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 ____________ .

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

habedo007

I've done same mistake.

while computing POS, take don't care also to get the minimize POS form.

POS is A'C

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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.

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

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 4

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

What if the given K map itself is converted into a maxterms K map?

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It should be $CA'$  brother, though the final answer won't change

@Sambhrant Maurya --> you will get $CA'$ itself as there are two octet of zero's .

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@Hemanth_13 corrected

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NOTE : a¯,a¯,c¯ is given so no need to use NOT gate for that

 It mean that if this was not given then minimum of nor gate required 2 ?
am i correct? 

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OR-AND realization is equal to NOR - NOR realization, so we should write the POS(product of sum) form rather than the SOP(sum of product) form.

SOP: $f(b,a,d,c) = c \cdot a' $

POS: $f(b,a,d,c) = c\cdot a’ $

So, only one NOR gate is required.

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Lakshman Patel RJIT Pls can you Explain more after finding POS as (c + a'). How to apply this POS expression from NOR Gate?
I follow the same approach rather find POS not SOP.
 

Also In this Below question suggest me which approach to follow, as Convert min term SOP to POS or any other? 
What is the minimum number of 2-input NOR gates required to implement a 4-variable function function expressed in sum-of-minterms form as f = Σ(0, 2, 5, 7, 8, 10, 13, 15)? Assume that all the inputs and their complements are available._________3.(Answer).

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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?