Test: Comparators - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test GATE Electrical Engineering (EE) 2024 Mock Test Series - Test: Comparators

Test: Comparators for Electrical Engineering (EE) 2024 is part of GATE Electrical Engineering (EE) 2024 Mock Test Series preparation. The Test: Comparators questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Comparators MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Comparators below.
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Test: Comparators - Question 1

Identify the 1 – bit comparator circuit 

Detailed Solution for Test: Comparators - Question 1

Comparator: 
A comparator is a combinational circuit that gives the output in terms of input signals.
It is used to compare numbers and represent their relationship with each other.
It compares the signals as A > B, A < B, and A = B.
1-bit comparator:
A 1-bit comparator circuit is shown below.

In this circuit, we can compare two input signals
The truth table for the 1-bit comparator is

The expression for A = B: A ⊙ B
The expression for A > B: A B̅ 
The expression for A < B: A̅ B

Test: Comparators - Question 2

A digital circuit which compares two numbers A3 A2 A1 A0, B3 B2 B1 B0 is shown in figure. To get output Y = 1, the correct pair of input numbers are

Detailed Solution for Test: Comparators - Question 2


Y = (A0 ⊕ B0) + (A1 ⊕ B1) + (A2 ⊕ B2) + (A3 ⊕ B3
Given that output is 1.
To get the output y = 1, all the inputs of OR gate should not be zero.
From the options, A3 A2 A1 A0 = 0010, B3 B2 B1 B0 = 1011 satisfies this condition.

Test: Comparators - Question 3

X = X1X0 and Y = Y1Y0 are 2-bit binary numbers. The Boolean function S that satisfies the condition “If X > Y, then S = 1”, in its minimized form, is

Detailed Solution for Test: Comparators - Question 3

x = x1 x0 y = y1 y0
The Boolean function S that satisfies condition,
If x > y, then s = 1.
We can represent the above condition through a truth table.

By using k-maps

Minimized form:
s = x11 + x1x00 + x010  

Test: Comparators - Question 4

Maximum conversion time in clock cycles for three types of 8 bit ADCs (i) Successive approximation, (ii) Dual slope and (iii) Parallel comparator are respectively

Detailed Solution for Test: Comparators - Question 4

Concept:
The conversion time of different types of n-bit ADC is shown :

  • From the above table, it is clear that the dual-slope is the slowest ADC and Flash Type is the fastest ADC.
  • The conversion type of Flash type is independent of the number of bits 

Application:
n = 8

  • Maximum conversion time in clock cycles of 8-bit Successive approximation type ADC  = n=  8
  • Maximum conversion time in clock cycles of 8 bit Dual slope type ADC = 2n+1 = 29 = 512
  • Maximum conversion time in clock cycles of 8 bit Successive approximation type ADC = 1
Test: Comparators - Question 5

The circuit shown in the figure is a –

Detailed Solution for Test: Comparators - Question 5

From the given circuit we can find the outputs expression as follows:

Now, we can observe that

So, the given circuit in question behaves like a comparator circuit. 

Test: Comparators - Question 6

Circuit for comparing 2 n-bit numbers has ____ entries in truth table

Detailed Solution for Test: Comparators - Question 6

In an n-bit, compare the n columns of bits in one binary number (let it be A) and n columns of bits of another number (let it be B).
For all possible values of bits in A and B truth table is taken for A > B, A < B and A = B.
So there are 2n inputs in the comparator.
For 2n inputs total possible combinations are 22n
So total number of entries is 22n.

Test: Comparators - Question 7

The given below logic circuit will work as

Detailed Solution for Test: Comparators - Question 7


By analysing the above truth table it is giving the expression for 1 bit comparator.

Test: Comparators - Question 8

The logic circuit of the figure is a

Detailed Solution for Test: Comparators - Question 8

Y = AB + A̅B̅
It represents a XNOR gate
The truth table of XNOR gate is


The output is high when both the inputs are equal.
It represents a 1-bit comparator (or) equality detector.

Test: Comparators - Question 9

The output Y of a 2-bit comparator is logic 1 whenever the 2-bit input A is greater than the 2-bit input B. The number of combinations for which the output is logic 1, is

Detailed Solution for Test: Comparators - Question 9

The only possible combinations are
A = 01 and B = 0 0
A = 10 and B = 00, 01
A = 11 and B = 00, 01, 10
So there are only 6 combinations
Tips and Tricks:

where n = 2 bit

Test: Comparators - Question 10

A 3 bit comparator circuit gives logic 1 output whenever the 3 bit input A is greater than the 3 bit input B. The output Boolean function y is represented by a K-map. The no. of 1 in the K-map is –

Detailed Solution for Test: Comparators - Question 10

Total no. of input combination of A & B are = 23 x 23 = 64
In 64 combination only in 8 combination A = B situation occurs. So the no. of combination for which A ≠ B is 64 - 8 = 56
Only of these 56 combinations only in 28 combinations A > B situation occur.
So, the no of 1 in K map will be 28

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