數(shù)字電子技術(shù)（佘新平 著） 華中科技大學(xué)出版社 課后答案-

1、第一章數(shù)制與編碼1.1 自測練習(xí)1.1.1、模擬量數(shù)字量1.1.2、(b1.1.3、(c1.1.4、(a是數(shù)字量,(b(c(d是模擬量1.2 自測練習(xí)1.2.1. 21.2.2. 比特bit1.2.3. 101.2.4. 二進制1.2.5. 十進制1.2.6. (a1.2.7. (b 1.2.8. (c 1.2.9. (b 1.2.10. (b 1.2.11. (b 1.2.12. (a 1.2.13. (c 1.2.14. (c 1.2.15. (c 1.2.16. 1001001 1.2.17. 111.2.18. 110010 1.2.19. 1101 1.2.20. 8進制1.2.21

2、. (a1.2.22. 0,1,2,3,4,5,6,71.2.23. 十六進制1.2.24. 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F 1.2.25. (b1.3 自測練習(xí)1.3.1. 1221.3.2. 675.521.3.3. 011111110.011.3.4. 521.3.5. 1BD.A81.3.6. 1110101111.11101.3.7. 38551.3.8. 28.3751.3.9. 100010.111.3.10. 135.6251.3.11. 570.11.3.12. 120.51.3.13. 2659.A1.4 自測練習(xí)1.4.1. BCD Bin

3、ary coded decimal 二十進制碼1.4.2. (a1.4.3. (b1.4.4. 8421BCD碼,4221BCD碼,5421BCD 1.4.5. (a1.4.6. 011001111001.10001.4.7. 111111101.4.8. 101010001.4.9. 111111011.4.10. 61.051.4.11. 01011001.01110101 1.4.12. 余3碼1.4.13. XS31.4.14. XS31.4.15. 1000.10111.4.16. 1001100000111.4.17. 521.4.18. 110101.4.19. 0101111.

4、4.20. (b1.4.21. ASCII1.4.22. (a1.4.23. ASCII American Standard Code for Information Interchange美國信息交換標(biāo)準(zhǔn)碼EBCDIC Extended Binary Coded Decimal Interchange Code 擴展二-十進制交換嗎1.4.24. 10010111.4.25. ASCII1.4.26. (b1.4.27. (b1.4.28. 110111011.4.29. -1111.4.30. +231.4.31. -231.4.32. -861.5 自測練習(xí)1.5.1 略1.5.2 11

5、0111011.5.3 010001011.5.4 11100110 補碼形式1.5.5 011111011.5.6 10001000 補碼形式1.5.7 11100010 補碼形式習(xí)題1.1 (a(d是數(shù)字量,(b(c是模擬量,用數(shù)字表時(e是數(shù)字量,用模擬表時(e是模擬量1.2 (a7, (b31, (c127, (d511, (e40951.3 (a,(b,(c(d522104108×+×+320410910×+×+×26108108×+×+321102105100×+×+×+21+1.4

6、 (a, (b, (c(d3212121×+×+984+12+12+×××4311212121×+×+×+212×64212+12+12+12+1××××1212+×2220110327.15310210710110510.=×+×+×+×+×3210-1-221011.0112+02+12+12+02+12=××××××210-18437.448

7、+38+78+48=××××10-1-2163A.1C316+A16+116+C16=××××,1.6 (a11110, (b100110,(c110010, (d10111.7 (a1001010110000, (b10010111111.8 110102 = 2610, 1011.0112 = 11.37510,57.6438 = 71.81835937510,76.EB16 = 118. 91796875101.9 1101010010012 = 65118 = D4916,0.100112 = 0.468

8、= 0.9816,1011111.011012 = 137.328 = 5F.68161.10 168 = 1410,1728 = 12210,61.538 = 49.671875,126.748 = 86.9375101.11 2A16 = 4210 = 1010102 = 528,B2F16 = 286310 = 1011001011112 = 54578,D3.E16= 211.87510 = 11010011.11102 = 323.78,1C3.F916 = 451.9726562510 =111000011.111110012 = 703.76281.12 (aE, (b2E, (

9、c1B3, (d3491.13 (a22, (b110, (c1053, (d20631.14 (a4094, (b1386, (c492821.15 (a23, (b440, (c27771.16 198610 = 111110000102 = 00011001100001108421BCD,67.31110 = 1000011.010012 =01100111.0011000100018421BCD, 1.183410 = 1.0010112 = 0001.00011000001101008421BCD ,0.904710 = 0.1110012 = 0000.10010000010001

10、118421BCD1.17 1310 = 000100118421BCD = 01000110XS3 = 1011Gray, 6.2510 = 0110.001001018421BCD =1001.01011000 XS3 = 0101.01Gray,0.12510 = 0000.0001001001018421BCD =0.010*XS3 = 0.001 Gray1.18 101102 = 11101 Gray,0101102 = 011101 Gray1.19 110110112 = 0010000110018421BCD,45610 = 0100010101108421BCD,1748=

11、0010011101008421BCD,2DA16 = 0111001100008421BCD,101100112421BCD = 010*BCD,11000011XS3 = 100100008421BCD1.20 0.0000原= 0.0000反= 0.0000補,0.1001原= 0.1001反= 0.1001補,11001原= 10110反= 10111補1.21 010100原= 010100補,101011原= 110101補,110010原= 101110補,100001原= 111111補1.22 1310 = 00001101補,11010 = 01101110補,-2510

12、= 11100111補,-90 =10100110補1.23 01110000補= 11210,00011111補= 3110,11011001補= -3910,11001000補= -56101.24 1000011 1000001 1010101 1010100 1001001 1001111 1001110 0100001 0100000 1001000 1101001 1100111 1101000 0100000 1010110 1101111 1101100 1110100 1100001 1100111 11001011.25 0100010 1011000 0100000 01

13、11101 0100000 0110010 0110101 0101111 1011001 01000101.26 BEN SMITH1.27 00000110 100001101.28 01110110 10001110第二章邏輯門2.1 自測練習(xí)2.1.1. (b 2.1.2. 162.1.3. 32, 6 2.1.4. 與2.1.5. (d 2.1.6. 162.1.7. 32, 6 2.1.8. 或2.1.9. 非2.1.10. 12.2 自測練習(xí)2.2.1. FAB=.2.2.2. (b2.2.3. 高2.2.4. 322.2.5. 16,52.2.6. 12.2.7. 串聯(lián)2.2.

14、8. (d2.2.9. 不相同2.2.10. 高2.2.11. 相同2.2.12. (a2.2.13. (c2.2.14. 奇2.3 自測練習(xí)2.3.1. OC,上拉電阻2.3.2. 0,1,高阻2.3.3. (b2.3.4. (c2.3.5. ,FAB=. 高阻2.3.6. 不能2.4 自測練習(xí)2.4.1. TTL,CMOS2.4.2. Transisitor Transistor Logic2.4.3. Complementary Metal Oxide Semicoductor2.4.4. 高級肖特基TTL,低功耗和高級低功耗肖特基TTL 2.4.5. 高,強,小2.4.6. (c2.4

15、.7. (b2.4.8. (c2.4.9. 大2.4.10. 強2.4.11. (a2.4.12. (a2.4.13. (b2.4.14. 高級肖特基TTL2.4.15. (c習(xí)題2.1 與,或,與2.2 與門,或門,與門2.3 (aF=A+B, F=AB (bF=A+B+C, F=ABC (cF=A+B+C+D, F=ABCD 2.4 (a0 (b1 (c0 (d02.5 (a0 (b0 (c1 (d02.6 (a1 (b1 (c1 (d12.7 (a4 (b8 (c16 (d322.8 (a3 (b4 (c5 (d6ABCF0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 0 0

16、 1 1 0 11111112.9 (a(b ABCDF1110 1 1 1 0 1 0 0 00 1 0 1 1 0 1 1 0 1 0 1 1 1 0 1 0 0 0 0 1 00 1 1 1 0 1 0 1 1 0 1 1 0 1 1 0 0 1 1 1 0 111111112.10 YABAC=+ 2.11ABCY0 0 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 0 1 0 1 11111112.122.13F1 = A(B+C, F2=A+BCA B C F1 F2 0 0 0 0 0 0 0 1 0 0 0 1 0 00 0 1 1 0 1 1 0 1 1 1 1

17、 0 0 0 1 1 1 0 1 1 111112.142.15 (a0 (b1 (c1 (d02.16 (a1 (b0 (c0 (d12.17 (a0 (b02.182.19 CDEF.2.20 CY ABDF.2.21 102.22 402.23 當(dāng)TTL反相器的輸出為3V,輸出是高電平,紅燈亮。當(dāng)TTL反相器的輸出為0.2 V時,輸出是低電平,綠燈亮。2.24 當(dāng)TTL反相器輸出高電平時三極管會導(dǎo)通,LED燈會點亮;當(dāng)TTL反相器輸出低電平時三極管不會導(dǎo)通,LED燈不會點亮。3.1自測練習(xí)答案1. 邏輯函數(shù)2. 邏輯表達式、真值表、邏輯電路圖、卡諾圖和波形圖3.表3-1 真值表ABC11

18、11111 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 04.ACD+BAC+5. (略3.2自測練習(xí)答案1. 與、或、非2. 代入規(guī)則、反演規(guī)則、對偶規(guī)則3. a和c4. a d5. a6. 、7. 、8. F3.3自測練習(xí)答案1. A2. AD3.4.5.6.7.8.9.3.4自測練習(xí)答案1. 標(biāo)準(zhǔn)與或表達式、標(biāo)準(zhǔn)或與表達式2. 1、3.4. 最大項5. 4,5,6,7,12,13,14,156.7.8.9.A B CF0 0 00 0 10 1 00 1 11 0 0 1 0 1 1 1 0 1 1 1 0111110.11.12. C3.5自測練習(xí)答案1. 12.3

19、. 格雷碼4.、*F5. m66. M17.(A=AF*=8.9.+10.第三章練習(xí)答案3.1、(3.2、(a1,0,0 (b1,1,1 (c0,1,03.3.略3.4.(aCBC+(bF3.5(a(b3.6 提示: 列出真值表可知: (1不正確, (2不正確, (3 正確, (4 正確3.7(a(bABC (cF(d(e(fAB(g(hF (iF=(j3.83.9 (a(b3.10 函數(shù)Y和函數(shù)Z互補, 即:Y 3.11、111 1 03.12 3.133.14 3.15 3.163.17 AB習(xí)題4.1寫出圖所示電路的邏輯表達式,并說明電路實現(xiàn)哪種邏輯門的功能。ABABA習(xí)題4.1圖解:該電路實現(xiàn)異或門的功能4.2分析圖所示電路,寫出輸出函數(shù)F。習(xí)題4.2圖解:4.3已知圖示電路及輸入A、B的波形,試畫出相應(yīng)的輸出波形F,不計門的延遲.習(xí)題4.3圖解:4.4由與非門構(gòu)成的某表決電路如圖