Question 1

Boolean simplification — minimize F = ABD + CD + ACD + ABC + ABCD Using Boolean algebra (no K-map required), find the most simplified SOP form from the options provided.

Accepted Answer

Introduction / Context: Algebraic simplification reduces hardware by removing redundant terms. For the given four-variable function, we apply absorption and factoring to reach a minimal or near-minimal sum-of-products (SOP) that matches one of the choices. Given Data / Assumptions: F = ABD + CD + ACD + ABC + ABCD. Variables A, B, C, D are Boolean. Operators: + is OR, concatenation is AND. Concept / Approach: Use absorption: X + XY = X, and redundancy removal where a more specific product term is covered by a more general one. Combine terms where a factorization can eliminate literals without altering coverage of minterms. Step-by-Step Solution: Start: F = ABD + CD + ACD + ABC + ABCD.Observe ABCD is contained in ABD (and also in ACD and ABC). Apply absorption: remove ABCD.Now F = ABD + CD + ACD + ABC.Notice CD + ACD = CD(1 + A) = CD. Absorb ACD into CD.Thus F simplifies to F = ABD + ABC + CD. Verification / Alternative check: Factor the remaining two AB terms: F = AB(C + D) + CD. Re-expanding yields ABC + ABD + CD, identical to the simplified expression. A truth table or logic simulator would confirm equivalence with the original function across all 16 input combinations. Why Other Options Are Wrong: F = CD + AD: AD covers minterms where B = 0 that are not in the original, so it overstates the function. F = BC + AB: Omits required D-dependent coverage and includes minterms not present. F = AC + AD: Again introduces minterms not guaranteed by the original and misses specific AB-dependent ones. Common Pitfalls: Dropping CD entirely after absorbing ACD; CD must remain. Over-aggressive factoring leading to terms like AD or AB alone, which add unwanted minterms. Final Answer: F = ABD + ABC + CD

Question 2

K-map grouping rules — which cell combinations are NOT valid to group? In Karnaugh map minimization, identify the combination below that cannot be grouped into valid power-of-two implicants.

Accepted Answer

Introduction / Context: Karnaugh maps (K-maps) simplify Boolean expressions by grouping adjacent 1-cells (or 0-cells for POS) into power-of-two blocks. Understanding what counts as adjacency is crucial to form correct groups and achieve minimal logic. Given Data / Assumptions: Adjacency is defined between cells that differ by exactly one variable (Gray code ordering). Wrapping is allowed—edges and corners can be adjacent across the map boundaries. Groups must be 1, 2, 4, 8, … cells in rectangular shapes. Concept / Approach: Valid groupings include horizontally or vertically adjacent cells, with edge wrap making opposite edges (and therefore corners) adjacent. Overlapping groups are permitted when they enable larger implicants. Diagonal cells are not adjacent (they differ in two variables) and therefore cannot be grouped together as a pair by themselves. Step-by-Step Solution: Check “corners in the same row/column”: with wrap-around, corner cells can join to form a 2-cell or 4-cell group.Check “overlapping combinations”: allowed if each group size is a power of two and improves simplification.Check “diagonal”: diagonally opposite cells differ in two variables; not adjacent; cannot form a valid 2-cell group. Verification / Alternative check: Label a 4-variable K-map in Gray code. Compare coordinates of diagonal cells; Hamming distance is 2, confirming non-adjacency. Contrast with edge-adjacent cells where Hamming distance is 1, validating permitted groupings. Why Other Options Are Wrong: Corners in same row/column: valid due to wrap-around adjacency. Overlapping: permitted and often necessary to achieve larger implicants. Common Pitfalls: Forgetting edge wrap makes opposite edges adjacent. Forming L-shaped or diagonal-only groups, which violate the rectangular, power-of-two rule. Final Answer: diagonal

Question 3

Converting W(X + YZ) to standard SOP form By applying which Boolean algebra law can the expression W(X + YZ) be expanded into a sum-of-products (SOP) form?

Accepted Answer

Introduction / Context: In digital logic design, we frequently convert mixed expressions into a standard form for implementation with basic gates or programmable logic. The two most common canonical forms are sum-of-products (SOP) and product-of-sums (POS). This question asks which Boolean law lets us expand W(X + YZ) into SOP, where each term is a product (AND) of literals and the overall expression is a sum (OR) of those products. Given Data / Assumptions: The expression is W(X + YZ). Variables W, X, Y, Z are Boolean (take values 0 or 1). Goal: express as SOP (sum of product terms). Concept / Approach: The key law is Boolean distribution, which lets a factor multiply (AND) into a sum (OR). The distributive law appears in two useful forms: A(B + C) = AB + AC A + BC = (A + B)(A + C) (the dual, for POS work) Here, we need the first form to push W across the parentheses so that the result is a sum of product terms suitable for SOP implementation. Step-by-Step Solution: Start with: F = W(X + YZ)Apply distributive law: W(X + YZ) = WX + W(YZ)Associate the second product (implicit): W(YZ) = (WY)Z = WYZTherefore, F = WX + WYZ, which is clearly SOP (sum of the products WX and WYZ). Verification / Alternative check: You can construct a quick truth table for W, X, Y, Z and verify that F = 1 exactly when either WX = 1 or WYZ = 1. A Karnaugh map grouping would also show the same on-cells, confirming that distribution produced an equivalent SOP expression without changing functionality. Why Other Options Are Wrong: Associative law changes grouping only, such as (A + B) + C = A + (B + C); it does not expand across a sum. Commutative law swaps order, such as AB = BA or A + B = B + A; it does not expand products over sums. Idempotent law (A + A = A, AA = A) simplifies duplicate literals, not distribute factors. “None of the above” is incorrect because distribution is exactly what is used. Common Pitfalls: Stopping at WX + W(YZ) and forgetting that W(YZ) is already a valid product term WYZ; no further expansion is needed for SOP. Accidentally turning W + (YZ) into (W + Y)(W + Z), which is a POS transformation, not SOP. Final Answer: distributive law

Question 4

Interpreting De Morgan’s theorem Given the theorem (A * B)' = A' + B', which pair of gate descriptions are logically equivalent according to this statement?

Accepted Answer

Introduction / Context: De Morgan’s theorems are fundamental identities used to transform logic expressions and convert between gate types. They allow designers to swap between NAND/NOR implementations and AND/OR structures with input or output inversions, which is crucial for technology mapping and minimizing parts in practical circuits. Given Data / Assumptions: Use positive logic (1 = HIGH, 0 = LOW). De Morgan’s theorem provided: (A * B)' = A' + B'. Symbol * denotes logical AND; + denotes logical OR; ' denotes inversion. Concept / Approach: The given identity states that the complement of an AND equals the OR of the complements. In gate terms, an AND followed by an inverter (i.e., a NAND) is equivalent to an OR gate whose inputs are both inverted. This is the classic translation used to realize functions with only NAND gates, or to interpret bubble-logic schematics correctly. Step-by-Step Solution: Start with: Y = (A * B)'.Apply De Morgan: Y = A' + B'.Gate interpretation: left side is NAND; right side is an OR gate with inverted inputs (bubbles on A and B).Thus, a NAND gate has identical logic behavior to an OR gate with input inversions. Verification / Alternative check: Build a 2-input truth table for both expressions. For all four input combinations of A and B, (A * B)' and A' + B' match exactly. A quick logic simulator test or simple gate-level substitution confirms the equivalence in practical circuits. Why Other Options Are Wrong: NOR equals an AND with inverted inputs per the other De Morgan identity (A + B)' = A' * B', not the one given here. “AND and NOR with inverted inputs” mismatches the provided form. “NOR and NAND with inverted inputs” is not a direct De Morgan pair for this identity. “NAND and AND with inverted inputs” confuses complement placement; an AND with inverted inputs corresponds to a NOR under the other identity. Common Pitfalls: Confusing which De Morgan form you are using. Remember: complement of AND becomes OR of complements; complement of OR becomes AND of complements. Misreading bubble logic: bubbles on the inputs of an OR create a NAND-equivalent; bubbles on the inputs of an AND create a NOR-equivalent. Final Answer: a NAND and an OR gate with inverted inputs

Question 5

The process of reduction or simplification of combinational logic circuits increases the cost of the circuit.

Accepted Answer

False

Question 6

By applying De Morgan's theorem to a NOR gate, two identical truth tables can be produced.

Accepted Answer

True

Question 7

The OR function is Boolean multiplication and the AND function is Boolean addition.

Accepted Answer

False

Question 8

CPLD software can be used to design original circuits that prove the Boolean rules and laws.

Accepted Answer

True

Question 9

In the commutative law, in ORing and ANDing of two variables, the order in which the variables are ORed or ANDed makes no difference.

Accepted Answer

True

Question 10

The systematic reduction of logic circuits is performed using Boolean algebra.

Accepted Answer

True

Question 11

In Boolean algebra, A + 1 = 1.

Accepted Answer

True

Question 12

The product-of-sums (POS) is basically the ORing of ANDed terms.

Accepted Answer

False

Question 13

Five-variable Karnaugh maps are impossible.

Accepted Answer

False

Question 14

The VHDL editor provided with a schematic editor development system will produce a file with the extension .vhd, which can be used by the simulator to test the output of the logic design.

Accepted Answer

True

Question 15

SOP stands for sum-of-powers.

Accepted Answer

False

Question 16

A Karnaugh map is similar to a truth table because it presents all the possible values of input variables and the resulting output of each value.

Accepted Answer

True

Question 17

A variable is a symbol used to represent a logical quantity that can have a value of 1 or 0.

Accepted Answer

True

Question 18

In Boolean algebra, = A.

Accepted Answer

True

Question 19

Boolean algebra simplifies logic circuits.

Accepted Answer

True

Question 20

If you need a NAND, an AND, and an inverter you would purchase a 7400 IC.

Accepted Answer

True

Boolean Algebra and Logic Simplification Questions