CANONICAL FORMS FOR BOOLEAN FUNCTION
A Boolean function specified by a
truth table can be expressed algebraically in many different ways. Two ways of
forming Boolean expression are canonical and noncanonical forms. Canonical forms express all
binary variables in every product (AND)
or sum (OR) term of the Boolean function. There are two types of canonical forms of a Boolean function. The first one is called sumof product and the second one is called productofsum. A’BC+A’BC’+ABC’+AB’C’+A’B’C’+ABC is a canonical sumofproduct form of a Boolean function, the given function has contained three inputs that means three binary variables. In each term of the given function has contained three variables.(X’+Y+Z).(X+Y’+Z).(X’+Y+Z).(X’+Y+Z’) is a canonical productofsum form of a Boolean function. A noncanonical form does not express all binary variables in every product or sum term of the Boolean function.
or sum (OR) term of the Boolean function. There are two types of canonical forms of a Boolean function. The first one is called sumof product and the second one is called productofsum. A’BC+A’BC’+ABC’+AB’C’+A’B’C’+ABC is a canonical sumofproduct form of a Boolean function, the given function has contained three inputs that means three binary variables. In each term of the given function has contained three variables.(X’+Y+Z).(X+Y’+Z).(X’+Y+Z).(X’+Y+Z’) is a canonical productofsum form of a Boolean function. A noncanonical form does not express all binary variables in every product or sum term of the Boolean function.
Minterms and Maxterms
A binary variable may appear
either in its normal form (X) OR in its complement form (X’). Now consider two
binary variables X and Y combined with an AND operation. Since each variable
may appear in either form, these are four possible combinations ( X’.Y’, X’.Y, X.Y’,
X.Y). Each of these four AND terms is called a minterm or a standard product. In a similar manner, n variables can be
combined to form 2^{n}
minterm. The 2^{n }different
minterms may be determined by a method similar to the one shown below for three
variables. The binary numbers from 0 to 2^{n} – 1 are listed under the n variable. Each minterm
is obtained from an AND term of the n variables, with each variable being
primed if the corresponding bit of the binary number is 0 and unprimed if a 1. In a similar fashion, n variables
forming an OR term, with each variable being primed or unprimed, provide 2^{n} possible combinations
called maxterms or standard sums. The eight maxterms for three
variables, together with their symbolic designation, are listed below. Any 2^{n} maxterms for n
variables may be determined similarly. Each maxterm is obtained from an OR term
of the n variables, with each variable being unprimed if the corresponding bit
is 0 and primed if it is a 1.
MINTERMS AND MAXTERMS FOR THREE VARIABLES
VARIABLES

MINTERMS

MAXTERMS


X

Y

Z

TERM

DESIGNATION

TERM

DESIGNATION

0

0

0

X’. Y’. Z’

m_{0}

X+Y+Z

M_{0}

0

0

1

X’. Y’. Z

m_{1}

X+Y+Z’`

M_{1}

0

1

0

X’. Y. Z’

m_{2}

X+Y’+Z

M_{2}

0

1

1

X’. Y. Z

m_{3}

X+Y’+Z’

M_{3}

1

0

0

X. Y’. Z’

m_{4}

X’+X+Z

M_{4}

1

0

1

X. Y’. Z

m_{5}

X’+Y+Z’

M_{5}

1

1

0

X. Y. Z’

m_{6}

X’+Y’+Z

M_{6}

1

1

1

X. Y. Z

m_{7}

X’+Y’+Z’

M_{7}

MINTERMS AND MAXTERMS FOR THREE VARIABLES
SUMOFPRODUCT
A sum0fproduct expression is a
product term (minterm) or several product terms (minterms) logically added
(ORed) together.
The following steps are used to
express a Boolean function in its sumof –product form:
 Construct a truth table for the given Boolean function.
 Form a minterm for each combination of the variables which produces a 1 in the function.
 The desired expression is the sum (OR) of all the minterms obtained in step 2.
For example, 011, 100, and 111,
there corresponding minterms are X’.Y.Z, X.Y’.Z’, X.Y.Z . Hence, taking the sum of all
these minterms, the given function can be expressed in its sumofproduct form
as:
F_{1}= X’.Y.Z + X.Y’.Z’ + X.Y.Z
Or F_{1}= m_{3}+m_{4}+m_{7}
Similarly, F_{2} = X’.Y.Z
+ X.Y’.Z + X.Y.Z’ + X.Y.Z can be easily expressed in its sumofproduct form:
F_{2} = X’.Y.Z + X.Y’.Z + X.Y.Z’
+ X.Y.Z
Or F_{2 }= m_{3}+m_{5}+m_{6}+m_{7}
X

Y

Z

F_{1}

F_{2}

0

0

0

0

0

0

0

1

0

0

0

1

0

0

0

0

1

1

1

1

1

0

0

1

0

1

0

1

0

1

1

1

0

0

1

1

1

1

1

1

This is the truth table for F_{1}
and F_{2}
It is sometimes convenient to
express a Boolean function in its sumofproduct form. If not in this form, it
can be made. So by first expanding the expression into a sum of AND terms. Each
term is then inspected to see if it contains all the variables. If it misses
one or more variables, it is ANDed with an expression of the form (X+X’), where
X is one of the missing variable.
Let us take an example, express the given
Boolean function into its sumofproduct form.
F = A’.(B+C’)
F = A’.B+A’.C’
The given function has three variables A, B,
and C. The first term of the function A’B is missing one variable, therefore
A’B = A’.B.(C+C’)
A’B = A’.B.C+A’B.C’
Similarly, the second term of the
function A’C’ is missing one variable, therefore
A’.C’= A’.C’.(B+B’)
A’.C’ = A’.C’.B+A’.C’.B’
So by combining all the terms we
get
F = A’.B.C+A’.B.C’+A’.C’.B+A’.B’.C’
Or F = A’.B.C+A’.B.C’+A’.B.C’+A’.B’.C’
But in the above expression, the
term A’.B.C’ appears twice and according to Boolean postulate we have X+X=X.
Hence it is possible to remove one of them.
F = A’.B.C+A’.B.C’+A’.B’.C’
Rearranging the minterms in
ascending order, we finally obtain:
F = A’.B’.C’+A’.B.C’+A’.B.C
F = m_{0}+m_{2}+m_{3}
F (A, B, C) = ∑ (0, 2, 3)
The summation symbol ‘∑’
stands for the ORing of terms. The numbers following it are the minterm of the
function.
PRODUCTOFSUM
A productofsum expression is a
sum term (maxterm) or several maxterms logically multiplied (ANDed) together.
For example, the expression (X’+Y).(X+Y’) is a productofsum expression.
The following steps are used to
express a Boolean function in its productofsum form:
1. Construct a truth table for
the given Boolean function.
2. Form a maxterm for each
combination of the variables which produces a 0 in the function.
3. The desired expression is the
sum (AND) of all the maxterms obtained in step 2.
For example, 000, 010, 011, 101,
110. The following five combinations of the variable produce a 0.
Their corresponding maxterms are:
(X+Y+Z), (X+Y’+Z), (X+Y’+Z’), (X’+Y+Z’), and (X’+Y’+Z)
If taking the product (AND) of
all these maxterms, we can write them as
F =
(X+Y+Z).(X+Y’+Z).(X+Y’+Z’).(X’+Y+Z’).(X’+Y’+Z)
Or F = M_{0}.M_{2}.M_{3}.M_{5}.M_{6}
Let us take an example:Express the Boolean
function in the productofsum form.
F = X.Y+X’.Y
At first convert the function
into OR term using the distributive law;
F = X.Y+X’.Z
= (X.Y+X’).(X.Y+Z)
= (X+X’).(Y+X’).(X+Z).(Y+Z)
= (X+Y’).(X+Z).(Y+Z)
The function has three variables
X, Y, and Z. Each OR term is missing one variable, therefor:
X’+Y = X’+Y+Z.Z’ =
(X’+Y+Z).(X’+Y+Z’) = (X’+Y+Z).(X’+Y+Z’)
X+Z = X+Z+Y.Y’ = (X+Z+Y).(X+Z+Y’) =
(X+Y+Z).(X+Y’+Z)
Y+Z = Y+Z+X.X’ = (Y+Z+X).(Y+Z+X’) = (X+Y+Z).(X’+Y+Z)
Combining all the terms and removing
those that appear more than one, we finally obtain:
F = (X+Y+Z).(X+Y’+Z).(X’+Y+Z).(X’+Y+Z’)
F = M_{0}.M_{2}.M_{4}.M_{5}
F (X,Y,Z) = π (0,2,4,5)
F (X,Y,Z) = π (0,2,4,5)
The product
symbol π
denotes the ANDing of the maxterms. The numbers following it are the maxterms
of the function.
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