lecture 4 4. boolean functions – advanced 4.1 the normal form and the canonical form 4.2...

Lecture 4

4. Boolean functions – advanced4.1 The normal form and the canonical form4.2 Fundamental product and sums4.3 Disjunctive and conjunctive canonical forms4.4 Logical operations of 2 variables4.5 Functionally complete sets4.6 Karnaugh map – another way of denotation4.7 Karnaugh map of 5 and 6 variables4.7 Don’t care terms

Boolean functions and logical operations

• Some fundamental terms, forms and definitions relating to Boolean functions and expressions will now be studied

• We shall also see how many other logical operations besides AND, OR and negation can be derived

4.1 The normal form

D1:

• A Boolean variable in true form or in complemented form is called a literal. Thus a,b,c, Na,Nb,Nc (a’,b’,c’) and so on are literals.

D2:

• The Boolean product of two or more literals is called a product term.

The normal form-2

D3:• The Boolean sum of two or more literals is

called a sum term. D4:• When a Boolean function appears as a

sum of several product terms, it is said to be expressed as a sum of products (SOP). The SOP form is also called the disjunctive normal form (DNF).

The normal form-3

D5:

• 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).

The normal form-4

Example 1

f(a,b,c,d)=N.a+b.Nc+c.d

is an expression in SOP form or DNF.

Example 2f(a,b,c,d)=(a+b).(b+c+d)is an expression in POS form or CNF.

The canonical form

D6:

• When each of the terms of a Boolean functions expressed either in SOP or POS form has all the variables in it it is said to be expressed in canonical form.

• Here also by idempotence law of Boolean algebra, the canonical form cannot have the same term more than once.

The canonical form-2

• The canonical SOP form is called the disjunctive canonical form (DCF) and the canonical POS form is called the conjunctive canonical form (CCF).

Example:Express the following functions in

canonical formsa) f1=a.Nb.c+b.Nc+a.cb) f2=(a+b).(b+Nc)

The canonical form-3Solutions:

a)f1=a.Nb.c+b.Nc+a.c=

=a.Nb.c+(a+Na).b.Nc+a.(b+Nb).c=

=a.Nb.c+a. b.Nc +Na.b.Nc+a.b.c+a.Nb.c

b)f2= (a+b)(b+Nc)=

= (a+b+c.Nc)(a.Na+b+Nc)=

= (a+b+c).(a+b+Nc).(a+b+Nc).(Na+b+Nc)=

= (a+b+c).(a+b+Nc).(Na+b+Nc)

f1 is expressed in DCF,

f2 is expressed in CCF.

4.2 Fundamental products and sums

D7:• A product term of n variables is called a minterm

of n variables. Thus Na.Nb.Nc and a.Nb.Nc are minterms of three variables.

• Since each variable can appear in a minterm in one of the two forms, true and complemented, the number of all possible minterms of n variables is 2 powered by n. A minterm is also called fundamental product.

Fundamental products and sums-2

• If a 0 is written for complemented variable and 1 for the uncomplemented variable, each minterm can be expressed as a binary number.

• Each minterm is then designated as mi, where the subscript i is the decimal value of the binary number.

• See following table for the three variable case:

Fundamental products and sums-3Minterms(n=3) Corresponding

binary number

representations

Corresponding

symbolic

representations

Na.Nb.Nc

Na.Nb.c

Na.b.Nc

Na.b.c

a.Nb.Nc

a.Nb.c

a.b.Nc

a.b.c

000

001

010

011

100

101

110

111

m0

m1

m2

m3

m4

m5

m6

m7


D8:• A sum term of n variables is called a maxterm of

n variables. Thus Na+b+c+Nd and a+b+c+d are maxterms of four variables.

• Like minterms, the number of maxterms is also 2 powered by n for n variables. A maxterm is also called fundamental sum.It is customary to designate each maxterm by Mi, where i is the decimal equivalent of the binary number representation of a maxterm, as shown in the following table for the three variable case.

Fundamental products and sums-5Maxterms(n=3) Corresponding

binary number

representations

Corresponding

symbolic

representations

a+b+c

a+b+Nc

a+Nb+c

a+Nb+Nc

Na+b+c

Na+b+Nc

Na+Nb+c

Na+Nb+Nc

000

001

010

011

100

101

110

111

M0

M1

M2

M3

M4

M5

M6

M7


• It should be noted, that in the maxterms we have written a 0 for a variable in the true form and 1 for a variable in the complemented form. This is the opposite of the procedure we followed in the case of minterms. It can easily be verified, that the complement of mi is Mi and vice versa, that is:

N(mi)=Mi and N(Mi)=mi


• From the definitions of minterm and maxterm it is evident, that the disjunctive canonical form of a Boolean function is a sum of minterms, and the conjunctive canonical form is a product of maxterms. Thus

f1=Na.Nb.Nc+Na.Nb.c+a.b.c =m0+m1+m7

and

f2=(a+b+c).(Na+Nb+c).(Na+Nb+Nc)=M0.M6.M7


• This has led to another convenient way of writing the canonical form wherein the m’s (minterms) and M’s (maxterms) are not written at all.

Thus

f1=sum(0,1,7)

and

f2=pí(0,6,7)


• Here the decimal numbers (which are subscripts of the m’s or M’s) identify the particular terms, sum indicates, that the terms are minterms and the function is a summation, and pí indicates, that the terms are maxterms and the function is a product.

4.3 Disjunctive and conjunctive canonical forms

• It has been mentioned earlier that the disjunctive canonical form is the same as the sum of minterms and the conjunctive canonical form is the same as the product of maxterms.

• It is interesting to note that if a function is expressed in the DCF, it can also be expressed in the CCF, and vice versa.

Disjunctive and conjunctive canonical forms-2

Theorem: A Boolean function expressed as a sum of minterms or as a product of maxterms can be converted into the other form as given by

sum(mi)=pí(Mj) and pí(Mi)=sum(mj)

Where the subsets i and j are two partitions of the entire set of 2 powered by n subscripts of either m’s or M’s.


From this theorem it can be concluded, that any Boolean function can be expressed in both the disjunctive and conjunctive canonical form.

Example1:

Express the following function in the other type of canonical form:

f1=Na.b.c+a.Nb.Nc+a.b.c


Solutionf1=f1(a,b,c)=m3+m4+m7=sum(mi)Subset i=(3,4,7)Entire set of subscripts: 2 powered by n (n=3) is 8

(0,1,2,3,4,5,6,7)Subset j=(0,1,2,5,6)f1=pí(Mj)=pí(M0,M1,M2,M5,M6)==M0.M1.M2.M5.M6=(a+b+c).(a+b+Nc).(a+Nb+c).(Na+b+Nc).(Na+Nb+c)


Example2:

Express the function y from the Example 1 from exercises in the both canonical forms:

y=f(S1,S2,S3)

y=sum (mi)=(i:(3,5,6))=NS1.S2.S3+ S1.NS2.S3+ S1.S2.NS3…….DCF (DSF)

y=pí(Mj)=(j:(0,1,2,4,7))=(S1+S2+S3).

.(S1+S2+NS3).(S1+NS2+S3).(NS1+S2+S3).

.(NS1+NS2+NS3)…..CCF (CSF)

4.4 Logical operations of 2 variables

• So far we have discussed mainly two logical operations, AND and OR of two variables and NOT(negation, inversion) of one variable.

• But there are altogether 16 various functions of two variables.

2 powered by 2 powered by 2=2 powered by 4=16 • Generally the number of various functions of n

binary variables is equal

2 powered by 2 powered by n

Logical operations of 2 variables -2

ab f1 f2 f3 f4 f5 f6 f7 f8 f9 f10

f11

f12

f13

f14

f15

f16

00 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

01 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1

10 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

11 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1


• f1…allways FALSE • f16…allways TRUE• f2….AND• f15……NAND• f8…..OR• f9…..NOR• f7….EXOR (nonequivalence)• f10…equivalence

Logical operations- of 2 variables-4

ab f1 f2 f3 f4 f5 f6 f7 f8 f9 f10

f11

f12

f13

f14

f15

f16

00 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

01 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1

10 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

11 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

negation


• f4…assertion (repetition) of a

• f13…negation of a

• f6…. assertion (repetition) of b

• f11…negation of b

• f3,f5,f12,f14 inclusion

(does not have direct technical meaning)

4.5 Functionally complete sets

• Definition: A set of operations is called a functionally

complete set, if and only if any Boolean function can be expressed by operations belonging to the set only.

• In fact, to prove any other set of operations to be functionally complete, it should be shown that its members can express the three operations of AND, OR, and NOT.

Functionally complete sets-2

• For example, the sets of operations{AND, OR,NOT}{AND, NOT}{NAND}{OR,NOT} {NOR}{XOR, AND}

are functionally complete sets.


• It can be seen that the NAND and NOR operations alone form a functionally complete set.

• It is therefore possible to implement any switching function using only one type of gate, either NAND or NOR.

• Hence these two operations are also known as universal operations, and the gates are used much more extensively than other gates in the synthesis of combinational logic circuits.


• How to implement switching function NOT (negation) using only NAND gate

&a

1Na

N(a . 1)=N(a)=Na

&a

aNa

N(a . a)=N(a)=Na


• How to implement switching function NOT (negation) using only NOR gate

>=1a

0Na

N(a + 0)=N(a)=Na

a

aNa

N(a + a)=N(a)=Na

>=1


• How to implement switching function AND using only NAND gate

&a

bN(a.b) &

N(a.b)a.b

N(a.b)


• How to implement switching function OR using only NOR gate

a

bN(a+b)

N(a+b)a+b

N(a+b)

>=1 >=1


• How to implement switching function OR using only NAND gate

&a

aNa

&b

bNb

&

N(Na.Nb)=

N(Na)+N(Nb)=

a+b


• How to implement switching function AND using only NOR gate

>=1a

aNa

b

bNb

N(Na+Nb)=

N(Na).N(Nb)=

a.b

>=1

>=1

4.6 Karnaugh map – another way of denotation

cd

ab

00 01 11 10

00 0

0000

1

0001

3

0011

2

0010

01 4

0100

5

0101

7

0111

6

0110

11 12

1100

13

1101

15

1111

14

1110

10 8

1000

9

1001

11

1011

10

1010

Karnaugh map – another way of denotation-2

cd

ab

0

0000

1

0001

3

0011

2

0010

4

0100

5

0101

7

0111

6

0110

12

1100

13

1101

15

1111

14

1110

8

1000

9

1001

11

1011

10

1010

Karnaugh map – another way of denotation -3

ab

cd

00 01 11 10

00 0

0000

4

0100

12

1100

8

1000

01 1

0001

5

0101

13

1101

9

1001

11 3

0011

7

0111

15

1111

11

1011

10 2

0010

6

0110

14

1110

10

1010


ab

cd

0

0000

4

0100

12

1100

8

1000

1

0001

5

0101

13

1101

9

1001

3

0011

7

0111

15

1111

11

1011

2

0010

6

0110

14

1110

10

1010


What is important:

• Index is a binary value of the same expression, e.g. „abcd“

• Property of neighbourhood

4.7 Karnaugh map of 5 and 6 variables

• The five-variable map is drawn as two separate four variable maps.

• Each of the two four variable maps is a map for the four variables e.g. x2,x3,x4,x5.

• They differ in the value of the first variable, x1. • In one of the maps x1=0, in the other x1=1.• Decimal values of the index can be computed

following the same procedure as in other maps.• Similarly, the six-variable map consists of four

four-variable maps.

Five-varibles Karnaugh map-2Example – values of indexes,

1st map for x1=0 x2x3

x4x5

00 01 11 10

00 0

00000

4

00100

12

01100

8

01000

01 1

00001

5

00101

13

01101

9

01001

11 3

00011

7

00111

15

01111

11

01011

10 2

00010

6

00110

14

01110

10

01010

Five-varibles Karnaugh map-2Example – values of indexes,

2nd map for x1=1 x2x3

x4x5

00 01 11 10

00 16

10000

20

10100

28

11100

24

11000

01 17

10001

21

10101

29

11101

25

11001

11 19

10011

23

10111

31

11111

27

11011

10 18

10010

22

10110

30

11110

26

11010

4.8 Don’t care terms

• Karnaugh map is a representation of a switching function expressed in the form of a truth table or as a sum of minterms.

• In all the maps that we discussed so far, in each cell we plotted either a 1(shown on the map) or a 0 (shown or not shown but meant by implication)

• This means, that for such functions the output of the circuit is clearly specified to be either 1 or 0.

• Such functions are called completely specified functions.

Don’t care terms-2

• There may be many practical cases, however, where the output of a circuit cannot be precisely specified.

• Very often this happens due to the simple fact that all input combinations may not occur in a practical situation

• For example, if the input variables of a function f(a,b) represent the values of limit switches for level in a container, we know that in practical case there is not possible to have high level and simultanously don’t have the low level too.


• The exact value of the function for this input combination cannot be specified.

• Such functions are known as incompletely specified functions

• The input combinations for which the function is not specified are shown by plotting a x on the corresponding cell of the map.

• It is obvious that while minimizing, the value of x can be assumed to be either 1 or 0.


• Since the circuit or software designer does not care for the exact value of these cells and chooses either 1 or 0 to suit the objective of obtaining the minimum solution, these terms are known as don’t –care terms.

• The minterms for which the function has the value x are also known as don’t –care minterms.

Don’t care terms-example -5

Z cd

ab

00 01 11 10

00

x 1 0 0

01

1 1 1 0

11

0 1 x 0

10

0 x x 0


Z cd

ab

00 01 11 10

00

x 1 0 0

01

1 1 1 0

11

0 1 x 0

10

0 x x 0


• Without using of „x“ fields the minimized function will be:

• Z= Na.b.Nc+Na.b.d+Na.Nc.d+b.Nc.d

• Using of „x“ fields as shown we get minimized function :

• Z=Na.Nc+ b.d

Literature

• Nripendra N. Biswas: Logic Design Theory,Prentice Hall International,1993,ISBN 0-13-010695-X

lecture 4 4. boolean functions – advanced 4.1 the normal form and the canonical form 4.2...

Documents

b ncthe canonical form

na b c nd

canonical sop form

canonical pos form

na b ncf1

complemented form

nc na

true form