2 votes

Simplify the following expression

AB’C + A’BC + A’B’C

Solution given is A’C + B’C can someone show me how?

AB’C + A’BC + A’B’C

Solution given is A’C + B’C can someone show me how?

4 votes

:AB'C + A'BC +A'B'C

=AB'C+A'C(B+B')

=AB'C+A'C x 1

=C(AB'+A')

=C[(A+A')(B'+A')] [`Distributive Law XY+Z==(X+Z)(Y+Z)`

]

=C x 1 x (B'+A')

=A'C+B'C

1 vote

Let $f(A,B,C)=A\cdot\overline{B}\cdot C+\overline{A}\cdot B\cdot C+\overline{A}\cdot\overline{B}\cdot C$

Take common from first and last term

$f(A,B,C)=\overline{B}\cdot C\cdot(A+\overline{A})+\overline{A}\cdot B\cdot C$

$f(A,B,C)=\overline{B}\cdot C\cdot1+\overline{A}\cdot B\cdot C$ ${\color{Red}{[X+\overline{X}=1]} }$

$f(A,B,C)=\overline{B}\cdot C+\overline{A}\cdot B\cdot C$

$f(A,B,C)=C(\overline{B}+\overline{A}\cdot B)$

$f(A,B,C)=C(\overline{B}+\overline{A})\cdot(\overline{B}+B)$ ${\color{Magenta} {[X+Y\cdot Z=(X+Y)\cdot(X+Z)]}}$

$f(A,B,C)=C(\overline{A}+\overline{B})\cdot 1$

$f(A,B,C)=\overline{A}\cdot C+\overline{B}\cdot C$

0 votes

$A\bar{B}C + \bar{A}BC + \bar{A}\bar{B}C$

Taking C common,

=$(A\bar{B} + \bar{A}B + \bar{A}\bar{B} ) . C$

=$((A \bigoplus B) + \bar{A}\bar{B} ) . C$

Remember this shortcut, [ $(P \bigoplus Q) + PQ => P + Q $], also we know that $A \bigoplus B = \bar{A} \bigoplus \bar{B}$, negating both inputs have no effect on output of XOR and XNOR. Hence we can write

$(A \bigoplus B) + \bar{A}\bar{B} = \bar{A}+\bar{B} $

Hence it directly becomes,

=$(\bar{A}+\bar{B}).C$

=$\bar{A}.C+\bar{B}.C$

Another method is to draw K-MAP from given min terms and then minimize :

$f(A,B,C) = A\bar{B}C + \bar{A}BC + \bar{A}\bar{B}C$

101 011 001

$f(A,B,C) = \sum \small m (1,3,5)$