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For the following function implement the SOP and POS circuit F(A,B,C,D)=?m (2,3,5,7)+?d(6,13,14,15)
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To find sum of products.

Using K-map [4 variable K-map, with 16 cells]

  • In K-map, Grey code representation is used to represent logical expression.
  • The minterms in the above K-map should be included atleast once.
  • The minterms (2,3,5,7) and don't care values (6,13,14,15) should be grouped to a rectangle, to minimize the no. of products in the final expression.
  • Make each rectangle as large as possible, to minimize the no. of literals in each term.
  • In the above K-map, we found 2 rectangles.
  • Each rectangle corresponds to one product term.
  • The product is determined by finding the common literals in that rectangle.

$\text{rectangle 1}=m(2,3,6,7)=A'C$

$\text{rectangle 2}=m(5,7,13,15)=BD$

In the first rectangle $m(2,3,6,7)$. The product term is $A'C$.

$\left. \begin{align*} &3\text{ represented as }0011 \rightarrow A'B'CD\\[2ex] &2\text{ represented as }0010\rightarrow A'B'CD'\\[2ex] &7\text{ represented as }0111\rightarrow A'BCD\\[2ex] &6\text{ represented as }0110\rightarrow A'BCD' \end{align*}\right\}$

Adding the above expression,

$\begin{align*} &\rightarrow A'B'CD+A'B'CD'+A'BCD+A'BCD'\\[2ex] &\rightarrow A'B'C[D+D']+A'BC[D+D']\\[2ex] &\rightarrow A'B'C+A'BC\\[2ex] &\rightarrow A'c[B+B']\\[2ex] &\rightarrow A'C\end {align*}$

$\because[x+x'=1]$

SOP - $A'C+BD$

To find POS:

Using the De-Morgan's Laws,

The complement of a SOP is always a POS and vice versa.

SOP: $F=A'C+BD$

$\begin {align*} (F)' &=\text{POS}\\[2ex] &=(A'C+BD)'\\[2ex] & =(A'C)'\cdot (BD)'\rightarrow (A+C')(B'+D')\end {align*}$

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