试按表121所列的数字集成电路的分类依据

1.1.1试按表1.2.1所列的数字集成电路的分类依据，指出下列器件属于何种集成度器件：(1)微处理器；(2)IC计算器；(3)IC加法器；(4)逻辑门；(5)4兆位存储器IC。解：(1)微处理器属于超大规模；(2)IC计算器属于大规模；(3)IC加法器属于中规模；(4)逻辑门属于小规模；(5)4兆位存储器IC属于甚大规模。1.1.2一数字信号的波形如图1.1.1所示，试问该波形所代表的二进制数是什么？解：010110101.2.2将下列十进制数转换为二进制数、八进制数、十六进制数和8421BCD码（要求转换误差不大于2-4）：(1)43(2)127(3)254.25(4)2.718解：(1)43D=101011B=53O=2BH；43的BCD编码为01000011BCD。(2)127D=1111111B=177O=7FH；127的BCD编码为000100100111BCD。(3)254.25D=11111110.01B=376.2O=FE.4H；001001010100.00100101BCD。(4)2.718D=10.10110111B=2.56O=2.B7H；0010.011100011000BCD。1.2.3将下列每一二进制数转换为十六进制码：(1)101001B(2)11.01101B解：(1)101001B=29H(2)11.01101B=3.68H1.2.4将下列十进制转换为十六进制数：(1)500D(2)59D(3)0.34D(4)1002.45D解：(1)500D=1F4H(2)59D=3BH(3)0.34D=0.570AH(4)1002.45D=3EA.7333H1.2.5将下列十六进制数转换为二进制数：(1)23F.45H(2)A040.51H解：(1)23F.45H=1000111111.01000101B(2)A040.51H=1010000001000000.01010001B1.2.6将下列十六进制数转换为十进制数：(1)103.2H(2)A45D.0BCH解：(1)103.2H=259.125D(2)A45D.0BCH=41024.046D2.1.3用逻辑代数证明下列不等式(1)BABAA由交换律))((CABABCA，得BABAAABAA))(((2)ACABCABCBAABCACABBCACBCACBCBBCACABCBAABC)()()((3)ECDAEDCCDACBAAECDAECDCDAEDCCDAAEDCCDACBAA_____)(2.1.4用代数法化简下列等式(1))(ABCABABABABCABCAB)((2)))((BABABABABA))(((3))(_______CBBCACABCCBCACBABCBCBACBBCA))(()(_______(4)BCCBBCAABCA_____CABCCBBCAABCA_____(5)____________________________BABABAAB0_____________________________________AABABABAAB(6)____________________________________________________________________________)()()()(BABABABABABABABBABABABABBABABABABABABABA)())(()()()()()()()()(___________________________________________________________________________________________________________(7)))((CBACBABACBACBA))(((8)CBAABCCBACBACACBCBACBCBAACBAABCCBACBA(9)_____________________________)(BAABBABABABAABBAAB))(()()(__________________________________(10)BACAABCBCAACBCAABCBBACAABCB(11)CBABCDDBCABDDABCDBCBABDACABDCADACBDBCBABDABCDCBABDABCCDCBABDABCCBABCDDBCABDDABC)()()()((12)__________________________________________________CABCBBCAACBCBCBCACBABCCBABCAABCCBACBBCAACCABCBBCAAC)())(()()()(__________________________________________________(13)__________________________________________________)(BABAABCBA0)()()(____________________________________________________________________________________________________________________________________________________________________ACBAACBABAABBCBABABAABCBA2.1.5将下列各式转换成与–或形式(1)__________________DCBA当0________BA，1__________DC时，真值为1。于是AB=01，CD=00或CD=11时，真值为1；AB=10，CD=00或CD=11时，真值为1。则有四个最小项不为0，即DCBA、BCDA、DCBA、CDBA当1________BA，0__________DC时，真值为1。AB=00，CD=10或CD=01时，真值为1；AB=11，CD=10或CD=01时，真值为1。则有四个最小项不为0，即DCBA、DCBA、DABC、DCAB)14,13,11,8,7,4,2,1(__________________mDCBA(2)__________________________________________________________________DADCDCBADBCACDCDBDBCADACDBADCDADCDCBADADCDCBA))(())(())((__________________________________________________________________(3)______________________________________________________________________ABBCBDACCADCDABCBBCABADCDACBBABACBDBCAABBCBDACABBCBDAC))(())((__________________________________________________________________________________________2.1.7利用与非门实现下列函数(1)L=AB+AC_______________________ACABL(2)_____________)(CADL_____________________________)(CADCADL(3)____________________))((DCBAL__________________________________________))((DCBADCBAL2.2.3用卡诺图法化简下列各式(1)CABCBBCAAC______________________CCABCCABCACCABCBBCACCABCBBCAAC____________________________________________________(2)CBADABADCABCDBACADABACDDABADADCABADADCBBADADCABBACBADABADCABCDBA)()((3)__________________)()()(BADCABDCDBBABCDABDCBCBADABBCDABDDCBCBAABDCABDDCBCBABADCABDCDBBA)()()()(__________________(4)__________)()()(CBADBCADCBDCDBA)14,12,11,9,6,5,4,1()()()(5414614129111__________mmmmmmmmmmCBADBCDABDCBCDBACBADBCADCBDCDBADBADCADB(5))15,14,13,12,10,9,8,7,6,5,4,3(),,,(mDCBALCDBADACDCADBCBAB(6))14,913,8,7,6,5,2,1,0(),,,(mDCBALDBCDCABCADCCB(7))15,11,7,5,3,2()13,9,6,4,1,0(),,,(dmDCBALDA(8))11,10,9,3,2,1()15,14,13,0(),,,(dmDCBALACADBA4.1.6试分析图题4.1.6所示逻辑电路的功能。CBASCBAABABCBAC)()(_____________________________________全加器4.1.7分析图题4.1.7所示逻辑电路的功能。000BAS000BAC0111CBAS011111)(CBABAC二位加法电路4.2.2试用2输入与非门和反相器设计一个4位的奇偶校验器，即当4位数中有奇数个1时输出为0，否则输出为1。DCBAL_______________________BABABABAL4.2.9某雷达站有3部雷达A、B、C，其中A和B功率消耗相等，C的功率是A的功率的两倍。这些雷达由两台发电机X和Y供电，发电机X的最大输出功率等于雷达A的功率消耗，发电机Y的最大输出功率是X的3倍。要求设计一个逻辑电路，能够根据各雷达的启动和关闭信号，以最节约电能的方式启、停发电机。ABCXY0000000101010100110110010101011100111111ABCCBACBAXCABmmmmmY765314.1.1解：_____________________________________________________________________________________________________________________________________________________________0123_____________________0123_____________________0123____________