1. 1 p.213 6(a) 8 words and 4 bits per word 0 1 1 0 1 0 0 0 0 1 1 0 1 1 0 1 1 1 1 0 0 0 0 1 What will be the values of F1, F2, F3, F4 if A=0, B=1, C=1? Answer: For ABC = 011 m3 is selected F1=1, F2= 0, F3=1, F4=0 Give the minterm expansion for F1, F2 F1 =  m(1,3,4,5) F2 =  m(0,4,5,7) F1 F2 F3 F4 A B C DECODERDECODER 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) m0 m1 m2 m3 m4 m5 m6 m7 m8 m9 m10 m11 m12 m13 m14 m15 A B C F1 F2 F3 AB 0000 0404 1 12 1818 00 01 11 10 C CD A D B 1 5 13 9 2 6 14 10 3 7 15 11

2. 2 X=A’BD+C’D+AB’+AB’C’D’ Y=A’BD+BCD+AB’ Z=A’BD+BCD+ABC+AB’C’D’ # ABCD XYZ 0 0000 000 1 0001 100 2 0010 000 3 0011 000 4 0100 000 5 0101 111 6 0010 000 7 0111 111 8 1000 111 9 1001 110 A 1010 110 B 1011 110 C 1100 000 D 1101 100 E 1110 001 F 1111 011 (a) (b)

3. 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) I0 I1 I2 I3 I4 I5 I6 I7 I8 I9 I10 I11 I12 I13 I14 I15 E’ 1 0 E’ 3 2 0 5 4 1 7 6 1 9 8 E’ 15 14 1 13 12 1 11 10 E’ 17 16 1 19 18 E’ 31 30 0 21 20 0 23 22 0 24 25 0 27 26 E 29 28 F Note: On line I4 we selected the don’t care 9 to simplify from E’ to 1. For logic one on this line we only need to connect to +5V. This leaves less of a current load on whatever gate supplies E’. A B C D S 3 S 2 S 1 S 0

4. 4 A B C S 2 S 1 S 0

5. 5

6. 6

7. 7 (a) (b)

8. 8 (OR is in the set)

9. 9