Lecture 논리 회로의 연산과 정리

Lecture • Views 771 • Comments 0 • Last Updated at 2 years ago  
  • 제어 설비
  • 논리 회로
  • 부울 대수
  • 드모르간의 정리

부울 대수의 주요 정리 및 법칙

항등 법칙

A+0=AA1=A \begin{align*} \mathrm{A} + 0 &= \mathrm{A} \\ \mathrm{A} \cdot 1 &= \mathrm{A} \end{align*}

소멸자

A+1=1A0=0 \begin{align*} \mathrm{A} + 1 &= 1 \\ \mathrm{A} \cdot 0 &= 0 \end{align*}

멱등 법칙

A+A=AAA=A \begin{align*} \mathrm{A} + \mathrm{A} &= \mathrm{A} \\ \mathrm{A} \cdot \mathrm{A} &= \mathrm{A} \end{align*}

보수 (역원)

A+A=1AA=0 \begin{align*} \mathrm{A} + \overline{\mathrm{A}} &= 1 \\ \mathrm{A} \cdot \overline{\mathrm{A}} &= 0 \end{align*}

이중 부정

A=A \overline{\overline{\mathrm{A}}} = \mathrm{A}

교환 법칙

A+B=B+AAB=BA \begin{align*} \mathrm{A} + \mathrm{B} &= \mathrm{B} + \mathrm{A} \\ \mathrm{A} \cdot \mathrm{B} &= \mathrm{B} \cdot \mathrm{A} \end{align*}

결합 법칙

A+(B+C)=(A+B)+CA(BC)=(AB)C \begin{align*} \mathrm{A} + \left( \mathrm{B} + \mathrm{C} \right) &= \left( \mathrm{A} + \mathrm{B} \right) + \mathrm{C} \\ \mathrm{A} \cdot \left( \mathrm{B} \cdot \mathrm{C} \right) &= \left( \mathrm{A} \cdot \mathrm{B} \right) \cdot \mathrm{C} \end{align*}

흡수 법칙

A(A+B)=AA+(AB)=A \begin{align*} \mathrm{A} \cdot \left( \mathrm{A} + \mathrm{B} \right) &= \mathrm{A} \\ \mathrm{A} + \left( \mathrm{A} \cdot \mathrm{B} \right) &= \mathrm{A} \end{align*}

분배 법칙

A(B+C)=AB+ACA+(BC)=(A+B)(A+C) \begin{align*} \mathrm{A} \cdot \left( \mathrm{B} + \mathrm{C} \right) &= \mathrm{A} \cdot \mathrm{B} + \mathrm{A} \cdot \mathrm{C} \\ \mathrm{A} + \left( \mathrm{B} \cdot \mathrm{C} \right) &= \left( \mathrm{A} + \mathrm{B} \right) \cdot \left( \mathrm{A} + \mathrm{C} \right) \end{align*}

드모르간의 정리

A+B=ABAB=A+B \begin{align*} \overline{ \mathrm{A} + \mathrm{B}} &= \overline{\mathrm{A}} \cdot \overline{\mathrm{B}} \\ \overline{ \mathrm{A} \cdot \mathrm{B}} &= \overline{\mathrm{A}} + \overline{\mathrm{B}} \end{align*}

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