數(shù)字邏輯第二版毛法堯課后題答案

1、第一章第一章 數(shù)制與碼制數(shù)制與碼制1.1 1.1 把下列各進(jìn)制數(shù)寫成按全展開的形式把下列各進(jìn)制數(shù)寫成按全展開的形式（1 1）（）（4517.2394517.239）1010 =4 =410103 3+5+510102 2+1+110101 1+7+710100 0+2+21010-1-1+3+31010-2 -2 +9 +91010-3-3（2 2）（）（10110.010110110.0101）2 2 =1 =12 24 4+0+02 23 3+1+12 22 2+1+12 21 1+0+02 20 0+0+02 2-1 -1 +1+12 2-2 -2 +0 +02 2-3-3+1+12 2

2、-4-4（3 3）（）（325.744325.744）8 8 =3 =38 82 2+2+28 81 1+5+58 80 0+7+78 8-1-1+4+48 8-2-2+4+48 8-3-3（4 4）（）（785.4af)16785.4af)16 =7 =716162 2+8+816161 1+5+516160 0+4+41616-1-1+a+a1616-2-2+f+f1616-3-31.2 1.2 完成下列二進(jìn)制表達(dá)式的運(yùn)算完成下列二進(jìn)制表達(dá)式的運(yùn)算（1 1）10111+101.101 10111+101.101 （2 2）1100-111.0111100-111.011（3 3）10.01

3、10.011.01 1.01 （4 4）1001.00011001.000111.10111.101 10111 10111+ 101.101+ 101.101= 11100.101= 11100.101 11000.000 11000.000- 00111.011- 00111.011= 10000.101= 10000.101 10.01 10.01 1.011.01 1001 1001 0000 0000 1001 1001 10.1101 10.1101 10.0110.0111101 )1001000111101 )10010001 1.5 1.5 如何判斷一個(gè)二進(jìn)制正整數(shù)如何判斷一

4、個(gè)二進(jìn)制正整數(shù)b=bb=b6 6b b5 5b b4 4b b3 3b b2 2b b1 1b b0 0能否被（能否被（4 4）1010整整除？除？解：解：b b1 1b b0 0同為同為0 0時(shí)能整除，否則不能。時(shí)能整除，否則不能。1.6 1.6 寫出下列各數(shù)的原碼、反碼和補(bǔ)碼。寫出下列各數(shù)的原碼、反碼和補(bǔ)碼。（1 1）0.1011 0.1011 （2 2）0.0000 0.0000 （3 3）-10110-10110解：解：0.10110.1011原原=0.1011=0.1011反反=0.1011=0.1011補(bǔ)補(bǔ)=0.1011=0.1011 0.0000 0.0000原原=0.0000=

5、0.0000反反=0.0000=0.0000反反=0.0000=0.0000 -10110 -10110原原=110110=110110 -10110 -10110反反=101001=101001 -10110 -10110反反=101010=101010 1.7 1.7 已知已知nn補(bǔ)補(bǔ)=1.0110=1.0110，求，求nn原原、nn反反和和n n解：解： nn原原=1.1010 n=1.1010 n反反和和=1.1001 n=-0.1010=1.1001 n=-0.10101.8 1.8 用原碼、反碼和補(bǔ)碼完成如下運(yùn)算用原碼、反碼和補(bǔ)碼完成如下運(yùn)算（1 1）0000101-0011010

6、0000101-0011010 解解（1 1）0000101-00110100000101-0011010原原=10010101=10010101 0000101-0011010=-0010101 0000101-0011010=-0010101 0000101-0011010 0000101-0011010反反=0000101=0000101反反+-0011010+-0011010反反 =00000101+11100101=11101010=00000101+11100101=11101010 0000101-0011010=-0010101 0000101-0011010=-0010101

7、 0000101-0011010 0000101-0011010反反=0000101=0000101補(bǔ)補(bǔ)+-0011010+-0011010補(bǔ)補(bǔ) =00000101+11100110=11101011=00000101+11100110=11101011 0000101-0011010=-0010101 0000101-0011010=-00101011.8 1.8 用原碼、反碼和補(bǔ)碼完成如下運(yùn)算用原碼、反碼和補(bǔ)碼完成如下運(yùn)算 （2 2）0.010110-0.1001100.010110-0.100110解解（2 2）0.010110-0.1001100.010110-0.100110原原=1

8、.010000=1.010000 0.010110-0.100110=-0.010000 0.010110-0.100110=-0.010000 0.010110-0.100110 0.010110-0.100110反反=0.010110=0.010110反反+-0.100110+-0.100110反反 = 0.010110+1.011001=1.101111= 0.010110+1.011001=1.101111 0.010110-0.100110=-0.010000 0.010110-0.100110=-0.010000 0.010110-0.100110 0.010110-0.10011

9、0補(bǔ)補(bǔ)=0.010110=0.010110補(bǔ)補(bǔ)+-0.100110+-0.100110補(bǔ)補(bǔ) = 0.010110+1.011010=1.110000= 0.010110+1.011010=1.110000 0.010110-0.100110=-0.010000 0.010110-0.100110=-0.0100001.9 1.9 分別用分別用“對(duì)對(duì)9 9的補(bǔ)數(shù)的補(bǔ)數(shù)“和和”對(duì)對(duì)1010的補(bǔ)數(shù)完成下列十進(jìn)制的補(bǔ)數(shù)完成下列十進(jìn)制數(shù)的運(yùn)算數(shù)的運(yùn)算（1 1）2550-1232550-123解解：（：（1 1）2550-1232550-1239 9補(bǔ)補(bǔ)=2550-0123=2550-01239 9補(bǔ)補(bǔ)=

10、2550=25509 9補(bǔ)補(bǔ)+-0123+-01239 9補(bǔ)補(bǔ) =02550+99876=02427=02550+99876=02427 2550-123=+2427 2550-123=+2427 2550-1232550-1231010補(bǔ)補(bǔ)=2550-0123=2550-01231010補(bǔ)補(bǔ)=2550=25501010補(bǔ)補(bǔ)+-0123+-01231010補(bǔ)補(bǔ) =02550+99877=02427=02550+99877=02427 2550-123=+2427 2550-123=+24271.9 1.9 分別用分別用“對(duì)對(duì)9 9的補(bǔ)數(shù)的補(bǔ)數(shù)“和和”對(duì)對(duì)1010的補(bǔ)數(shù)完成下列十進(jìn)制的補(bǔ)數(shù)完成下

11、列十進(jìn)制數(shù)的運(yùn)算數(shù)的運(yùn)算 （2 2）537-846537-846解解：（：（2 2）537-846537-8469 9補(bǔ)補(bǔ)=537=5379 9補(bǔ)補(bǔ)+-846+-8469 9補(bǔ)補(bǔ) =0537+9153=9690=0537+9153=9690 537-846=-309 537-846=-309 537-846537-8461010補(bǔ)補(bǔ)=537=5371010補(bǔ)補(bǔ)+-846+-8461010補(bǔ)補(bǔ) =0537+9154=9691=0537+9154=9691 537-846=-309 537-846=-3091.10 1.10 將下列將下列8421bcd8421bcd碼轉(zhuǎn)換成十進(jìn)制數(shù)和二進(jìn)制數(shù)碼轉(zhuǎn)換

12、成十進(jìn)制數(shù)和二進(jìn)制數(shù)(1)011010000011(1)011010000011(2)01000101.1001(2)01000101.1001解解：(1)(1)（011010000011011010000011）8421bcd8421bcd=(683)=(683)d d=(1010101011)=(1010101011)2 2 (2)(01000101.1001) (2)(01000101.1001)8421bcd8421bcd=(45.9)=(45.9)d d=(101101.1110)=(101101.1110)2 21.11 1.11 試用試用8421bcd8421bcd碼、余碼、余3

13、 3碼和格雷碼分別表示下列各數(shù)碼和格雷碼分別表示下列各數(shù)(1)578)(1)578)1010 (2)(1100110)(2)(1100110)2 2解解：( (578)578)1010 =(010101111000)=(010101111000)8421bcd8421bcd =(100010101011) =(100010101011)余余3 3 = =（10010000101001000010）2 2 = =（11011000111101100011）g g 解解：（11001101100110）2 2 = =（10101011010101）g g = =( (102)102)1010 =(

14、000100000010)=(000100000010)8421bcd8421bcd =(010000110101) =(010000110101)余余3 3 1.12 1.12 將下列一組數(shù)按從小到大順序排序?qū)⑾铝幸唤M數(shù)按從小到大順序排序(11011001)(11011001)2 2,(135.6),(135.6)8 8,(27),(27)1010,(3af),(3af)1616,(00111000),(00111000)8421bcd8421bcd(11011001)(11011001)2 2=(217)=(217)1010 (135.6) (135.6)8 8=(93.75)=(93.7

15、5)1010 (3af) (3af)1616=(431)=(431)1010 (00111000) (00111000)8421bcd8421bcd=(38)=(38)1010按從小到大順序排序?yàn)椋喊磸男〉酱箜樞蚺判驗(yàn)椋?27)(27)10 10 , (00111000), (00111000)8421bcd 8421bcd ,(135.6),(135.6)8 8,(11011001),(11011001)2 2 (3af) (3af)1616, ,)1111,1101,1100,0111,0100() 1 (cabbdf)1111,1101,1011,1001,0111,0101,0011,

16、0001()()()2(ddbabadbabababaf時(shí)，即：時(shí)，為或11101101110010101001100001110110010101000010000100001010, 0)()()3(fabdcdcbadcbadcadacdbadcaafcabacaab)( ) 1 (cabacbcabacabacaab）（證明：)(1)2(bababaab1aabababaab證明：cabcbacbaabca) 3(cabacbaaabca)(證明：cabacabcbacbacabcbacbaabcacbaabccabccabacacbbacacbba)()()(證明：cacbbacba

17、abc)4(babcbaabc)5(bababccbbacababcbacbabcbaabc)(證明：cdbbdacacbbacadca)()()6(cdbbdacadcbabcdacdbacdbadcbadcabcadcabdcbacdbadcbabcdacdbacadcbdbacdacacbbacadcacbbacadca)()()(證明：)()() 1 (cbcafcbcafcbcaf)()()()()()2(dcacbbafdcacbbafdcacbbaf)()()()3(gfedcbafgfedcbafgfedcbafedcbafedcbafedcbaf)4()()()()()5(c

18、abbafcabbacabbafcabbaf11)2(bababababaabbaafbabbabcdbbaf) 1 (dbadcadbadadcadbadbaabdcadbadabf)3(adeacedceadacdcedcaadcedcacbaaf)()()()()4(cabcbbcaacf)5(bccabccbbaccbacbbaccbacbbcaaccabcbbcaacf)()()()()()(bcbacbcbacbbaccbacbbcaaccabcbbcaacf)()()()()()5(edcadcedcadcedcaadcedcacbaaf)()()()()()4()12, 8 ,

19、 4 , 3 , 2 , 1 , 0()(),()3()13,12, 9 , 8 , 7 , 6 , 5 , 4(),()2()7 , 6 , 5 , 4 , 3()()(),() 1 (7 . 2mdcabadcbbcadcbafmdcbbcdcabbadcbafmcbacabacabacbaf)7 , 6 , 5 , 3(),()3()6 , 5 , 3 , 0() ,()2()4 , 2 , 1 , 0(),() 1 (8 . 2mcbafmcbafmcbafdcbdcacdbdcbadcacdbdcadcbadcabcdbcddacbadccbcdbfff)()9 . 221)()()

20、() 1 (cbcabacabbafcacbf) 1 ()(bacffabcf最簡(jiǎn)或與表達(dá)式：與”表達(dá)式或”表達(dá)式和最簡(jiǎn)“或并寫出最簡(jiǎn)“與，用卡諾圖化簡(jiǎn)下列函數(shù)2.10cbacdcabadcbaf),()2(abcd00011110000111101110111111111000cbacbadcbaf),()2()(cbacbaffbcacbaf最簡(jiǎn)或與表達(dá)式：baddbcbadcbddbcdcbaf)(),() 3(dbf)3(abcd000111100001111011101111111110001111000000001111abcd0001111000011110dacdcdadbdc

21、baf),() 1 (的關(guān)系、和、用卡諾圖判斷函數(shù)d)cbg(ad)cbf(a2.110000111111110000abcd0001111000011110abddcacddbdcbag),() 1 (gf ) 1 (abccbaabccbacaabccbacacbaabccbacacbcabaabccbacacababccbacacbbaabccbaacbcabdcbag)()()()()()()()()()(),()2(1 11 11 11 1abcd00011110000111101 11 11 11 11 11 11 11 1abcd0001111000011110abccbacba

22、cbacbabacbacbacbabacbabadcbaf)()()(),()2(dcacf1(2)adbcac0112. 2cbbdcacbfa時(shí)時(shí)）當(dāng)（)10, 8 , 6 , 5 , 4 , 3 , 1 ()15,13, 7 , 2 , 0(),(113. 2dmdcbaf）（abcd00011110000111101 1d d0 0d dd dd d1 10 0d d1 11 10 01 1d d0 0d dbdaf)15,13,10, 8 , 7 , 4 , 2 , 0(),(213. 21mdcbaf）（)10, 8 , 7 , 6 , 5 , 2 , 1 , 0(),(2mdcb

23、af)7 , 4 , 3 , 2(),(3mdcbaf1 11 11 11 11 11 11 11 1ababcdcd000001011111101000000101111110101 11 11 11 11 11 11 11 1ababcdcd000001011111101000000101111110101 11 11 11 1ababcdcd00000101111110100000010111111010abdbcdadcbadbf12）（dcabcdadcadbf2bcdadcbacbaf33.1將下列函數(shù)化簡(jiǎn)，并用將下列函數(shù)化簡(jiǎn)，并用“與非與非”、 “或非或非” 門畫門畫出邏輯電路圖

24、。出邏輯電路圖。)7 , 3 , 2 , 0(),() 1 (mcbaf)6 , 3(),()2(mcbafcbcadcabadcbaf),()3(cdbcaabdcbaf),()4(解解（1 1）bccabccafcbcafcbcacbcaff)()6 , 3(),()2(mcbaf解解（1 1）accabaccabfbcacabfcbacbacbacbaff)(cbcadcabadcbaf),()3(解解: :bacbcabacbcafcbcadcabadcbaf),()3(解解: :abccbafcbacbacbacbaff)(bacdbcabacdbcaabdcbaf),()4(解解：

25、abf &abfbabaf1abf1cdbcabadcbaf),()4(解解：daabcacbfdabacacbdabacacbff)()()()(cbabaabcbaf)(),() 1 (3.2 3.2 將下列函數(shù)化簡(jiǎn)，并用將下列函數(shù)化簡(jiǎn)，并用“與或非門與或非門”畫出邏輯電路圖。畫出邏輯電路圖。)15,14,13,10, 9 ,78, 6 , 2 , 1 (),()2(mdcbaf解解：bcacbaabcbabaabcbaf)(),() 1 (bcacabbcacabf)15,14,13,10, 9 ,78, 6 , 2 , 1 (),()2(mdcbaf解解：cbadcbdcadcbdcb

26、adcbdcadcbdfabcabcz z0000000 00010011 10100100 00110111 11001000 01011010 01101101 11111111 13.3 3.3 解：由時(shí)間圖得真值表如下：解：由時(shí)間圖得真值表如下：)7 , 6 , 3 , 1 (),(mcbafbacafbacabacaff)(3.43.4解：解：cbaabccbabcbcaccbbcacbcbcacbcbaf)()()()(3.5 13.5 1）解：）解：cbafbccabag一位二進(jìn)制數(shù)全減器：一位二進(jìn)制數(shù)全減器：2 2）全加器）全加器3. 6 3. 6 解：解：babaf1baba

27、f2babababaf3當(dāng)當(dāng)a=ba=b時(shí)等效時(shí)等效f f1 1=f=f2 2=f=f3 3=0=0xyxy y y3 3y y2 2y y1 1y y0 00000 000000000101 000100011010 010001001111 100110013.7 3.7 解解(1)(1)byybabayababy012302xy byyabaababayababy01230&ab3y2y1y0y11xyxy y y4 4y y3 3y y2 2y y1 1y y0 00000 00000000000101 00001000011010 01000010001111 11011110113

28、.7 3.7 解解(2)(2)byabyyayababy0123403xy &ab4y2y1y0y3y1c c1 1 c c0 0 x xy y0000000 00010010 00100100 00110111 11001001 11011010 01101101 11111111 13.8 3.8 解：真值表如下：解：真值表如下：xcxcxcxcf0101)7 , 6 , 4 , 3(),(01mxccfy&11cxy0c&3.9 3.9 依題意得真值表如下：依題意得真值表如下：b b8 8b b4 4b b2 2b b1 1f f7 7f f6 6f f5 5f f4 4f f3 3f

29、f2 2f f1 1f f0 00 0 0 00 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 10 0 0 1 0 0 0 0 0 1 0 10 0 0 0 0 1 0 10 0 1 00 0 1 0 0 0 0 1 0 0 0 00 0 0 1 0 0 0 00 0 1 10 0 1 1 0 0 0 1 0 1 0 10 0 0 1 0 1 0 10 1 0 00 1 0 0 0 0 1 0 0 0 0 00 0 1 0 0 0 0 00 1 0 10 1 0 1 0 0 1 0 0 1 0 10 0 1 0 0 1 0 10 1 1 00 1 1 0

30、 0 0 1 1 0 0 0 00 0 1 1 0 0 0 00 1 1 10 1 1 1 0 0 1 1 0 1 0 10 0 1 1 0 1 0 11 0 0 01 0 0 0 0 1 0 0 0 0 0 00 1 0 0 0 0 0 01 0 0 11 0 0 1 0 1 0 0 0 1 0 10 1 0 0 0 1 0 1依真值表得依真值表得:07f86bf 45bf 24bf 03f12bf 01f10bf 3.10 3.10 依題意得真值表如下：依題意得真值表如下：y y1 1y y0 0 x x1 1x x0 0z z1 1 z z0 0000000001111000100010

31、101001000100101001100110101010001001010010101011111011001100101011101110101100010001010100110011010101010101111101110110101110011001010110111011010111011101010111111111111)15,14,13,12,10, 9 , 8 , 5 , 4 , 0(1mz)15,11,10, 7 , 6 , 5 , 3 , 2 , 1 , 0(0mz)15,14,13,12,10, 9 , 8 , 5 , 4 , 0(1mzabcddcbacadad

32、cz0abcddcbacacbbaz1)15,11,10, 7 , 6 , 5 , 3 , 2 , 1 , 0(0mz3.113.11依題意得真值表如下：依題意得真值表如下：b b1 1b b0 0b b1 1b b0 0f f000000001 1000100010 0001000100 0001100111 1010001000 0010101011 1011001101 1011101110 0100010000 0100110011 1101010101 1101110110 0110011001 1110111010 0111011100 0111111111 1)15,12,10,

33、 9 , 3 , 5 , 3 , 0(mf1234bbbbf=1=1=1=1=1=1=1=14b3b2b1b1f bcd 碼 0 0 1 1 余 3 碼 s4 s3 s2 s1 c4 c0 a4 a3 a2 a1 b4 b3 b2 b1 bcd 碼 0 0 1 1 余 3 碼 s4 s3 s2 s1 c4 c0 a4 a3 a2 a1 b4 b3 b2 b1 6.16.1：用兩個(gè)：用兩個(gè)4 4位二進(jìn)制并行加法器實(shí)現(xiàn)兩位十進(jìn)制位二進(jìn)制并行加法器實(shí)現(xiàn)兩位十進(jìn)制8421bcd8421bcd碼到余碼到余3 3碼的轉(zhuǎn)換碼的轉(zhuǎn)換高位高位低位低位a3a2a1a0b3b2b1b0abab a=b abab a

34、=b ab74ls85(2)0 0a a7 7a a6 6a a5 50 0b b7 7b b6 6b b5 5a a4 4a a3 3a a2 2a a1 1b b4 4b b3 3b b2 2b b1 16.26.2：用兩塊：用兩塊4 4位數(shù)值比較器芯片實(shí)現(xiàn)兩個(gè)位數(shù)值比較器芯片實(shí)現(xiàn)兩個(gè)7 7位二進(jìn)制的比較位二進(jìn)制的比較6.36.3：用：用3-83-8線譯碼器線譯碼器7413874138和必要邏輯門實(shí)現(xiàn)下列函數(shù)和必要邏輯門實(shí)現(xiàn)下列函數(shù)yxxyzyxfyxzyxfzxyyxzyxf),(),(),(321解：解：6106101),(mmmmmmzxyzyxzyxzxyyxzyxf7632107

35、632102),(mmmmmmmmmmmmxyzzxyyzxzyxyzxzyxzyxzyxyxzyxf761076103),(mmmmmmmmxyzzxyzyxzyxyxxyzyxfa074ls138y0a1a2g2ag1g2by1y2y3y4y5y6y7&f1xyz 100&f2f3變量多數(shù)表決電路）（全加器用四選一電路實(shí)現(xiàn)函數(shù)32) 1 (:4 . 61111is) 1 (iiiiiiiiiiiicbacbacbacba：由全加器的真值表可得解：iiicba輸出，輸出輸出端，、制端輸入從四路選擇器的選擇控2ysy1i 12d 2d 2d 02d 1d1d 1d1d312110121130可得：iiiicccc1111iiiiiiiiiiiiicbacbacbacbac1a1aiaib01d11d21d31d02d12d22d32dy2y2isic 12d