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1、Chapter7solutionENQTQTFF:QT*=EN∙Q’+EN’∙QDFF:QD*=DDCLKQQN?QD*=QT*→EN=F(D,Q)→EN==QT*EN0110→EN=EN017.15画出表7-9的状态图S0xyx’y+xy’S1x’y+xy’X’y’S2x’y+xy’S3x’y+xy’X’y’X’y’X’y’xyxyxy7.17构造一个与图x7-17等效的状态和输出表X=1ForstateAXy=01→B→Eelse→AwhenXY00011110ZSS*ABCDEFGH1110000101001011AEBBBBDDCGAADDC
2、CEFFEFFBBGHHGHHDD7.21(c)列出状态图中的二义性yz0001111000011110WxNextstateofstateAA,BABBA,BCCAAAAA下一状态不唯一，不满足互斥性。下一状态未定义，不满足完备性。ForstateC:Z+z’=1Z∙z’=0同时满足互斥和完备性。ForstateB:Z+xz’+x’z’=z+z’=1Z∙(xz’)=0;z(x’z’)=0;同时满足互斥和完备性。A,BABBA,BCCAAAAAAAAA7.41DCQQNSRQQN0DC001D’D’1D’DDD’Q1Q2XZ0100011000111
3、01011010011001110Q1*Q2*1010001000Q1*=D1=X∙Q2’+X’∙Q1’∙Q2Q2*=D2=X’+Q1’∙Q2+Q1∙Q2’=(X’+Q1+Q2)(x’+Q1+Q2)Z=Q1∙Q2按照Q2Q1编码：Q2*=D2=X∙Q1’+X’∙Q2’∙Q1按照Q2Q1编码：Q1*=D1=X’+Q2’∙Q1+Q2∙Q1’xy0001111000011110Q1Q2Q2*7.52按照00,01,10,11的顺序进行状态赋值，比较所得电路与教材表7-9中的“1”计数器的激励方程的成本。Q2*=D2=Q2x’y’+Q2’x’y+Q2xy+Q
4、2’xy’Q1*=D1=Q1x’y’+Q1’xy+Q1’Q2y+Q1’Q2x+Q1Q2’x’+Q1Q2’y’xy0001111000011110Q1Q2Q1*01001100001111011101000010010111D1=Q2x’y+Q1x’y+Q1’x’y’+Q2xy’D2=Q1’x’y+Q1’xy’+Q2x’y+Q2’xy’7.52的赋值方法比格雷码赋值方法，实现的D1多用2个三输入与门,第二级或门增加2个输入端。补充题1解答：激励方程:D1=Q1⊕Q2，D2=Q/1+Q/2转移方程：Q1*=D1=Q1⊕Q2，Q2*=D2=Q/1+Q/2输
5、出方程：Z=Q1•Q2Analyzethesequential-circuitasshowninfigure1.Writeouttheexcitationequations,transitionequationsandoutputequation.AssumetheinitialstateisQ2Q1=00,completethetimingdiagramforQ2,Q1andZ.(Don’tneedconsiderpropagationdelayofeachcomponent).Q1Q2Z00010011101100110110Q1*Q2*Desi
6、gnaMealysequentialdetectorwithoneinputxandoneoutputz.Ifandonlyifxcontinuestobe1111or1001,theoutputzis1.Otherwise,theoutputzis0.Theoverlapispermitted.Pleasedescribethestatemeaningandfinishthestate/outputtable.Example：x：010111100110011111z：000000100100010011补充题2XStatemeaningS01In
7、itialAA,0A1,0Received1A1A10,0A11,0Received10A10A100,0A1,0Received11A11A10,0A111,0Received100A100A,0A1001,1Received111A111A10,0A1111,1Received1001A1001A10,0A11,0Received1111A1111A10,0A1111，1S*,ZXStatemeaningS01InitialAA,0A1,0Received1A1A10,0A11,0Received10A10A100,0A1,0Received11
8、A11A10,0A111,0Received100A100A,0A1001,1Received111A111
