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I HAVE my digital circuits xam tommorow.. cnany1 plz help me out wid this problem Find the complement of the expression: (A'B + CD)E' + E
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no help :(
Simply use D Morgan Theorem (A+B')(C'+D')E'
I think it will be:- (A'B+CD)E'+E USING DEMORGAN >>[(A'B+CD)E'+E]' >>[(A'B+CD)'+E"".E'] i.e E"=E >>{[(A"+B')(C'+D')]+0} i.e E.E'=0 >>(A+B')(C'+D')
get the dual of the expression and then compliment each of the variable , that is the shot cut to check ur result dual expression : (A'+B)(C+D)+EE' when each variable is complimented : ((A')' + B' ) (C' +D')+1'= (A+B')(C'+D')+0= (A+B')(C'+D')
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