Question 1

Diagnosing a shorted input on a shared node of a load gate What observable indication best characterizes a short on the input of a load (receiving) gate connected to a shared signal node?

Accepted Answer

Introduction / Context:In multi-fanout digital nets, several load (receiving) gate inputs share one driving source. If one load input becomes shorted to ground (or otherwise abnormally low impedance), the entire node can be clamped, preventing valid logic transitions and propagating failure across otherwise healthy loads. Recognizing this signature enables quick isolation of the faulty part.Given Data / Assumptions:A single signal node fans out to multiple gate inputs.One load input is shorted (e.g., to ground inside the IC).Technology: TTL/CMOS with standard logic thresholds.Concept / Approach:A shorted input presents a very low resistance to ground (or to Vcc, depending on the fault). The shared node voltage is pulled hard toward that rail. As a result, the driver cannot change the node level effectively; all other loads on the node lose their signal drive. With a ground short, the node appears permanently LOW, which then forces every downstream stage to interpret a constant 0 level.Step-by-Step Solution:Observe symptom: no transitions seen on the shared node despite a healthy driver.Check level: node remains at a solid LOW (or HIGH for a Vcc short) regardless of input stimuli.Conclude impact: every gate connected to that node receives a stuck logic level → signal loss.Isolate: disconnect suspected loads one at a time to find which input releases the node to normal operation.Verification / Alternative check:Use an ohmmeter (with power removed) to measure from node to ground/Vcc; an abnormally low resistance indicates a short. In-circuit, monitor current draw or use a thermal camera to spot the culprit IC.Why Other Options Are Wrong:Only the output of the defective gate is affected: false; the whole node is compromised.Signal loss only, without stating stuck level: incomplete; the dominant symptom is also a stuck LOW for a ground short.Stuck LOW only, without noting global signal loss: incomplete; it misses the network-wide consequence on all connected loads.Common Pitfalls:Assuming the driver is weak or faulty; forgetting that any single shorted load can overpower the driver and clamp the net; overlooking that a Vcc short would clamp HIGH instead.Final Answer:There is a signal loss to all gates on the node, and the affected node will be stuck in the LOW state.

Question 2

Logic probe diagnostics — dim output and no response to input pulsing A logic probe on a gate’s output shows a dim indicator. A pulser applied to each input terminal does not change the output indication. What is the most probable fault?

Accepted Answer

Introduction / Context:Logic probes commonly display three states: LOW, HIGH, and mid/indeterminate (often a dim or flicker). When an output node is floating (open circuit), the probe may show a weak, unstable or dim indication. If input stimulation fails to move the output, the evidence points away from valid gate operation and toward a physical discontinuity at the output pin or inside the device.Given Data / Assumptions:Probe shows a dim (indeterminate/mid-level) reading on the gate output.Pulsing each input (with a pulser) does not affect the output indication.Assume the probe and pulser are working and ground reference is sound.Concept / Approach:An open output results when the output driver is not electrically connected to the node (broken bond, lifted pin, cracked trace) or when an internal transistor pair has failed open. With no solid path to Vcc or ground, the node floats and couples small leakage or capacitive signals, yielding a dim or ambiguous probe reading. Stimulating inputs will not change a truly open output node.Step-by-Step Solution:Recognize symptom pattern: dim output + no response → suspect non-driven node.Eliminate alternatives: if Vcc were low, many nodes would misread, not just one output.Confirm hypothesis: gently press the IC, reflow, or continuity-check between pin and trace; substitute the IC if socketed.Optionally add a pull-up/pull-down resistor temporarily; if the level goes strongly HIGH/LOW, the output was indeed floating.Verification / Alternative check:Use an ohmmeter (power off) to check continuity from the output pin to the connected net; inspect under magnification for cracked solder joints. Compare with a known-good identical gate on the board.Why Other Options Are Wrong:Supply voltage low: would likely affect multiple points and be diagnosed at the power pins, not just one output.Bad probe ground: would corrupt all readings, including known references; also typically gives erratic, not consistently dim, behavior.Tristate device: a high-impedance output can float, but input pulsing would toggle the enable if the device were functioning; the stated behavior fits a fault (open) rather than a valid high-Z condition.Common Pitfalls:Confusing dim indication with a valid mid-level analog signal; overlooking cracked solder joints or lifted pads; failing to verify probe/calibration on a known-good logic level.Final Answer:The output of the gate appears to be open.

Question 3

Decoder sizing for full output decoding How many 1-of-16 decoders are required to decode a full 7-bit binary input (128 unique outputs), assuming you may use an enable scheme to cascade devices?

Accepted Answer

Introduction / Context:Address decoders translate binary input lines into one-hot outputs. A 1-of-16 decoder asserts exactly one of sixteen outputs based on 4 binary inputs (since 2^4 = 16). To decode N input bits fully into 2^N unique outputs, multiple decoder devices are cascaded using enable inputs. This question examines how to cover 7 input bits (128 outputs) using only 1-of-16 parts.Given Data / Assumptions:Each 1-of-16 decoder handles 4 select inputs and has an enable input.Total inputs to decode: 7 bits → 128 distinct outputs.Cascading via enables is allowed (higher-order bits select which device is active).Concept / Approach:Partition the 7 bits into higher-order bits that choose which decoder is enabled and lower-order bits that feed the 4 select lines of each decoder. With 3 high-order bits you can choose among 2^3 = 8 devices; with 4 low-order bits you select one of 16 output lines on the enabled device. This yields 8 * 16 = 128 unique outputs.Step-by-Step Solution:Choose bit split: use the 3 MSBs as device enables and the 4 LSBs as decoder select inputs.Number of devices required = 2^3 = 8 (one per MSB combination).Each device's 4 select inputs are driven in parallel by the same 4 LSBs.Enable logic activates exactly one 1-of-16 at a time, producing the correct one-hot among its 16 outputs.Verification / Alternative check:Count total outputs: 8 devices × 16 outputs/device = 128 outputs, exactly matching 2^7 possibilities. Timing and fanout requirements are satisfied because only one device is active at a time.Why Other Options Are Wrong:5, 6, 7 devices: insufficient, yielding at most 80, 96, or 112 outputs, respectively.More than 8 devices are unnecessary; 8 is minimal with 1-of-16 parts because 2^7 / 16 = 8.Common Pitfalls:Attempting to cascade in series on the 4-bit inputs (which does not increase total outputs), or forgetting to reserve higher-order bits for device enables. Also, mixing up the roles of MSBs and LSBs can lead to duplicated or missing address ranges.Final Answer:8

Question 4

Digital fault diagnosis — output short to ground at a driving gate What is the most likely indication observed in a digital system when the output node of a driving logic gate is shorted directly to ground (0 V), considering its fan-out to multiple …

Accepted Answer

Introduction / Context:A common hardware fault in digital circuits is a “hard short” from a gate’s output node to ground. Because this node often fans out to several inputs, understanding the observable symptoms helps technicians localize and fix the failure quickly. This question tests practical fault-analysis intuition for TTL/CMOS outputs driving multiple loads.Given Data / Assumptions:The driving gate's output pin is shorted to 0 V.Downstream devices receive this node as their input (fan-out > 1).Standard logic thresholds apply; no special wired-OR bus is intended unless explicitly designed.No series isolation resistors or current-limiting devices are assumed at the node.Concept / Approach:A solid short to ground forces the node to a persistent LOW level regardless of the gate's intended state. Every load input tied to that node now “sees” a LOW. Thus the functional symptom is loss of the intended signal at all of those loads. Side effects can include excessive current draw, heating of the damaged device, and potential logic contention if the system had multiple drivers (which is bad practice outside of wired-logic designs).Step-by-Step Solution:Identify fault: output node is physically tied to 0 V.Evaluate logic level: node is forced LOW at all times (stuck-at-0).Propagate effect: every load connected to this node receives LOW, so the original waveform is lost across the fan-out.System symptom: “signal loss to all load gates.”Verification / Alternative check:Measuring the node with an oscilloscope shows a flat line near 0 V regardless of upstream toggling. Disconnecting the short (or isolating the driver) immediately restores the waveform at the node, confirming the diagnosis.Why Other Options Are Wrong:Only defective gate affected: false; all loads on that net are affected.Node stuck HIGH: a short to ground forces LOW, not HIGH.Stuck either HIGH or LOW: a defined short to ground yields LOW specifically.Only nearest load fails: the net is common to all loads, so all are impacted.Common Pitfalls:Confusing “short to ground” with “short to Vcc”.Overlooking multiple loads on a single net; failures propagate to every connected input.Final Answer:There is a signal loss to all load gates.

Question 5

Parity bit calculation — append an even parity bit to a 6-bit word Given the data word 110010, determine the 7-bit result after adding an even parity bit (total number of 1s must be even).

Accepted Answer

Introduction / Context:Parity is a lightweight error-detection method used in memories, serial links, and legacy buses. With even parity, the parity bit is chosen so that the total count of 1s in the transmitted word (data + parity) is even. This question checks your ability to compute and append an even parity bit correctly.Given Data / Assumptions:Data word: 110010 (6 bits).Even parity requirement: total number of 1s across all 7 bits must be even.Parity bit is appended as an extra bit; its exact position (MSB vs LSB) is implementation-defined, but options indicate it is prefixed to create a 7-bit result.Concept / Approach:Count the 1s in the data and add 1 if the count is odd (to make it even). If the count is already even, add 0. Here we prepend the computed parity bit (consistent with the presented choices).Step-by-Step Solution:Count ones in 110010: 1 + 1 + 0 + 0 + 1 + 0 = 3 (odd).Even parity requires an additional 1 to make the total count even.Place parity bit as the new MSB: 1 110010 ⇒ 1110010.Verification / Alternative check:Recount 1s in 1110010: there are 4 ones, which is even. If the system appended the bit at the LSB instead, the 7-bit sequence would be 1100101; however, the given options reflect the MSB placement, so 1110010 is the correct choice here.Why Other Options Are Wrong:1111001, 001101, 0110010: each yields the wrong total parity or scrambles the original bit order.110010: contains no parity bit and still has an odd count of ones.Common Pitfalls:Losing track of where the parity bit is appended; match the convention implied by the answer choices.Confusing even and odd parity definitions.Final Answer:1110010

Question 6

PLD design methodology — naming the building-block approach In programmable logic device (PLD) development, what is the design concept called where complex systems are constructed from reusable circuit “blocks” arranged in levels?

Accepted Answer

Introduction / Context:When targeting PLDs, CPLDs, or FPGAs, engineers routinely assemble larger functions from smaller, well-tested modules. This modular thinking supports clarity, reuse, and verification at multiple abstraction layers, and it is central to scalable hardware development flows.Given Data / Assumptions:Goal: name the design paradigm built around reusable blocks.Context: PLD projects, but the concept applies broadly in digital design.Blocks can be RTL modules, IP cores, or schematic macros.Concept / Approach:Hierarchical design organizes a system into layers: leaf modules implement primitive functions; mid-level modules compose these into subsystems; a top-level integrates all subsystems. This hierarchy enables focused simulation, easier timing closure, and team-based parallel development, while also easing incremental synthesis and floorplanning in larger devices.Step-by-Step Solution (Reasoning):Identify key features: modular blocks, reuse, and layered composition.Map to terminology: these features define hierarchical design.Rule out distractors that describe languages or vague categories rather than a methodology.Verification / Alternative check:Industry practice: vendor reference designs and IP catalogs are organized hierarchically (e.g., UART block inside a communication subsystem inside a SoC). Synthesis and static timing tools exploit hierarchical boundaries to manage complexity.Why Other Options Are Wrong:Architectural design: addresses system-level choices, not the specific building-block method.Digital design: overly broad; not a methodology name.Verilog: a hardware description language; not a design paradigm.Behavioral synthesis: a tool technique, not the general building-block concept.Common Pitfalls:Confusing language (Verilog/VHDL) with methodology (hierarchy, top-down/bottom-up).Final Answer:hierarchical design.

Question 7

Magnitude comparator vs selector: Does a magnitude comparator output the highest or lowest numeric value of its inputs based on control signals, or is that the role of a different device?

Accepted Answer

Introduction / Context:Combinational building blocks have specialized roles. A magnitude comparator takes two binary numbers and asserts flags indicating A > B, A = B, or A Given Data / Assumptions:Inputs are two binary words A and B.Magnitude comparator outputs are relational flags only.Selectors (MUX/min–max units) use flags or control lines to route values.Concept / Approach:A magnitude comparator does not drive out a numeric operand. It asserts logic-high on one or more of its relational outputs: A>B, A=B, AB is true, select A, else select B. Therefore, the statement that a comparator “outputs the highest or lowest value depending on control signals” incorrectly attributes selector behavior to the comparator itself.Step-by-Step Solution:Identify primitive: comparator ⇒ relational flags only.Identify selection function: MUX/min–max ⇒ routes operand to output.Map the claim: it describes a selector, not a comparator.Therefore, the statement is incorrect.Verification / Alternative check:Review standard ICs: 74x85 outputs A>B, A=B, AWhy Other Options Are Wrong:“Correct” contradicts device definitions. Bit-width or two’s-complement representation do not change the role separation; tri-state buses are unrelated to comparator core behavior.Common Pitfalls:Assuming any block that “compares” will also select; in modular design, comparison and selection are distinct steps wired together by control logic.Final Answer:Incorrect

Question 8

Decoder outputs vs inputs: For a binary decoder, does having 2 select inputs imply only 2 outputs for the decoded value, or should it produce 2^n outputs?

Accepted Answer

Introduction / Context:Decoders are one-hot generators: they convert an n-bit binary input into one of 2^n active outputs. They are used for address decoding, memory selection, and instruction demultiplexing. The stem claims a 2-input decoder has only 2 outputs; we will verify the general n-to-2^n rule and correct this misconception.Given Data / Assumptions:Ideal binary decoder with n input (select) lines.Exactly one output is asserted for each input code when enabled.Active-high outputs assumed (active-low variants behave analogously).Concept / Approach:The defining property is exponential fan-out: an n-to-2^n decoder. For n = 2, outputs = 2^2 = 4. Typical parts: 2-to-4, 3-to-8, 4-to-16 decoders. Therefore, the claim of “two outputs” is incorrect. Confusing decoders with demultiplexers or simple selectors can cause wiring mistakes in address maps and control logic.Step-by-Step Solution:Use the rule: outputs = 2^n.Substitute n = 2 ⇒ outputs = 4.Hence, a 2-input decoder has four outputs (00, 01, 10, 11 one-hot).Therefore, the statement is incorrect.Verification / Alternative check:Look up a standard 74HC139/138 family: for n = 3, 3-to-8 outputs; for n = 2, 2-to-4 outputs. Datasheets reaffirm the 2^n relation.Why Other Options Are Wrong:Enable polarity, Gray coding, or encoder terminology do not change the fundamental output count. “Correct” contradicts the definition.Common Pitfalls:Mixing up decoder (n→2^n) with demultiplexer (routes one input to 1 of 2^n outputs but still has 2^n outputs). Ensure device roles and counts are clear before schematic capture.Final Answer:Incorrect

Question 9

Multiplexer function clarity: Does a multiplexer (MUX) choose which output to send an input to, or does it choose which input is forwarded to a single output?

Accepted Answer

Introduction / Context:Multiplexers and demultiplexers are duals but not identical. A multiplexer selects exactly one of many inputs and forwards it to a single output based on select lines. A demultiplexer takes a single input and routes it to one of many outputs. The stem incorrectly describes a demultiplexer as if it were a multiplexer. Clarifying this prevents common wiring errors in datapaths and I/O selection networks.Given Data / Assumptions:MUX: many-to-one (N inputs, 1 output).DEMUX: one-to-many (1 input, N outputs).Select lines determine which path is active.Concept / Approach:A k-select-line MUX chooses one of 2^k inputs to appear at a single output. Conversely, a k-select-line DEMUX activates one of 2^k outputs to receive the single input. Hence the statement “chooses which output to send an input to” describes a DEMUX, not a MUX. Therefore, as a definition of a multiplexer, the statement is incorrect.Step-by-Step Solution:Recall device roles: MUX = many→one; DEMUX = one→many.Map the stem: “choose which output an input goes to” ⇒ DEMUX.Therefore, the stem does not define a MUX correctly.Conclude: the statement is incorrect for a multiplexer.Verification / Alternative check:Typical parts: 74HC157 (2:1 quad MUX) selects between inputs for a single output per channel. 74HC138 (3:8 decoder/DEMUX) routes one enable/input to one of eight outputs—distinct functions.Why Other Options Are Wrong:Limiting to 2:1 or invoking tri-state buses does not change the core definition. “True for demultiplexers” acknowledges the dual but confirms the statement is not correct for MUX.Common Pitfalls:Using the wrong symbol in schematics; confusing enable pins with select lines; forgetting that MUX outputs are singular per channel.Final Answer:Incorrect

Question 10

Error-detection capability of single-parity schemes: Can simple parity checking detect only an odd number of bit errors in a codeword (and miss even numbers of bit flips)?

Accepted Answer

Introduction / Context:Parity bits are the simplest error-detection mechanism used in memory systems, serial links, and legacy protocols. A single parity bit forces the total number of 1s in a codeword to be either even (even parity) or odd (odd parity). Understanding exactly what error patterns parity detects—and which it cannot—guides the selection of stronger codes when needed, such as checksums, CRCs, or Hamming codes.Given Data / Assumptions:Single parity bit appended to a data word.Even or odd parity convention is known to both transmitter and receiver.Independent bit-flip error model (no burst-correction mechanisms assumed).Concept / Approach:Let w be the number of 1s in the transmitted data. The parity bit is set so that w + p has a fixed parity (even or odd). If an odd number of bits flip during transmission, the parity of the received word toggles, producing a mismatch flag. If an even number of bits flip, parity returns to the original parity class and the simple parity check fails to detect the error. Thus, parity detects any odd-weight error pattern (1, 3, 5, … flips) and misses all even-weight error patterns (2, 4, 6, … flips). The minimum Hamming distance of a single-parity code is 2, enabling detection of all single-bit errors but not correction and not detection of double-bit errors.Step-by-Step Solution:Define parity rule: choose p so that total ones count has fixed parity.Consider one-bit error: parity flips ⇒ detected.Consider two-bit errors: parity flips twice ⇒ original parity restored ⇒ undetected.Generalize to any odd vs even number of flips.Verification / Alternative check:Compute Hamming distance: with one parity bit, any valid codeword differs from any other by at least 2, confirming detect-1 capability and failure to detect all double-bit errors.Why Other Options Are Wrong:Odd vs even parity choice does not affect the odd-error detection rule; both detect odd numbers of flips. Hamming distance > 2 requires more redundancy than single parity. Applicability is general to binary data, not restricted to ASCII.Common Pitfalls:Assuming parity can correct errors (it cannot); assuming it detects double-bit errors (it does not); mixing parity with CRC behavior, which has different detection guarantees.Final Answer:Correct

Question 11

Digital design — encoder fundamentals: in combinational logic, is it accurate that an encoder circuit is specifically designed to generate a coded output (e.g., binary or BCD) corresponding to which single input line is active?

Accepted Answer

Introduction / Context:An encoder is a standard combinational logic block used in digital systems to compress multiple input lines into a smaller number of output bits. Typical examples include 8-to-3 binary encoders and keyboard encoders. This question checks the core definition: does an encoder generate a specific code on its outputs based on which input line is asserted?Given Data / Assumptions:One-hot or priority-encoded inputs are considered.Exactly one valid input is active at a time unless a priority encoder is used.Outputs are binary-coded representations (binary/BCD etc.).Concept / Approach:An N-to-m encoder maps N input lines to m outputs, where m = ceil(log2 N). The function is the inverse of a decoder: instead of expanding a coded input into one-of-N outputs, an encoder compresses one-of-N inputs into a code. Priority encoders add logic to resolve multiple simultaneous inputs according to a defined priority, ensuring a valid coded result.Step-by-Step Solution:1) Identify the definition: encoder = device that converts active input line number into a code.2) Note the coding scheme (e.g., 8 inputs → 3-bit binary output).3) Recognize enhancements (priority, enable, valid) that keep the core behavior the same.4) Conclude the statement is an accurate summary of encoder purpose.Verification / Alternative check:Truth tables for standard 4-to-2 or 8-to-3 encoders show unique coded outputs for each asserted input line. Priority encoder datasheets confirm the same with added priority/valid flags.Why Other Options Are Wrong:Does not apply: contradicts the device definition. Only for parity encoders: parity generation is a different function (error checking), not line-to-code encoding. Applies only when all inputs are active: encoders assume one active input; all-active yields invalid unless priority or masking is used.Common Pitfalls:Confusing encoders with decoders or multiplexers; ignoring the need for priority when simultaneous inputs can occur.Final Answer:Applies

Question 12

Decoder realization with minimal gates: can a single AND gate plus two inverters act as a functional “basic decoder,” i.e., uniquely select one-of-N outputs from coded inputs?

Accepted Answer

Introduction / Context:Decoders expand a coded input (e.g., 2 bits) into one-of-N outputs (e.g., 4 lines). This item probes whether an extremely small gate set—one AND gate and two inverters—can truly perform the generic decoder function, which requires producing uniquely asserted output lines for each input code.Given Data / Assumptions:“Basic decoder” refers to standard one-of-N selection (e.g., 2-to-4, 3-to-8).Inputs are the coded lines; outputs are mutually exclusive selects.No ROM or PLA structures are assumed.Concept / Approach:A k-to-2^k decoder uses product terms (minterms) of the inputs and their complements. For k = 2, four AND terms are required, each AND-ing either the true or complemented inputs. Two inverters provide complements, but a single AND gate cannot generate all required minterms. Thus, the proposed gate count is insufficient for one-of-N decoding.Step-by-Step Solution:1) For 2-to-4 decode: outputs D0 = ~A * ~B, D1 = ~A * B, D2 = A * ~B, D3 = A * B.2) Count required gates: four 2-input AND gates plus two inverters.3) Given only one AND gate and two inverters, you can realize at most one minterm.4) Therefore the gate set cannot implement a full decoder.Verification / Alternative check:Truth table inspection shows that a single AND gate yields only one of the minterms; additional AND gates are needed to cover all code combinations.Why Other Options Are Wrong:Applies: overclaims capability. Applies only with enable: an enable simply gates all outputs; it does not create missing minterms. Applies only for one-to-two decoding: even 1-to-2 decoding typically uses a buffer and an inverter (or two ANDs with enable), not a lone AND plus two inverters to realize “decoder” semantics.Common Pitfalls:Confusing “some decoding” (one minterm) with a full decoder; underestimating the number of product terms needed.Final Answer:Does not apply

Question 13

Multiplexer function check: is a multiplexer (MUX) a device that generates a fixed sequence of binary states on each clock pulse, regardless of its data inputs?

Accepted Answer

Introduction / Context:A multiplexer is often called a “data selector.” It forwards one of several input lines to a single output based on select inputs. This question tests whether a MUX inherently produces a fixed output sequence on each clock pulse, which would actually describe a counter or sequence generator, not a MUX.Given Data / Assumptions:Standard n-to-1 MUX with data inputs D0…Dn−1 and select lines S.No internal clocked memory in a simple combinational MUX.Any clocking pertains to external logic driving the select lines, not the MUX core.Concept / Approach:A MUX is purely combinational: Output Y = Dk where k is the index set by S. If a clock drives the select lines through a counter, the observed output may change in a pattern—but the MUX itself is not the sequence generator. The “fixed sequence on each clock” description fits counters or shift registers, not multiplexers.Step-by-Step Solution:1) Define MUX behavior: selects one input to pass to output.2) Recognize sequence generation requires state (flip-flops, counters).3) Conclude the statement confuses the roles of selector vs sequencer.4) Therefore, it does not apply to a MUX by itself.Verification / Alternative check:Datasheets for basic MUX ICs (e.g., 74HC151) list propagation delay and truth table only—no clock, no internal state.Why Other Options Are Wrong:Applies: misidentifies the device function. “Only if select lines are clocked”: even then, the pattern comes from the external counter, not from the MUX.Common Pitfalls:Assuming “clocked” equals “sequential”; forgetting a MUX is combinational.Final Answer:Does not apply

Question 14

Decoder fundamentals: does the general function of a decoder involve asserting one or more specific outputs when its input lines match a particular digital state (code)?

Accepted Answer

Introduction / Context:Decoders are the counterpart to encoders: they expand a compact code into one-of-N (or sometimes multiple) outputs. They are used in address decoding, display driving, and memory chip selection. The question checks the canonical role of decoders in activating outputs when an input code is recognized.Given Data / Assumptions:K input lines represent a binary code.Up to 2^K distinct outputs are possible (or grouped outputs).Active-high or active-low conventions may apply.Concept / Approach:A decoder generates minterms of the inputs and asserts the output(s) corresponding to the input pattern. Examples include the 3-to-8 decoder (one of eight outputs goes active) and 4-to-16 address decoders. Some variants include enable inputs and multiple active outputs for special coding schemes, but the essence remains detection and activation based on input states.Step-by-Step Solution:1) Model input code as a K-bit vector.2) Implement minterms with AND gates and input complements.3) Assert the matching output line(s) for the given code.4) Optional: gate the entire decoder with an enable pin.Verification / Alternative check:Standard truth tables (e.g., 74HC138) show exactly one active output for each 3-bit input when enabled, confirming the definition.Why Other Options Are Wrong:Does not apply: contradicts core functionality. Active-low only / all-ones-only: output polarity and specific codes are implementation details; the function is general detection and activation.Common Pitfalls:Confusing decoders with demultiplexers; overlooking enable and polarity conventions.Final Answer:Applies

Question 15

Hexadecimal decoder input width: does a hexadecimal (1-of-16) decoder select one of sixteen outputs based on an 8-bit binary input?

Accepted Answer

Introduction / Context:Hexadecimal decoders are common in address decoding and display drivers. A “hex” quantity spans sixteen distinct symbols (0–F), which directly maps to 4 bits of binary information. The question probes whether such a decoder requires 8-bit inputs.Given Data / Assumptions:Hexadecimal implies 16 distinct codes.Binary representation uses 4 bits for 16 states.One-of-16 output selection is assumed.Concept / Approach:Because 2^4 = 16, only 4 input bits are needed to uniquely select one of 16 outputs. An 8-bit input would encode 256 states, vastly exceeding the 16 outputs of a basic hexadecimal decoder. Some devices may include enables or polarity options, but the main data input width for hex selection is 4 bits.Step-by-Step Solution:1) Compute required width: m outputs need log2(m) input bits.2) For m = 16, log2(16) = 4, so four inputs are sufficient.3) Therefore, 8-bit input is not required for a 1-of-16 decoder.4) Conclude the statement is incorrect.Verification / Alternative check:Datasheets for 4-to-16 decoders (e.g., 74HC154) specify four address inputs A–D and sixteen outputs.Why Other Options Are Wrong:Applies: misstates the necessary input width. BCD and extra enables do not change that 4 bits encode 16 states; enables gate functionality but are not primary address bits.Common Pitfalls:Confusing “hexadecimal display decoder” (which may include segment mapping) with the address-width requirement; conflating control lines with data width.Final Answer:Does not apply

Question 16

MUX select-line count: for a sixteen-input (16-to-1) multiplexer, how many data-select control lines are required to address all inputs?

Accepted Answer

Introduction / Context:Select-line count is a fundamental sizing rule for multiplexers. A 16-to-1 MUX forwards one of sixteen inputs to the output, so the number of address lines must be sufficient to encode sixteen distinct choices. This question reinforces the log2 relationship used throughout digital design.Given Data / Assumptions:The device is a single 16-to-1 multiplexer.Binary addressing is used for the select lines.Optional enable pins do not replace address bits.Concept / Approach:With M inputs, the number of select lines s must satisfy 2^s ≥ M. For M = 16, 2^4 = 16, so s = 4. Fewer than four lines cannot uniquely address all inputs; more than four would be redundant for basic selection, though sometimes extra lines exist for hierarchical selection or banking.Step-by-Step Solution:1) Set 2^s = 16 → s = 4.2) Confirm that codes 0000 through 1111 map to inputs D0…D15.3) Recognize that enables are orthogonal to select-line count.4) Conclude four select lines are required.Verification / Alternative check:Commercial 16:1 MUXes (e.g., 74HC4067 analog MUX) use four address lines S0–S3 to select channels 0–15.Why Other Options Are Wrong:Two or three lines: 2^2 = 4 and 2^3 = 8, insufficient. Five lines: 2^5 = 32, more than needed for 16 inputs.Common Pitfalls:Counting enable pins as address lines; mixing up multiplexer and decoder sizing.Final Answer:Four select lines

Question 17

Priority encoder definition check: A priority encoder asserts an output code for the highest-priority active input only. Evaluate the claim that it “encodes the highest and lowest value inputs simultaneously.”

Accepted Answer

Introduction / Context: Encoders map one-of-N inputs to a binary-encoded output. When multiple inputs might be active, a priority encoder resolves conflicts by selecting the highest-priority input according to a predefined order. Understanding this behavior is essential for interrupt controllers and bus arbitration.Given Data / Assumptions:Priority encoders produce a code for one active input at a time.They typically include a valid output and enable signals.“Highest and lowest simultaneously” contradicts single-winner resolution.Concept / Approach: A priority encoder masks lower-priority inputs when any higher-priority input is active. Its truth table ensures only the top-ranked active line determines the output code, never a combination of highest and lowest simultaneously.Step-by-Step Solution:Assume inputs In7..In0 with In7 highest priority.If In7=1 and any lower Ink=1, output encodes 7, not both 7 and k.Therefore, “highest and lowest simultaneously” is not how a priority encoder operates.Verification / Alternative check:Inspect a 74148 8-to-3 priority encoder datasheet: the output corresponds to the highest-priority asserted input.Why Other Options Are Wrong:Correct / code variants: No mainstream encoder simultaneously reports highest and lowest values on one code output.Thermometer/Gray references: These are coding schemes, not priority mechanisms.Common Pitfalls:Confusing a plain encoder with a priority encoder.Expecting multi-winner outputs when the device is designed for single-winner arbitration.Final Answer:Incorrect

Question 18

Carry-generation techniques: The look-ahead carry (carry lookahead) method is designed to reduce, not incur, long ripple propagation delays. Judge the accuracy of the claim that it “suffers from propagation delays.”

Accepted Answer

Introduction / Context: Ripple-carry adders suffer from cumulative carry propagation delay, which scales with bit-width. Carry lookahead (CLA) logic addresses this by computing carry signals using generate/propagate terms in parallel, drastically shortening the critical path compared to ripple structures.Given Data / Assumptions:Ripple-carry: carry-out of each stage waits on the previous stage.CLA: uses G (generate) and P (propagate) to precompute carries.All digital logic still has finite gate delays; the question targets the concept of “suffering from” ripple-like delays.Concept / Approach: CLA replaces serial carry dependency with parallel equations such as c1 = G0 + P0c0, c2 = G1 + P1G0 + P1P0c0, etc. While there is nonzero combinational delay in CLA networks, they specifically mitigate the linear propagation characteristic that plagues ripple adders.Step-by-Step Solution:Define Gk = Ak*Bk and Pk = Ak + Bk (for a bit k).Compute carries using sums of products rather than waiting stage-by-stage.Observe reduced depth vs. ripple; hierarchical CLAs further minimize fan-in/fan-out issues.Hence the claim that CLA “suffers from propagation delays” (as ripple does) is inaccurate.Verification / Alternative check:Compare timing: ripple delay ~ O(n); CLA delay ~ O(log n) or bounded by gate levels with hierarchy.Why Other Options Are Wrong:Correct / Only true for 1-bit / low Vcc: None reflect the fundamental timing advantage of CLA over ripple.Common Pitfalls:Equating “nonzero delay” with “suffers from propagation delay” in the ripple sense.Ignoring fan-in limitations and practical buffering, which are engineering details but do not negate CLA’s purpose.Final Answer:Incorrect

Question 19

Form classification check: Consider the Boolean expression (A + B) * (C + D’). Decide whether classifying it as a product-of-sums (POS) expression is correct.

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Introduction / Context: Boolean expressions are commonly written in canonical or standard forms. Two key forms are sum-of-products (SOP) and product-of-sums (POS). Being able to recognize these patterns is essential for mapping to AOI/OAI structures and for minimization with K-maps or algebraic methods.Given Data / Assumptions:Expression under review: (A + B) * (C + D’).Operator conventions: + denotes OR, * denotes AND, ’ denotes NOT.Each parenthesized term is a sum (OR) of literals.Concept / Approach: A product-of-sums (POS) is an AND of sum terms. Each sum term is an OR of literals. In this case, (A + B) is a sum term, and (C + D’) is another sum term; their AND constitutes a POS expression. In contrast, SOP is an OR of product terms such as AB + CD’.Step-by-Step Solution:Identify sums: S1 = (A + B), S2 = (C + D’).Combine by product: S1 * S2 — an AND of sums.Definition alignment: AND of sums ⇒ POS.Therefore, the classification as POS is correct.Verification / Alternative check:Distribute to SOP equivalence if needed: (A + B)(C + D’) = AC + AD’ + BC + BD’, confirming it is an AND of sum terms before distribution.Why Other Options Are Wrong:Incorrect / SOP / Neither: They contradict the structural definition of POS as an AND of sum terms.Common Pitfalls:Confusing operator precedence or notation conventions.Assuming complements invalidate form classification; complements within sums are allowed.Final Answer:Correct

Question 20

Karnaugh map adjacency rule: In a K-map, the cell in the top row is considered adjacent to the corresponding cell directly below it in the bottom row due to wraparound. Assess this statement.

Accepted Answer

Introduction / Context: Karnaugh maps (K-maps) visualize adjacency in Boolean spaces, enabling grouping of 1-cells to simplify expressions. The map wraps around at the edges so that cells differing in only one variable are considered neighbors even across boundaries.Given Data / Assumptions:K-map uses Gray-coded ordering along rows and columns.Edge adjacency follows toroidal wraparound (left-right and top-bottom).Adjacency implies one-bit Hamming distance between cells.Concept / Approach: Because the variable ordering is Gray coded, the first and last rows (and columns) differ by only one variable for corresponding columns (or rows). Therefore, the top-row cell and the bottom-row cell directly beneath/above it are adjacent, allowing valid grouping across the border.Step-by-Step Solution:Consider a 4-variable K-map with rows AB and columns CD in Gray code.Row AB = 00 (top) and AB = 01, 11, 10 (middle), and AB = 00 wraps to AB = 10 (bottom) with one-bit difference.Thus, top/bottom cells align as neighbors for grouping 2, 4, 8, etc. cells.This enables larger implicants and greater simplification.Verification / Alternative check:Draw a 3- or 4-variable K-map and confirm that opposite edges correspond to single-bit changes; edges are conceptually “touching.”Why Other Options Are Wrong:Incorrect: Ignores the Gray code and toroidal adjacency.Only true for 2-variable / don’t-care cells: Adjacency rules are general; don’t-cares can be used, but they are not required for edge adjacency.Common Pitfalls:Treating the K-map as a flat grid without wraparound, which prevents optimal grouping.Misaligning Gray code order, breaking single-bit adjacency.Final Answer:Correct

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