1

p.213 6(a) 8 words and 4 bits per word

0 1 1 01 0 1 00 0 0 11 0 1 01 1 0 11 1 1 00 0 0 00 1 0 1

What will be the values of F1, F2, F3, F4if A=0, B=1, C=1?Answer: For ABC = 011 m3 is selectedF1=1, F2= 0, F3=1, F4=0

Give the minterm expansion for F1, F2F1 = m(1,3,4,5)F2 = m(0,4,5,7)

F1 F2 F3 F4

A

B

C

DECODER

p. 214 6(d ) Functions in Fig. 9-3 p. 218 to be realized by ROM

F1=AB+ACD, F2=ACD+ABC’+A’CD, F3=AB+A’CD-must express functions in minterm form -- can use a k-map or algebraic expansion to do this, e.g. AB = m(12,13,14,15) ACD = m(11,15)

m0m1m2m3m4m5m6m7m8m9m10m11m12m13m14m15

A

B

C

F1 F2 F3

AB 00

04

112

18 00

01

11

10 C

CD

A

D

B

1 5 13 9

2 6 14 10

3 7 15 11

2

X=A’BD+C’D+AB’+AB’C’D’Y=A’BD+BCD+AB’Z=A’BD+BCD+ABC+AB’C’D’

# ABCD XYZ0 0000 0001 0001 1002 0010 0003 0011 0004 0100 0005 0101 1116 0010 0007 0111 1118 1000 1119 1001 110A 1010 110B 1011 110C 1100 000D 1101 100E 1110 001F 1111 011

(a)(b)

3

9.17(a) p. 245 F(A,B,C,D,E) = m(0,2,6,7,8,10,11,12,13,14,16,18,19,29,30)

+ d(4,9,21)

I0I1I2I3I4I5I6I7I8I9I10I11I12I13I14I15

E’ 1 0E’ 3 20 5 41 7 6

1 9 8

E’ 15 14 1 13 121 11 10

E’ 17 161 19 18

E’ 31 30

0 21 200 23 220 24 25

0 27 26E 29 28

F

Note: On line I4 we selected thedon’t care 9 to simplify fromE’ to 1.For logic one on this line weonly need to connect to +5V.This leaves less of a current loadon whatever gate supplies E’.

A B C D

S3 S2 S1 S0

4

A B C

S2 S1 S0

5

6

7

(a)

(b)

8

(OR is in the set)

9