1
ECE 15A
Fundamentals of Logic Design
Lecture 5
Malgorzata Marek-Sadowska
Electrical and Computer Engineering Department
UCSB
2
Last time
±
Minterm and maxterm representations of Boolean
functions
±
Any Boolean function can be expressed as a sum of 1-
minterms (dnf)
∑
=
)
,
,
(
2
1
n
x
x
x
F
L
(1_minterms)
Example
x
y
z
minterms notation
F
0
0
0
x’y’z’
m0
0
0
0
1
x’y’z
m1
0
0
1
0
x’yz’
m2
0
0
1
1
x’yz
m3
1
1
0
0
x y’z’
m4
0
1
0
1
xy’z
m5
1
1
1
0
xyz’
m6
1
1
1
1
xyz
m7
1
∑
=
)
,
(
y
x
F
(m3,m5,m6,m7)
3
Last time
±
Any Boolean function can be expressed as a product of its 0-maxterms
(cnf)
∏
=
)
,
,
(
2
1
n
x
x
x
F
L
(0_maxterms)
Example
x
y
z
maxterms notation
F
0
0
0
x+y+z’
M0
0
0
0
1
x+y+z’
M1
0
0
1
0
x+y’+z
M2
0
0
1
1
x’+y+z
M3
1
1
0
0
x’+y+z
M4
0
1
0
1
x’+y+z’
M5
1
1
1
0
x’+y’+z
M6
1
1
1
1
x’+y’+z’
M7
1
∏
=
)
,
(
y
x
F
(M0,M1,M2,M4)
4
Canonical representations of Boolean
functions
±
Canonical forms are unique representations of Boolean functions.
±
dnf and cnf are canonical representations of Boolean functions
±
To convert between dnf and cnf, we interchange the sum and
product symbols and list those numbers which were absent in the
original form.
∑
=
)
,
(
y
x
F
(m3,m5,m6,m7)
∏
=
)
,
(
y
x
F
(M0,M1,M2,M4)
Example:
5
Boolean functions in standard forms
±
Sum-of-products (SOP)
²
Boolean expression that contains AND terms (called
products or implicants) which are ORed together
²
Example:
F1(x,y,z,w)=xy+z’+x’w’w’
±
Product of sums (POS)
²
Boolean expression that contains OR terms (called sum
terms or implicants) which are ANDed together
²
Example: F2=(x+z’+w)(x’+y+w’)
±
These normal forms are not unique: the same
functions can be expressed in many different ways
±
Standard forms usually contain fewer terms and
literals than the corresponding dnfs or cnfs.
6
XOR operation
0
1
0 0
1
1
1
0
A
B
= B
A
(commutative)
(A
B)
C
= A
(B
C)
(associative)
(AB)
(AC)
= A (B
C)
multiplication distributive over
L = (A
B)(A
C)
=
A
(BC)= R
A
B
C
A
B
A
C
L
BC
R
0
0
0
0
0
0
0
0
0
0
1
0
1
0
0
0
0
1
0
1
0
0
0
0
0
1
1
1
1
1
1
1
1
0
0
1
1
1
0
1
1
0
1
1
0
0
0
1
1
1
0
0
1
0
0
1
1
1
1
0
0
0
1
0
A
C
= B
B
C
= A
A
B
C
=
0
If
A
B = C, then