Test: K Map & Logic Gates - Computer Science Engineering (CSE) MCQ

We need to map min terms of 
Hence as highlighted g₂ matches with option C
Edit : we have to map h(x) with g(x)...mod 16 is used in g(x) only because since we have 4 bits ,mamximum possible no that can be represented is 15..so after 15 we shouldn't get 16 and go back to 0.. thats why..
now mapping is simple...we just have to map such that h(x)---->g(x+1)
means if h represents gray code of 0 then g will represent gray code of 1
if h represents gray code of 1 then g will represent for 2 and so on....
for last number h(15) ,mod 16 actually comes into picture which will make it to represent as g(0)
so draw table as mentioned above... now write g as a function of f ..simply how we do minimisation...see the min terms..
be careful only in one thing here...
example... for 2..gray code representation is 0011 meaning 3 in decimal..so if we select this row as a min term(just an example) then we have selected 3 and not 2 ..means row numbers are not representing min terms...rest everything is fine!

1). In the question it is stated that -> Thus, for n=3, the ordering (000, 100, 101, 111, 110, 010, 011, 001) is a Gray code.
2). We have to find orderings such that they start with 0ⁿ , must contain all bit strings of length n and successive strings must differ in one bit.
Option c&d are clearly wrong.
Now, consider n=1. The only Gray code possible is {0,1}. Hence no of Gray code = odd for n=1.
For n=2 only two Gray code exists {00, 10, 11, 01} and {00, 01, 11, 10}. Thus no of Gray code = even for n=2.
Thus it is not that gray codes could be only even or only odd.
Hence option e is correct. 

00000(code1),01011(code2),10101(code3),11110(code4)
Haming distance = min of all hamming distances.
which is 3 b/w (code1) and (code2) so p = 3
Now to correct d bit error we need hamming distance = 2d+1
so 2d+1 = 3 will gives d= 1.
A is answer

answer = option D

Sign bit is 1 -> number is negative
Exponent bits- 10000011
Exponent is added with 127 bias in IEEE single precision format.
So, Actual exponent = 10000011 - 127 = 131 - 127 = 4
Mantissa bits- 101000000000000000000000
In IEEE format, an implied 1 is before mantissa, and hence the actual number is
-1.101 * 2⁴ 
=- (11010)2 = - 26

(B) is the answer. In IEEE uses normalized representation and hence an implicit '1' is used before the decimal point. So, if mantissa is 0000..0
it would be treated as 1.000..0
and hence the exponent need to be -1 for us to get 0.1 which is the binary representation of 0.5. 
More into IEEE floating point representation:

A. can't be ans. ans wants without solving ??
Two Max Terms are: (a+b)(a+c) so "a" is common here only possibility is option D. ----------------------------------------------------------------------------------------------------------------------------------
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Kmap will give min terms = a[4 1's circle] + bc[2 1's box] Option D ckt will give  = (a+b) (b+c) = a+bc
D will be answer

If all the empty cells are filled with '0' and then the POS expression is calculated as:

Then the POS expression of f = (D+B)(D'+B') = BD' + B'D

CD+AD

Essential prime implicants are prime implicants that cover an output of the function that no combination of other prime implicants is able to cover.

m1,m3,m9,m11 form one quad xz'
m4,m6,m12,m14 form one quad x'z So f=x'z+z'x 

essential prime implicants which r grouped only by only one method or way so in above question cornor's ones r grouped by only one method
d) will be the answer

ad'   [fill minterm in K-map in front for a and d' ]

similarly fill all minterms  for ad'+a'c' +bc'd , resulting K-map will be 

option a) c'd'+ ad' + abc' + a'c'd

s equivalent to given expression option b) a'c' + c'd' + ad' + abc'd

is equivalent to given expression.
option c) a'c' + ad' + abc' + c'd

is not equivalent to given expression.
option d) b'c'd' + acd' + a'c' + abc'

is equivalent to given expression.
So, answer is C.

2 Quads are getting formed

Value for First one is a'd' and value for 2nd one is b'd'.

2 quads are getting formed.
Value for First one is b'd' and value for 2nd one is b'c'.So,Ans is option B

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 

NOR gate, NAND gate, Multiplexers and Half adders can also be used to realise all digital circuits.

So, Ans D)

ROM memory size =2^m x n
m=no. of address lines  n= no. of data lines
given  4K x 16
= 2² x 2¹⁰ x 16
= 2¹² x 16 address lines =12
data lines= 16 

Answer should be (B) As according to question .. truth table will be like
A B C D    f    
0 0 0 0      0
0 0 0 1     0
0 0 1 0     0
0 0 1 1     0
0 1 0 0    0
0 1 0 1    1
0 1 1 0    1
0 1 1 1    1
1 0 0 0     1
1 0 0 1     1
1 0 1 0     dont care
1 0 1 1    dont care
1 1 0 0     dont care
1 1 0 1    dont care
1 1 1 0    dont care
1 1 1 1    dont care 
using this truth table we get  3 sub cube which are combined with following minterms  A (8,9,10,11,12,13,14,15) , BD( 5,13,7,15) and BC(6,7,14,15)
 SO f = A+ BD +BC= A+ B(C+D)
SO minimum gate required 2 OR gate and 1 AND gate = 3 minimum gate ...

given boolean function is
 f = AB + C
  = (A+C) . (B +C)  
   =((A+C)' +(B+C)')' 

therefore 3 NOR gate required .

The K-map would be

So, the minimized expression would be x'z + xz'.
So, option B.