Question 1: Simplify the following Boolean expressions using De Morgan law and Boolean Algebra rules. Show process steps and your work clearly. Boolean expression below can be simplified as two variables and AND gate. [15 Marks] Q = A.B.(\overline{B} + C) + B.C + (\overline{(\overline{B} + \overline{C})} + \overline{A})
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B.(B + C) + B.C + ((B + C) + A) Show more…
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Simplify the Boolean expression $(\overline{\bar{A} \cdot B})+(\overline{\bar{A}+B})$ by using de Morgan's laws and the rules of Boolean algebra. Applying de Morgan's law to the first term gives: $\overline{\bar{A} \cdot B}=\overline{\bar{A}}+\bar{B}=A+\bar{B}$ since $\overline{\bar{A}}=A$ Applying de Morgan's law to the second term gives: $$ \overline{\bar{A}+B}=\overline{\bar{A}} \cdot \bar{B}=A \cdot \bar{B} $$ Thus, $(\overline{\bar{A} \cdot B})+(\overline{\bar{A}+B})=(A+\bar{B})+A \cdot \bar{B}$ Removing the bracket and reordering gives: $A+A$ : $\bar{B}+\bar{B}$ But, by rule 15 , Table $11.7, A+A \cdot B=A$, It follows that: $A+A \cdot \bar{B}=A$ Thus: $(\overline{A \cdot B})+(\overline{\bar{A}+B})=A+\bar{B}$
Simplify the Boolean expression $(\overline{A \cdot \bar{B}+C}) \cdot(\bar{A}+\overline{B \cdot \bar{C}})$ by using de Morgan's laws and the rules of Boolean algebra. pplying de Morgan's laws to the first term gives: $$ \begin{aligned} \overline{A \cdot \bar{B}+C} &=\overline{A \cdot \bar{B}} \cdot \bar{C}=(\bar{A}+\overline{\bar{B}}) \cdot \bar{C} \\ &=(\bar{A}+B) \cdot \bar{C}=\bar{A} \cdot \bar{C}+B \cdot \bar{C} \end{aligned} $$ pplying de Morgan's law to the second term gives: $$ \bar{A}+\overline{B \cdot \bar{C}}=\bar{A}+(\bar{B}+\overline{\bar{C}})=\bar{A}+(\bar{B}+C) $$ hus $(\overline{A \cdot \bar{B}+C}) \cdot(\bar{A}+\overline{B \cdot \bar{C}})$ $$ \begin{aligned} &=(\bar{A} \cdot \bar{C}+B \cdot \bar{C}) \cdot(\bar{A}+\bar{B}+C) \\ &=\bar{A} \cdot \bar{A} \cdot \bar{C}+\bar{A} \cdot \bar{B} \cdot \bar{C}+\bar{A} \cdot \bar{C} \cdot C \\ &\quad+\bar{A} \cdot B \cdot \bar{C}+B \cdot \bar{B} \cdot \bar{C}+B \cdot \bar{C} \cdot C \end{aligned} $$ But from Table $11,7, \bar{A} \cdot \bar{A}=\bar{A}$ and $\bar{C} \cdot C=B \cdot \bar{B}=0$ Hence the Boolean expression becomes: $$ \begin{aligned} \bar{A} & \cdot \bar{C}+\bar{A} \cdot \bar{B} \cdot \bar{C}+\bar{A} \cdot B \cdot \bar{C} \\ &=\bar{A} \cdot \bar{C}(1+\bar{B}+B) \\ &=\bar{A} \cdot \bar{C}(1+B) \\ &=\bar{A} \cdot \bar{C} \end{aligned} $$ Thus: $\overline{(A \cdot \bar{B}+C}) \cdot(\bar{A}+\overline{B \cdot \bar{C}})=\bar{A} \cdot \bar{C}$
Simplify $A \cdot \bar{C}+\bar{A} \cdot(B+C)+A \cdot B \cdot(C+\bar{B})$ using the rules of Boolean algebra.With reference to Table $11.7$ Reference $$ \begin{aligned} A \cdot \bar{C}+\bar{A} \cdot(B+C) \\ &=A \cdot \bar{C}+\bar{A} \cdot B+\bar{A} \cdot C \\ &=A \cdot \bar{C}+\bar{A} \cdot B+\bar{A} \cdot C \\ &=A \cdot \bar{C}+\bar{A} \cdot B+\bar{A} \cdot C+A \cdot B \cdot C \\ &=A \cdot(\bar{C}+B \cdot C)+\bar{A} \cdot B+\bar{A} \cdot C \\ &=A \cdot(\bar{C}+B)+\bar{A} \cdot B+\bar{A} \cdot C \\ &=A \cdot \bar{C}+A \cdot B+\bar{A} \cdot B+\bar{A} \cdot C \\ &=A \cdot \bar{C}+B \cdot(A+\bar{A})+\bar{A} \cdot C \\ &=A \cdot \bar{C}+B \cdot 1+\bar{A} \cdot C \\ =& A \cdot \bar{C}+B+\bar{A} \cdot C \end{aligned} $$ 14 17
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