Question

Simplify the following expressions, then implement them using digital logic gates. (a) f = A + AB + AC (b) f = AB + AC + BC (c) f = A + B + AB + BC (d) f = ABC + A + C (e) f = A B + BC + AB (f) f = A BD + ABD + BCD + ACD

Answer 1

Explanation:

(a)

$f=A+A B+A C\\=A[1+B+C]=A \quad[1+x=1]\\F=A$

No gate is required to implement this function

(b)

$\begin{aligned}&\ f=A B+\bar{A} C+B C\\&\therefore f=A B+A C\end{aligned}$                                  $\begin{array}{l}(A B+\bar{A} C+B E=A B+\bar{A} C \\B C \text { is redendant })\end{array}$

Note: Refer the first image.

(c)

$\begin{aligned}f &=\overline{A+B}+A \bar{B}+B \bar{C} \\&=(\bar{A} \bar{B})+A \bar{B}+B \bar{C} \\&=\bar{B}[A+\bar{A}]+B \bar{C} \\& F=\bar{B}+B \bar{C} =\bar{B}+\bar{C}\end{aligned}$    

Note: Refer the second image      

(d)

$\begin{aligned}f=& A B \bar{c}+\overline{A+\bar{c}} \\=& A B \bar{c}+\bar{A} \bar{c}=\bar{A} B \bar{c}+\bar{A} c \\f=& \bar{A}[c+B \bar{c}] . \\& f=\bar{A} B+\bar{A} c=\bar{A}(B+c)\end{aligned}$

Note: Refer the third image

(e)

$\begin{aligned}f=& A \bar{B}+\bar{B} C+A \bar{B} \\&=\bar{B}[A+\bar{A}+c] \\&=\bar{B}[1+C]\end{aligned}$

       $f=\frac{}{B}$

(f)

$\begin{aligned}f &=A B C+A B D+A B C \\&=A B[C+C]+A B D \\&=A B+A B D \\&=B[A+A D] \\&=B[A+D] \\\therefore & A=B[A+D]\end{aligned}$

Note: Refer the fourth image