39 votes

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 ____________ .

49 votes

Best answer

0

can someone explain me with diagram how is the k-map reduced ? My answer is coming like : ca' + ac' which is by considering the two quad squares. Where am I wrong ? Why aren't we using the dont cares in first and fourth row ?

12

*There isn't any requirement that **do**n't cares MUST be used.* If with the use of don't cares we can reduce the term size, only then a don't care need be used.

3

This function has same $SOP$ and $POS$ form - $c.a$'

For $SOP$ - $c.a'$ is treated as single term.

For $POS$ - $c.a'$ is treated as 2 terms each of which is composed of a single variable.

For $SOP$ - $c.a'$ is treated as single term.

For $POS$ - $c.a'$ is treated as 2 terms each of which is composed of a single variable.

0

Don't we need one nor gate for inverting the input "C" as well? So 2 NOR gates should be required in my opinion.

0

No need NOR gate for inverting the input because it's available in the question.

And one more thing, OR-AND realization is equal to NOR-NOR realization, so it's better we write Product of sum (POS) form.

17 votes

**Answer : 1 Only**

Here we should take note that * all inputs (a,b,c,d)* and their

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

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

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

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

0

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 .

0

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?

1

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.

0

*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).**

0