| written 9.5 years ago by | • modified 9.5 years ago |
Mumbai University > COMPS > Sem 3 > Digital Logic Design and Analysis
Marks: 8 M
Year: May 2015
| written 9.5 years ago by | • modified 9.5 years ago |
Mumbai University > COMPS > Sem 3 > Digital Logic Design and Analysis
Marks: 8 M
Year: May 2015
| written 9.5 years ago by |
$A+\overline{B}\overline{C}+AB\overline{D}+ABCD \\ =A(B+\underline{B})(C+\overline{C})(D+\overline{D})+\overline{B}\overline{C}+(A+\overline{A})(D+\overline{D})+AB\overline{D}(C+\overline{C})+ABCD \\ = A(B+\underline{B})(CD+C\overline{D}+\overline{C}D+\overline{C}\overline{D})+\overline{B}\overline{C}(AD+A\overline{D}+\overline{A}D+\overline{A}\overline{D})ABC\overline{D}+AB\overline{C}\overline{D}+ABCD \\ =A\bigg(BCD+BC\overline{D}+B\overline{C}D+B\overline{C}\overline{D}+\overline{B}CD+\overline{B}C\overline{D}+\overline{B}\overline{C}D+\overline{B}\overline{C}\overline{D}+A\overline{B}\overline{C}D+\overline{A}\overline{B}\overline{C}D+\overline{A}\overline{B}\overline{C}\overline{D}+ABC\overline{D}+AB\overline{C}\overline{D}+ABCD\bigg) \\ =ABCD+ABC\overline{D}+AB\overline{C}D+AB\overline{C}\overline{D}+A\overline{B}CD+A\overline{B}C\overline{D}+A\overline{B}\overline{C}D+A\overline{B}\overline{C}\overline{D}+A\overline{B}\overline{C}D+\overline{A}\overline{B}\overline{C}D+\overline{A}\overline{B}\overline{C}\overline{D}+ABC\overline{D}+AB\overline{C}\overline{D}+ABCD \\ =ABCD+ABC\overline{D}+A\overline{B}\overline{C}D+AB\overline{C}\overline{D}+A\overline{B}\overline{C}\overline{D} AB\overline{C}D+A\overline{B}CD+A\overline{B}C\overline{D}+\overline{A}\overline{B}\overline{C}D$

