Digital Logic Design NTA NET Questions

Answer: II
Explanation:
Given,
(D.C)₁₆
Calculation,
To obtain octal equivalent for a given number , first convert into base 2 representation and then make a group of 3 bits.
why 3 bits? Because octal representation have base 8 and log₂8 = 3bits.
(D)₁₆ = (13)₁₀ = (1101)₂
(C)₁₆ = (12)₁₀ = (1100)₂
(D.C)₁₆ = (1101.1100)₂
make a pair of 3 bits
(D.C)₁₆ = (001 101.110 000)₂ = (15.60)₈ = (15.6)₈
So, Option(II) is correct.

Answer: III
Explanation:
➡ Some Boolean algebra laws:
i) Identity Law
x + 0 = x , x .1 = x
ii) Absorption Law
x + xy = x , x.(x + y) = x
iii) Idempotent Law
x + x = x , x . x = x
iv) Domination Law
x + 1 = 1 , x . 0 = 0
v) Commutative Law
x + y = y + x , x . y = y . x
vi) Associative Law
x + (y + z) = (x + y) + z , x .(y . z) = (x . y). z
vii) Distributive Law
x.(y + z) = x.y + x.z , x + (y.z) = (x + y).(x + z)
viii) Complementation Law
x + x’ = 1 , x . x’ = 0
ix) DeMorgan’s Law
(x . y)’ = x’ + y’ , (x + y)’ = x’ . y’

So, Option(III) is correct.

Answer: III
Explanation:
➡ ODD Function: X-OR is an ODD Function Because it gives High output for Odd number of High inputs.
Le us check for two input A and B
Logical equation of X-OR gate.
Y = AB = AB’ + A’B

➡ Universal Gate: The NAND Gate and NOR Gate are said to be Universal Gate because any Logic circuit can be implemented with it.
➡ Self Complementing Code: In Self Complementing code, the code of a digit and the code of 9’s complement of the digit are 1’s complement to each other.
2421 code
let’s take decimal number 2,
9’s complement of the digit 2 = 7
the 2421 code of a digit 2 = 0010 ………….Code(1)
the 2421 code of a digit 7 = 1101 ………….Code(2)
Code (1) and code(2) are 1’s complement to each other. That’s why 2421 code is self complement code.
➡Buffer: Amplification
So, Option(III) is correct.

Answer: II
Explanation:
Combinational Circuits
• Combinational circuits are defined as time independent circuits which do not depends upon previous inputs to generate any output.
• Output depends upon only present input.
• As combinational circuit don’t have clock, they don’t require triggering.


A. Output at any time is function of inputs at that time. (True)
Outputs depend on present inputs only.
B. Contains memory elements. (False)
• Combinational Circuits does not have any memory elements because it depends on present inputs only.
C. Do not have feedback paths. (True)
D. Clock is used to trigger the circuits to obtain outputs. (False)
So, Option(II) is correct.

Answer: II
Explanation:
Given,
Number of registers = 8
Number of bits in each register = 16bits
Formula,
Number of multiplexer in the bus = number of bits in each register
Size of multiplexer : number of register X 1
Calculation,
Number of multiplexer in the bus:
Number of Multiplexer = number of bits in each register
Here, number of bits in each register = 16
Therefore, Number of register = 16
Size of multiplexer:
Size of multiplexer : number of register X 1
Here, number of register = 8
Therefore, Size of multiplexer = 8×1
So, Option(II) is correct.

Answer: IV
Explanation:
Two K-maps can be constructed from given Boolean Function:
F(A,B,C,D)=Σ(0,1,2,3,6,12,13,14,15)



Boolean Expression of K-maps are: (A’B’ + AB + A’CD’) and (A’B’ + AB + BCD’)
So, Option(IV) is correct.

Answer: I
Explanation:
Conjunctive normal form(CNF) or clausal normal form is a conjunction of one or more than one clauses, where a clause is a disjunction of literals; or it is a product of sums or an AND of ORs.
These are in conjunctive normal form:
• (A + B’ + C’) . (D’ + E + F)
• (A + B ) . (C)
• (A + B )
• (A)

Confusing Point: Literals vs variables
Suppose, we have a function F(A,B)
The variables of function are: A, B
The literals of function are: A, A’, B, B’ (a literal is a variable either in complemented form or uncomplemented form)

So, Option(I) is correct.

Answer: I
Explanation:
Given,
Hamming Distance =5
Formula,
For error correction, The hamming distance between any two codeword is 2t+1, where t is error correcting bits.
Calculation,
=> 2t+1 = 5
=> t =2 bits
Now, let’s explore each option and see which codeword have hamming distance 2.
Hamming distance is number of 1’s in X-OR of two code.
Option(I):

Hamming distance = 2
Option(II):

Hamming distance = 3
Option(III):

Hamming distance = 8
Option(IV):

Hamming distance = 7
Only option(I) have hamming distance 2.
So, Option(I) is correct.

Answer: II
Explanation:
In Self Complementing code, The code of a digit and the code of 9’s complement of the digit are 1’s complement to each other.

Let’s take each one,
(a) 8421 code
let’s take decimal number 2,
9’s complement of the digit 2 = 7
the 8421 code of a digit 2 = 0010 ………….Code(1)
the 8421 code of a digit 7 = 0111 ………….Code(2)
Code(1) and code(2) are not 1’s complement to each other. That’s why 8421 code is not self complement code.

(b) 2421 code
let’s take decimal number 2,
9’s complement of the digit 2 = 7
the 2421 code of a digit 2 = 0010 ………….Code(1)
the 2421 code of a digit 7 = 1101 ………….Code(2)
Code (1) and code(2) are 1’s complement to each other. That’s why 2421 code is self complement code.

(c) excess-3 code
let’s take decimal number 2,
9’s complement of the digit 2 = 7
the excess-3 code of a digit 2 = 0101 ………….Code(1)
the excess-3 code of a digit 7 = 1010 ………….Code(2)
Code (1) and code(2) are 1’s complement to each other. That’s why excess-3 code is self complement code.

(d) excess-3 gray code
let’s take decimal number 2,
9’s complement of the digit 2 = 7
the excess-3 gray code of a digit 2 = 0111 ………….Code(1)
the excess-3 gray code of a digit 7 = 1111 ………….Code(2)
Code (1) and code(2) are not 1’s complement to each other. That’s why excess-3 code is not self complement code.
So, Option(II) is correct.

Answer: II
Explanation:
=> AB+AB’+A’C+AC
=> A.(B+B’) + (A’+A).C
=> A.1 + 1.C
=> A + C
The Boolean AB+AB’+A’C+AC is unaffected by the value of the Boolean variable B.
So, Option(II) is correct.

Answer: IV
Explanation:
(142)_b + (112)_b-2 = (75)₈
convert into base 10
=> ( 1*b² + 4*b¹ + 2*b⁰ ) + ( 1*(b-2⁾² + 1*(b-2)¹ + 2*(b-2)⁰ ) = 7*8¹ + 5*8⁰
=> ( b² + 4b +2 ) + ( b² + 4 -3b) = 61
=> 2b² + b – 55 = 0
Formula,
if an equation is ax² + bx +c =0



b = (-1+21)/4, (-1-21)/4
b = 5, -5.5
Base can’t be negative number.
So, Option(IV) is correct.

Answer: I
Explanation:
Total number of input sequence with n variable = 2ⁿ
Each input sequence have boolean value i.e 0 or 1.


Therefore, Total Boolean functions of degree n =
So, Option(I) is correct.

Answer: II
Explanation:
(146)_b + (313)_b-2 = (246)₈
convert into base 10
=> ( 1*b² + 4*b¹ + 6*b⁰ ) + ( 3*(b-2)² + 1*(b-2)¹ + 3*(b-2)⁰ ) = 2*8² + 4*8¹ + 6*8⁰
=> ( b² + 4b +6 ) + ( 3b² + 13 -11b) = 166
=> 4b² – 7b – 147 = 0
Formula,
if an equation is ax² + bx +c =0




b = (7+49)/8, (7-49)/8
b = 7, -6
Base can’t be negative number.
So, Option(IV) is correct.

Answer: III
Explanation:
ROM memory size = 2^m x n
where, m is address lines and n is data line.
Given,
ROM memory size = 16K x 16 = 2¹⁴ x 16
Address line = 14
Data line = 16
So, Option(III) is correct.

Answer: III
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Answer: II
Explanation:
➡ MOD-2 coumter followed by MOD-5 counter, it denotes cascading of two counters. Therefore,
equivalent counter = MOD (2 x 5) counter = MOD 10 counter.

➡ Hence, the number of states when a MOD-2 counter is followed by a MOD-5 counter is 10.

➡ MOD-10 counter is also called as BCD counter or Decade counter.
So, Option(II) is correct.

Answer: III
Explanation:
Formula,
|AՍB| = A + B – |AՈB|
where,
A = number of bit strings of length ten start with 1
B = number of bit strings of length ten end with 00
|AՈB| = number of bit strings of length ten start with 1 and end with 00
|AՍB| = number of bit strings of length ten either start with 1 or end with 00

Calculation,
➡ number of bit strings of length ten start with 1
first bit out of 10 bits is fixed to 1, and remaining 9 bits have 2 option either 1 or 0
Total possible number of bit strings of length ten start with 1 = 2⁹ = 512

➡ number of bit strings of length ten end with 00
last 2 bits out of 10 bits are fixed to 00, and remaining 8 bits have 2 option either 1 or 0
Total possible number of bit strings of length ten end with 00 =2⁸ = 256

➡ number of bit strings of length ten start with 1 and end with 00
first bit out of 10 bits is fixed to 1 and last 2 bits out of 10 bits are fixed to 00, and remaining 7 bits have 2 option either 1 or 0
Total possible number of bit strings of length ten start with 1 and end with 00 =2⁷ = 128

|AՍB| = A + B – |AՈB|
= 512 + 256 – 128
= 640

Therefore, Number of bit strings of length ten either start with 1 or end with 00 is 640.
So, Option(III) is correct.

Answer: IV
Explanation:
Priority Encoder: Priority Encoder is a combinational circuit that have 2ⁿ inputs and n outputs. The size of Priority Encoder is 2ⁿ x n.
Priority Decoder: Priority decoder is a combinational circuit that have n inputs and 2ⁿ outputs. The size of Priority Decoder is n x 2ⁿ.

Calculation,
It is given, a parallel bus arbiter has 5 bus arbiter.
so, number of bits to represent 5 bus = Гlog₂5Ⴈ = 3 bits
➡ Priority Encoder
n = 3 bits
Number of inputs to priority encoder = 2ⁿ = 2³ = 8
Number of outputs to priority encoder = n = 3
Therefore, size of priority encoder = 8 x 3
➡ Priority Decoder
n = 3 bits
Number of inputs to priority decoder = n = 3
Number of outputs to priority decoder = 2ⁿ = 2³ = 8
Therefore, size of priority decoder = 3 x 8

So, Option(IV) is correct.

Answer: III
Explanation:
In computers, subtraction is generally carried out by 2’s complement.
So, Option(III) is correct.

Answer: I
Explanation:
=> A’⋅B+A.B’+A.B
=> A’⋅B+A.(B’+B)
=> A’⋅B+A {since, B’+B = 1}
=> (A’ + A) . (B + A)
=> 1 . (B + A)
=> (B + A)
=> A + B
So, Option(I) is correct.

Answer: II
Explanation:
➡ (i)If x ≤ y and y ≤ z, then x ≤ z (True)
it is by transitive rule.
➡ (ii)If x ≤ y and y ≤ x, then x=y (True)
As x ≤ y , it means x ∨ y = y
and
As y ≤ x , it means y ∨ x = x ∨ y = x
Therefore, x ∨ y = y = y ∨ x = x
➡ (iii)If x < y and y < z, then x ≤ y (False)
As x < y(x less than y), then x ≤ y (x less than equal to y) is incorrect
➡ (iv)If x < y and y < z, then x < y (True)
As x < y, then x < y which is same in both cases.
So, Option(II) is correct.

Answer: I
Explanation:
Let’s take each boolean equation:
➡ (i) wx + w(x + y) + x(x + y)=x+wy
= wx + wx + wy + xx + xy
= wx + wy + x + xy
= x(1 + w + y) + wy
= x + wy
Therefore, equation(i) is True.

➡(ii) (wx’(y+xz’)+w’x’)y=x’y
= (wx’y + wx’xz’ + w’x’)y
= (wx’y + w’x’)y
= wx’yy + w’x’y
= wx’y + w’x’y
= (w + w’)x’y
= x’y
Therefore, equation(iI) is True.
So, Option(I) is correct.

Answer: I
Explanation:

= [(A.B)’ + (A + B)’]’
= [A’ + B’ + A’B’]’
= [A’ + B'(1 + A’)]’
= [A’ + B’]’
= A.B
So, Option(I) is correct.

Answer: IV
Explanation:
IEEE-754 32-bit representation
In the standard format for a single-precision binary number, the sign bit (S) is the left-most bit, the exponent (E) includes the next eight bits, and the mantissa or fractional part (F) includes the remaining 23 bits, as shown next.

The eight bits in the exponent represent a biased exponent, which is obtained by adding 127 to the actual exponent.
Calculation,
Given, (-40.1)₁₀
convert the decimal number into binary number
(-40.1)₁₀ = (- 101000 . 00011 0011 0011 0011)₂
Normalize above binary number into implicit normalization
(- 1 . 01000 00011 0011 0011 0011 x 2⁵)₂
here,
sign bit = 1
because it is negative number
Exponent = 5 + 127 = (132)₁₀ = (1000 0100)₂
Mantissa = 01000 00011 0011 0011 0011 0

(1100 0010 0000 0110 0110 0110 0110)₂
Convert into base 16 representation
0xC206666
So, Option(IV) is correct.

Answer: II

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Answer: IV
Explanation:
To guarantee correction of upto t errors, the minimum Hamming distance d_min = t+1
So, Option(I) is correct.

Answer: I

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Answer: IV

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Answer: II

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Answer: II

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Answer: I

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Answer: II

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Answer: I

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Answer: IV

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Answer: I

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Answer: IV

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Answer: I

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Answer: II

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Answer: I , II Both(Marks to all)

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Answer: II

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Answer: II

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Answer: II

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Answer: IV

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Answer: I

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Answer: III

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Answer: II

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Answer: III

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Answer: IV

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Answer: III

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Answer: I

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Answer: IV

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Answer: II

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Answer: IV

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Answer: III

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Answer: III

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Answer: II

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Answer: I

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Answer: IV

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Answer: III

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Answer: II

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Answer: IV

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Answer: III

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Answer: IV

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Answer: IV

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Answer: III

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Answer: II

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Answer: III

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Answer: II

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Answer: IV

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Answer: III

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Answer: IV

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Answer: I, IV
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