精品北科大数字电子技术老师留的作业答案

数字电子技术（V10）作业讲解ALLCHAP03P1011CHAP03P1027、10107CHAP03P10315CHAP03P10419CHAP03P10424CHAP03P10425CHAP03P10537CHAP03P10537CHAP12P5225CHAP12P5227BCDEATHEPULSEGOESTHROUGHTHREEGATESINTHESHORTESTPATHCHAP12P52211G2ISOVERLOADEDBECAUSEITHAS12UNITLOADSCHAP12P52417HIGHFLOATINGHIGHHIGHZCHAP12P52419LOWLOWLOWCHAP04P15610DCHAP04P15611BCHAP04P15721CHAP04P15833DEVELOPTRUTHTABLESCBACACABBAAZYXWZZYXBCHAP04P15836CDCABDCBADBCADCBACDBADCBADCBAXSTANDARDSOPSTANDARDPOSDCBADCBADCBADCBADCBADCBADCBADCBADCBAXCHAP04P15944CDCBADCABDCABDCBADCBADCBADCDCABDCDCBADCACB11111CHAP04P15946、474746CHAP04P15946CHAP04P15947CHAP05补充1COMBINATIONALLOGICANALYSISWRITETHEBOOLEANEXPRESSIONOFTHECIRCUIT,DRAWTHETRUETABLEANDTHENDESCRIBETHEFUNCTIONOFTHISCIRCUIT21YCBAABCYBCACABY2CHAP05补充1ABCY1Y2000001010011100101110111FULLADDER21YCBAABCYBCACABY2CHAP05P1893/5FCBBAXABCX00001111001100110101010100011101CHAP05P19111/17BCAABCBAX1117CHAP05P19114CHAP05P19225ECDABCDABCDABXECHAP05P19225FBDADBCACBDADBCACDCBAORDCBADCBADCBAXFCHAP05补充2DESIGNAHALLLIGHTCIRCUITTOTHEFOLLOWINGSPECIFICATIONTHEREISASWITCHATEITHERENDOFAHALLTHATCONTROLSASINGLELIGHTIFTHELIGHTISOFF,CHANGINGTHEPOSITIONOFEITHERSWITCHCAUSESTHELIGHTTOTURNONSIMILARLY,IFTHELIGHTISON,CHANGINGTHEPOSITIONOFEITHERSWITCHCAUSESTHELIGHTTOTURNOFFWRITEYOURASSUMPTIONS,DERIVEATRUTHTABLE,ANDDESCRIBEHOWTOIMPLEMENTTHISFUNCTIONINTERMSOFLOGICGATESCHAP05P19330FORTHEINPUTWAVEFORMSINTHEFIGURE,WHATLOGICCIRCUITWILLGENERATETHEOUTPUTWAVEFORMSHOWNINPUTSOUTPUTABCX000001010011100101110111CBACBAABCXCHAP05P19330ACBACBACBAABCX另一种方法是将输出为1所对应的最小项输入求和优点缺点CHAP05P19434DESIGNALOGICCIRCUITTOPRODUCEAHIGHOUTPUTONLYIFTHEINPUT,REPRESENTEDBYA4BITBINARYNUMBER,ISGREATERTHAN12ORLESSTHAN3CHAP05P19434CHAP05P19434321320321320XAAAAAAAAAAAA321320321320AAAAAAAAAAAA323203213201AAAAAAAAAAAA变换CHAP06P2486（修正）00000电路实现两个二进制加法功能0111111B反码输入，最低位进位输入为1电路完成A（B的反码1）操作，即AB的补码实现两个二进制减法功能涉及计算机文化基础中负数表示方法以及减法计算方法，可参考教材P28SIGNMAGNITUDEFORM和P33ARITHMATICOPERATIONSWITHSIGNEDNUMBERS相关预备知识带符号的二进制数表示方法,负数以补码（2SCOMPLEMENT）形式存在计算机中带符号数的减法用加法实现，即加上减数的补码，补码反码（1SCOMPLEMENT）1CHAP06P2498CHAP06P24910CHAP06P25014CHAP06P25016（D）1000CHAP06P25117/18（A）17110118CHANGETHEANDGATESTONANDGATESCHAP06P25122CHAP06P25230USEGATESTOIMPLEMENTCHAP06P25437FULLADDERUSENANDGATESTOIMPLEMENTCHAP06P25437USEMULTIPLEX74LS151TOIMPLEMENT用译码器和数据选则器实现逻辑函数时，用最小项形式表达更方便CHAP06P2543737USE74LS1383LINE8LINEDECODERTOIMPLEMENT见仿真CHAP06补充2用一片38线译码器和必要的门电路实现一个全加器。4210765365307421MMMMMMMMCMMMMMMMMSOUTABCINHOMEWORKIMPLEMENTA32LINETO5LINEENCODERUSING74LS148SANDEXTERNALLOGICHINTS1用几块芯片2）芯片间如何连接3）编码的结果应该有几位4）应该如何编码HOMEWORKIMPLEMENTA32LINETO5LINEENCODERUSING74LS148SANDEXTERNALLOGIC低3位编码同16LINETO4LINE扩展HOMEWORKIMPLEMENTA32LINETO5LINEENCODERUSING74LS148SANDEXTERNALLOGIC高两位编码和各芯片有有效输入情况关系A4A3第四片有有效输入11第三片有有效输入10第二片有有效输入01第一片有有效输入00EOISLOWWHENEIISLOWANDNONEOFTHEINPUTSISACTIVEGSISLOWWHENEIISLOWANDANYOFTHEINPUTSISACTIVE高两位编码和各芯片扩展端输出关系如下INPUTSOUTPUTSGS4GS3GS2GS1A4A3011111101110110101111000（4）片内有有效信号（3）片内有有效信号（2）片内有有效信号（1）片内有有效信号INPUTSOUTPUTSGS4GS3GS2GS1A4A3011111101110110101111000高两位编码和各芯片扩展端输出关系如下34344GSGSGSGSA24243GSGSGSGSAEOISLOWWHENEIISLOWANDNONEOFTHEINPUTSISACTIVEGSISLOWWHENEIISLOWANDANYOFTHEINPUTSISACTIVE（4）片内有有效信号（3）片内有有效信号（2）片内有有效信号（1）片内有有效信号34344GSGSGSGSA24243GSGSGSGSA完整电路参见仿真用KMAP化简INPUTSOUTPUTSEO4EO3EO2EO1A4A3111111011110001101000100如用EO端扩展，高两位编码和各芯片输出关系如下34EOA2343EOEOEOA（4）片内有有效信号（3）片内有有效信号（2）片内有有效信号（1）片内有有效信号用KMAP化简，有补充作业同学都不写设计思路，直接画图。正确的做法是简明扼要地介绍自己的设计思路；输入和输出信号都是低电平有效，所以规范的画图应该是小圈不可以少。设计思路（1）四个芯片的8个输入作为总体的32根输入线，在5线的输出中，有3根线基本上由4个芯片的3线输出经过一定的逻辑运算得到，关键是另外2根输出线的组成。很显然，应该由扩展输出端组成。整理4个芯片的工作情况是000111101011011101111110ZZYYYY34EXEXEXEX43213121EXEX3EXEX4YYZYYZ用KMAP化简CHAP07P3018CHAP07P30112CHAP07P30214CHAP07P30425NNQDQ1CHAP07P30426NABNABAAQKQJKJ1CHAP07P28126NABNABAAQKQJKJ1NNNQKQJQJK1触发器特性方程NANBNANBNANBNANAQQQQQQQQ11画出下图中触发器的状态量Q1、Q2与输出Y、Z波形。CHAP07补充4画出该电路的状态量Q1、Q2和输出Z在时钟脉冲作用下的波形。CHAP07补充5112112122111NNNQRQCLKZQDQQDQCHAP08P3334CCONNECTAMODULUS13COUNTERBY74LS93CHAP08P3334CCONNECTAMODULUS13COUNTERBY74LS93CHAP08补充1CONNECTAMODULUS13COUNTERBY74HC163HIGHCHAP08P3336SHOWTHECOMPLETETIMINGDIAGRAM5BITSYNCHRONOUSBINARYCOUNTERCHAP08P33517DESIGNACOUNTERTOPRODUCETHEFOLLOWINGSEQUENCE1,4,3,5,7,6,2,1,USE3FLIPFLOPSPRESENTSTATENEXTSTATEQ2Q1Q0Q2Q1Q0010111000101111011100101110001011100111001NEXTSTATETABLECHAP08P33517TRANSITIONTABLEOUTPUTSTATETRANSITIONSPRESENTSTATETONEXTSTATEFLIPFLOPINPUTSQ2Q1Q0J2K2J1K1J0K00TO11TO00TO11TO11TO11TO00TO00TO00TO11TO00TO11TO11TO11TO01TO00TO11TO11TO11TO00TO00TO11X1XXX0X1X001X01X1XXXXX1X001X1XXX011X001XXCHAP08P33517212121021021210202QQQQQQKQQJQKQJQKQJCHAP08补充2设计一个自动售饮料机的逻辑电路。每次只允许投入一元硬币或5元纸币，累计投入达三元自动给出一罐饮料，并找零钱。要求设计的电路能自启动。输入变量A1元，A1投入，A0未投B5元，B1投入，B0未投输出变量X饮料，X0不给，X1给YZ找钱，YZ00不找钱，YZ01找两元，YZ10找三元，YZ11找四元CHAP08补充2状态变量S0累计0元S1累计1元S2累计2元BASNSN1/XYZ00011011S0S0/000S1/000S0/101/S1S1/000S2/000S0/110/S2S2/000S0/100S0/111/状态分配选用2个触发器Q1、Q0。Q1Q000S0；Q1Q001S1；Q1Q010S2；CHAP08补充2001000010010000100000111001000011000000101000Q1N1Q0N1XYZCHAP08补充20101001101011QBZBQBQYAQBXQBAQQAQQBAAQQNNCHAP08P33621CHAP08P33623CHAP09P3706DETERMINETHEDATAOUTPUTWAVEFORMCHAP09P37112DEVELOPTHEDATAOUTPUTWAVEFORMINRELATIONTOTHEINPUTSDETERMINETHESTATEOFTHEREGISTERAFTEREACHCLOCKPULSEFORTHECONTROLWAVEFORMGIVENINITIALLYSTORING76D01001100BLSBRIGHTMOSTPOSITIONCHAP09P372191001100001001100001001100001001100100110010011000010011001001100001001100100110010011000CHAP09P37021USETWO74HC194TOCREATEAN8BITBIDIRECTIONALSHIFTREGISTERCHAP09P37022DETERMINETHEQOUTPUTSOFA74HC194WITHTHEINPUTSSHOWNINPUTSD0、D1、D2ANDD3AREALLHIGHCHAP09P3702323HOWMANYFLIPFLOPSAREREQUIREDTOIMPLEMENTEACHOFTHEFOLLOWINGINAJOHNSONCOUNTERCONFIGURATIONAMODULUS62N6N3BMODULUS102N10N5CMODULUS142N14N7DMODULUS162N16N8CHAP09P37026DEVISEARINGCOUNT