Question 1

Exclusive-NOR (XNOR) gate behavior In digital electronics, the exclusive-NOR gate outputs logic 1 when both inputs have the same logic level. For a 2-input XNOR gate, which input combination will produce an output of 1?

Accepted Answer

Introduction / Context: Exclusive-NOR (XNOR) is a fundamental digital logic gate used for equality detection. It outputs logic 1 when both inputs are equal (both 0 or both 1) and outputs logic 0 when the inputs differ. Understanding this truth relationship is essential for comparators, parity checkers, and error-detection circuits. Given Data / Assumptions: Gate: 2-input XNOR. We test listed input pairs to determine which yields output 1. Logic levels: 1 = HIGH, 0 = LOW. Concept / Approach: The XNOR operation implements logical equality. For inputs A and B, XNOR can be expressed as A XNOR B = (A AND B) OR (A' AND B'). Equivalently, it is the complement of XOR: A XNOR B = (A XOR B)'. Thus, output is 1 when A = B and 0 otherwise. Step-by-Step Solution: Test A = 0, B = 0: inputs are equal, so output = 1.Test A = 1, B = 0: inputs differ, so output = 0.Test A = 0, B = 1: inputs differ, so output = 0.Therefore, among the provided options, only A = 0, B = 0 gives output 1. Verification / Alternative check: Construct the full truth table: equal pairs (0,0) and (1,1) produce 1; unequal pairs produce 0. Since 1,1 is not listed, 0,0 is the only valid choice here. Why Other Options Are Wrong: A = 1, B = 0: unequal inputs yield 0, not 1. A = 0, B = 1: unequal inputs yield 0, not 1. none of the above: incorrect because 0,0 is valid. Common Pitfalls: Confusing XOR with XNOR; assuming XNOR is 1 only when both inputs are 1 (that is AND), or overlooking that 0,0 also satisfies equality and therefore produces 1 for XNOR. Final Answer: A = 0, B = 0

Question 2

Implementing an inverter using NAND logic In standard digital logic design, how many 2-input NAND gates are required to construct a NOT (inverter) function from a single input?

Accepted Answer

Introduction / Context: NAND is a universal gate. Any Boolean function, including the basic NOT, AND, and OR operations, can be implemented using only NAND gates. Knowing minimal gate counts is key for cost, area, and delay optimization in logic design and discrete implementations. Given Data / Assumptions: Available gate: a 2-input NAND. Goal: realize a NOT function (inverter) from a single input. Concept / Approach: The NAND truth function is Z = (A * B)'. If both inputs of a NAND are tied together to the same signal X, then Z = (X * X)' = (X)' = NOT X. Therefore, a single 2-input NAND can implement an inverter by shorting its two inputs. Step-by-Step Solution: Start with a 2-input NAND gate: Z = (A * B)'.Tie inputs: set A = B = X.Compute output: Z = (X * X)' = X' = NOT X.Thus, 1 NAND gate realizes an inverter. Verification / Alternative check: Truth table for X vs. Z shows X = 0 gives Z = 1; X = 1 gives Z = 0, confirming NOT behavior. This is a standard identity in universal gate constructions. Why Other Options Are Wrong: 2, 3, 4 NANDs: More than necessary; a single NAND suffices when its inputs are tied together. Common Pitfalls: Forgetting to tie both NAND inputs together; confusing NAND-based realizations with NOR-based realizations (NOR is also universal but uses a different wiring for NOT). Final Answer: 1

Question 3

8-input XOR output evaluation An 8-input XOR gate produces Y = 1 when an odd number of its inputs are at logic 1. For the input vector ordered A to H, which of the following bit patterns will make Y = 1?

Accepted Answer

Introduction / Context: An exclusive-OR (XOR) gate outputs logic 1 if and only if the number of 1s at its inputs is odd. Multi-input XORs are common in parity generators/checkers and error-detection coding. Recognizing the odd-parity rule enables quick mental evaluation without writing full truth tables. Given Data / Assumptions: Gate: 8-input XOR. Rule: output Y = 1 when count_of_ones is odd; otherwise Y = 0. Input ordering is A through H; options are 8-bit strings. Concept / Approach: Count the number of 1s in each candidate bit pattern. Odd count implies Y = 1; even count implies Y = 0. This is equivalent to computing parity of the vector. Step-by-Step Solution: Option A: 10111100 has five 1s (odd) => Y = 1.Option B: 10111000 has four 1s (even) => Y = 0.Option C: 11100111 has six 1s (even) => Y = 0.Option D: 00011101 has four 1s (even) => Y = 0. Verification / Alternative check: Pairwise XOR accumulation confirms the same results: XOR of an odd number of 1s yields 1; XOR of an even number of 1s yields 0. Using parity intuition gives identical outcomes. Why Other Options Are Wrong: B, C, D each contain an even number of 1s, so an 8-input XOR outputs 0. Common Pitfalls: Miscounting the ones; confusing XOR with XNOR (which outputs 1 for an even number of 1s). Always confirm the odd vs. even parity rule for XOR vs. XNOR respectively. Final Answer: 10111100

Question 4

Timing of XOR output for overlapping pulses Two positive pulses are applied to a 2-input XOR gate. Pulse X1 has width 75 µs starting at t = 0 µs. Pulse X2 has width 15 µs starting at t = 20 µs (thus ending at t = 35 µs). Describe the XOR output timi…

Accepted Answer

Introduction / Context: The exclusive-OR (XOR) gate outputs logic 1 when its inputs are different and logic 0 when they are the same. When pulses overlap in time, the XOR output toggles according to intervals of mismatch vs. match. Timing analysis of overlaps is a core skill for digital timing diagrams. Given Data / Assumptions: Input X1: HIGH from t = 0 µs to t = 75 µs (tw = 75 µs). Input X2: HIGH from t = 20 µs to t = 35 µs (tw = 15 µs), starting 20 µs after X1 rises. XOR rule: output = 1 when exactly one input is HIGH; otherwise 0. Concept / Approach: Partition the timeline into three intervals based on X2: before overlap, during overlap, and after overlap (while X1 is still HIGH). Evaluate XOR in each interval using the one-HIGH-only condition. Step-by-Step Solution: Interval 1 (0–20 µs): X1 = 1, X2 = 0 → XOR = 1 → produces a 20 µs output pulse.Interval 2 (20–35 µs): X1 = 1, X2 = 1 → XOR = 0 → this is a 15 µs separation (LOW gap) between output pulses.Interval 3 (35–75 µs): X1 = 1, X2 = 0 → XOR = 1 → produces a 40 µs output pulse.Therefore, the output is a 20 µs pulse, a 15 µs LOW gap, and then a 40 µs pulse. Verification / Alternative check: Constructing a quick timing diagram or tabulating values at the boundary points t = 0, 20, 35, 75 confirms the pulse widths and the separation. No other transitions occur since inputs are constant outside these boundaries. Why Other Options Are Wrong: Option B: Claims a 15 µs pulse in the output; during 20–35 µs both inputs are HIGH, so XOR must be 0 (a gap), not a pulse. Option C: Omits the initial 20 µs pulse that occurs before the overlap. Option D: Adds an extra 15 µs output pulse that does not exist; the 15 µs is the LOW gap during overlap. Common Pitfalls: Confusing the overlap interval as producing an XOR high; miscounting widths; or forgetting to include the post-overlap tail while X1 remains HIGH. Final Answer: The exclusive-OR output is a 20 µs pulse followed by a 40 µs pulse, with a separation of 15 µs between the pulses.

Question 5

Logic design using only NOR gates How many 2-input NOR gates are required to implement a 2-input NAND gate (using NOR gates only)?

Accepted Answer

Introduction / Context: Universal gates such as NOR and NAND can be combined to realize any Boolean function. This question tests gate-level synthesis skills by asking how to build a 2-input NAND strictly from 2-input NOR gates, an essential technique when constrained to a single gate family. Given Data / Assumptions: Available building block: 2-input NOR gates only. Target function: 2-input NAND, i.e., Y = (A * B)' = (A * B)' which equals (A * B)'. Allowed trick: use a NOR gate as an inverter by tying its inputs together (X NOR X = X'). Concept / Approach: Apply DeMorgan's law to express NAND in terms of OR and complements: (A * B)' = A' + B'. If we can generate A' and B' with NOR inverters and then realize an OR using NOR, we can obtain NAND with only NOR devices. Step-by-Step Solution: 1) Invert A with a NOR inverter: A1 = A NOR A = A'.2) Invert B with a NOR inverter: B1 = B NOR B = B'.3) Form the OR of A1 and B1 using NOR then invert: T = A1 NOR B1 = (A1 + B1)'.4) Invert T with a NOR inverter: Y = T NOR T = ((A1 + B1)')' = A1 + B1 = A' + B' = (A * B)' (the desired NAND). Verification / Alternative check: Truth-table check confirms that the composite four-NOR network outputs 0 only when A = B = 1, matching a NAND gate. A quick simulation or hand evaluation for all four input pairs verifies correctness. Why Other Options Are Wrong: 1, 2, or 3 gates: Insufficient because you need at least two inverters (for A and B) plus a NOR-based OR (which itself requires a NOR and a final inversion).5 gates: Possible but not minimal; the construction above achieves the function with four NORs. Common Pitfalls: Forgetting that OR with NOR requires an additional inversion, or attempting to realize AND directly with NOR without proper inversions, which yields a different function (e.g., NOR rather than NAND). Final Answer: 4

Question 6

Negative logic viewpoint – can a NAND gate implement the behavior of a negative-OR (i.e., OR with inverted signal senses)?

Accepted Answer

Introduction / Context: Universal gates such as NAND are powerful because they can realize a variety of logic functions under different signal conventions. In negative logic, the logical meaning of 0/1 is swapped (low = asserted, high = deasserted), which changes how we interpret familiar gates. This question asks whether a NAND gate corresponds to a negative-OR function under that convention. Given Data / Assumptions: Inputs and/or outputs may be interpreted under negative logic (active-LOW). Device is an ideal NAND gate with standard Boolean behavior. No analog limitations (fan-in, thresholds) affect the logical equivalence. Concept / Approach: In positive logic, NAND(A,B) = NOT(A * B). Under negative logic interpretation, complementing input and output senses maps NAND to OR. Intuitively, if low level “means” true, then a NAND’s output low occurs precisely when either input is low-true, aligning with OR semantics in the negative domain. This is a classic duality used in logic families and active-LOW signal design. Step-by-Step Solution: 1) Start with Y = NOT(A * B) (NAND).2) Define negative-logic variables: A_n = NOT(A), B_n = NOT(B), Y_n = NOT(Y).3) Substitute: Y_n = NOT(Y) = NOT(NOT(A * B)) = A * B.4) In terms of negative variables: A = NOT(A_n), B = NOT(B_n) → A * B = NOT(A_n) * NOT(B_n) = NOT(A_n + B_n).5) Therefore Y_n = NOT(A_n + B_n) and Y = A_n + B_n in negative sense, which corresponds to OR. Verification / Alternative check: De Morgan’s law confirms NOT(A * B) = NOT(A) + NOT(B). Interpreting inputs as active-LOW means the gate acts like OR on the active meanings. Textbooks present NAND as negative-OR and NOR as negative-AND under signal inversion conventions. Why Other Options Are Wrong: Claims about Schmitt triggers, low frequency, or CMOS-only are hardware distractions; the equivalence is logical and independent of device technology or speed regime. Common Pitfalls: Mixing physical voltage levels with logical meaning. Remember that changing the active sense (positive vs negative logic) changes the interpretation of the same hardware function. Final Answer: Correct

Question 7

Gate universality – is an AND gate itself considered a universal gate capable of implementing any Boolean function by composition?

Accepted Answer

Introduction / Context: A universal gate is one that can be combined in multiple copies to implement any Boolean function without requiring additional primitive types. This item checks whether an AND gate by itself qualifies as universal, which is a common misconception among beginners who equate “fundamental” with “universal.” Given Data / Assumptions: “Universal” means functionally complete using only a single gate type (plus signal constants if needed). Known universal gates include NAND and NOR. We assume ideal Boolean behavior and unlimited composition of identical gates. Concept / Approach: AND alone cannot reproduce inversion (NOT) or OR without external inversion sources. Universality requires the ability to synthesize NOT and one of {AND, OR}. NAND can create NOT (tie inputs together) and then build AND/OR; NOR can similarly produce NOT and then build other functions. Pure AND lacks a built-in way to invert, so it is not functionally complete by itself. Step-by-Step Solution: 1) Requirement: a universal set must generate NOT and AND/OR.2) Using only AND gates: Y = A * B * ... cannot produce NOT(A) without an inverter or complemented constant.3) Therefore, AND fails the universality criterion; NAND/NOR succeed because inversion is intrinsic through De Morgan transformations.4) Conclusion: an AND gate by itself is not universal. Verification / Alternative check: Standard logic design texts list functional completeness properties: {NAND} and {NOR} are complete; {AND} or {OR} alone are not; {AND, NOT} or {OR, NOT} are complete pairs. This external reference pattern confirms the conclusion. Why Other Options Are Wrong: “Correct” contradicts functional completeness theory. “Only with complemented inputs and outputs” smuggles in extra NOTs, violating the single-gate-type constraint. “Only for two-level SOP” still needs inversion. “Only when fan-in is unlimited” does not fix the absence of inversion. Common Pitfalls: Equating frequent use in SOP forms with universality. Remember: you must be able to generate NOT internally to claim universality. Final Answer: Incorrect

Question 8

Two-input AND gate behavior – if one input is HIGH (logic 1), does the output mirror the other input?

Accepted Answer

Introduction / Context: Understanding how an AND gate behaves under partial assertion is essential for gating and enabling signals. This question asks whether forcing one input HIGH causes the output to equal the remaining input, a useful identity in practical designs. Given Data / Assumptions: Standard 2-input AND gate with ideal Boolean behavior. Active-HIGH logic convention: 1 = asserted, 0 = deasserted. No hazards or metastability considered for the logical statement. Concept / Approach: The AND operation is multiplicative in Boolean algebra: Y = A * B. If A = 1, then Y = 1 * B = B, so the output mirrors B. This is the enabling property of AND: a HIGH input acts like an enable that passes the other signal through unchanged (subject to normal gate delays and thresholds). Step-by-Step Solution: 1) Write the Boolean relation: Y = A * B.2) Substitute A = 1 → Y = 1 * B = B.3) Therefore, with one input HIGH, the output equals the other input.4) Similarly, if an input is LOW, Y = 0 * B = 0, which effectively blocks the signal. Verification / Alternative check: Truth table rows confirm the identity: (1,0) → 0 and (1,1) → 1, so when A = 1 the output follows B exactly. This is why AND is used for clock enables, conditional data paths, and masking logic where a control line gates another signal. Why Other Options Are Wrong: Technology (CMOS vs TTL) and propagation delay do not change the logical identity; tying inputs together is irrelevant to the stated condition. Common Pitfalls: Confusing AND enabling with OR enabling; for OR, tying one input HIGH forces the output HIGH regardless of the other input. With AND, HIGH enables and LOW disables. Final Answer: Correct

Question 9

AOI realization – does complementing the output of an AND–OR network produce an AND–OR–Invert (AOI) circuit?

Accepted Answer

Introduction / Context: AOI (AND–OR–Invert) gates are common library elements in standard-cell ASICs and TTL/CMOS logic families. They are formed when a two-level AND–OR structure is followed by a single inversion at the output. This item checks whether you recognize that adding a final inversion to an AND–OR network yields an AOI function. Given Data / Assumptions: Network consists of first-level AND gates feeding a second-level OR gate. We apply a single inversion to the overall output. Ideal Boolean behavior is assumed. Concept / Approach: Let the AND–OR network compute S = (P1) + (P2) + ... where each Pi is a product term (e.g., A * B, C * D * E). Adding an inversion gives Y = NOT(S), which by definition is an AOI structure. Using De Morgan’s law, Y = NOT(P1 + P2 + ...) = NOT(P1) * NOT(P2) * ..., demonstrating the dual NAND–NOR interpretation and motivating why AOI (and OAI) forms are efficient in transistor networks. Step-by-Step Solution: 1) Start with S = (A * B) + (C * D) (example two-term SOP).2) AOI adds a final inverter: Y = NOT(S).3) Apply De Morgan: Y = NOT(A * B + C * D) = NOT(A * B) * NOT(C * D).4) The result is exactly an AND–OR–Invert function, often implemented compactly in CMOS using series–parallel transistor networks. Verification / Alternative check: Standard cell libraries list AOI21, AOI22, etc., where the suffix indicates how many inputs feed the first-level ANDs. Datasheets and timing libraries confirm they are implemented by inverting the output of an AND–OR combination. Why Other Options Are Wrong: Input inversion is not required to form AOI (that would be OAI variants or different polarities). Limiting to “two-level SOP only” is misleading—AOI refers specifically to two levels plus inversion, which is exactly the structure described. Wired-logic is unrelated to the functional definition. Common Pitfalls: Confusing AOI with OAI (OR–AND–Invert). Also, forgetting that a single output inverter transforms SOP to POS complements via De Morgan’s law. Final Answer: Correct

Question 10

Truth-table complement – is a NOR gate's output the logical complement of an OR gate's output for the same inputs?

Accepted Answer

Introduction / Context: OR and NOR are complementary logic functions. Recognizing how one relates to the other is essential when switching between active-HIGH and active-LOW conventions and when reading symbols that include inversion bubbles. Given Data / Assumptions: Inputs are identical for both OR and NOR comparisons. Active-HIGH logic levels and ideal Boolean behavior are assumed. No consideration of propagation delay or analog drive strength is required. Concept / Approach: By definition, NOR(A,B,...) = NOT(OR(A,B,...)). Therefore, for every input combination, the NOR output bit is exactly the inversion of the OR output bit. This is what it means for the truth table to be the “opposite” or complement—each row's output is flipped between the two gates. Step-by-Step Solution: 1) Write OR relation: S = A + B (+ C ...).2) Define NOR: Y = NOT(S) = NOT(A + B (+ C ...)).3) For any input row, compute S (OR) and invert it to obtain Y (NOR).4) Conclude that the tables are complements row-by-row. Verification / Alternative check: Consider two inputs: OR truth table is 00→0, 01→1, 10→1, 11→1; NOR is 00→1, 01→0, 10→0, 11→0—a perfect complement across all rows. Symbols also show an inversion bubble at the NOR output, signaling the negation. Why Other Options Are Wrong: “Incorrect” contradicts the formal definition. Conditions about open-collector outputs or DC operation are irrelevant to the Boolean complement relationship. Common Pitfalls: Misreading schematic bubbles or assuming that hardware packaging changes logic definition. Hardware details affect electrical behavior, not truth-table identity. Final Answer: Correct

Question 11

NAND behavior – is the output LOW only when all inputs are simultaneously HIGH?

Accepted Answer

Introduction / Context: NAND gates are ubiquitous universal primitives. Knowing the exact condition for a LOW output is crucial when designing fault indicators, active-LOW control signals, and safety interlocks. This question verifies the canonical NAND condition for zero output. Given Data / Assumptions: N-input NAND with ideal Boolean behavior. Active-HIGH convention for logic 1. No analog issues (timing or fan-in limits) change the logic definition. Concept / Approach: NAND is the negation of AND: Y = NOT(A * B * ... * N). The AND product equals 1 only when every input is 1. Therefore, Y becomes 0 only in that all-HIGH case. For any input being 0, the product is 0 and the NAND output is 1. Step-by-Step Solution: 1) Define Y = NOT(∏ inputs).2) If all inputs are HIGH, ∏ = 1 → Y = NOT(1) = 0.3) If any input is LOW, ∏ = 0 → Y = NOT(0) = 1.4) Conclude: NAND outputs LOW only for the all-HIGH input row. Verification / Alternative check: Truth tables for 2-input, 3-input, and 4-input NANDs all share this property. Timing diagrams show brief hazards during transitions in real hardware, but the steady-state condition remains exact. Why Other Options Are Wrong: The gate count (two vs many inputs), technology family (CMOS vs TTL), and pull-ups do not alter Boolean function; they affect electrical characteristics only. Common Pitfalls: Confusing NAND with NOR (which is the complement of OR). Also, overlooking that any single LOW forces NAND HIGH, which designers exploit to build robust inhibit signals. Final Answer: Correct

Question 12

In digital electronics, an exclusive-OR (XOR) gate produces a HIGH output when its inputs are different and a LOW output when its inputs are the same. Evaluate the statement: "An exclusive-OR gate's output is HIGH when its inputs are equal."

Accepted Answer

Introduction / Context: The exclusive-OR, commonly abbreviated as XOR, is a fundamental combinational logic function used in parity generation, adders, and data comparison. Understanding precisely when XOR outputs a HIGH (1) versus a LOW (0) is essential for circuit design, troubleshooting, and test writing. Given Data / Assumptions: Two-input XOR gate is considered (extension to more inputs follows the same odd-parity idea). Inputs are binary levels: 0 (LOW) and 1 (HIGH). The claim under evaluation says “XOR output is HIGH when inputs are equal.” Concept / Approach: By definition, XOR outputs 1 if and only if the number of HIGH inputs is odd; for a two-input XOR, this happens when exactly one input is HIGH and the other is LOW. Therefore, when the inputs are equal (both 0 or both 1), the XOR output is 0. The related function XNOR (exclusive-NOR) outputs 1 when inputs are equal and 0 when they differ, which is the logical complement of XOR. Step-by-Step Solution: 1) List two-input combinations: 00, 01, 10, 11.2) XOR rule: output = 1 when inputs differ, 0 when equal.3) Evaluate: XOR(0,0)=0; XOR(0,1)=1; XOR(1,0)=1; XOR(1,1)=0.4) Observe that equal pairs (00 and 11) give 0, not 1. Verification / Alternative check: Truth tables in any digital logic text confirm the above mapping. Hardware realizations using sums-of-products or via gates (e.g., (A AND NOT B) OR (NOT A AND B)) also show the output is 1 for unequal inputs only. Why Other Options Are Wrong: “Correct” contradicts XOR definition. “HIGH only when both inputs are 1” describes AND, not XOR. “HIGH only when both inputs are 0” describes NOR for two inputs (with inversion nuances), not XOR. Common Pitfalls: Confusing XOR with XNOR; assuming “exclusive” means “both or none.” Also, mixing up XOR with simple OR, which is 1 when any input is 1 (including 11). Final Answer: Incorrect

Question 13

Consider the AND logic function used in combinational circuits. Evaluate the claim: "The output of an AND gate is HIGH when any input is HIGH."

Accepted Answer

Introduction / Context: The AND gate is one of the most basic logic functions. Designers use it for enabling, masking, and multi-signal qualification. Misunderstanding its HIGH-output condition can cause design errors and test misinterpretations. Given Data / Assumptions: Standard n-input AND gate. Inputs are binary levels: 0 and 1. We are evaluating a statement that attributes OR-like behavior to AND. Concept / Approach: By definition, AND outputs 1 only when all inputs are 1 simultaneously. If any single input is 0, the AND output becomes 0 regardless of the other inputs. The incorrect statement confuses AND with OR, which outputs 1 when any input is 1. Step-by-Step Solution: 1) For a two-input AND, enumerate: 00, 01, 10, 11.2) AND rule: output = 1 only for 11; otherwise output = 0.3) Therefore, “any input HIGH” causing output HIGH is false; that rule belongs to OR. Verification / Alternative check: Truth tables and Boolean algebra confirm AND = logical multiplication. In algebraic form for two inputs: Y = A * B, which equals 1 only if A = 1 and B = 1 together. Why Other Options Are Wrong: “Correct” is wrong because it restates the claim. “HIGH only when all inputs are HIGH” is actually true for AND but is not the requested evaluation of the claim; it is included as a distractor. “LOW only when all inputs are LOW” is a property of OR's complement situations, not AND in general. Common Pitfalls: Confusing AND with OR; assuming “any” instead of “all.” Forgetting that one LOW blocks an AND chain entirely. Final Answer: Incorrect

Question 14

Evaluate the statement: "A NOR gate is a universal gate," meaning it alone can be composed to realize any Boolean function.

Accepted Answer

Introduction / Context: Universal gates are gate types from which any Boolean function can be constructed without requiring additional primitive gates. Understanding which families are universal informs design choices, minimization, and library constraints. Given Data / Assumptions: NOR is defined as NOT(OR). We can freely interconnect multiple NOR gates. Inverter behavior can be synthesized from a NOR gate by tying its inputs together. Concept / Approach: NOR is universal because you can build NOT, OR, and AND using only NOR gates. Once those primitives are available, arbitrary combinations (SOP, POS, and multi-level networks) can be constructed to realize any Boolean expression. The dual universal gate is NAND, which similarly suffices for complete logic construction. Step-by-Step Solution: 1) NOT from NOR: Y = NOR(A, A) = NOT(A OR A) = NOT A.2) OR from NOR: Use De Morgan transformations; OR can be achieved via inverted inputs to a NOR, implemented with additional NOR inverters.3) AND from NOR: A AND B = NOT(NOT A OR NOT B). Using NOR-based inverters and one NOR, we realize AND.4) With NOT/AND/OR available, arbitrary Boolean functions follow. Verification / Alternative check: Standard logic textbooks provide canonical constructions for NOT, OR, and AND from NOR, proving universality. Why Other Options Are Wrong: “Incorrect” contradicts established theory. “Only universal with XOR present” is unnecessary; XOR itself can be built from NORs. “Universal only for two-level SOP” is too restrictive; multi-level implementations are also possible with NOR-only networks. Common Pitfalls: Assuming universality requires a special gate like XOR; forgetting that tying inputs together yields inversion with NOR. Final Answer: Correct

Question 15

In VHDL-based design, a "component" declaration introduces a reusable interface (entity-like) that can be instantiated within an architecture. Assess the statement: "A VHDL component is a predefined logic function."

Accepted Answer

Introduction / Context: Hardware description languages such as VHDL separate interface specification (entity/component) from behavior/structure (architecture). Misidentifying a component as a “predefined logic function” blurs the distinction between declarations, implementations, and libraries. Given Data / Assumptions: VHDL uses entities/architectures to describe modules. Components declare a local interface used for instantiation. Libraries/packages may contain predefined functions or primitives, but that is distinct from a component declaration itself. Concept / Approach: A VHDL component is not inherently a predefined function. Rather, it is a declaration that mirrors an entity interface (ports and generics) so that instances can be placed inside an architecture. The actual logic behavior resides in an entity/architecture pair or is bound via configuration. A design may bind a component to different entities/architectures, enabling reuse and swapping implementations without changing the instantiation sites. Step-by-Step Solution: 1) Write an entity/architecture pair that defines the module behavior.2) In a higher-level architecture, declare a component that matches the entity interface (port/generic list).3) Instantiate the component; later, bind it to the specific entity/architecture via direct entity instantiation or configuration.4) Observe that the component is a placeholder for an implementation, not itself a predefined gate or function. Verification / Alternative check: Review VHDL reference manuals: components serve as instantiation templates; predefined functions (e.g., numeric_std arithmetic) exist in packages, separate from component declarations. Why Other Options Are Wrong: “Correct” mistakes declaration for implementation. “Correct only for IEEE library gates” is inaccurate—IEEE packages provide types and functions, not generic “component = logic function.” “Correct when synthesized to LUTs” confuses implementation technology with the HDL construct’s meaning. Common Pitfalls: Equating component with entity behavior; assuming components imply fixed, built-in logic; forgetting about binding/configuration. Final Answer: Incorrect

Question 16

Digital logic driving capability: A gate can drive a certain maximum number of standard load inputs. This limit is specified by which parameter?

Accepted Answer

Introduction / Context: When interconnecting logic gates, each output has a limit to how many inputs it can reliably drive without violating logic-level specifications. Recognizing the correct parameter ensures signal integrity and reliable operation in larger designs. Given Data / Assumptions: Logic families specify VOH/VOL, VIH/VIL, and input/output currents. Multiple receiving inputs load a single driving output. Standard loads are defined per family (e.g., TTL unit load). Concept / Approach: Fan-out is the maximum number of standard inputs that a gate output can drive while keeping output levels within guaranteed ranges. It is derived from the ratio of the driver’s available source/sink current to the required input current of the loads. Fan-in, by contrast, counts the number of inputs a gate possesses; supply voltage and noise margin are important parameters but do not count loads directly. Step-by-Step Solution: 1) Identify the logic family (e.g., TTL, CMOS) and its unit load definitions.2) Obtain driver I_OH/I_OL and receiver I_IH/I_IL from datasheets.3) Compute allowable loads so that VOH/VOL remain within spec; this is reported as “fan-out.”4) Ensure total capacitive loading and wiring do not degrade edges excessively, even if static fan-out is met. Verification / Alternative check: Cross-check with family datasheets or design guides that list typical fan-out values and conditions. Why Other Options Are Wrong: “Supply voltage” sets operating range, not load count. “Noise margin” measures robustness against noise, not how many loads are driven. “Fan-in” is the count of inputs on a gate, unrelated to how many loads an output can drive. Common Pitfalls: Ignoring dynamic effects (capacitance) that reduce practical fan-out; mixing fan-in with fan-out terminology. Final Answer: fan-out

Question 17

You have a 3-input NAND gate available but only need a 2-input NAND function. How should the unused input be handled to preserve correct two-input NAND behavior?

Accepted Answer

Introduction / Context: Real designs frequently adapt available parts to required functions. Correctly fixing unused inputs on multi-input gates prevents unintended logic behavior and noise susceptibility. Given Data / Assumptions: Device: 3-input NAND; required behavior: 2-input NAND. Inputs are standard digital levels with family-appropriate thresholds. No desire to alter the intended truth table for the two active inputs. Concept / Approach: A NAND outputs LOW only when all inputs are HIGH. To reduce a 3-input NAND to an effective 2-input NAND, force the third input to a benign value that preserves the 2-input truth table. Setting the unused input to logic 1 yields Y = NOT(A * B * 1) = NOT(A * B), which is exactly a 2-input NAND of A and B. Leaving inputs open is bad practice; tying to 0 changes behavior to Y = NOT(A * B * 0) = NOT(0) = 1 (trivial constant HIGH), which is incorrect. Step-by-Step Solution: 1) Desired: Y = NOT(A * B).2) With a third input C fixed at 1: Y = NOT(A * B * 1) = NOT(A * B).3) Therefore tie the unused input to HIGH to preserve function. Verification / Alternative check: Check full truth table: with C = 1, the 3-input NAND reduces to the 2-input NAND truth table for A and B. Why Other Options Are Wrong: “connected to ground” (0) forces output always HIGH. “left open” risks undefined/floating behavior and noise. “any of the above” is incorrect because only tying HIGH is functionally correct and robust. Common Pitfalls: Leaving inputs floating; assuming tying LOW is harmless; ignoring family-specific recommendations about unused inputs. Final Answer: connected to a HIGH

Question 18

If both inputs of a 2-input NOR gate are tied together and driven by the same signal A, what single-function behavior does the device implement?

Accepted Answer

Introduction / Context: Gate reconfiguration through tying inputs is a classic technique for reducing parts and creating needed primitives. With NOR and NAND (universal gates), tying inputs can yield inversion, enabling construction of other logic functions. Given Data / Assumptions: 2-input NOR gate with inputs connected together: A, A. Standard binary logic levels 0 and 1. No additional components assumed. Concept / Approach: NOR is NOT(OR). With identical inputs, OR(A, A) equals A. Therefore, NOR(A, A) equals NOT(A). Thus, the tied-input NOR behaves exactly as an inverter. This property is exploited when building universal-gate-only designs: NOT, OR, and AND can all be composed using NOR gates. Step-by-Step Solution: 1) Start from the NOR definition: Y = NOT(A OR B).2) Tie inputs: B = A, so Y = NOT(A OR A) = NOT(A).3) Conclude that the gate acts as an inverter on signal A. Verification / Alternative check: Truth table check: if A = 0, output = 1; if A = 1, output = 0, matching an inverter behavior exactly. Why Other Options Are Wrong: “OR gate” is the opposite function before inversion. “AND gate” is unrelated to tying NOR inputs. “any of the above” cannot be true because only one behavior results from the identity OR(A, A) = A. Common Pitfalls: Assuming tied inputs average or produce undefined levels; overlooking that OR with identical inputs reduces to the input itself, making NOR a straightforward inverter. Final Answer: inverter

Question 19

NAND as a Universal Gate — Using only NAND gates, which standard logic function(s) can be implemented without needing any other gate type?

Accepted Answer

Introduction: The NAND gate is called a universal gate because, by appropriate interconnection, it can realize every basic logic function. This question checks whether you know that NAND alone can synthesize NOT, AND, and OR, and therefore any combinational logic circuit. Given Data / Assumptions: Only 2-input NAND gates are available. Binary logic levels are ideal (0 and 1). No constraints on the number of stages, other than logical feasibility. Concept / Approach: A universal gate set can implement all other gates. With NAND: NOT is obtained by tying a gate’s inputs together; AND is obtained by NAND followed by NOT; OR is obtained via De Morgan by inverting inputs into a NAND (i.e., OR = NOT(A) NAND NOT(B)). Step-by-Step Solution: 1) Inverter: NOT A = A NAND A.2) AND: A AND B = NOT(A NAND B) = (A NAND B) NAND (A NAND B).3) OR: A OR B = NOT(NOT A AND NOT B) = (A NAND A) NAND (B NAND B). Verification / Alternative check: Build truth tables or simulate the three constructions; each matches the target gate across all input pairs. Why Other Options Are Wrong: OR gate: True that NAND can make OR, but the option claims only OR. NAND can do more.AND gate: Same issue—NAND can make AND but is not limited to it.Inverter (NOT): Also derivable, but not the only possibility. Common Pitfalls: Forgetting the NOT-with-tied-inputs trick (A NAND A).Missing De Morgan’s law when deriving OR from NAND. Final Answer: Any of the above

Question 20

Circuit Topology — In digital and analog circuit analysis, what is the correct definition of a node?

Accepted Answer

Introduction: The concept of a node underpins circuit analysis techniques such as nodal analysis and Kirchhoff’s current law. Accurately identifying nodes is essential for writing equations and understanding how signals and currents distribute in a network. Given Data / Assumptions: Passive and active components are connected by conductors. Ideal wires have negligible impedance for topological definitions. Analysis follows standard circuit theory conventions. Concept / Approach: A node is defined topologically, not functionally: it is any point of interconnection among two or more circuit elements. All points on the same continuous conductor (with no intervening elements) belong to the same node and share the same electrical potential at a given instant. Step-by-Step Solution: 1) Identify continuous conducting paths between element terminals.2) Group all physically connected points with no intervening elements as one node.3) Assign a reference (ground) node; express other node voltages relative to it. Verification / Alternative check: Use a continuity check (mentally or in CAD) to verify that points designated as a node are indeed directly connected. Why Other Options Are Wrong: A circuit implemented strictly as a sum-of-products: That describes a realization style, not a topological point.The complete set of output signals: That refers to signals/ports, not a connection point.A permanently shorted input terminal: A fault condition, not a definition of a node. Common Pitfalls: Confusing nodes with components or ports.Splitting a single conductor into multiple “nodes” when no element lies in between. Final Answer: A common point in a circuit where two or more elements connect

Combinational Logic Analysis Questions