic gate.
Compute the Boolean Function represented by the
contact network
Calculate the switching functions realized by the
contact network shown below
Boolean Functions & Logical Operations
A Boolean variable in the true form or in the
complemented form is called a literal.
Eg: a,a’,b,b’
The Boolean product of two or more literals is called
a product term
Eg: ab’c , a’bc’
The Boolean sum of two or more literals is called a
sum term
Eg: (a+b’+c’) , (a’+b+c’)
Disjunctive Normal Form
When a Boolean function appears as a sum of
several product terms, it is said to be expressed as a
sum of product(SOP) form.
The SOP form is also called the Disjunctive normal
form(DNF).
Eg: f(abcd)=a’b+bc’+abcd
Conjunctive Normal Form
When a Boolean function appears as a product of
several sum terms, it is said to be expressed as a
product of sums(POS).
The POS form is also called the conjunctive normal
form(CNF).
F(abcd) = (a+b)(b’+c’+d)
Express the following function in DNF
F = (a+b)(b+c+d)
Express the following Boolean function in CNF
F= ab+c’d
F= abc+b’d
Canonical Forms
When each of the terms of a Boolean function
expressed either in SOP or POS form which has all
the variables in it, it is said to be in canonical form.
The canonical SOP form is called the disjunctive
canonical form(DCF) and the canonical POS form is
called the conjunctive canonical form(CCF)
Express the following function in a canonical form
F = ab’+bc’+ac
F= a’+bc
Express the following function in a canonical form
F = (a+b)(b+c’)
F(a,b,c)= a+b
Fundamental Products and Sums
A product term of n variables is called a minterm of n
variables.
Eg: ab’c’ , ab’c’
The no.of all possible minterms of n variables is 2n.
A minterm is also called a fundamental product.
A sum term of n variables is called a maxterm of n variables.
Eg: (a+b+c+d), (a+b’+c+d’)
The no.of all possible maxterms of n variables is 2n.
A maxterm is also called a fundamental sum.
Disjunctive canonical normal form(DCNF) of a Boolean function is a sum of minterms and the conjunctive canonical normal form(CCNF) is a product of maxterms.
F=a’b’c’ + a’b’c + abc
= m0+m1+m7
= ∑(0,1,7)
F = (a+b+c)(a’+b’+c)(a’+b’+c’)
= M0M6M7
= ∏(0,6,7)
∑ Indicates that the
terms are minterms and
the function is a
summation
∏ Indicates that the
terms are maxterms and
the function is a product
Properties
The Sum of all 2n minterms of n variables is 1.
The product of any two n-variable minterms, which are
different is 0.
mimj=0 when i!=j
Complement of mi is Mi
mi’=Mi & Mi’=mi
Convert the given DCNF to CCNF
F=a’bc + ab’c’ + abc
Express the following function in both canonical
forms
F= a’+bc
F(abcd) = (a+b)(b’+c’+d)
Theorem
(Conversion b/w minterms & maxterms)
A boolean function expressed as a sum of minterms
or as a product of maxterms can be converted into
the other forms as given by
∑mi = ∏Mj &
∏Mi=∑mj
Where the subsets i and j are two partitions of the
entire set of 2n subscripts of either m’s or M’s
Express the function f = a’c’d’ + a’bc’+ a’bcd’ in both
types of canonical form
Express the function
f(a,b,c) = (a’+b+c)(a’+b’)(a+b+c) in both types of
canonical form
Express the following function in CCF
F(a,b,c) = ab’ +c’
Self Dual Function
If a Boolean function and its dual are the same, then
the function is called a self-dual function
Check whether the given function is self dual or not
f= ab+ac+bc
Check whether f = b’(a’c’+ac) + b(a’c +ac’) is self
dual or not
Logical Operations
Like AND,OR etc. , there are three more significant
boolean operators. They are NAND, NOR and
EXCLUSIVE-OR
Properties of NAND operation
SHOW That A↑B= B↑A
Check whether NAND operation is associative or not
Properties of XOR operation
XOR is commutative
XOR is also associative
Karnaugh Maps(Veitch Diagram)
The number of gates required and the number of
input terminals for the gates for the implementation
of a Boolean function get reduced considerably if the
Boolean function can be simplified.
Karnaugh Map method gives a systematic approach
for simplifying a Boolean expression.
K-map contains boxes called cells. Each of the cell
represents one of the 2n possible products(of the
variables) that can be formed from n variables.
Thus a 3-variable map contains 8 cells, 2 variable-4
cells, 4 variable-16 cells.
Each nearby cells represents only one bit change.
Numbering of each cells will be done based on the
gray code sequence.
K-map representation
Plotting SOP form in K-map
If a Boolean function is given as a sum of its
minterms(ie in its DCF form), then the function is
plotted on the map by writing 1’s on those cells
which represent the minterms that are present in the
function.
Plot the Boolean function
Y = ABC’ + ABC + A’B’C on the K-map
Plot the boolean function
Y = A’BC’D’ + AB’C
Plot the boolean function Y = AB + BC
Plot f(abc) = ∑(0,2,6,7) in the K-map
Plot f(abcd) = ∑(1,3,4,7-10,12,15)
Plotting POS form in K-map
If a Boolean function is given as a product of
maxterms (ie in CCF), then the function is plotted on
the map by writing 0’s on those cells which represent
the maxterms that are present in the function.
Represent f = ∏(0,3,6,7)
Plot function f(a,b,c)= a +b in K-map
Plot f= c’d’ + a’bd
Plot f = b’c + d + abc’
K-map minimization
Two cells are said to be adjacent if there is only one
bit change among the cells.
Group those cells and eliminate variables which is in
true form and complementary form.
Minimum Sum Of Products
Minimize the expression
Y = A’B’C’+A’BC+AB’C’+ABC
Y= AB’C + A’B’C + A’BC + AB’C’ + A’B’C’
Minimize the expression
Y= A’BC’D’ + A’BC’D + ABC’D’ + ABC’D + AB’C’D +
A’B’CD’
Minimum Product of Sums
Minimize the expression
Y = (A+B+C’) (A+B’+C’) (A’+B’+C’)(A’+B+C) (A+B+C)
Write each Maxterms. Plot 0 for the Maxterm cells
Minimize the expression
F= ∏(0,2,3,8,9,12,13,15)
5-variable K-map
Simplify
F = ∑(0,2,4,6,9,11,13,17,21,25,27,29,31)
Don’t care terms
The combinations for which the values of the
expression are not specified are called don’t care
combinations. The output is a don’t care term for
these invalid combinations
Minimize the boolean expression of the below circuit
logic
A B C Y
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
Prime Cubes
Each cells in the k-map corresponds to minterms is
called a cube.
A cube of a given function that can not grow larger
by expanding into other cubes of the function is
called a prime cube(PC) or a prime implicant (PI) of
the function.
Find the prime implicants/ prime cubes of the
function
f = ∑(2,5,6)
Essential Prime Cubes
If among the minterms subsuming a prime cube,
there is at least one that is covered by this and only
this prime cube, then the prime cube is called an
essential prime cube (EPC) or an essential prime
implicant(EPI)
Redundant Prime Cubes
If each of the minterms subsuming a prime cube is
covered by other essential prime cubes, then the
prime cube is a redundant prime cube(RPC) or a
redundant prime implicant
Find the EPC and RPC of the following boolean
function
∑(1,3,6,7)
Quine-Mc Cluskey Minimization
wrt to minimization, K-maps Only good for:
Small functions (<6variables)
Single output function at a time
Not Implementable on Computer
Subjective Interpretation, Different Coverings
Q-M (Tabular Minimization) Solves These Problems
Basic Definitions
Any single or group of ones that can be
combined is called an implicant.
A prime implicant of a function F is a product
term implicant which is no longer an implicant if
any literal is deleted from it.
Essential prime implicants are the implicant
which will definitely in the final expression.
Procedure
F (w,x,y,z)= (0,1,2,3,5,7,8,10,14,15)
Write the binary equivalent of all the minterms.
Group all the minterms by the no of 1s they contain.
Step 1
0 0000
1 0001
2 0010
8 1000
3 0011
5 0101
10 1010
7 0111
14 1110
15 1111
Combine the members of new groups to create more
new groups.
Combined group must differ by one bit and must
have – in the same position.
Combine as much as possible.
Select prime implicants to cover all the ones.
Finding Prime Implicants
0 0000
1 0001
2 0010
8 1000
3 0011
5 0101
10 1010
7 0111
14 1110
15 1111
Step 1 Step 2
(0,1) 000-
(0,2) 00-0(0,8) -000
(1,3) 00-1
(1,5) 0-01(2,3) 001-
(2,10) -010
(8,10) 10-0
(3,7) 0-11
(5,7) 01-1
(10,14) 1-10
(7,15) -111
(14,15) 111-
Step 3
(0,1,2,3) 00--
(0,2,1,3) 00--
(0,2,8,10) -0-0
(0,8,2,10) -0-0
(1,3,5,7) 0--1
(1,5,3,7) 0--1
6 Prime Implicants
1-10
-111
111-
00--
-0-0
0--1
(0,1) 000-
(0,8) -000
(2,3) 001-
Finding Prime Implicants
0 0000
1 0001
2 0010
8 1000
3 0011
5 0101
10 1010
7 0111
14 1110
15 1111
Step 1 Step 2
(0,1) 000-
(0,2) 00-0(0,8) -000
(1,3) 00-1
(1,5) 0-01
(2,3) 001-
(2,10) -010
(8,10) 10-0
(3,7) 0-11
(5,7) 01-1
(10,14) 1-10
(7,15) -111
(14,15) 111-
Step 3
(0,1,2,3) 00--
(0,2,1,3) 00--
(0,2,8,10) -0-0
(0,8,2,10) -0-0
(1,3,5,7) 0--1
(1,5,3,7) 0--1
6 Prime Implicants
1-10
-111
111-
00--
-0-0
0--1
(0,1) 000-
(0,8) -000
(2,3) 001-
Find Essential Prime Implicants –
Prime Cube table
Prime
ImplicantCovered
minterms
1-10
-111
111-
00--
-0-0
0--1
Minterms
0 1 2 3 5 7 8 10 14 15
10,14
7,15
14,15
0,1,2,3
0,2,8,101,3,5,7
X XX X
X X
X X X X
X X X XX X X X
Essential Prime implicant
Here essential prime cubes are x’z’ & w’z
Selective Prime Cube Table
• In an SPC table, If a row I has crosses in
all columns where another row J has
crosses, and there is at least one more
column with a cross under I but not under
J, row I is said to dominate row J
• A row in an SPC table that is dominated
by another row of the same or smaller cost
can be deleted from the SPC table
14 15
A(10,14) *
B(7,15) *
C(14,15) * *
• As per the theorem , Here C can be
selected.