Question 1

Assertion (logic design): “A popular waveform generator is the Johnson shift counter.” Select the most appropriate evaluation.

Accepted Answer

Introduction / Context:This statement mixes up roles. Johnson (twisted-ring) counters generate sequences of states and are excellent for timing/sequence generation, but the commonly cited popular general-purpose waveform generator in basic electronics is the 555 timer or function generator ICs.Given Data / Assumptions:“Popular waveform generator” generally means a device used to create analog waveforms like square/triangle/sawtooth.Johnson counter is a digital sequential circuit.Concept / Approach:Johnson counters circulate a pattern through flip-flops to produce non-overlapping timing signals. They are not standard analog waveform generator ICs. Thus the claim is not correct in the usual sense taught at this level.Step-by-Step Solution:Step 1: Define “waveform generator” in context (often analog/function generation).Step 2: Identify Johnson counter as a digital counter/sequencer.Step 3: Conclude the statement is incorrect.Verification / Alternative check:Compare typical lab instruments/ICs: 555 timer, XR2206, ICL8038, or DDS devices are taught as waveform generators, not Johnson counters.Why Other Options Are Wrong:Correct: Conflicts with the standard classification.Partially correct under special cases: Even though counters can drive digital patterns, this does not make them “popular waveform generators.”Ambiguous statement / Insufficient information: The wording is sufficiently clear to judge as incorrect.Common Pitfalls:Assuming any periodic digital sequence source is a “waveform generator.” The term is broader and usually analog-focused at this level.Final Answer:Incorrect

Question 2

Fundamentals of logic gates: “A logic gate has one or more output terminals and a single input terminal.” Select the most appropriate evaluation.

Accepted Answer

Introduction / Context:Basic logic gates are combinational devices with one output and one or more inputs. The given statement flips this conventional description.Given Data / Assumptions:Typical gates: AND, OR, NAND, NOR, XOR, XNOR, NOT.Standard I/O structure: multiple inputs (except NOT) and single output.Concept / Approach:Review the canonical definition: A logic gate implements a Boolean function. In almost all common cases, there is exactly one output; the number of inputs can be one (NOT) or more (AND with 2, 3, 4 inputs, etc.).Step-by-Step Solution:Step 1: Recall the I/O pattern for common gates.Step 2: Compare with the statement that claims one input and multiple outputs.Step 3: Conclude the statement is incorrect.Verification / Alternative check:Datasheets for 7400-series or CMOS 4000-series gates show single-output symbols with varying input counts, confirming the standard arrangement.Why Other Options Are Wrong:Correct: Contradicts standard definitions.Ambiguous / Depends on implementation: While complex ICs can have multiple outputs, basic “logic gate” definition remains single-output.Not applicable to combinational logic: Gates are combinational by definition.Common Pitfalls:Generalizing from complex devices (like decoders) back to a single logic gate symbol.Final Answer:Incorrect

Question 3

Power characteristics of IC families: “As a rule, CMOS has the lowest power consumption among common IC logic families.” Select the most appropriate evaluation.

Accepted Answer

Introduction / Context:Different logic families (CMOS, TTL, ECL, etc.) exhibit distinct power behaviors. This question probes recognition of CMOS power advantages in typical operating regimes.Given Data / Assumptions:Comparison is across common logic families.“As a rule” means typical, not pathological edge cases.Concept / Approach:Static CMOS consumes very little static power because complementary transistors ideally do not both conduct in steady states. Dynamic power is roughly proportional to C * V^2 * f. Compared to TTL or ECL at similar V and f, CMOS is generally lower in power.Step-by-Step Solution:Step 1: Recall the CMOS static power advantage.Step 2: Note dynamic power scales with switching activity, but is still typically lower than bipolar families at like-for-like conditions.Step 3: Conclude the statement is correct.Verification / Alternative check:Datasheet comparisons (HC/HCX vs. LS-TTL vs. ECL) show CMOS families with lower typical supply current in steady state.Why Other Options Are Wrong:Incorrect: Conflicts with widely observed datasheet values.Only true at high frequency: High f increases dynamic losses; claim is not limited to high f.Only true for bipolar families: Statement is about CMOS vs. others; bipolar families are typically higher power.Insufficient information: Standard comparison suffices.Common Pitfalls:Overlooking dynamic power and assuming it negates CMOS advantages; it does not for typical use.Final Answer:Correct

Question 4

Troubleshooting practice: “Good troubleshooting is done by looking at the input signal and how it interacts with the circuits.” Select the most appropriate evaluation.

Accepted Answer

Introduction / Context:Effective troubleshooting requires observing signals at inputs, intermediate nodes, and outputs while considering expected behavior. The statement narrows the practice to the input and “how it interacts,” which is incomplete and potentially misleading.Given Data / Assumptions:We consider general troubleshooting across digital and analog circuits.Good process includes hypotheses, measurement at multiple points, and comparison with known-good references.Concept / Approach:A robust method traces signal flow: stimulus → input conditioning → core function → output stage. Restricting observation to the input and vaguely to “interaction” omits verification at outputs and key internal nodes, timing relationships, and power/ground integrity checks.Step-by-Step Solution:Step 1: Define expected behavior at each stage (block diagram).Step 2: Probe strategically: input, critical internal nodes, and output.Step 3: Compare measurements with the model; isolate the faulty block.Verification / Alternative check:Service manuals advocate “divide and conquer” or “half-split” methods that explicitly require checks beyond the input.Why Other Options Are Wrong:Correct: Too narrow to be best practice.Only correct for linear analog circuits: Even there, observing outputs and supply rails is essential.Ambiguous / Insufficient information: The claim is sufficiently concrete to evaluate as incomplete.Common Pitfalls:Focusing solely on inputs and ignoring timing, loading, power integrity, or thermal issues.Final Answer:Incorrect

Question 5

Learning digital logic effectively: “It is important to memorize logic symbols, Boolean equations, and truth tables for logic gates.” Select the most appropriate evaluation.

Accepted Answer

Introduction / Context:This statement emphasizes memorization. In digital design education and practice, understanding how to derive equations and truth tables from function specifications is more valuable than rote memorization, especially because symbols can vary slightly by standard.Given Data / Assumptions:We focus on enduring competence, not short-term recall.Standard gate behaviors can be reconstructed logically.Concept / Approach:While familiarity with common symbols and basic truth tables is useful, long-term proficiency comes from reasoning: reducing expressions, drawing timing diagrams, and building or analyzing circuits from specifications. Therefore, plain memorization is not “important” in the sense of being primary.Step-by-Step Solution:Step 1: Note that symbols/truth tables are derivable from definitions.Step 2: Emphasize skill in derivation and application over rote memory.Step 3: Conclude the statement, as written, is not correct as a primary learning strategy.Verification / Alternative check:Curricula encourage Boolean algebra, Karnaugh maps, and derivations; quick reference tables exist if needed, reducing the need to memorize everything.Why Other Options Are Wrong:Correct: Overstates memorization as the main objective.Only correct for exam cramming: Even then, understanding improves accuracy and transfer.Ambiguous / Insufficient information: The statement is clear enough to evaluate.Common Pitfalls:Relying on memory without understanding leads to errors on unfamiliar problems.Final Answer:Incorrect

Question 6

Digital test equipment – is a logic pulser used to detect floating logic levels on a node?

Accepted Answer

Introduction / Context: A logic pulser and a logic probe are complementary handheld tools used in digital electronics labs and troubleshooting. This question checks whether you understand what each tool actually does: injecting pulses versus sensing logic states and floating nodes.Given Data / Assumptions:A 'logic pulser' can drive a digital node with a short HIGH/LOW pulse sequence.A 'floating' node means it is not driven to a valid logic 0 or logic 1, so its level is indeterminate and noise-susceptible.We are discussing standard TTL/CMOS logic circuits on a bench or in-system.Concept / Approach: A logic pulser is a stimulator. It forces transitions on a node to see if downstream logic responds. A logic probe is a sensor. It reads whether a node is LOW, HIGH, toggling, or floating/undefined (some probes include a Hi-Z or pulse detect indicator). Therefore, using a pulser to determine floating level conflates roles; probing determines state, pulsing provides stimulus.Step-by-Step Solution:1) Identify the claim: “Logic pulser is used to determine floating level.”2) Recall roles: pulser = injects pulses; probe = measures logic state (including float indicators on some probes).3) Because measuring float requires a sensing function, a pulser is not the correct instrument.4) Conclusion: the statement is incorrect.Verification / Alternative check: Many tool kits are sold as “logic pulser + logic probe”. Manuals describe the pulser’s drive capability (for example, sourcing a brief 1 pulse) versus the probe’s threshold-based detection (LEDs for LOW/HIGH, pulse detect, sometimes Hi-Z). The division of labor confirms the conclusion.Why Other Options Are Wrong: “Correct” is wrong because a pulser injects transitions, it does not read a floating state. “Depends on the clock frequency” is irrelevant; floating detection is not about clock speed. “Not enough information” is unnecessary; the tool roles are standard. “Only for analog circuits” is off-topic; these tools target digital logic.Common Pitfalls: Confusing names: both start with “logic”, but probe senses; pulser stimulates. Another pitfall is assuming any tool connected to a node can infer float; without threshold sensing, you cannot classify a node's logic state reliably.Final Answer: Incorrect

Question 7

Exclusive-NOR (XNOR) truth: is the output HIGH when inputs are unequal?

Accepted Answer

Introduction / Context: Exclusive-NOR (XNOR) is a fundamental logic gate widely used in equality detectors, comparators, and parity checking. This item tests the core truth condition for XNOR versus XOR.Given Data / Assumptions:Standard two-input XNOR gate.Logic levels use conventional binary 0 and 1.No timing hazards or analog details; we rely on ideal Boolean behavior.Concept / Approach: XOR is HIGH when inputs are different (unequal). XNOR is the complement of XOR, so it is HIGH when inputs are the same (equal). Therefore, the claim that XNOR is HIGH for unequal inputs contradicts its definition.Step-by-Step Solution:1) Recall: XOR(A,B) = 1 when A != B.2) Therefore, XNOR(A,B) = NOT XOR(A,B) = 1 when A == B.3) The statement says “HIGH when inputs are unequal,” which is the XOR condition, not XNOR.4) Conclusion: the statement is incorrect.Verification / Alternative check: Build the truth table: 00 → 1, 01 → 0, 10 → 0, 11 → 1 for XNOR. Unequal pairs (01,10) give 0. Equal pairs (00,11) give 1. This independently confirms the reasoning.Why Other Options Are Wrong: “Correct” would describe XOR, not XNOR. “Only when both inputs are LOW” is incomplete; XNOR is also HIGH when both are HIGH. “Only when both inputs are HIGH” similarly misses the 00 case. “Only with open-collector outputs” confuses output structure with logic function; collector configuration does not change truth conditions.Common Pitfalls: Learners often swap XOR and XNOR semantics or forget that XNOR is an equality detector. Another pitfall is thinking output stage hardware affects the Boolean function—it does not for ideal logic definitions.Final Answer: Incorrect

Question 8

Inverter (NOT gate) behavior – does the output equal the logical complement of the input?

Accepted Answer

Introduction / Context: The inverter, also called a NOT gate, is the simplest single-input logic element. Understanding its complementing action is foundational for all combinational and sequential design.Given Data / Assumptions:Ideal Boolean logic levels: 0 and 1.One input, one output logic device.No analog edge cases (e.g., undefined thresholds) considered for the basic truth statement.Concept / Approach: By definition, a NOT gate outputs the logical complement of its input: output = 1 when input = 0, and output = 0 when input = 1. This is the Boolean operation NOT(A).Step-by-Step Solution:1) State the function: Y = NOT(A).2) Truth mapping: A=0 → Y=1; A=1 → Y=0.3) The statement claims the output is the complement; that matches the formal definition.4) Conclusion: the statement is correct.Verification / Alternative check: Consider an inverter chain: two inverters in series yield Y = NOT(NOT(A)) = A, proving each stage performs a complement. Oscillator and Schmitt variants still implement logical inversion around thresholds in normal operation.Why Other Options Are Wrong: “Incorrect” contradicts the definition. “Only at low frequencies” is false; frequency limits affect signal integrity but not the function. “Only for CMOS devices” is false; TTL and others also implement NOT. “Depends on the pull-up resistor” applies to open-collector/open-drain topologies, but the logical function remains inversion once levels are valid.Common Pitfalls: Confusing physical signal inversion with analog artifacts (slew, thresholds). While non-idealities exist, the logical operation is still complementing for valid 0/1 levels.Final Answer: Correct

Question 9

AND gate truth – if all inputs are HIGH, is the output LOW?

Accepted Answer

Introduction / Context: The AND function is one of the core Boolean primitives. Knowing the output when all inputs are asserted is essential for designing enable signals, decoders, and control logic.Given Data / Assumptions:Standard AND gate with two or more inputs.Active-HIGH logic levels.No inversion bubbles at the output (i.e., not a NAND).Concept / Approach: For an AND gate, the output is 1 only if every input is 1. Therefore, if all inputs are HIGH, the output must be HIGH. The statement claims the output would be LOW in that case, which is the opposite and would correspond to a NAND gate, not an AND gate.Step-by-Step Solution:1) Recall AND definition: Y = A * B * C * ...2) When all inputs are 1, the product 1*1*... = 1.3) The claim says Y is LOW with all HIGH inputs, which describes NAND (NOT of AND).4) Conclusion: the statement is incorrect for an AND gate.Verification / Alternative check: Consult truth tables: AND(1,1) = 1; AND(1,1,1) = 1. Only NAND gives 0 for all-HIGH inputs. Schematic symbols with an inversion bubble at the output indicate NAND, not AND.Why Other Options Are Wrong: “Correct” conflicts with the AND definition. “Only for three or more inputs” is irrelevant; any number of inputs follows the same rule. “Only with active-LOW outputs” confuses electrical polarity with logic function. “Cannot be determined” is incorrect; the AND definition is precise.Common Pitfalls: Mixing up AND and NAND due to similar symbols. Always check for the small inversion bubble to identify NAND, which inverts the AND result.Final Answer: Incorrect

Question 10

OR gate behavior – is the output HIGH only when all inputs are HIGH?

Accepted Answer

Introduction / Context: The OR gate forms the basis of sum-of-products design. Understanding its activation condition is crucial for alarms, enables, and multiplexing logic where any triggering input should set the output.Given Data / Assumptions:Standard OR gate with two or more inputs.Active-HIGH logic convention.No inversion at inputs or output (i.e., not NOR).Concept / Approach: OR outputs 1 if at least one input is 1. Requiring “all inputs HIGH” would turn OR into an AND function, which contradicts the definition. Therefore, the claim is false.Step-by-Step Solution:1) Define OR: Y = A + B + C + ...2) If any input equals 1, Y = 1.3) The statement demands all-HIGH for HIGH output, which is the AND condition.4) Conclusion: the statement is incorrect.Verification / Alternative check: Truth table check: OR(1,0)=1; OR(0,1)=1; OR(1,1)=1. Only AND requires all 1s. NOR is the inverted OR, giving 0 when any input is 1 and 1 only when all are 0, but that is a different gate entirely.Why Other Options Are Wrong: “Correct” contradicts the OR definition. “Only for wired-OR buses” conflates wiring technique with logic function. “Only with positive logic” is irrelevant; OR semantics remain with consistent polarity. “Only for two-input devices” is false; the rule scales to N inputs.Common Pitfalls: Confusing OR with AND or NOR due to symbol similarities. Remember: “+” denotes OR in Boolean algebra; “*” denotes AND.Final Answer: Incorrect

Question 11

Exclusive-OR (XOR) truth – is the output HIGH when the inputs are unequal?

Accepted Answer

Introduction / Context: The XOR gate is widely used for parity generation, adders (sum bit), and change detection. This item verifies the canonical XOR condition in two-input form.Given Data / Assumptions:Two-input XOR gate under ideal Boolean behavior.Active-HIGH convention for logic 1.No timing hazards considered.Concept / Approach: XOR evaluates to 1 if and only if its inputs differ. It is often read as “either A or B, but not both.” Therefore, stating that the output is HIGH when inputs are unequal is accurate.Step-by-Step Solution:1) Recall definition: XOR(A,B) = (A * NOT(B)) + (NOT(A) * B).2) For unequal inputs (01 or 10), one product term is 1 → output = 1.3) For equal inputs (00 or 11), both product terms are 0 → output = 0.4) Conclusion: the statement is correct.Verification / Alternative check: Truth table: 00→0, 01→1, 10→1, 11→0. In full adders, XOR generates the sum bit, further reinforcing the “unequal → 1” behavior.Why Other Options Are Wrong: “Incorrect” would describe XNOR, not XOR. The options “Only when both inputs are LOW/HIGH” refer to equality cases, which produce 0 for XOR. “Only for TTL families” confuses logic family with function; XOR behavior is universal across families.Common Pitfalls: Mixing XOR with inclusive OR. Inclusive OR is 1 for 01, 10, and 11; XOR excludes the 11 case, hence “exclusive.”Final Answer: Correct

Question 12

NOR versus OR – do they operate in exactly the same way?

Accepted Answer

Introduction / Context: OR and NOR are related but distinct logic operations. Recognizing the inversion bubble on NOR is vital for correct reading of schematics and active-LOW signal conventions.Given Data / Assumptions:Standard two-input (extendable to N input) gates.Active-HIGH logic convention.Ideal Boolean behavior.Concept / Approach: NOR is the logical negation of OR: NOR(A,B) = NOT(OR(A,B)). Their truth tables are complements. Therefore, they do not operate in the same way; NOR only outputs 1 when all inputs are 0, while OR outputs 1 if any input is 1.Step-by-Step Solution:1) OR truth: any input HIGH → output HIGH.2) NOR truth: any input HIGH → output LOW; only all LOW → output HIGH.3) Since outputs differ for all input combinations except the all-LOW case's complement, they are not identical functions.4) Conclusion: the statement is incorrect.Verification / Alternative check: Draw symbols: NOR shows an inversion bubble at the output, signaling logical negation. Truth tables confirm complementarity. De Morgan's laws also express relationships: NOR equals AND of inverted inputs: NOR(A,B) = NOT(A + B) = (NOT A) * (NOT B).Why Other Options Are Wrong: “Correct” is false by definition. “Only with wired-OR lines” and “Only with tri-state outputs” confuse hardware interconnects with logical functions. “Only when inputs are all LOW” cherry-picks a single row and ignores other combinations.Common Pitfalls: Ignoring inversion bubbles in schematic symbols; that bubble changes the function dramatically. Always check polarity indicators.Final Answer: Incorrect

Question 13

Programmable logic arrays – is an OR array programmed by blowing fuses to remove selected connections?

Accepted Answer

Introduction / Context: Classic programmable logic devices include PROM, PLA, and PAL. Understanding which array (AND or OR) is user-programmable and how programming occurs (e.g., fuse blowing) is key to legacy digital design knowledge and reverse engineering older boards.Given Data / Assumptions:PROM: typically fixed AND (minterm generator), programmable OR.PLA: both AND and OR arrays programmable.PAL: usually programmable AND, fixed OR (device-dependent variants exist).Historical devices used fusible links to make or break connections.Concept / Approach: In PROM-style architectures, programming the OR array involves blowing fuses to disconnect certain minterm lines from an output OR plane, effectively removing product-term connections so that only the desired minterms contribute to each output function. Describing this as “eliminating selected variables/connections from the output functions” is a common summary for beginners, even though what is actually eliminated are product-term connections rather than literal algebraic variables.Step-by-Step Solution:1) Recall device families: PROM → programmable OR plane via fuses.2) Programming = selectively blowing fuses to open undesired connections.3) Result: output OR sums only the intended product terms.4) Therefore, the statement captures the essence of OR-plane programming.Verification / Alternative check: Textbook block diagrams show a fixed decoder (AND) generating all minterms and a programmable OR matrix with fusible links to outputs. Modern non-volatile cells (EPROM/EEPROM/Flash/antifuse) replace literal fuses but the logical idea of connect/disconnect remains.Why Other Options Are Wrong: “Incorrect” would deny well-documented PROM behavior. “Only the AND array is programmable” applies to many PALs, not PROMs. “Only EEPROM cells are used” is historically false; many devices used metal fuses. “Depends solely on fan-in limits” is off-topic; fan-in affects capacity, not the programming principle.Common Pitfalls: Taking the word “variables” literally; in hardware you program connections to product terms, which correspond to minterms in Boolean expressions. Another pitfall is assuming all PLDs have the same programmable plane; families differ.Final Answer: Correct

Question 14

74LS08 / 7408 quad 2-input AND gate – are VCC and GND on pins 14 and 7 powering all four gates?

Accepted Answer

Introduction / Context: Standard 14-pin TTL logic ICs (e.g., 74xx series) use a common power pinout convention. Knowing supply pins is essential for prototyping and troubleshooting power issues across all internal gates.Given Data / Assumptions:Device: 7408 (or 74LS08) quad two-input AND gate.Typical DIP-14 package.Standard TTL convention: VCC at pin 14, GND at pin 7.Concept / Approach: The IC houses four independent AND gates that share the same supply rails. Supplying VCC to pin 14 and GND to pin 7 powers the entire die, not just one gate. Individual gates are accessed through their respective input/output pins but all rely on the same VCC/GND pins.Step-by-Step Solution:1) Identify pinout: pin 14 = VCC, pin 7 = GND for 7408.2) The internal four gates reside on a common substrate powered by these rails.3) Therefore, connecting these pins powers all four AND gates simultaneously.4) Conclusion: the statement is correct.Verification / Alternative check: Datasheets and pinout diagrams consistently show the same assignment across many 74xx devices. Schematic symbols group four gates (A, B, C, D) but share the same power pins, simplifying board routing.Why Other Options Are Wrong: “Incorrect” contradicts datasheets. “Only two of the four gates are powered” has no basis in the design. “Depends on package subtype” is misleading; DIP and SOIC keep the same logical assignment. “Only in CMOS versions” is false; TTL and CMOS families alike follow the convention for 7408 variants.Common Pitfalls: Confusing per-gate pins with power pins or mixing up devices (e.g., thinking pin 8 is GND like on some 8-pin parts). Always verify with the device's pinout.Final Answer: Correct

Question 15

Boolean notation – does the “+” symbol represent the OR operation in Boolean equations?

Accepted Answer

Introduction / Context: Boolean algebra uses compact symbols to denote logic operations. Recognizing that “+” stands for OR is essential for reading and manipulating sum-of-products (SOP) and related forms in digital design.Given Data / Assumptions:Standard Boolean notation with “+” for OR and “*” (or adjacency) for AND.NOT is indicated by a bar, prime, or explicit NOT().We are not performing numeric arithmetic but logical operations.Concept / Approach: In Boolean algebra, “+” is logical OR, “*” is logical AND, and complement is inversion. These symbols are conventional and appear in textbooks, datasheets, and design tools. Therefore, the statement is correct.Step-by-Step Solution:1) Map symbols: OR → “+”, AND → “*”, NOT → overbar/prime.2) Example: Y = A + B means Y is 1 if A=1 or B=1 (or both).3) Compare with XOR: A ⊕ B is distinct and notated differently; “+” is not XOR.4) Conclusion: “+” represents OR in Boolean equations.Verification / Alternative check: Truth table for Y = A + B matches OR: 00→0, 01→1, 10→1, 11→1. Karnaugh maps and SOP forms also rely on “+” as the sum (OR) of product terms.Why Other Options Are Wrong: “Incorrect” denies widely accepted notation. “It means XOR” is wrong; XOR has its own symbol. “Logical addition only in arithmetic circuits” confuses arithmetic with logic, though the symbol resembles addition. “It denotes enable” is a signal label, not an operator.Common Pitfalls: Assuming “+” always implies numeric addition; in Boolean context it denotes logical OR. Context determines meaning: Boolean algebra vs integer arithmetic.Final Answer: Correct

Question 16

In digital electronics, a periodic signal (waveform) can be gated—i.e., enabled so it passes through, or disabled so it is blocked—by using either an AND gate with a suitably chosen enable input or an OR gate with the appropriate active-low/active-h…

Accepted Answer

Introduction / Context:In digital electronics, waveform gating is the practice of allowing or blocking a periodic signal using a logic control input. AND and OR gates are both used for this purpose, depending on whether the control is active-high or active-low.Given Data / Assumptions:A periodic input signal (the waveform) must be either enabled (passed) or disabled (blocked).We have access to AND and OR gates and, if needed, the inverted form of the control signal.Active-high means control = 1 enables; active-low means control = 0 enables.Concept / Approach:Use an AND gate for active-high enabling (signal AND enable). Use an OR gate for active-low enabling by ORing the signal with the complement of the disable control, or equivalently by ORing the inverted waveform with an active-high control before reinverting. The choice depends on available polarities and required logic levels.Step-by-Step Solution:1) AND gating (active-high enable E): output = signal * E.2) OR gating (active-low enable En): output = signal OR 0 when En = 0; to block, force output to 1 or hold the line as needed with proper polarity planning.3) With appropriate inversion, both gates can implement enable or disable behavior.Verification / Alternative check:Truth tables for two-input AND/OR confirm these behaviors when one input is the control or its complement.Why Other Options Are Wrong:“Only AND…” and “Only OR…” ignore polarity transformations; “Incorrect” contradicts standard gating practice using either gate with proper inversion.Common Pitfalls:Forgetting control polarity; neglecting that an inverter may be needed; confusing blocking a waveform with forcing a constant level.Final Answer:Correct

Question 17

Consider the NAND logic function. The output of a NAND gate goes LOW only in the single case when every one of its inputs is HIGH. Evaluate the statement.

Accepted Answer

Introduction / Context:A NAND gate is the logical negation of AND. Understanding its truth condition for a LOW output is foundational for digital design and troubleshooting.Given Data / Assumptions:NAND is defined as NOT(AND).Inputs can be 0 or 1; output depends on all inputs.Concept / Approach:Since AND outputs 1 only when all inputs are 1, NAND will output 0 exactly in that same case (because it inverts the AND result). Otherwise, NAND outputs 1.Step-by-Step Solution:1) For all inputs = 1, AND = 1, so NAND = 0.2) If any input = 0, AND = 0, so NAND = 1.3) Therefore, “LOW only if all inputs are HIGH” is correct.Verification / Alternative check:Construct a two-input truth table to verify the unique LOW case (1,1).Why Other Options Are Wrong:“Incorrect” and claims about LOW on mixed or any 0 contradict the definition of NAND.Common Pitfalls:Confusing NAND with NOR or with AND; forgetting that NAND inverts the AND output.Final Answer:Correct

Question 18

A truth table is intended to show how a logic gate’s output responds to every possible combination of its inputs. Does the statement that it shows “how the input level responds to output combinations” hold true?

Accepted Answer

Introduction / Context:Truth tables are core to digital logic because they map input combinations to the resulting output. Misstating the direction of dependency causes conceptual errors in design and debugging.Given Data / Assumptions:Truth tables enumerate all input combinations for a function.Each row lists inputs first and the corresponding output after.Concept / Approach:The output is a function of the inputs. A valid truth table shows output for each input pattern; it does not show inputs responding to output states.Step-by-Step Solution:1) Define the gate (e.g., AND, OR, NAND, NOR).2) List all input combinations in rows.3) Compute the output for each row by the gate’s rule.4) Conclude that the statement reversing input/output dependency is incorrect.Verification / Alternative check:Check any standard logic text: outputs are determined by inputs, not vice versa.Why Other Options Are Wrong:“Correct” contradicts the definition; limiting the claim to NAND or equal inputs has no basis.Common Pitfalls:Thinking of logic in reverse causality; assuming symmetry where none exists.Final Answer:Incorrect

Question 19

Evaluate the statement about a NOR gate: “The output of a NOR gate is LOW if any of its inputs is LOW.” Is this statement valid for NOR logic?

Accepted Answer

Introduction / Context:NOR is the negation of OR. Understanding when NOR outputs 0 versus 1 is vital when designing active-low logic and combined functions.Given Data / Assumptions:NOR = NOT(OR).Inputs may be 0 or 1.Concept / Approach:OR outputs 1 when any input is 1. Consequently, NOR outputs 0 when any input is 1, and NOR outputs 1 only when all inputs are 0. The given statement wrongly references any input being LOW (0).Step-by-Step Solution:1) If any input = 1, OR = 1, so NOR = 0.2) If all inputs = 0, OR = 0, so NOR = 1.3) Therefore, the claim about “any input LOW” causing LOW output is incorrect.Verification / Alternative check:Compare two-input truth tables for OR and NOR to confirm the complement relationship.Why Other Options Are Wrong:“Correct” contradicts NOR behavior; restrictions to input count or equality are irrelevant to NOR’s definition.Common Pitfalls:Confusing NOR with NAND; mixing up the OR condition (any 1) with the wrong trigger (any 0).Final Answer:Incorrect

Question 20

TTL troubleshooting scenario: An inverter (NOT gate) drives one input of a 2-input AND gate. If the inverter’s output lead is open (disconnected due to a fault) and logic pulses are applied at the other AND input (point B), what will appear at the A…

Accepted Answer

Introduction / Context:Floating nodes behave differently in TTL than in CMOS. In classic TTL, an unused or open TTL input tends to read as logic HIGH due to internal bias currents. This behavior influences fault outcomes when a driving output lead is broken.Given Data / Assumptions:Standard two-input TTL AND gate.One AND input is driven by an inverter whose output lead is open (disconnected).The other AND input (point B) carries logic pulses.No external pull resistors changing the default bias.Concept / Approach:With the inverter’s output lead open, the AND input it should drive is effectively floating. In TTL, a floating input biases HIGH. Therefore, the AND gate effectively sees one input fixed at HIGH, and the output follows the other input.Step-by-Step Solution:1) Open output leaves the receiving AND input floating.2) Floating TTL input defaults to HIGH.3) AND truth: X * 1 = X.4) With point B pulsing, the AND output reproduces the pulses.Verification / Alternative check:Consult standard TTL input characteristics showing that an unconnected input is read as HIGH, unlike CMOS which is high-impedance and undefined without biasing.Why Other Options Are Wrong:“a steady LOW/HIGH” ignore that the pulsing input propagates when the other input is effectively HIGH; “an undefined level” is more typical of CMOS or analog float, not classic TTL inputs.Common Pitfalls:Assuming CMOS-like undefined behavior; overlooking that the failure is the driver lead, not the AND gate pin itself.Final Answer:pulses

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