Question

Simplify the following expression by k maps method
1. A'BC'+ABC'+A'B'C'
2. AB'C+ABC'+A'BC+ABC+AB'C'+A'B'C'

Answer 1

Answer:

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Explanation:

Simplification with Karnaugh Map

It is always desirable to simplify a given Boolean function (as either a Boolean expression or a Truth Table) so that the hardware for realizing the function will be minimized in terms of the number of logic gates and the number of inputs to these gates necessary for representing the function.

Deduction method could be used to apply various axioms and theorems to the given expression so that a simpler form can be derived. However, this is very heuristic and you need luck as well as skill to succeed.

Example 0:

\begin{displaymath}AB+AB'=A(B+B')=A\;1=A \end{displaymath}

\begin{displaymath}ABC+ABC'=AB(C+C)=AB\;1=AB \end{displaymath}

In general

\begin{displaymath}(ABC\cdots)X'+(ABC\cdots)X=ABC\cdots \end{displaymath}

Example 1:

$\displaystyle f(A,B)$ $\textstyle =$ $\displaystyle A'B + AB' + AB$

$\textstyle =$ $\displaystyle A'B + A = ?$

$\textstyle =$ $\displaystyle (A' + A)(B + A)\;\;\;\;\;

\mbox{(If you remember this axiom)}$

$\textstyle =$ $\displaystyle B + A$

$\displaystyle or$ $\textstyle =$ $\displaystyle A'B + \underline{AB} + AB' + \underline{AB}$

$\textstyle =$ $\displaystyle B(A'+A) + A(B'+B)$

$\textstyle =$ $\displaystyle B + A$

Example 2:

$\displaystyle f(A,B,C)$ $\textstyle =$ $\displaystyle A'BC'+A'BC+A'B'C+ABC$

$\textstyle =$ $\displaystyle A'B+A'B'C+ABC$

$\displaystyle or$ $\textstyle =$ $\displaystyle A'BC'+\underline{A'BC}+A'B'C+\underline{A'BC}+ABC+\underline{A'BC}$

$\textstyle =$ $\displaystyle A'B+A'C+BC$

A function can be more systematically simplified by Karnaugh Map.

Two variables

Consider Example A above:

\begin{displaymath}

\begin{tabular}{c\vert cc\vert c} \hline

& A & B & f \hl...

...& 1 \\

2 & 1 & 0 & 1 \\

3 & 1 & 1 & 1 \hline

\end{tabular}\end{displaymath}