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Consider numbers represented inbe the Gray code representation of a number n and let be the Gray code of value of the number .Which one of the following functions is correct?
We need to map min terms of
Hence as highlighted g_{2} 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!
For any natural number , an ordering of all binary strings of length n is a Gray code if it starts with 0^{n}, and any successive strings in the ordering differ in exactly one bit (the first and last string must also differ by one bit). Thus, for n = 3, the ordering (000, 100, 101, 111, 110, 010, 011, 001) is a Gray code. Which of the following must be TRUE for all Gray codes over strings of length n?
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 0^{n} , 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.
A half adder is implemented with XOR and AND gates. A full adder is implemented with two half adders and one OR gate.
The propagation delay of an XOR gate is twice that of an AND/OR gate. The propagation delay of an AND/OR gate is 1.2 microseconds. A 4bitripplecarry binary adder is implemented by using four full adders. The total propagation time of this 4bit binary adder in microseconds is ______.
Consider the binary code that consists of only four valid codewords as given below: 00000, 01011, 10101, 11110
Let the minimum Hamming distance of the code p and the maximum number of erroneous bits that can be corrected by the code be q. Then the values of p and q are
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
In the IEEE floating point representation the hexadecimal value corresponds to
answer = option D
The following bit pattern represents a floating point number in IEEE 754 single precision format
1 10000011 101000000000000000000000
The value of the number in decimal form is
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 * 2^{4}
= (11010)2 =  26
The decimal value 0.5 in IEEE single precision floating point representation has
(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 Boolean function f is to be realized only by NOR gates. Its Kmap is given below:
The realization is
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. 

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
What is the equivalent Boolean expression in productofsums form for the Karnaugh map given in Fig
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
Which of the following expressions is in the productofsums form?
Which of the following functions implements the Karnaugh map shown below?
CD+AD
The prime implicant which has at least one element that is not present in any other implicant is known as
Essential prime implicants are prime implicants that cover an output of the function that no combination of other prime implicants is able to cover.
Given the following karnaugh map, which one of the following represents the minimal SumOfProducts of the map?
Minimum sum of product expression for f(w,x,y,z) shown in Karnaughmap below
m1,m3,m9,m11 form one quad xz'
m4,m6,m12,m14 form one quad x'z So f=x'z+z'x
The boolean function for a combinational circuit with four inputs is represented by the following Karnaugh map.
Which of the product terms given below is an essential prime implicant of the function?
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
Consider the following expression
Which of the following Karnaugh Maps correctly represents the expression?
Consider the following expression
Which of the following expressions does not correspond to the Karnaugh Map obtained for the given expression?
ad' [fill minterm in Kmap in front for a and d' ]
similarly fill all minterms for ad'+a'c' +bc'd , resulting Kmap 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.
In the Karnaugh map shown below, X denotes a don’t care term. What is the minimal form of the function represented by the Karnaugh map?
2 Quads are getting formed
Value for First one is a'd' and value for 2nd one is b'd'.
What is the minimal form of the Karnaugh map shown below? Assume that X denotes a don’t care term
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
Consider the Karnaugh map given below, where X represents "don't care" and blank represents 0.
Assume for all inputs the respective complements are also available. The above logic is implemented using 2input NOR gates only. The minimum number of gates required is ____________ .
From Kmap simplification we get the minterm 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
Choose the correct alternatives (more than one may be correct) and write the corresponding letters only:
All digital circuits can be realized using only
NOR gate, NAND gate, Multiplexers and Half adders can also be used to realise all digital circuits.
Which one of the following circuits is NOT equivalent to a 2input XNOR (exclusive NOR) gate?
So, Ans D)
The Boolean function obtained by adding an inverter to each and every input of an gate is:
The capacity of a memory unit is defined by the number of words multiplied by the number of bits/word. How many separate address and data lines are needed for a memory of ?
ROM memory size =2^{m} x n
m=no. of address lines n= no. of data lines
given 4K x 16
= 2^{2} x 2^{10} x 16
= 2^{12} x 16 address lines =12
data lines= 16
A circuit outputs a digit in the form of 4 bits. 0 is represented by 0000, 1 by 0001, …, 9 by 1001. A combinational circuit is to be designed which takes these 4 bits as input and outputs 1 if the digit > 5, and 0 otherwise. If only AND, OR and NOT gates may be used, what is the minimum number of gates required?
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 ...
What is the minimum number of NAND gates required to implement a 2input EXCLUSIVEOR function without using any other logic gate?
What is the minimum number of gates required to implement the Boolean function (AB+C) if we have to use only 2input NOR gates?
given boolean function is
f = AB + C
= (A+C) . (B +C)
=((A+C)' +(B+C)')'
therefore 3 NOR gate required .
Consider the following Boolean function of four variables:
The function is
The Kmap would be
So, the minimized expression would be x'z + xz'.
So, option B.
Consider the following Boolean function of four variables
The function is
The simplified SOP (Sum of Product) from the Boolean expression
is
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