Question

• Given P(a,b,c,d)=∑(1,3,8,9,10,11,12,14,15).
  • Determine P as a most simplified SOP
  • Determine P’ as a most simplified SOP
• Generate gate diagrams of P and P’ from the first bullet
• Implement P’, by converting each of the gate diagrams from above to only 2-input Nand gates. Do realize that a Nand equivalent of invertor must be added to the output for P to create P’.
• (its is one question)

Answer 1

Given Function is

P (a, b, c, d) = m (1, 3, 8, 9, 10, 11, 12, 14, 15)

Above Function in K-map as follows

Simplified K-map as follows

The most Simplified SOP of P (a, b, c, d) = ad’+b’d+ac

Given Function is

P (a, b, c, d) = m (1, 3, 8, 9, 10, 11, 12, 14, 15)

The complement of the above Function is nothing but all the minterms other than the above minterms

So the complement of P is P'

P' (a, b, c, d) = m (0, 2, 4, 5, 6, 7, 13)

Above Function in K-map as follows

Simplified K-map as follows

The most Simplified SOP of P' (a, b, c, d) = a’b+bc’d+a’d’

Given P  = ad’+b’d+ac

[P']'  = [(ad’+b’d+ac)']' { We know that (P')'= P }

P  = [(ad’)' (b’d)' (ac)']'   { We know that (P+Q)'= P' Q' }

P  = [(ad’)' (b’d)' (ac)']'

Simplified Circuit: Using NAND Gates

Given P' = a’b+bc’d+a’d’

[(P')']' = [(a’b+bc’d+a’d’)']' { We know that (P')'= P }

P' = [(a’b)' (bc’d)' (a’d’)']' { We know that (P+Q)'= P' Q' }

P' = [(a’b)' (bc’d)' (a’d’)']'

Simplified Circuit: Using NAND Gates