数字电子技术基础-2-逻辑代数化简

逻辑代数化简

• List of uesd equation or theorem
• Helpful principles
• Drill problems

List of uesd equation or theorem

• $De.Morgantheorem$
• 恒真式$A+A^{\prime }=1$
• 加法的分配律$A+BC=(A+B)(A+C)$

Helpful principles

• If one term contains parts of other two terms, then we usually times $A+A^{\prime }=1$
  for example: $Y=AB+AC+B{C}^{\prime }$, it’s obvious that AB contains parts of other two terms, then we transfer like below
  $$\begin{array}{rl}Y& =AB(C+C^{\prime })+AC+B{C}^{\prime }\\ & =(ABC+AC)+(AB{C}^{\prime }+B{C}^{\prime })\\ & =AC(B+1)+B{C}^{\prime }(A+1)\\ & =AC+B{C}^{\prime }\end{array}$$
• when you see a notation of notation, then it true, not notation
  for example: $Y=((A+B{)}^{\prime }{)}^{\prime }=A+B$

Drill problems

1. try to simplify expression followed: $$Y=AC+B^{\prime }C+B{D}^{\prime }+C{D}^{\prime }+A(B+C^{\prime })+A^{\prime }BC{D}^{\prime }+A{B}^{\prime }DE$$
   Solution
   $$\begin{array}{rl}Y& =A(C+C^{\prime })+B^{\prime }C+B{D}^{\prime }+C{D}^{\prime }(1+A^{\prime }B)+AB+A{B}^{\prime }DE\\ & =A+AB+A{B}^{\prime }DE+B^{\prime }C+B{D}^{\prime }+C{D}^{\prime }\\ & =A+B^{\prime }C+B{D}^{\prime }+(B+B^{\prime })C{D}^{\prime }\\ & =A+B^{\prime }C(1+D^{\prime })+B{D}^{\prime }(1+C)\\ & =A+B^{\prime }C+B{D}^{\prime }\end{array}$$