# Canonical SOP and POS

```CpE 71 INTRODUCTION TO LOGIC DESIGN
Σ
Canonical/Standard Forms
Π
Product Term : A function that is defined to be the logical AND of a set
of variables that are either primed(not) or unprimed, i.e. if f(A,B,C,D)is a function, the product
terms of the function can be A . B , B . C ’ . D , A . B . C ’ . D , and etc.
Minterm : A product term in which all variables (primed or unprimed) appear once and only
n
once, i.e. if a function has n variable, then there are 2 possible unique minterms. Each
n
minterms has a value of “1” for only one of the 2 possible values and “Zero” for all other value.
A symbol for each minterm is of the form m k where k denotes the decimal equivalent of the binary.
Example: The truth table below illustrates the form of all possible minterms for 3-variables.
W
0
0
0
0
1
1
1
1
X
0
0
1
1
0
0
1
1
Y
0
1
0
1
0
1
0
1
TERM
W’.X’.Y’
W’.X’.Y
W’.X.Y’
W’.X.Y
W.X’.Y’
W.X’.Y
W.X.Y’
W.X.Y
DESIGNATION
m0
m1
m2
m3
m4
m5
m6
m7
Sum Term : A function that is defined to be the logical OR of a set of variables that are either
primed(not) or unprimed, i.e. if f(A,B,C,D)is a function, the sum terms of the function can be
(A+B), (B+C’+D) , (A+B+C’+D) , and etc.
Maxterm : A sum term in which all variables (primed or unprimed) appear once and only
n
once, i.e. if a function has n variable, then there are 2 possible unique maxterms. Each
n
maxterms has a value of “0” for only one of the 2 possible values and “one” for all other value. A
symbol for each maxterm is of the form Mk where k denotes the decimal equivalent of the binary.
Example: The truth table below illustrates the form of all possible maxterms for 3-variables.
W
0
0
0
0
1
1
1
1
X
0
0
1
1
0
0
1
1
Y
0
1
0
1
0
1
0
1
TERM
DESIGNATION
W+X+Y
M0
W+X+Y’
M1
W+X’+Y
M2
W+X’+Y’
M3
W’+X+Y
M4
W’+X+Y’
M5
W’+X’+Y
M6
W’+X’+Y’
M7
CANONICAL REPRESENTATION OF
LOGIC FUNCTION
Σ
Π
SUM OF PRODUCTS: A boolean function can be expressed
algebraically from a given truth table by forming a minterm for
each combination of the variables which produces a “1” in the
function, and then taking OR (sum) of all those terms.
Example:
Decimal
0
1
2
3
4
5
6
7
A
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
C F(A,B,C)
0
1
1
0
0
0
1
1
0
1
1
0
0
0
1
0
F(A,B,C) = Σm(0,3,4)
= m0 +m3 +m4
= A’.B’.C’+A’.B.C+A.B’.C’
The function is said to be represented in the
canonical sum of products form.
PRODUCT OF SUMS: A boolean function can be expressed
algebraically from a given truth table by forming a maxterm for
each combination of the variables which produces a “0” in the
function, and then taking AND (product) of all those terms.
Example 1:
Decimal
0
1
2
3
4
5
6
7
A
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
C F(A,B,C)
0
1
1
0
0
0
1
1
0
1
1
0
0
0
1
0
F(A,B,C) = ΠM(1,2,5,6,7)
= M1 .M2 .M5 .M6 .M7
=(A+B+C’) (A+B’+C) (A’+B+C’) (A'+B'+C) (A'+B'+C')
The function is said to be represented in the
canonical product of sums form.
CONVERSION BETWEEN CANONICAL FORM
Interchane the symbols Σm and ΠM and list those number missing from the original form
Examples:
F(W,X,Y) = Σm(0,1,3,5,7) = W’X’Y’ + W’X’Y + W’XY + WX’Y + WXY
= ΠM (2,4,6) = (W+X’+Y) (W’+X+Y) (W’+X’+Y)
F(A,B)
= ΠM (0,1,2)
= Σm(3)
= (A+B) (A+B’) (A’+B)
= A.B
EXPRESS THE BOOLEAN FUNCTION IN A
CANONICAL SUM OF PRODUCT (MINTERMS)
AND/OR PRODUCT OF SUMS(MAXTERMS)
Σ
Π
Expand each individual term to incorporate all variables,
i.e. use X+X’ = 1or X.X’=0.
Example 1: Express F(A,B,C,D) = A’B + ACD
into the canonical sum of products (Σm) and product of sum (ΠM)
F(A,B,C,D) = A’B + ACD
= A’B(C+C’) + ACD(B+B’)
= A’BC+A’BC’+ABCD+AB’CD
= A’BC(D+D’)+A’BC’(D+D’)+ ABCD+AB’CD
= A’BCD+A’BCD’+A’BC’D+A’BC’D’+ ABCD+AB’CD
= Σm(7,6,5,4,15,11)
= ΠM(0,1,2,3,8,9,10,12,13,14)
= (A+B+C+D).(A+B+C+D’).(A+B+C’+D).(A+B+C’+D’).(A’+B+C+D)
.(A’+B+C+D’).(A’+B+C’+D). (A’+B’+C+D). (A’+B’+C+D’). (A’+B’+C’+D)
Example 2: Express F(A,B,C,D) = (A’+B’+C).(A+C’+D).(B+C+D’)
into the canonical product of sum (ΠM) and sum of products (Σm).
F(A,B,C,D) = (A’+B’+C).(A+C’+D).(B+C+D’)
= (A’+B’+C+D.D’).(A+B.B’+C’+D).(A.A’+B+C+D’)
= (A’+B’+C+D)(A’+B’+C+D’).(A+B+C’+D) (A+B’+C’+D)
.(A+B+C+D’) (A’+B+C+D’)
= ΠM(12,13,2,6,1,9)
= Σm(0,3,4,5,7,8,10,11,14,15)
= A’B’C’D’+ A’B’CD+ A’BC’D’+ A’BC’D+ A’BCD+ AB’C’D’+ AB’CD’+
AB’CD+ ABCD’+ ABCD