南京郵電大學數(shù)字電路第二章習題2.1有A、B、C三個輸入信號，試列出下列問題的真值表，并寫出其最小項表達式m()。

(1)如果A、B、C均為0或其中一個信號為1時，輸出F=1，其余情況下F=0。(2)若A、B、C中出現(xiàn)奇數(shù)個0時輸出為1，其余情況下輸出為0。(3)若A、B、C中有兩個或兩個以上為1時，輸出為1，其余情況下輸出為0。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.2/2/20232ABCF1F2F3000110001100010100011011100100101011110011111001解：真值表和最小項表達式如下：Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.2/2/202332.4試直接寫出下列各式的反演式和對偶式。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.2/2/20234Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.2/2/202352.10試寫出下列各函數(shù)表達式F的F和F′的最小項表達式。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.2/2/20236Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.2/2/202372.11試用公式法把下列各表達式化簡為最簡與或式。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.2/2/20238Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.2/2/202392.12用卡諾圖法把下列函數(shù)化簡為最簡與或式。011111011010110100ABCEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.2/2/20231011?10?1?111?101?110010110100ABCDEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.2/2/2023111011111??01????0010110100ABCDEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.2/2/2023122.13用卡諾圖法將下列函數(shù)化簡為最簡或與式。011111011010110100ABCEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.2/2/2023132.14已知Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.2/2/202314?1101111?010110100ABC?1?11?01?010110100ABCA10?11?101010110100BC⊕﹦Evaluationonly.