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Produce the truth table of the combination of four NAND gates arranged as shown in Fig.Name the gate so formed. |
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Answer» Solution :The TRUTH table of COMBINATION of gates is shown below, `|:{:(A,B,y'(=bar(A.B)),y_(1)(=bar(A.y')),y_(2)(=bar(B.y)),y=bar(y_(1).y_(2))),(0,0,1,1,1,0),(1,0,1,0,1,1),(0,1,1,1,0,1),(1,1,0,1,1,0):}:|-= |{:(A,B,y),(0,0,0),(1,0,1),(0,1,1),(1,1,0):}|` It means a EXOR gate is produced. Explaination Using Boolean expression and De morgan's THEOREM[i.e., `bar(C.D)=barC+barD and bar(C+D)=barC.barD]` `y^(')=bar(A.B)=barA+barB` `y_(1)=bar(A.(barA+barB))=barA+bar((barA+barB))=barA+(bar(barA).bar(barB))=barA+(A.B)` `y_(2)=bar(B.(barA+barB))=barB+bar((barA+barB))=barB+(bar(barA).bar(barB))=barB+(A.B)` `y = bar(y_(1).y_(2)) =bar([bar(A)+(A.B)].[bar(B)+(A).B))] =bar([bar(A)+(A+B)]) =barbar(A).bar(A+B)+barbar(B).bar(A+B)` `=A.(barA+barB)+B.(barA+barB)=A.barA+A.barB+B.barA+B.barB=A.barB+B.barA( :'A.barA=B.barB=0)` which is for EXOR gate. Thus, arrangement WORKS as EXOR gate. |
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