COURSE: ECSE – 323
MIDTERM EXAM
WINTER
2009
2
YOUR
NAME:
ANSWER
KEY
________________________________________________________________________
Question 1:
Booelan Logic Theory
(10 points)
The prime implicants of
the function
F(A,B,C,D) =
m(0,1,5,7,8,9,12,14,15) with
“Don’t cares” D = (3,11,13) are given in the following table
Group of Minterms
Prime implicant
A
B
C
D
P1:(1,3,9,11,7,15,5,13)



1
P2:(0,1,8,9)

0
0

P3:(1,3,5,7)
0


1
P4:(8,12,9,13)
1

0

P5:(12,13,14,15)
1
1


(5 points)
a)
Using the covering table produce
ALL
minimal sumofproducts
forms of
F.
(5 points)
b)
Repeat part a) but using this time the Petrick function.
________________________________________________________________________
ANSWER
a)
Group of Minterms
Prime implicant
A
B
C
D
0
1
5
7
8
9
12
14
15
essential
essential
P1:(1,3,9,11,7,15,5,13)



1
XXX X
X
P2:(0,1,8,9)

0
0

XX
P3:(1,3,5,7)
0


1
XXX
P4:(8,12,9,13)
1

0

X X X
P5:(12,13,14,15)
1
1


XXX
P2, P5 are essential primeimplicant and , therefore, they have to appear in every minimal
form
These prime implicants cover all minterms expect 5 and 7 which can be covered with
either
P1 or P3. P1 should be prefered because it contains only one variable, while
P3
requires is product term with two variables.Therefore, there is only one minimal sumof
products form, namely , F = P2 + P5 + P1 =
B
C
+
A B + D
b) The Petrick function
is
P = (P2)(P1+P2+P3)(P1+ P3)(P1+P3)(P2 + P4) )(P1+P2+P4)
(P4 + P5)(P5)(P1 + P5)
= (P2)(P5)(P1 + P3) = P2 P5 P1 + P2 P5 P3
Apparently, the Petrick function gives us two minimal forms, each one containing three
primeimplicants. Closer analysis of each one of these solutions shows that the solution
P2P5P1 contains fewer number of variables than P2P5P3, therefore,
P2P5P1 forms the
minimal sumofproducts of F,namely, F = P2 + P5 + P1 =
B
C
+
A B + D, as we
found in part a)