《数字电路与系统设计》第4章习题答案

1578606928217第1页共9页4.1分析图4.1电路的逻辑功能解：（1）推导输出表达式（略）(2)列真值表（略）（3）逻辑功能：当M=0时，实现3位自然二进制码转换成3位循环码。当M=1时，实现3位循环码转换成3位自然二进制码。4.2分析图P4.2电路的逻辑功能。123456ABCD654321DCBATitleNumberRevisionSizeBDate:3-Mar-2002SheetofFile:E:\DesignExplorer99SE\Library\YangHengXin\MyDesign.ddbDrawnBy:=1=11&&1BCAFF12解：(1)从输入端开始，逐级推导出函数表达式。（略）(2)列真值表。（略）(3)确定逻辑功能。假设变量A、B、C和函数F1、F2均表示一位二进制数，那么，由真值表可知，该电路实现了一位全减器的功能。A、B、C、F1、F2分别表示被减数、减数、来自低位的借位、本位差、本位向高位的借位。ABCF1F2－被减数减数借位差4.3分析图4.3电路的逻辑功能解：实现1位全加器。4.4设ABCD是一个8421BCD码，试用最少与非门设计一个能判断该8421BCD码是否大于等于5的电路，该数大于等于5，F=1；否则为0。解：逻辑电路如下图所示：123456ABCD654321DCBATitleNumberRevisionSizeBDate:3-Mar-2002SheetofFile:E:\DesignExplorer99SE\Library\YangHengXin\MyDesign.ddbDrawnBy:&&&DBCAF&4.5试设计一个2位二进制数乘法器电路。解：为了使电路尽量简单，希望门数越少越好，本电路是四输出函数，圈卡诺圈时要尽量选择共有的卡诺圈以减少逻辑门的数量。电路图略。1578606928217第2页共9页4.6试设计一个将8421BCD码转换成余3码的电路。解：电路图略。4.7在双轨输入条件下用最少与非门设计下列组合电路：解：略4.8在双轨输入信号下，用最少或非门设计题4.7的组合电路。解：将表达式化简为最简或与式：(1)F=(A+C)(A+B+C)=A+C+A+B+C(2)F=(C+D)(B+D)(A+B+C)=C+D+B+D+A+B+C(3)F=(A+C)(A+B+D)(A+B+D)=A+C+A+B+D+A+B+D(4)F=(A+B+C)(A+B+C)=A+B+C+A+B+C4.9已知输入波形A、B、C、D，如图P4.4所示。采用与非门设计产生输出波形如F的组合电路。解：F=AC+BC+CD电路图略4.10电话室对3种电话编码控制，按紧急次序排列优先权高低是：火警电话、急救电话、普通电话，分别编码为11，10，01。试设计该编码电路。解：略4.11试将2/4译码器扩展成4/16译码器解：A3A2A1A0Y0Y1Y2Y3Y4Y5Y6Y7Y8Y9Y10Y11Y12Y13Y14Y15A1ENY3A02/4Y2译码器Y1Y0ENA12/4(1)A0Y0Y1Y2Y3ENA12/4(2)A0Y0Y1Y2Y3ENA12/4(3)A0Y0Y1Y2Y3ENA12/4(4)A0Y0Y1Y2Y31578606928217第3页共9页4.12试用74138设计一个多输出组合网络，它的输入是4位二进制码ABCD，输出为：F1：ABCD是4的倍数。F2：ABCD比2大。F3：ABCD在8～11之间。F4：ABCD不等于0。解：电路如下图所示：4.13试将八选一MUX扩展为六十四选一MUX。解：方法一：123456ABCD654321DCBATitleNumberRevisionSizeBDate:4-Mar-2002SheetofFile:E:\DesignExplorer99SE\Library\YangHengXin\MyDesign.ddbDrawnBy:YYYYYYYYAAAEEE0112A012345677413822BYYYYYYYYAAAEEE0112A012345677413822B&&&0001ABCDF1F2F3F4413123456ABCD654321DCBATitleNumberRevisionSizeBDate:4-Mar-2002SheetofFile:E:\DesignExplorer99SE\Library\YangHengXin\MyDesign.ddbDrawnBy:YYYYYYYYAAAEEE0112A012345677413822BYYYYYYYYAAAEEE0112A012345677413822B&&&0001ABCDF1F2F3F44131578606928217第4页共9页方法一电路图ENA2A1A0D0D174151(1)YD2D3D4D5D6D7ENA2A1A0D0D174151(2)YD2D3D4D5D6D7ENA2A1A0D0D174151(7)YD2D3D4D5D6D7ENA2A1A0D0D174151(8)YD2D3D4D5D6D71A2Y0A1Y1A0Y274138Y3E1Y4E2AY5E2BY6Y7100A5A4A3A2A1A0D0D1D7D8D9D15D48D49D55D56D57D63Y0Y1Y6Y7Y1578606928217第5页共9页方法二：方法二电路图ENA2A1A0D0D174151(1)YD2D3D4D5D6D7ENA2A1A0D0D174151(2)YD2D3D4D5D6D7ENA2A1A0D0D174151(7)YD2D3D4D5D6D7ENA2A1A0D0D174151(8)YD2D3D4D5D6D7A2A1A0D0D1D7D8D9D15D48D49D55D56D57D63Y0Y1Y6Y7ENA2A1A0D0D174151(1)YD2D3D4D5D6D7A5A4A3Y1578606928217第6页共9页4.14试用74151实现下列函数：。)7,4,2,1(),,,()1(mDCBAF解：(1)电路图如下所示：123456ABCD654321DCBATitleNumberRevisionSizeBDate:4-Mar-2002SheetofFile:E:\DesignExplorer99SE\Library\YangHengXin\MyDesign.ddbDrawnBy:DDDD013412D74151DAAA20ENY567DDABCFDD(2)F(A,B,C)=AB+AB+C解：(3)F(A,B,C,D)=ABC+BCD+ACD解：。)8,7()14,13,12,3,0(),,,()4(mDCBAF解：令A=A2、B=A1、C=A0则：D0=D7=D,D1=D,D6=1,D2=D3=D4=D5=0。相应的电路图如下图所示：ENA2A1A0D0D174151YD2D3D4D5D6D7ABC01111101FENA2A1A0D0D174151YD2D3D4D5D6D7ABC00D001DDF1578606928217第7页共9页123456ABCD654321DCBATitleNumberRevisionSizeBDate:4-Mar-2002SheetofFile:E:\DesignExplorer99SE\Library\YangHengXin\MyDesign.ddbDrawnBy:DDDD013412D74151DAAA20ENY567DDABCFDDD14.14(4)(5)F(A,S,C,D,E)=ABCD+ABCE+BCDE解：电路图略。4.15用½74153实现下列函数：。)15,7,4,2,1(),,,()1(mDCBAF解：电路图如下：123456ABCD654321DCBATitleNumberRevisionSizeBDate:5-Mar-2002SheetofFile:E:\DesignExplorer99SE\Library\YangHengXin\MyDesign.ddbDrawnBy:YAADENDD012301D7415312_AB=1=&FCD（2）F(A,B,C)=m(1,2,4,7)解：ENA1A0D0YD1D2D3ABCCCCF74153211578606928217第8页共9页4.16试在图4.2.31的基础上增加一片7485，构成25位数据比较器。解：4.17设A=A3A2A1A0,B=B3B2B1B0均为8421BCD码。试用74283设计一个A、B的求和电路。（可用附加器件）解：设COS3S2S1S0为A、B的二进制和，则当CO=1或S3S2S1S01001时，须加0110修正项进行调整，计算结果为C4C3C2C1C0。4.18用74283将8421BCD码转换为余3BCD码。解：电路图如右所示：4.20用74283将8421BCD码转换为5421BCD码。解：123456ABCD654321DCBATitleNumberRevisionSizeBDate:5-Mar-2002SheetofFile:E:\DesignExplorer99SE\Library\YangHengXin\MyDesign.ddbDrawnBy:SBBB120174283AAA3ASS021B2303SCOCI000118421BCD余3BCD=A3A2A1A0B3B2B1B0(AB)i(A=B)i7485(AB)iFABFA=BFABA20B20A24A23A22A21B24B23B22B21=A3A2A1A0B3B2B1B0(AB)i(A=B)i7485(AB)iFABFA=BFABA5B5A9A8A7A6B9B8B7B6A3A2A1A0B3B2B1B0(AB)i(A=B)i7485(AB)iFABFA=BFAB0=A3A2A1A0B3B2B1B0(AB)i(A=B)i7485(AB)iFABFA=BFABA0B0A4A3A2A1B4B3B2B1=A3A2A1A0B3B2B1B0(AB)i(A=B)i7485(AB)iFABFA=BFABA10B10A14A13A12A11B14B13B12B11=A3A2A1A0B3B2B1B0(AB)i(A=B)i7485(AB)iFABFA=BFABA15B15A19A18A17A16B19B18B17B16FABFA=BFABA3A2A1COA0S3CI74283S2B3S1B2S0B1B0A3A2A1A0B3B2B1B01A3A2A1COA0S3CI74283S2B3S1B2S0B1B0S3S2S1S000C4C3C2C1C0123456ABCD654321DCBATitleNumberRevisionSizeBDate:5-Mar-2002SheetofFile:E:\DesignExplorer99SE\Library\YangHengXin\MyDesign.ddbDrawnBy:SBBB120174283AAA3ASS021B2303SCOCI000118421BCD余3BCD5421BCD00123456ABCD654321DCBATitleNumberRevisionSizeBDate:19-Feb-2002SheetofFile:C:\ProgramFiles\DesignExplorer99SE\Library\YangHengXin\MyDesign.ddbDrawnBy:F(A=B)BBB1201ii(AB)7485(AB)AAA3iAFFABABA=BB2308421BCD0100123456ABCD654321DCBATitleNumberRevisionSizeBDate:5-Mar-2002SheetofFile:E:\