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(Joyce) #1
PROBLEMS 331

x

y

z

f

Figure P6.1.28

z

y

x

f

Figure P6.1.29

xyzf

0001
0010
0101
0110
1001
1010
1100
1110

6.1.32With the use of a K map, simplify the following
Boolean expressions and draw the logic diagram.


(a) F 1 =A·B+B ̄·C+A·B·D+A·C·D

(b) F 2 =(X+Y)·(X ̄+Z)·(Y+Z) ̄

(c) F 3 =A·C+B·C ̄+A·B·C

(d) F 4 =(X ̄+Y)·(X+Z) ̄ ·(Y+Z)

6.1.33The K map of a logic function is shown in Figure
P6.1.33.
(a) Obtain a POS expression and its correspond-
ing realization.
(b) For the purpose of comparison, obtain the
corresponding SOP circuit, and comment on
the number of gates needed.


*6.1.34Given the K map of a logic function as shown
in Figure P6.1.34, in whichds denote don’t-care
conditions, obtain the SOP expression.
6.1.35The K map of a logic function is shown in Figure
P6.1.35, in whichds denote don’t-care conditions.
Obtain the SOP expressions.
6.1.36Obtain a minimum two-level NAND–NAND
realization for the following Boolean expressions.
(a)F(A,B,C)=


mi( 1 , 6 )+


di( 2 , 4 , 5 )
(b)F(A,B,C,D)=


∑ mi(^0 ,^4 ,^5 ,^7 ,^13 )+
di( 2 , 6 , 8 , 10 , 11 )
Note that


di()denotes the sum of minterms
corresponding to don’t-care outputs.
6.1.37(a) Show the equivalent NOR realizations of the
basic NOT, OR, and AND gates.
(b) Show the equivalent NAND realization of the
basic NOT, AND, and OR gates.
6.1.38Using a minimum number of NAND gates, realize
the following Boolean expression:∑ F(A,B,C)=
mi( 0 , 3 , 4 , 5 , 7 ).
6.1.39Figure P6.1.39 shows afull adderwith the idea of
addingCito the partial sumS′, which is the same
logic process as addition with pencil and paper.
(a) Draw the truth table for the full adder.
(b) Add decimal numbersA=7 andB=3in
binary form, showing the values ofA, B, S′,

AB
00
01
11
10

00 01 11 10
0000
0000
0001
0100

CD

Figure P6.1.33

AB
00
01
11
10

00 01 11 10
0111
d 1 0 d
100 d
010 d

CD

Figure P6.1.34

AB
00
01
11
10

00 01 11 10
1100
0010
0 d dd
0 1 dd

CD

Figure P6.1.35
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