GATE CSE 2017 | Set 1 | Question: 21

Question

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 ____________ .

Comments

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final exp:- AC'

Only one NOR gate is needed

 

Answer 1

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 .

Comments

if complemented inputs are not available then 2 NOR gates are required for $CA’$?? please verify.

$\overline{\overline {C\bar A}}\implies\overline{\bar C+A}$

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@Hira Thakur yes, you’re correct.

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If complements were not available in that case the answer would be 2 , 1 extra bcz of comlementation of c

Answer 2

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.

Comments

the expression which comes is only ca'  how did u get cb'a'+cba' ????

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ca' is the minimized form .
Check my answer again.

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@Bikram sir check pls

if  question is max number of nor gate :

 (c' + a )'

term with a single literal  (c' ) requires an inverter 

(instead using inverter we can also  use nor gate to get this ) in this case max nor gate =2

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See this  Shivam Chauhan

in this case max nor will be 2

Answer 3

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

Comments

Actually the above method is more suitable when POS is given. But we got $a'+c$ which is sum term.

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 $f(a,b,c,d) = \sum{(0,2,5,7,8,10,13,15)}$ Sum of Minterms(SOP)

$f(a,b,c,d) = \Pi{(0,2,5,7,8,10,13,15)}$ Product of Maxterms(POS)

Using $k-$map,we get

$f(a,b,c,d) = B'D'+BD$ (SOP)

$f(a,b,c,d) = (B'+D)\cdot(B+D')$ (POS)

Using the POS,we can easily make NOR gate.

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We know,

Pos = (sop)'

Here sop expression is ca'

So converting to pos we get

(ca')' = (c'+a)

How you are applying double complement over the sop expression can u please explain?

Answer 4

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

Answer 5

Answer is 1 NOR gate,

on Solving the K-map for a POS expression we get,

f= (A'+C') (A+C)

which can be written as, f= ( (A'+C')'  + (A+C)' )'

this requires 3 NOR gates for implementation,

 

on solving K-map for SOP expression we get,
 

f= C+A'

which can be written as, f= (C' + A)'
 

this requires only 1 NOR gate for implementation and hence it is the answer

Answer 6

Look at this Explanation