Test Series

Question

The minimum number of two input NOR gates required to implement the simplified expression of the following function are $\qquad$ $\mathrm{f}(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D})=\Sigma \mathrm{m}(0,2,6,10,14)$

• A. One
  
• B. Two
  
• C. Three
  
• D. Four

Comments

Is it C?

---

Yes please tell in detail

---

We should know that an expression in POS form is always easier to be realized by NOR gates.

Given is the SOP ∑m(0,2,6,10,14) so we should convert it to POS.

Hence piM(1,3,4,5,7,8,9,11,12,13,15) [I couldn't find the "pi" symbol :P]

So the expression is D'.(B'+C).(A'+C)

Now we have to convert this expression in such a way that we can realize it with NOR

( ( D'.(B'+C).(A'+C) )' )' = (  (D')' + (B'+C)' + (A'+C)' )' = (D + BC' + AC' )'

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

= (D + ( C + (A+B)' )' )' 

X=(A+B)'  --> 1 NOR

Y=(C+ X)'  --> 1 NOR

(D+Y)' --> 1 NOR

---

By k-map, SOP format ===> CD' + A' B' D' ===> D' (C+A'B') ==> D' . ( C + (A+B)' ) ===> ( D + ( C + (A+B)' )'  ) '

TOTAL = 3 NOR GATES

---

Yeah.. Shaik's one is faster.. usually converting to POS works better..but in this case it's not

Answer 1

3 nor gates is the ans right?