Question 1 :Boolean Logic Theory (10 points)
(5 points)
a) Poof that the Shannon Expansion Theorem
F(A,B,C,…,Z) =
f8e5
A F(0,B,C,…,Z)
+
A F(1,B,C,…,Z)
can be generalized by replacing the
+ operator in the above expression by the
⊕
operator, namely, that
F(A,B,C,…,Z) =
f8e5
A F(0,B,C,…,Z)
⊕
A F(1,B,C,…,Z)
is true.
(5 points)
b) By using conveniently ONLY the ExOR and AND operators, convert
F(A,B,C) = AB +
f8e5
C
into an all positive function, i.e. a function that does
not have any complemented variable.
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ANSWER
a)
For A = 0,
F(A,B,C,…,Z) =
f8e5
A F(0,B,C,…,Z)
⊕
A F(1,B,C,…,Z)
=
1 . F(0,B,C,…,Z)
⊕
0
=
F(0,B,C,…,Z)
⊕
0
=
F(0,B,C,…,Z).
For A = 1,
F(A,B,C,…,Z) =
f8e5
A F(0,B,C,…,Z)
⊕
A F(1,B,C,…,Z)
=
0
⊕
1.
F(1,B,C,…,Z)
=
0
⊕
F(1,B,C,…,Z)
=
F(1,B,C,…,Z).
The same result is obtained substituting A = 0 and A = 1 in the expression of the Shannon
Expansion Theorem, F(A,B,C,…,Z) =
f8e5
A F(0,B,C,…,Z)
+
A F(1,B,C,…,Z). Thus, since the
Shannon Expansion Theorem is true, its generalization obtained by replacing the
+ operator, in
the above expression, by the
⊕
operator is also true.
b)
We expand
F (A,B,C) = AB +
f8e5
C
with respect to
C using the generalized Shannon
Expansion Theorem. We obtain
F(A,B,C) =
f8e5
C [ 1 ]
⊕
C [AB].
Replacing
f8e5
C
by 1
⊕
C, we obtain
F(A,B,C) = 1
⊕
C
⊕
CAB.
ECSE323 Digital Systems Design

Midterm Exam
 Fall 2007
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