91Ó°ÊÓ

Construct a switching circuit to meet the requirements of the Boolean expression: \(Z=A \cdot \bar{C}+\bar{A} \cdot B+\bar{A} \cdot B \cdot \bar{C}\). Construct the truth table for this circuit. The three terms joined by or-functions, \((+)\), indicate three parallel branches,

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: \((\overline{A \cdot B})+(\overline{\bar{A}+B})=A+\bar{B}\)