数电第四章习题答案

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1577894046631第1页共11页第四章习题答案4.1分析图4.1电路的逻辑功能解:(1)推导输出表达式Y2=X2;Y1=X1X2;Y0=(MY1+X1M)X0(2)列真值表MX2X1X0Y2Y1Y00000000100100011010001010110011110001001101010111100110111101111000001011010110111101100000001011010111110100101(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。F1=A⊕B⊕CF2=A(B⊕C)+BC=ABC+ABC+ABC+ABC(2)列真值表。ABCF1F20000010100111001011101110011110110000011(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。1577894046631第2页共11页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:(1)F1=ABC;F2=(AB)C+AB(2)真值表:ABCF2F10000010100111001011101110001011001101011(3)逻辑功能: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:(1)列真值表1011010101001000001110111000100000100000FABCDØ1110Ø1101Ø1100Ø1011Ø1111Ø10101100111000FABCD(2)写最简表达式CDAB000111100001111111011F=A+BD+BC=A·BD·BC1577894046631第3页共11页(3)画逻辑电路,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:(1)设被乘数AB,乘数CD,乘积结果为F4F3F2F1。真值表为:ABCDF4F3F2F100000001001000110100010101100111100010011010101111001101111011110000000000000000000000010010001100000010010001100000001101101001根据真值表可直接得到:F4=ABCDCDAB00011110CDAB000111100000010111111111110111011F3=ACABCDF2=ADABCD+BCABCDCDAB00011110000111111110F1=BD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。1577894046631第4页共11页4.6试设计一个将8421BCD码转换成余3码的电路。解:(1)列真值表ABCDF4F3F2F100000001001000110100010101100111100010010011010001010110011110001001101010111100(2)化简输出表达式CDAB00011110CDAB0001111000001110111101111111011101F4=A+BD+BCF3=BD+BC+BCDCDAB00011110CDAB0001111000110011011101111111101101F2=CD+CDF1=D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:(1)F(ABC)=m(1,3,4,6,7)解:F=AC+AC+BCBCA000111100111111(2)F(ABCD)=m(0,2,6,7,8,10,12,14,15)解:F=BD+AD+BCCDAB00011110001101111111110111577894046631第5页共11页)10,7,4,0()12,9,8,6,5,2(),,,()3(mDCBAF解:函数的卡诺图如下所示:ABCDØ11101111Ø1Ø011Ø0010110100F=ABC+CD+AB+AD=ABC·CD·AB·AD画逻辑电路,如下图所示:123456ABCD654321DCBATitleNumberRevisionSizeBDate:3-Mar-2002SheetofFile:E:\DesignExplorer99SE\Library\YangHengXin\MyDesign.ddbDrawnBy:&&&&&AAABBCCDDF(4)F(ABC)=AB+BC+AC解:F(ABC)=AB+BC+ACBCA00011110011111114.9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:直接画卡诺图为:CDAB00011110001101111111101111F=AC+BC+CD电路图如上。ACBCCDF1577894046631第6页共11页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:设火警为A,急救为B,普通为C,列真值表为:ABCF1F2000001010011100101110111011010111111110001111001111111ABCF1=A+B000111100111111ABCF2=BA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Y154.12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。A1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Y31577894046631第7页共11页解:由题意,各函数是4变量函数,故须将74138扩展为4-16线译码器,让A、B、C、D分别接4-16线译码器的地址端A3、A2、A1、A0,可写出各函数的表达式如下:)12,8,4,0(),,,(1mDCBAF=m0m4m8m12=Y0Y4Y8Y12)2,1,0(),,,(2mDCBAF=m0m1m2=Y0Y1Y2)11,10,9,8(),,,(3mDCBAF=m8m9m10m11=Y8Y9Y10Y1104),,,(mDCBAF=Y0实现电路如下图所示:4.14试用74151实现下列函数:。)7,4,2,1(),,,()1(mDCBAF解:(1)函数有4个输入变量,而74151的地址端只有3个,即A2、A1、A0,故须对函数的卡诺图进行降维,即降为3维。10111101110010110100ABCD00001DDDD010110100ABC令A=A2、B=A1、C=A0则:D0=D3=D,D1=D2=D,D4=D5=D6=D7=0相应的电路图如下所示:123456ABCD654321DCBATitleNumberRevisionSizeBDate:4-Mar-2002SheetofFile:E:\DesignExplorer99SE\Library\YangHengXin\MyDesign.ddbDrawnBy:YYYYYYYYAAAEEE0112A012345677413822BYYYYYYYYAAAEEE0112A012345677413822B&&&0001ABCDF1F2F3F4413123456ABCD654321DCBATitleNumberRevisionSizeBDate:4-Mar-2002SheetofFile:E:\DesignExplorer99SE\Library\YangHengXin\MyDesign.ddbDrawnBy:YYYYYYYYAAAEEE0112A012345677413822BYYYYYYYYAAAEEE0112A012345677413822B&&&0001ABCDF1F2F3F44131577894046631第8页共11页123456ABCD654321DCBATitleNumberRevisionSizeBDate:4-Mar-2002SheetofFile:E:\DesignExplorer99SE\Library\YangHengXin\MyDesign.ddbDrawnBy:DDDD013412D74151DAAA20ENY567DDABCFDD。)8,7()14,13,12,3,0(),,,()4(mDCBAF解:函数有4个输入变量,而74151的地址端只有3个,即A2、A1、A0,故须对函数的卡诺图进行降维,即降为3维。Ø1011111Ø01110010110100ABCD1D00100DD010110100ABC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:123456ABCD654321DCBATitleNumberRevisionSizeBDate:4-Mar-2002SheetofFile:E:\DesignExplorer99SE\Library\YangHengXin\MyDesign.ddbDrawnBy:DDDD013412D74151DAAA20ENY567DDABCFDDD14.14(4)4.15用½74153实现下列函数:。)15,7,4,2,1(),,,()1(mDCBAF解:(1)函数有4个输入变量,而½74153的地址端只有2个,即A1、A0,故须对函数的卡诺图进行降维,即降为2维。101111101110010110100ABCD0D001DDDD010110100ABC0C⊕D0C⊙D0CD11AB电路图如下:1577894046631第9页共11页123456ABCD654321DCBATitleNumberRevisionSizeBDate:5-Mar-2002SheetofFile:E:\DesignExplorer99SE\Library\YangHengXin\MyDesign.ddbDrawnBy:YAADENDD012301D7415312_AB=1=&FCD4.18用74283将8421BCD码转换为余3BCD码。解:由于同一个十进制数码的余3BCD码比相应的8421BCD码大3,故用一片74283既可以实现,电路图如右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