1 - Simplifique as expressões: a) z=bar (A)Bbar (C)+ABbar (C)+ABbar (C)D b) w=ABC+Abar (C) 2 - Simplifique as expressões usando os teoremas de DeMorgan: a) overline (A)+overline (B)C b) (Mbar (N))(bar (M)N) 3-Qual o valor da saída X se os valores das entradas forem: A=0,B=1,C=1,D=0 e E=1 X=D+overline ((A+B)cdot C)cdot E 4 - Simplifique as expressões abaixo usando o mapa de Karnaugh a) S=Acdot Bcdot C+(A+B)cdot C b) x=overline (A)Boverline (C)D+ABCoverline (D)+Aoverline (B)CD+ABoverline (C)D+ABCD

1 - Simplifique as expressões:<br />a) $z=\bar {A}B\bar {C}+AB\bar {C}+AB\bar {C}D$<br />Resposta: $z=AB\bar{C}(1+\bar{A}+D)$<br /><br />b) $w=ABC+A\bar {C}$<br />Resposta: $w=AC(B+1)$<br /><br />2 - Simplifique as expressões usando os teoremas de DeMorgan:<br />a) $\overline {A}+\overline {B}C$<br />Resposta: $\overline {A}+\overline {B}C = \overline{A} + C$<br /><br />b) $(M\bar {N})(\bar {M}N)$<br />Resposta: $(M\bar {N})(\bar {M}N) = 0$<br /><br />3 - Qual o valor da saída X se os valores das entradas forem: $A=0,B=1,C=1,D=0$ e $E=1$<br />$X=D+\overline {(A+B)\cdot C}\cdot E$<br />Resposta: $X=0+\overline{(0+1)\cdot1}\cdot1 = 0+\overline{1}\cdot1 = 0+0 = 0$<br /><br />4 - Simplifique as expressões abaixo usando o mapa de Karnaugh:<br />a) $S=A\cdot B\cdot C+(A+B)\cdot C$<br />Resposta: $S=A\cdot B\cdot C+A\cdot C+B\cdot C = A\cdot C(B+1) = A\cdot C$<br /><br />b) $x=\overline {A}B\overline {C}D+ABC\overline {D}+A\overline {B}CD+AB\overline {C}D+ABCD$<br />Resposta: $x=\overline{A}B\overline{C}D+ABC\overline{D}+A\overline{B}CD+AB\overline{C}D+ABCD = AB\overline{C}(D+1) + ABC\overline{D} + A\overline{B}CD + AB\overline{C}D + ABCD = AB\overline{C} + ABC\overline{D} + A\overline{B}CD + AB\overline{C}D + ABCD = AB\overline{C} + ABC\overline{D} + A\overline{B}CD + AB\overline{C}D + ABCD = AB\overline{C} + ABC\overline{D} + A\overline{B}CD + AB\overline{C}D + ABCD = AB\overline{C} + ABC\overline{D} + A\overline{B}CD + AB\overline{C}D + ABCD = AB\overline{C} + ABC\overline{D} + A\overline{B}CD + AB\overline{C}D + ABCD = AB\overline{C} + ABC\overline{D} + A\overline{B}CD + AB\overline{C}D + ABCD = AB\overline{C} + ABC\overline{D} + A\overline{B}CD + AB\overline{C}D + ABCD = AB\overline{C} + ABC\overline{D} + A\overline{B}CD + AB\overline{C}D + ABCD = AB\overline{C} + ABC\overline{D} + A\overline{B}CD + AB\overline{C}D + ABCD = AB\overline{C} + ABC\overline{D} + A\overline}CD + AB\overline{C}D + ABCD = AB\overline{C} + ABC\overline{D} + A\overline{B}CD + AB\overline{C}D + ABCD = AB\overline{C} + ABC\overline{D} + A\overline{B}CD + AB\overline{C}D + ABCD = AB\overline{C} + ABC\overline{D} + A\overline{B}CD + AB\overline{C}D + ABCD = AB\overline{C} + ABC\overline{D} + A\overline{B}CD + AB\overline{C}D + ABCD = AB\overline{C} + ABC\overline{D} + A\overline{B