K-Map MCQ Quiz in தமிழ் - Objective Question with Answer for K-Map - இலவச PDF ஐப் பதிவிறக்கவும்

From Truth Table:

D(A, B, C) = A̅.B.C + A.B.C̅

D(A, B, C) = B.(A̅.C + A.C̅)

D(A, B, C) = B.(A + C).(A̅  + C̅) 

Alternate:

K – Map:

D(A, B, C) = B.(A + C).(A̅ + C̅) 

The correct answer is A'B'C' + A'B'C + A'BC + A'BC' + AB'C

Key PointsThe given Boolean function F(A, B, C) = Σ(0, 1, 2, 3, 5) represents a sum of products (SOP) expression. The numbers in the Σ notation indicate the binary values for which the function is equal to 1.

Let's convert the given binary values (0, 1, 2, 3, 5) to their respective binary representations:

• 0 = A'B'C'
• 1 = A'B'C
• 2 = A'BC'
• 3 = A'BC
• 5 = AB'C

So Correct answer is A'B'C' + A'B'C + A'BC + A'BC' + AB'C

\(f = \bar AD + \bar AC + A\bar C\bar D + AB\bar C\)

\(f = \bar AD + \bar AC + A\bar C\bar D + B\bar CD\)

Explanation :

Given:

f(w, 0, 0, z)= 1

f(1, x, 1, z)= x+z

f(w, 1, y, z)= wz+y

Only thing makes this Question complicated is how you fill the K map cells.

1. f(w, 0, 0, z) = 1 ; means whenever x = 0 and y = 0 (total 4 cells) then put 1 in k-map cell.

2. f(1, x, 1, z) = x+z ; means whenever w=1 and y=1 (total 4 cells) then evaluate the expression “x+z” as per each cell (only those where w=1 and y=1) and fill the result in that particular cell.

3. f(w, 1, y, z) = wz+y ; means whenever x=1( total 8 cells) then evaluate the expression “wz+y” as per each cell (only those where x=1) and fill the result in that particular cell.

1. In the Below K map first we mark x = 0 and y = 0

  1001, 0000 , 0001 , 1000 these 4 cells will be marked as 1 .

  (wx’y’z , w’x’y’z’ , w’x’y’z , wx’y’z’) likewise we can fill the case where w=1 and y=1.

2. Now x=1( total 8 cells)

                     0100,     0101,           0110,          0111,          1100,        1101,        1110,        1111

                     w’xy’z’,      w’xy’z,      w’xyz’,        w’xyz,        wxy’z’,      wxy’z,        wxyz’,       wxyz

Likewise we can fill the K map as given in Image.

So after minimum sum of product expression will be x’y’ + xy + wz ; so total 6 literals. Answer will be 6. 

Concept: We can avoid Static -1 Hazard if all adjacent 1’s is covered by an implicant.

Analysis: For the given Question, expect Option (1), all other options have pairs of adjacent 1’s which are not covered by an implicant.

So, there is the possibility of Static -1 Hazard.

∴ Option (1) is not having Static -1 Hazard.

F (P, Q, R, S) = ∑ 0, 2, 5, 7, 8, 10, 13, 15

Don’t care min terms are 2, 7, 8, 13

By plotting the K-map, the minimal SOP (sum of products) can be found.

Explanation –

While putting the terms to k-map following things happen,

• 3^rd and 4^th columns are swapped
• 3^rd and 4^th rows.
• term 2 is going to (0, 3) column instead of (0, 2)
• 8 is going to (3, 0) instead of (2,0)

Solving, the above K-map, we get Q̅S̅ + QS

The correct answer is C' + D'

Key PointsTo simplify the Boolean function F(A, B, C, D) using the Karnaugh Map (K-map) method, we first need to construct a 4-variable K-map and then fill in the values based on the given minterms (0, 1, 2, 4, 5, 6, 8, 9, 10, 12, 13, 14). The minterms can be represented in binary form to determine their positions on the K-map.

The K-map for F(A, B, C, D) would look like this:

simplified expression =C’ + D’

Hence the correct answer is C’ + D’

Ans: The correct answer is option 1

Concept:

K-map:

We want to find expressions with minimum variables so K-map helps to minimize Boolean expression very easily without using any Boolean algebra theorem. we can solve it by using two forms, the first one is Sum of Product (SUM) and another one is Product of Sum (POS). We fill the grid of K-map with 0's and 1's then solve it by making groups.

EXAMPLE:\(\)

F(A, B, C) = ∑ (0,5,15,10,7,13,8)

This diagram shown

• 4 adjacent square is grouped and this gives the logical sum of 4 minterms but we can not group diagonal adjacent square and also not diagonal corners.
• we can group corners in the same column
• we can group corners in the same row, and also overlapping combinations.

F = P̅.Q̅0+ P̅.Q.1 + P.Q̅.R  + P.Q.R̅. 

F = P̅.Q + P.Q̅.R + P.Q.R̅

F = P̅.Q(R̅ + R ) + P.Q̅.R + P.Q.R̅

F = P̅.Q.R̅ + P̅.Q.R + P.Q̅.R + P.Q.R̅

F = ∑ (2, 3, 5, 6)

K – Map:

F = P̅ Q + Q R̅ + P Q̅ R

The correct answer is \(\rm {A\bar{B}C} {D}+{B} \bar{C}+{B}\bar {D}+\bar{A}\bar {D}+\bar{C} \bar{D}\)

Concept:
Given :
 F(A, B, C, D) = ∑m(0, 1, 2, 3, 5, 7, 8, 9, 11, 14) 
 

The minimized equation for the following logic function using K-map is \(\rm {A\bar{B}C} {D}+{B} \bar{C}+{B}\bar {D}+\bar{A}\bar {D}+\bar{C} \bar{D}\)

Important Points 

•  Karnaugh Maps (K-maps) are a graphical method used to simplify Boolean algebra expressions.
  • When minimizing a Boolean function using K-maps, there are several key points to keep in mind:
• Representation: Each cell in the K-map represents a unique combination of input values for the variables in the Boolean expression.
• Adjacency: Cells that are adjacent in the K-map differ by only one variable assignment (binary digit).
• Grouping: The main technique for simplifying a Boolean expression in a K-map involves identifying groups of adjacent 1s (or 0s)
  • that can be combined to form larger groups.
• Group Size: The goal is to form the largest groups possible without violating adjacency rules.
• Minimal Cover: The simplified expression should cover all the 1s (or 0s) in the original function using the fewest possible terms.
• Prime Implicants: A prime implicant is a group of cells in the K-map that cannot be combined
  • with any other adjacent group to form a larger group. Prime implicants are essential for finding the minimal cover.
• Essential Prime Implicants: These are prime implicants that cover at least one minterm
  • that no other prime implicant covers. Essential prime implicants must be included in the final simplified expression .