Logic and Computer Design Fundamentals
Chapter 2 – Combinational
Logic Circuits
Part 1 – Gate Circuits and Boolean Equations
2-3 Standard Forms
 There are four types of standard form
•
•
•
•
Sum of Minterm (SOM)
Product of Maxterm (POM)
Sum of Product (SOP)
Product of Sum (POS)
 SOM and POM are useful to specify
Boolean functions in a form that:
• Allows comparison for equality.
• Has a correspondence to the truth tables
Chapter 2 - Part 1
2
Minterms
 Given a function: L( D, X , A)  DX  A
Inputs
D
Outputs
X
L(D,X,A)
A
 Input variables are D, X, and A
 Minterms are AND terms with every variable
present in either true or complemented form.
DXA
D’XA’
A’XD
DAX’
…..
DX
D’A’
X’D
A
…..
These are NOT Minterm because …..
they are just called products
This is still a products: A
= A.1
Minterm examples
Chapter 2 - Part 1
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Minterms
 For function L(D,X,A) How many minterms do we
have?
 Let us give a binary number
to each minterm, such for
each variable we use 0 or 1,
depending on whether it is
complemented or not:
D X’A
(1 0 1)2 = (5)10
D’X’A’
D’X’A
D’X A’
D’X A
D X’A’
D X’A
D X A’
D XA
 In that case we can see that we will have 23 = 8 Minterms
in total (see table) going from D’X’A’ (000) to DXA (111).
Minterm examples
Chapter 2 - Part 1
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Minterms for three variables
 By convention, Minterms are named as m0, m1, …. m7
are named as shown in table below.
 Also note that m0 = 1 if and if x=0 and y=0 and z= 0
 Similarly for all other Minterms
Chapter 2 - Part 1
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Maxterms
L( D, X , A)  DX  A
 Maxterms are OR terms with every
variable in true or complemented form.
D’+X+A’
D+X+A
D+A+X’
A’+X+D
…..
D+X
D’+A’
X’+D
A
…..
These are NOT Maxterm because …..
they are just called SUMs
This is still a SUM : A
= A+0
Maxterm examples
Chapter 2 - Part 1
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Maxterms for three variables
Chapter 2 - Part 1
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Minterm and Maxterm Relationship
 Review: DeMorgan's Theorem
x · y  x  y and x  y  x ×y
 Two-variable example:
M 2  x  y and m 2  x·y
Thus M2 is the complement of m2 and vice-versa.
 Since DeMorgan's Theorem holds for n variables,
the above holds for terms of n variables
 giving:
M i  m i and m i  M i
Thus Mi is the complement of mi.
Chapter 2 - Part 1
8
Functions in terms of Minterms
 Recall any function is defined by a truth table.
 From the table we can say that L=1 if (D=0 and X=0 and
A=0) or (D=0 and X=0 and A=1) or (D=1 and X=0 and
A=0)
Any function can be expressed as
a sum of Minterms from its truth
table.
L=1 if
D’X’A’
D’X’A
D’X A’
D’X A
D X’A’
D X’A
D X A’
D XA
(D=0 and X=0 and A=0) or
(D=0 and X=0 and A=1) or
(D=1 and X=0 and A=0)
Thus in Boolean
L = (D’ . X’ . A’) + (D’ . X’ . A) + (D . X’ . A’)
L=
m0
+
m1
+
m4
Short hand notation: F1 =m (0,1,4)
Chapter 2 - Part 1
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Minterms and Maxterms










L = m0 + m1 + m4
What about L’ ?
Also L = (L’ ) ’
From the table we can say that
L’ = m2+ m3+ m5+ m6 + m7
D’X’A’
D’X’A
D’X A’
Thus L = (L’ ) ’
D’X A
D X’A’
L = (m2 + m3+ m5 + m6+ m7 ) ’
D X’A
D X A’
Using Demorgans Law:
D XA
L = m2 ’ . m3 ’ . m5 ’ . m6 ’ . m7 ’
Since mi ’ = Mi
L = M2 . M3 . M5 . M6 . M7
L = m0 + m1 + m4 = M2 . M3 . M5 . M6 . M7
L = m (0,1,4) = ΠM(2,3,5,6,7)
Chapter 2 - Part 1
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Conversion of Minterm and
Maxterm
 m(0, 2, 5, 7) = ΠM(1,3,4,6)
  m(1, 3, 4, 6) = ΠM(0,2,5,7)
F  XY Z  XYZ  XY Z  XYZ  m0  m2  m5  m7 
F  XY Z  XYZ  XY Z  XYZ  m1  m3  m4  m6
F = m (0,2,5,7) = ΠM(1,3,4,6)
F = m (1,3,4,6) = ΠM(0,2,5,7)
F = m (1,3,4,6) = ΠM(0,2,5,7) = m (0,2,5,7) = ΠM(1,3,4,6)
Chapter 2 - Part 1
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Alternatively: use Truth Table
 For the function table, the maxterms used
are the terms corresponding to the 0's.
F(A,B,C)= A+A’B’
ABC
000
001
010
011
100
101
110
111
F
1
1
0
0
1
1
1
1
m0
m1
M2
M3
m4
m5
m6
m7
F = m0. m1. m4. m5. m6. m7
F = M2.M3
= (A+B’+C)(A+B’+C’)
5 variable Example
 F(A, B, C, D, E) = m2 + m9 + m17 + m23
 F(A, B, C, D, E) write in standard form:
A’B’C’DE’ + A’BC’D’E + AB’C’D’E + AB’CDE
m2
m9
m17
m23
 Sum of Minterm(SOM) expression:
• F = Σm(2, 9, 17, 23)
F = DX + XA
ΠM(0,2,3,5,6)
Convert non-SOP expression to
SOP or POS expression
F = AB + C(D+E) = AB + CD + CE
NOT in SOP form
NOR SOP form
IN in SOP form
F = AB + C(D+E) = {A + C(D+E) } . {B + C(D+E) }
F = {(A+C) . (A+(D+E)) } . {(B+C) . (B+(D+E)) }
F = (A+C) . (A+D+E) . (B+C) . (B+D+E)
IN in POS form
Chapter 2 - Part 1
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Convert non-SOP expression to SOP
expression
F  AB  C ( D  E )  AB  CD  CE
 The decision whether to use a two-level or multiple-level
implementation is complex.
 no. of gates
No. of gate inputs
 amount of time delay
Chapter 2 - Part 1
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Converting a from SOP to
SOM form: F(A,B,C) = A+B’C
 Write the function as a canonical SOM
 There are three variables, A, B, and C which we take to
be the standard order.
 To add the missing variables:
“ANDing” any term that has a missing variable
with a term 1=( X + X’).
 F=A+B’C = A(B+B’)(C+C’) + B’C(A+A’)
= ABC + ABC’ + AB’C + AB’C’ + AB’C + A’B’C
= ABC + ABC’ + AB’C + AB’C’ + A’B’C
= m7 + m6 + m5 + m4 + m1
= m1 + m4 + m5 + m6 + m7
Converting a from SOP to POM
form: F(A,B,C) = A+B’C
 Any Boolean Function can be expressed as a
Product of of Maxterms (POM).
• For an expression, apply the second
distributive law , then “ORing” terms
missing variable x with a term equal to
0=(x.x’) and then applying the distributive
law again.
F(A,B,C)= A+A’B’
Apply the distributive law:
F= A+A’B’ = (A+A’)(A+B’)
= 1.(A+B’)
Add missing variable C:
F= A+B’+CC’
= (A+B’+C)(A+B’+C’)
= M2.M3
A Simplification Example
 Simplify F F( A, B, C)  m(1,4,5,6,7)
 Writing the minterm expression:
F = A’ B’ C + A B’ C’ + A B C’ + AB’C + ABC
 Simplifying using Boolean algebra (add AB’C):
F=
Simplification of two-level implementation of
SOP expression
 The two implementations for F are shown
below – it is quite apparent which is simpler!
A
B
C
A
B
C
A
B
C
A
B
C
A
B
C
A
F
B
C
F
Is there an easier way to simplify.
Yes, using k-maps
Chapter 2 - Part 1
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