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Consider the Karnaugh map given below, where $X$ represents "don't care" and blank represents $0$.

 

Assume for all inputs $\left ( a,b,c,d \right )$, the respective complements $\left ( \bar{a}, \bar{b}, \bar{c}, \bar{d} \right )$ are also available. The above logic is implemented using $2$-input $\text{NOR}$ gates only. The minimum number of gates required is ____________ .

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61 votes
61 votes

From K-map simplification we get the min-term as $CA'$. So We can simplyfy it for NOR gate expression

I.e. C' NOR $A = (C'+A)' = CA'$
Now complemented inputs are also given to us so, for $2$ input NOR gate we need only 1 NOR gate.

1 is correct answer .

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21 votes

Answer : 1 Only

Here we should take note that all inputs (a,b,c,d) and their respective complements are available.

And now when we solve the K-map the we get minterms like :

- >    cb'a'+ cba' 

- >    ca'(b+b')

- >    ca' 

If we give input to a NOR gate as c' and a the output will be (c' + a )' =  ca'.

So, only 1 NOR gate is Required.

6 votes
6 votes
Applying logic of k-map the simplified expression id ca'

now it is given that all inverted inputs are also available

so we can express ca' in terms of nor gate

(c'+a)'=ca'

so only 1 NOR gate required

1 is answer

ps: more editing will come
6 votes
6 votes

The expression of $K-map$ is $c.\bar a$

so the expression should be in $POS$ form for $NOR$ gate 

$\overline{\overline{c.\bar a}} = \overline{\bar c + a} $

Now the expression is in $POS$ form

NOTE : $\bar a , \bar a , \bar c \text { is given so no need to use NOT gate for that}$

 

 

So only 1 $NOR$ gate required

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