Digital electronics miscellaneous
- Number of NAND gates are required to realize the function—
F = ( X + Y). (Z + W)
-
View Hint View Answer Discuss in Forum
In order to solve such type of problem first reduce any from into to SOP (sum of product from).
F = ( X + Y). (Z + W)
or F = XY(Z + W)
or F = XYZ + XYW
So, the number of NAND gate requirement is equal to 4.
Correct Option: B
In order to solve such type of problem first reduce any from into to SOP (sum of product from).
F = ( X + Y). (Z + W)
or F = XY(Z + W)
or F = XYZ + XYW
So, the number of NAND gate requirement is equal to 4.
- By inspecting the Karnaigh map plot of the switching function F (X1, X2, X3) = Σm (1, 3, 6, 7), one can say that the redundant prime implicant is—
-
View Hint View Answer Discuss in Forum
Given, F (X1, X2, X3) = ∑ m (1, 3, 6, 7)
From K-map. The redundant prime implicant is X2, X3.
Correct Option: B
Given, F (X1, X2, X3) = ∑ m (1, 3, 6, 7)
From K-map. The redundant prime implicant is X2, X3.
- Which the Boolean function
F (X1, X2, X3) = Σ(0,1, 2, 3) + ∑d (4, 5, 6, 7) is minimized?
-
View Hint View Answer Discuss in Forum
Given, F (X1, X2, X3) = Σ(0,1, 2, 3) + ∑d (4, 5, 6, 7)
From K-map, we get F (X1, X2, X3 )= 1
Correct Option: A
Given, F (X1, X2, X3) = Σ(0,1, 2, 3) + ∑d (4, 5, 6, 7)
From K-map, we get F (X1, X2, X3 )= 1
- For the logic circuit given. Which is the simplified Boolean function?
-
View Hint View Answer Discuss in Forum
X = A + (A + B) (C + B)
or X = A + AC + BC + AB + B
or X = A (1 + C) + BC + B (A + 1)
or X = A + BC + B
or X = A + B (1 + C)
or X = A + B [Since 1 + C = 1]
Correct Option: B
X = A + (A + B) (C + B)
or X = A + AC + BC + AB + B
or X = A (1 + C) + BC + B (A + 1)
or X = A + BC + B
or X = A + B (1 + C)
or X = A + B [Since 1 + C = 1]
- The minimized expression for the given K-map (X = don't care) is—
-
View Hint View Answer Discuss in Forum
From given K-map F = A + BC
Correct Option: A
From given K-map F = A + BC
