课后习题答案 第2章 逻辑代数及其化简(3)

2018-11-21 22:21

(5) F5=(6) F6=解答:

??m(1,3,4,6,7,9,11,12,14,15) m(0,2,4,7,8,9,12,13,14,15)

(1) F1? (3,5,6,7)?m卡诺图:

BC A 0 1

由卡诺图可知:F1?

(2) F2?00 0 0 01 0 1 11 1 1 10 0 1 (3,5,6,7)?AC?AB?BC ?m

(4,5,6,7,8,9,10,11,12,13) ?m卡诺图:

CD AB 00 01 11 10 由卡诺图可知:F2?

(3) F3?00 0 1 1 1 01 0 1 1 1 11 10 0 1 0 1 0 1 0 1 (4,5,6,7,8,9,10,11,12,13)?AB?AB?AC ?m(2,3,6,7,10,11,12,15) ?m卡诺图:

CD AB 00 01 11 10

00 0 0 1 0 01 0 0 0 0 11 10 1 1 1 1 11

1 1 0 1

(2,3,6,7,10,11,12,15)?ABCD?AC?BC?CD ?m由卡诺图可知:F3?

(4) F4? (1,,34,5,8,9,13,15)?m卡诺图:

CD AB 00 01 11 10

由卡诺图可知:F4?(5) F5?00 0 1 0 1 01 1 1 1 1 11 10 1 0 1 0 0 0 0 0 (1,,34,5,8,9,13,15)??mABD?ABC?ABD?ABC

(1,,34,6,7,9,11,12,14,15) ?m卡诺图:

CD AB 00 01 11 10

由卡诺图可知:F5?

(6) F6?00 0 1 1 0 01 1 0 0 1 11 10 1 1 1 1 0 1 1 0 (1,,34,6,7,9,11,12,14,15)?BD?BD?CD ?m(0,,24,7,8,9,12,13,14,15) ?m卡诺图:

CD AB 00 01 11

00 1 1 1 01 0 0 1 11 10 0 1 1 12 1 0 1 10

由卡诺图可知:

F6?1 1 0 0 (0,,24,7,8,9,12,13,14,15)?AB?AC?CD?ABC?BCD ?m2-13 对具有无关项AB+AC=0的下列逻辑函数进行化简: (1) F1=AC+AB (2) F2=AC+AB

(3) F3=ABC+ABD+ABD+ABCD (4) F4=BCD+ABCD+ABC+ABD (5)F5=ACD+ABCD+ABD+ABCD (6) F6=BCD+ABCD+ABCD 解答:

(1) F1?AC?AB

F1?AC?AB?AC?AB?AB?AC?AC?B?AC

(2) F2?AC?AB

解:

F2?AC?AB?AC?AB?AB?AC?B?C

(3) F3?ABC?ABD?ABD?ABCD

F3?ABC?ABD?ABD?ABCD?AB?AC?ABC?AB?ABCD?AB?AC

?ABC?B?ABCD?AC?AC?B?ACD?AC?B?C?ACD?B?C?AD

(4) F4?BCD?ABCD?ABC?ABD

13

F4?BCD?ABCD?ABC?ABD?BCD?ABCD?ABC?ABD?AB?AC

?BCD?ACD?ABC?ABD?AB?AC?ABCD?ACBD?ABC?AB?AC?AB?CD?AC?BD?ABC?CD?BD?ABC

(5) F5?ACD?ABCD?ABD?ABCD

F5?ACD?ABCD?ABD?ABCD?ACD?ABCD?ABD?ABCD?AB?AC?ACD?ABD?ABD?ABCD?AB?AC ?ACD?AD?ABCD?AB?AC?AD?ABCD?AB?AC?AD?BCD?AB?AC?AD?BCD

(6) F6?BCD?ABCD?ABCD

F6?BCD?ABCD?ABCD?BCD?ABCD?ABCD?AB?AC

?BCD?AB?BCD?AC?ABD?BCD?AB?AD?BCD?AC?BCD?BCD?AD

2-14 化简下列具有无关项?的逻辑函数: (1) F1=(2) F2=(3) F3=(4) F4=(5) F5=(6) F6=解答:

(1)F1?邋m(0,1,3,5,8)+ (10,11,12,13,14,15)

(10,11,12,13,14,15) (5,6,8,9,10,11) (1,5,6,9,10,11,12) (1,2,3,9,10,11) (5,7,13,15)

邋m(0,1,2,3,4,7,8,9)+邋m(2,3,4,7,12,13,14)+邋m(0,2,7,8,13,15)+邋m(0,4,6,8,13)+邋m(0,2,6,8,10,14)+?m(0,1,3,5,8)???(10,11,12,13,14,15)

14

卡诺图如图所示:

CD AB 00 01 11 10

00 1 0 Φ 1 01 1 1 Φ 0 11 10 1 0 Φ Φ 0 0 Φ Φ 由卡诺图可知:F1?ABD?BCD?BCD

(2)F2??m(0,1,2,3,4,7,8,9)???(10,11,12,13,14,15)

CD AB 00 01 11 10 00 1 1 Φ 1 01 1 0 Φ 1 11 10 1 1 Φ Φ 1 0 Φ Φ 卡诺图如图所示:

由卡诺图可知:F2?B?CD?CD

(3)F3??m(2,3,4,7,12,13,14)???(5,6,8,9,10,11)

CD AB 00 01 11 10 00 0 1 1 Φ 01 0 Φ 1 Φ 11 10 1 1 0 Φ 1 Φ 1 Φ 卡诺图如图所示:

由卡诺图可知:F3?AC?AC?BD

(4)F4??m(0,2,7,8,13,15)???(1,5,6,9,10,11,12)

卡诺图如图所示:

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