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Test: K-Map - 1 - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test Analog and Digital Electronics - Test: K-Map - 1

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

Minimize the following expression using k-Map
Y = ∑m (0,1,5,9,13,14,15) + d (3,4,7,10,11)

Detailed Solution for Test: K-Map - 1 - Question 1


∴ Y = AC+ D + (AC)'
= D + ((AC)' + AC)
 = D +  A̅ C̅ + AC 
OR
= D + (A ⊙ C)

Test: K-Map - 1 - Question 2

The minimized sum of products expression for f(a,b,c,d) = Ʃm(0,1,5,6,7,8,9) with don’t care Ʃm(10,11,12,13,14,15) is ___________.

Detailed Solution for Test: K-Map - 1 - Question 2

4 variable K-Map:
Rules for the pairing of boolean expressions:

  1. Octet will be paired first, then quadrant, then doublet, and at last single term will be used.
  2. If needed, don't pair elements that can be paired with other elements to form an octet, quad, or doublet.
  3. All elements in any pair can not be don't care.
  4. Any two pairs must have at least one element uncommon in both of them.


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Test: K-Map - 1 - Question 3

There are 4 variables in the Boolean function and the value of the function is 1. Find the number of cells in the K-Map which will contain a 1 when SOP expression is used.

Detailed Solution for Test: K-Map - 1 - Question 3

Concept:
The K-map is a graphical method that provides a systematic method for simplifying and manipulating the Boolean expressions or to convert a truth table to its corresponding logic circuit in a simple, orderly process.
In an 'n' variable K map, there are 2n cells.

Application:
For 4 variables there will be 24 = 16 cells
The K map will give an output of 1, when all the cells have a 1, i.e. if all the input combinations give an output of 1, the maximum number of inputs can be simplified to give an output of 1.
This is explained with the following K map:

 

Since the K map forms a pair of 16, it can be eliminated giving an output:
Y = 1
Since the output contains no input variables (A, B, C, or D), all the four variables are simplified/eliminated.

Test: K-Map - 1 - Question 4

The simplification in minimal sum of product (SOP) of Y = F(A, B, C, D) = ∑m (0, 2, 3, 6, 7) + ∑d (8, 10, 11, 15) using K-maps is

Detailed Solution for Test: K-Map - 1 - Question 4

Analysis:
The K-map for the given SOP representation is drawn as:

Taking the don't care that helps in the minimization, we get the simplified expression as:
F(A, B, C, D) = A̅ C + B̅ D̅ 

Test: K-Map - 1 - Question 5

A K-map of 3 variables contains _______ cells.

Detailed Solution for Test: K-Map - 1 - Question 5

Karnaugh map (K-map):

  • The Karnaugh map (K-map) is a method of simplifying Boolean algebra expressions.
  • The Karnaugh map reduces the need for extensive calculations.
  • Karnaugh map can be explained as An array that contains 2k number of cells, where k is the number of variables in the Boolean expression that is to be reduced or optimized. 

Number of cells in 2 variable k-map = 22 = 4
Number of cells in 3 variable k-map = 23 = 8
Number of cells in 4 variable k-map = 24 = 16

2 variable K-maps:
There are 4 cells in the 2-variable k-map as shown,

3 variable K-maps:

  • For a 3-variable Boolean function, there is a possibility of 8 output minterms.
  • The general representation of all the minterms using 3-variables is shown below.

Test: K-Map - 1 - Question 6

The output expression for the Karnaugh map shown below is:

Detailed Solution for Test: K-Map - 1 - Question 6

Given K-map:

It can be grouped as follows:

Output expression in the form of SOP (sum of products) = BD̅ + ABC

Test: K-Map - 1 - Question 7

Digital input signals A, B, C with A as the MSB and C as the LSB are used to realize the Boolean function F = m0 + m2 + m3 + m5 + m7, where mi denotes the ith minterm. In addition, F has a don’t care for m1. The simplified expression for F is given by:

Detailed Solution for Test: K-Map - 1 - Question 7

F = m0 + m2 + m3 + m5 + m7
m1 is a don’t care.
 

Test: K-Map - 1 - Question 8

A problem detector system produces an alarm in the factory when one of the three conditions occurs. The system is designed as such tha only one condition can occur at a time. If the three conditions are defined as q, r, and s respectively, the output logic for the system is given as

Detailed Solution for Test: K-Map - 1 - Question 8

Concept:
K maps
This is a table that represents both minterms and max terms in it and it is used for the simplification of the boolean algebraic expressions easily.
For an 'n' variable k-map total 2n cells will be present in it.
The different k-maps structures are shown below:
2 variable k-map

3 variable k-map

Calculation:

  • Given conditions for the system are
  • A problem detector system produces an alarm in the factory when one of the three conditions occurs.
  • The system is designed as such only one condition can occur at a time.
  • conditions are defined as q, r, and s

The truth table is shown as:

The output function minterms according to the truth table are:
output = ∑ (1, 2, 4)
k-map for the circuit is:

Test: K-Map - 1 - Question 9

Consider the following sum of products expression, F

The equivalent product of sums expression is

Detailed Solution for Test: K-Map - 1 - Question 9

Concept:
The SOP representation of the circuit is:
F = Σm (minterms)
Minterm: a minterm of n variables is a product of the variables in which each appears exactly once in true or complemented form.
The POS representation of the circuit:
F = ΠM (max terms)
Maxterm: a maxterm of n variables is a sum of the variables in which each appears exactly once in true or complemented form.

Calculation:
Given,

For the function, we form the K-map as:

Hence, the function in the form of minterms is expressed as:
f(A, B, C) = Σm (0,1,3,5,7)
Now, we put 0 in each block of the K-map excluding the blocks corresponding to the terms in the above function.

Grouping the 0’s in K-map, we obtain the max terms as
F = πM(2, 4, 6)

Test: K-Map - 1 - Question 10

The Boolean expression  can be minimized to

Detailed Solution for Test: K-Map - 1 - Question 10


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