Question 1

Applications of shift registers in digital systems: which uses are valid for a general-purpose shift register in data handling and timing?

Accepted Answer

Introduction / Context: Shift registers are chains of flip-flops clocked to move data one stage per tick. Their versatility makes them core building blocks in serializers, deserializers, and timing circuits for communication and signal processing. Given Data / Assumptions: Devices can accept parallel or serial inputs based on configuration. Outputs may be tapped at intermediate stages for timing purposes. Clocking moves bits predictably along the chain. Concept / Approach: Parallel-to-serial: load multiple bits at once, then shift them out bit by bit. Serial-to-parallel: accept a bit stream over successive clocks, then read out all bits together. Digital delay line: each stage introduces one clock period of delay, creating a precise multi-cycle delay. Step-by-Step Solution: Recognize the data movement property inherent to shift registers. Map movement to serialization/deserialization functions. Use the per-stage propagation as a controllable delay line. Verification / Alternative check: Textbook examples and datasheets (e.g., 74xx/40xx families) routinely demonstrate all three uses, confirming the inclusive option. Why Other Options Are Wrong: Each of A, B, and C alone is correct but incomplete; the question asks for all valid uses. “None” is invalid because multiple valid applications exist. Common Pitfalls: Ignoring load and shift control signals; misaligning clock domains when using as delay lines can cause metastability. Final Answer: all of the above

Question 2

Convert the binary value 100011 (base 2) into its octal (base 8) representation. What is the correct octal result?

Accepted Answer

Introduction / Context: Binary-to-octal conversion is efficient because 1 octal digit maps exactly to 3 binary bits. Group the binary number into triples starting from the right (least significant bit), then translate each group to an octal digit. Given Data / Assumptions: Binary value: 100011₂. We will group bits in sets of three from the right. No fractional parts; integer conversion only. Concept / Approach: Write 100011₂ as grouped triples: 100 011. Convert each triple to octal: 100₂ = 4₈ and 011₂ = 3₈. Concatenate digits to obtain the octal number 43₈. Step-by-Step Solution: Group bits: 100011 → 100 011. Translate: 100₂ → 4; 011₂ → 3. Combine: result = 43₈. Verification / Alternative check: Decimal cross-check: 100011₂ = 32 + 2 + 1 = 35; 43₈ = 4*8 + 3 = 32 + 3 = 35. Both paths yield 35, so the conversion is correct. Why Other Options Are Wrong: 47₈ = 39₁₀; 49₈ is not valid (octal digits 0–7 only); 50₈ = 40₁₀. “None” is invalid because 43₈ is correct. Common Pitfalls: Failing to pad leftmost bits to a full group of three; using invalid octal digits (8 or 9). Final Answer: 43 (base 8)

Question 3

Hexadecimal conversion: what is the hexadecimal equivalent of the binary number 10101111 (base 2)?

Accepted Answer

Introduction / Context: Binary-to-hex conversion is convenient because each hex digit corresponds to 4 binary bits. This mapping avoids lengthy decimal conversions and is widely used in debugging, memory dumps, and instruction encoding. Given Data / Assumptions: Binary value: 10101111₂. We will group the bits in nibbles (4-bit groups). No sign or fractional parts are involved. Concept / Approach: Split into nibbles: 1010 1111. Translate: 1010₂ = A₁₆ and 1111₂ = F₁₆. Thus, the hexadecimal representation is AF₁₆. Step-by-Step Solution: Group: 1010 1111. Convert: 1010 → A; 1111 → F. Concatenate: AF. Verification / Alternative check: Decimal check: 10101111₂ = 175₁₀; AF₁₆ = 10*16 + 15 = 175, confirming equivalence. Why Other Options Are Wrong: 9E and 8C decode to different binary values (10011110₂ and 10001100₂). “All of the above” cannot be correct because hex representations are unique. Common Pitfalls: Misgrouping bits from the left; forgetting that 1111 equals F, not E. Final Answer: AF

Question 4

TTL families comparison: among the listed transistor–transistor logic (TTL) subfamilies, which is generally the fastest due to reduced storage delay?

Accepted Answer

Introduction / Context: TTL has multiple subfamilies optimized for speed, power, or noise immunity. Speed improvements often come from device physics and circuit tricks that shorten transistor saturation/storage delays. Schottky TTL integrates Schottky diodes to prevent deep saturation, achieving faster switching. Given Data / Assumptions: We compare general subfamilies: standard, high-speed, Schottky, and low-speed TTL. Process variations aside, Schottky clamping reduces storage time. We ignore later advanced variants (e.g., ALS/AS/FAST) not listed in the options. Concept / Approach: Schottky diodes across base–collector junctions keep transistors out of deep saturation, minimizing charge storage that must be removed during turn-off. This directly lowers propagation delay compared to standard/high-speed TTL without Schottky clamps. Step-by-Step Solution: Identify the subfamily using Schottky clamping. Relate clamping to reduced storage delay. Choose Schottky TTL as the fastest among listed options. Verification / Alternative check: Datasheets show lower typical propagation delays for S-TTL than standard or generic high-speed TTL families of the same era. Why Other Options Are Wrong: Standard/low-speed: designed for general use or power savings, not peak speed. High-speed (without Schottky) still trails Schottky-based variants in many classic lines. “None” is invalid because a known fastest choice is present. Common Pitfalls: Confusing Schottky TTL with CMOS families; CMOS speed depends on different factors and is not part of this comparison. Final Answer: schottky TTL

Question 5

Data routing primitives: which combinational logic device takes data from a single input source and routes it to one of several outputs (one-to-many) under control selection?

Accepted Answer

Introduction / Context: Multiplexers (MUX) and demultiplexers (DEMUX) are complementary. A MUX selects one of many inputs to a single output (many-to-one). A DEMUX does the reverse: from a single input, it drives one of many outputs (one-to-many) based on select lines. Given Data / Assumptions: Single data input, multiple output lines. Select lines determine which output carries the input data. Combinational (no internal memory). Concept / Approach: By definition, a demultiplexer fans out a single input to one of several outputs under digital control. This makes it ideal for time-division distribution, address decoding variants, and signal routing in communication buses. Step-by-Step Solution: Identify the direction: one input to many outputs. Match to device names: DEMUX matches one-to-many; MUX is many-to-one. Select demultiplexer. Verification / Alternative check: Truth tables for DEMUX devices (e.g., 1-to-4, 1-to-8) confirm one-hot output selection per select code with input replicated to the chosen output. Why Other Options Are Wrong: Decoder: converts binary codes to one-hot outputs but does not pass data through. Encoder: many inputs to fewer outputs. Multiplexer: many-to-one, the opposite direction. “None”: invalid because DEMUX fits exactly. Common Pitfalls: Confusing decoders and demultiplexers; while related, a decoder does not forward the data stream, it only asserts selection lines. Final Answer: demultiplexer

Question 6

Digital IC drive capability: how is the fan-out of a logic gate or building block defined in practical digital design?

Accepted Answer

Introduction / Context: Fan-out quantifies how many standard inputs a logic output can reliably drive without violating voltage levels or timing specifications. It is a key parameter for designing robust digital systems and avoiding signal integrity issues. Given Data / Assumptions: Standardized input load per gate family (e.g., TTL, CMOS). Output must meet logic-high/low thresholds under load. Capacitive loading and DC current levels affect fan-out. Concept / Approach: For TTL, fan-out historically references the ratio of available output current to required input current. For CMOS, static current is tiny, so fan-out is largely limited by capacitive loading and timing (rise/fall times). In both cases, the definition remains: how many inputs can one output reliably drive. Step-by-Step Solution: Identify output drive limits (current/voltage and timing). Divide by standard input load to estimate maximum count. State the definition accordingly. Verification / Alternative check: Datasheets list “fan-out” explicitly as the maximum number of standard inputs permitted per output for a given family and supply condition. Why Other Options Are Wrong: B: Refers to thermal management, not fan-out. C: Describes fan-in, not fan-out. D: Power dissipation is related but not the definition of fan-out. Common Pitfalls: Ignoring dynamic effects (capacitance) in CMOS; assuming static definitions apply identically across families without timing considerations. Final Answer: the number of inputs that one output can transmit to

Question 7

Convert the binary value 1100001 (base 2) to hexadecimal (base 16). What is the correct hex representation?

Accepted Answer

Introduction / Context: Binary-to-hex conversion proceeds by grouping bits into 4-bit nibbles from the right. Each nibble maps directly to a single hexadecimal digit (0–F), making the conversion both fast and exact. Given Data / Assumptions: Binary value: 1100001₂. We will left-pad with zeros if needed to complete a nibble. No sign or fractional bits are involved. Concept / Approach: Group from the right into nibbles: 1 1000 01 → pad to 0110 0001. Convert each nibble: 0110₂ = 6₁₆ and 0001₂ = 1₁₆. Thus, 1100001₂ = 0x61. Step-by-Step Solution: Pad and group: 1100001 → 0110 0001. Nibble conversions: 0110 → 6; 0001 → 1. Combine: 61₁₆. Verification / Alternative check: Decimal check: 1100001₂ = 97₁₀; 0x61 = 6*16 + 1 = 97; both match. Why Other Options Are Wrong: 57₁₆ = 87₁₀; 51₁₆ = 81₁₀; 43₁₆ = 67₁₀; all differ from 97₁₀. “None” is invalid because 61₁₆ is correct. Common Pitfalls: Forgetting to pad the leftmost group to 4 bits; misreading 0110 as 5 instead of 6. Final Answer: 61 (base 16)

Question 8

Digital logic fundamentals: How many output signals can a single logic gate (e.g., AND, OR, NOT, NAND, NOR, XOR) normally provide in a standard digital circuit design?

Accepted Answer

Introduction / Context: A logic gate is the basic building block of digital electronics. In textbooks, datasheets, and schematic symbols, a gate performs a Boolean operation on one or more inputs and presents the result as an output. Understanding how many outputs a single gate provides helps distinguish simple gates from multi-output devices such as decoders or demultiplexers and prevents design confusion when reading logic diagrams. Given Data / Assumptions: The term “logic gate” refers to elementary gates (AND, OR, NOT, NAND, NOR, XOR, XNOR) as singular functional blocks. We are considering the output count of one gate instance, not the number of gates inside an IC package. We are not discussing derived or complex combinational modules like encoders or decoders. Concept / Approach: By definition, a gate computes a single Boolean function value at a time and presents that value on one output node. While a physical integrated circuit may contain several such gates (e.g., a “quad 2-input NAND” has four independent NAND gates), each individual gate symbol in a schematic has a single output pin. Multi-output behaviors are characteristic of higher-level combinational circuits (e.g., a 3-to-8 decoder has eight outputs), not simple gates. Step-by-Step Solution: Identify the scope: one gate equals one Boolean function instance.Recall standard symbols: NOT, AND, OR, NAND, NOR, XOR—all with one output pin.Conclude that a single gate normally supplies one output signal. Verification / Alternative check: Examine standard TTL/CMOS datasheets (e.g., 74HC08 quad AND); each subgate has exactly one output even though the package contains multiple gates. Schematic libraries and EDA tools also model gates as single-output primitives. Why Other Options Are Wrong: “More than one” and “two only” describe multi-output modules (e.g., decoders), not single gates. “Both (a) and (b)” can’t be correct because they contradict the standard single-output definition. “None” fails because a correct choice exists. Common Pitfalls: Confusing an IC package that houses multiple gates with one gate that has multiple outputs; conflating demultiplexers/decoders (multi-output) with basic gates (single-output). Final Answer: one.

Question 9

Number systems in computing: The hexadecimal (base-16) number system is widely used when analyzing and programming which class of components?

Accepted Answer

Introduction / Context: Hexadecimal (base-16) provides a compact, human-friendly way to represent binary values. Because one hex digit maps exactly to four binary bits, programmers and hardware engineers rely on hex for addresses, instruction opcodes, and data inspection. This is especially common in low-level development and debugging for microprocessors and microcontrollers. Given Data / Assumptions: Hex digits span 0–9 and A–F, matching 4-bit nibbles. Microprocessors manipulate binary states; their memory addresses and registers are naturally represented in hex. The question asks which category most strongly aligns with hex use for analysis and programming. Concept / Approach: Microprocessor programming involves examining instruction streams, memory-mapped I/O, and register contents. Hexadecimal shortens long binary sequences (e.g., 0x3F7A) and aligns with instruction boundaries and byte groupings. While “registers” (option A) and “chips” (option B) are related, they are sub-elements within microprocessor systems. Hex is the de facto notation across assembler listings, linker maps, debuggers, and embedded monitors that target microprocessors. Step-by-Step Solution: Recognize that hex conveniently compresses 4-bit groups.Note that microprocessor addresses, opcodes, and data are commonly reported in hex by tools.Select “microprocessors” as the most appropriate, domain-specific answer. Verification / Alternative check: Examine any assembly listing or debugger (e.g., GDB, J-Link) to see addresses and opcodes displayed in hex by default. Datasheets specify registers and instruction encodings in hex as well. Why Other Options Are Wrong: “Registers” and “chips” are generic terms; hex is indeed used there, but the question targets the typical context of programming and analysis—microprocessors. “Vacuum tube” is irrelevant to modern hex notation usage. Common Pitfalls: Assuming hex is used only for memory addresses; it also covers constants, masks, and bitfields throughout firmware code. Final Answer: microprocessors.

Question 10

Base conversion: Convert the binary number 101110₂ to its octal equivalent and select the correct result.

Accepted Answer

Introduction / Context: Converting between binary and octal is efficient because octal is a power-of-two base. Each octal digit corresponds to exactly three binary bits, enabling quick grouping without intermediate decimal steps. This is a staple skill in digital electronics and computer architecture courses. Given Data / Assumptions: Binary value: 101110₂. Goal: express it in octal (base-8). Method: group bits into 3-bit chunks from the right. Concept / Approach: Binary to octal mapping is: 000→0, 001→1, 010→2, 011→3, 100→4, 101→5, 110→6, 111→7. Group the binary number into triads, padding the leftmost group with leading zeros if necessary; then translate each triad directly to an octal digit. Step-by-Step Solution: Write 101110₂ and group from the right: 101 110.Convert: 101₂ → 5₈; 110₂ → 6₈.Concatenate octal digits: 56₈. Verification / Alternative check: Decimal cross-check: 101110₂ = 32 + 8 + 4 + 2 = 46₁₀. Octal 56₈ = 5*8 + 6 = 40 + 6 = 46₁₀. Values match, confirming the conversion. Why Other Options Are Wrong: 35₈ converts to 29₁₀; 46₈ equals 38₁₀; 50₈ equals 40₁₀—none equal 46₁₀. Common Pitfalls: Grouping from the left instead of the right or forgetting to pad leading zeros can yield incorrect digit boundaries. Always group 3 bits per octal digit starting at the LSB side. Final Answer: 56₈.

Question 11

Functional completeness in Boolean logic: Which of the following gate sets is NOT functionally complete for constructing arbitrary Boolean functions?

Accepted Answer

Introduction / Context: A set of logic primitives is called functionally complete if any Boolean function can be constructed using only gates from that set. Recognizing complete versus incomplete sets is critical for hardware minimization and theoretical foundations of digital logic design. Given Data / Assumptions: NAND alone is known to be functionally complete. NOR alone is also functionally complete. AND and OR together, without inversion, cannot generate complements. Concept / Approach: To synthesize arbitrary Boolean functions, the gate set must be able to express AND, OR, and NOT operations (directly or indirectly). NAND and NOR each can create NOT and, with combinations, produce both AND and OR. However, the pair AND + OR lacks intrinsic inversion; without NOT, it cannot create functions that require complementing variables. Step-by-Step Solution: List candidate sets and check whether NOT can be synthesized.Verify that NAND can build NOT, AND, and OR; similarly for NOR.Conclude that “AND, OR” is not functionally complete because it cannot produce NOT by itself. Verification / Alternative check: Standard proofs show NOT(A) = NAND(A, A) or NOT(A) = NOR(A, A). No such identity exists using only AND and OR without a complemented input available. Why Other Options Are Wrong: NAND and NOR each form complete bases. AND, OR, NOT is trivially complete by definition. Common Pitfalls: Assuming complemented literals are “given” externally; functional completeness is judged without external inversion sources. Final Answer: AND, OR.

Question 12

Binary adders in digital design: A binary adder is a combinational logic circuit that can add how many binary numbers at a time?

Accepted Answer

Introduction / Context: Binary adders implement arithmetic at the hardware level. The canonical adder (half adder/full adder and multi-bit versions) is designed to add two binary operands, producing a sum and a carry. Distinguishing how many numbers are added simultaneously clarifies adder use versus more complex arithmetic units like accumulators or multi-operand adders. Given Data / Assumptions: A half adder adds two 1-bit inputs (A and B) producing sum and carry. A full adder adds A, B, and a carry-in but the carry-in is not a third independent operand; it is a chain signal. Ripple-carry and carry-look-ahead adders extend the bit width but still add two operands. Concept / Approach: Standard datapaths allocate two input buses to the adder: Operand X and Operand Y. The carry chain enables multi-bit addition but does not equate to adding a third independent number at the same time. Summation of three or more operands is handled by cascaded adders, carry-save adders, or tree structures (e.g., Wallace tree), not by a single basic adder that inherently accepts more than two numbers. Step-by-Step Solution: Identify the basic full adder inputs: A, B, and carry-in.Recognize that carry-in represents prior stage carry, not an independent operand.Conclude that a binary adder adds two numbers at a time. Verification / Alternative check: Examine ALU block diagrams: add instruction routes two register sources to the adder. Multi-operand sums use additional cycles or specialized adders. Why Other Options Are Wrong: “Hundreds/Thousands” are clearly incorrect; “One” misunderstands how addition requires at least two operands; “None” is invalid since a correct option exists. Common Pitfalls: Misinterpreting the carry-in as a third operand; confusing multi-cycle accumulation with single-cycle addition. Final Answer: Two.

Question 13

Computing terminology: Names, numbers, measurements, and other raw facts that a program uses or processes to solve a problem are collectively called _____

Accepted Answer

Introduction / Context: Solving problems on a computer requires three elemental components: data (the raw facts), instructions (the steps to process those facts), and the program (a structured collection of instructions). Distinguishing these terms is essential in computer science, databases, and software engineering. Given Data / Assumptions: The query asks for the correct term describing names, numbers, and facts that are processed. We are not being asked about the algorithms or control logic, only the raw inputs and stored facts. Standard computing definitions apply. Concept / Approach: Data represents raw facts, measurements, symbols, or values—such as temperatures, IDs, or text—before they are interpreted. An instruction is a directive telling the CPU what to do with the data (e.g., add, compare, move). A program is an ordered set of instructions that implements an algorithm. Controls typically refer to signals, UI elements, or control flow, not the raw facts themselves. Step-by-Step Solution: Isolate the definition sought: “names, numbers, and other facts.”Match to standard terminology: these are data.Select “data.” Verification / Alternative check: Any programming example shows variables holding data values; instructions manipulate those values. Database schemas store data; queries process it. Why Other Options Are Wrong: Program = set of instructions; instruction = single operation; controls ≠ raw facts. Common Pitfalls: Confusing data with information; information is data interpreted within context, while the question targets the raw inputs. Final Answer: data.

Question 14

Why hexadecimal is popular: The main advantage of hexadecimal numbers is the ease of conversion from hexadecimal to _____ and vice versa.

Accepted Answer

Introduction / Context: Hexadecimal (base-16) and binary (base-2) are tightly related because 16 is a power of 2. This relationship allows clean grouping of bits into nibbles (4-bit chunks), making conversions direct and error-resistant compared to decimal conversions. Given Data / Assumptions: A hex digit equals exactly 4 binary bits. We seek the system that offers the easiest two-way conversion with hex. Standard computing notation and grouping rules apply. Concept / Approach: Each hex digit maps to 0000–1111 in binary. Converting hex to binary is just replacing each hex digit by its 4-bit equivalent; converting binary to hex is grouping binary digits into sets of four (padding with leading zeros when needed) and translating each group to a hex digit. Decimal and BCD require arithmetic; ASCII/“ASCE” is a character encoding, not a numeric base. Step-by-Step Solution: Recognize that hex ↔ binary is a 1-to-1 mapping per nibble.Note that no division or multiplication is required; only table lookup or mental mapping is needed.Therefore, select “binary.” Verification / Alternative check: Example: hex 0x3A = binary 0011 1010. Reverse by regroupping bits into nibbles → 3 and A. Why Other Options Are Wrong: Decimal conversion involves arithmetic and is not as direct. “ASCE” (likely ASCII) is unrelated to numeric base conversion. BCD is a storage encoding of decimal digits, not a natural power-of-two base. Common Pitfalls: Forgetting to pad the leftmost binary group to 4 bits can cause digit misalignment. Final Answer: binary.

Question 15

Combinational logic identification: Which combinational circuit generates a specific binary word (code) as its output corresponding to an active input line?

Accepted Answer

Introduction / Context: Encoders and decoders are dual combinational circuits. An encoder accepts one-of-N active inputs and outputs the corresponding binary code (word). A decoder performs the inverse, taking a binary code and asserting one of several outputs. Understanding this distinction is critical in address decoding, keyboard scanning, and interrupt handling. Given Data / Assumptions: We seek the circuit that outputs a binary word representing which input is active. Only one input is assumed active at a time in a basic encoder. Standard device roles: MUX selects inputs to a single line; DEMUX routes one input to many outputs. Concept / Approach: An n-to-2^n decoder takes an n-bit code and activates exactly one of 2^n outputs. Conversely, a 2^n-to-n encoder maps an asserted input line to its n-bit index code. Priority encoders extend this to handle multiple simultaneous requests by outputting the highest-priority index. Neither multiplexer nor demultiplexer inherently “generates” a binary word from a one-hot input set. Step-by-Step Solution: Identify the required behavior: produce a binary word from an active input line.Match that behavior to the encoder definition.Select “encoder.” Verification / Alternative check: Datasheets for 74HC147 (10-to-4 priority encoder) illustrate input lines mapped to 4-bit outputs. Why Other Options Are Wrong: Decoder generates one-of-N outputs given a code. Multiplexer selects among data inputs, not a code generator. Demultiplexer routes one input to multiple outputs. Common Pitfalls: Confusing the direction of data flow between encoders and decoders due to similar naming. Final Answer: encoder.

Question 16

Logic level conventions: In a positive-logic system, which electrical level represents logic 1 (HIGH)?

Accepted Answer

Introduction / Context: Digital logic conventions define how electrical levels map to logical values. The most common convention is positive logic, where a higher voltage corresponds to logic 1 (HIGH) and a lower voltage corresponds to logic 0 (LOW). Clarity on this convention is vital when interfacing components and interpreting datasheets. Given Data / Assumptions: Assume a positive-logic system (e.g., 0 V as LOW, VCC as HIGH). Voltage thresholds (VIH/VIL) define valid logic level ranges. Negative logic is the inverse mapping and is not assumed here. Concept / Approach: In positive logic, logic 1 corresponds to a higher voltage near the supply rail (e.g., 3.3 V or 5 V, subject to VIH), and logic 0 corresponds to a lower voltage near ground (below VIL). Designers select logic families with compatible thresholds to guarantee noise margins and reliable communication between devices. Step-by-Step Solution: Identify system type: positive logic → 1 is HIGH.Translate to electrical terms: HIGH = higher voltage level within specified VIH–VOH constraints.Choose “higher voltage level.” Verification / Alternative check: CMOS/TTL datasheets list VIH and VIL that align with this mapping; logic analyzers default to showing HIGH as a higher voltage trace. Why Other Options Are Wrong: Zero or lower voltage level indicate LOW in positive logic. Negative voltage is unrelated unless using special bipolar families or negative supplies with different conventions. Common Pitfalls: Mixing positive/negative logic conventions or forgetting that thresholds, not ideal rails, determine valid logic recognition. Final Answer: higher voltage level.

Question 17

Multiplexing concept: A multiplexer (MUX) is also commonly known as a _____ because it selects one of many inputs to pass to a single output line.

Accepted Answer

Introduction / Context: Multiplexers are standard combinational circuits that route one of several data inputs to a single output, controlled by select lines. Because their primary job is to choose among inputs, they are often called “data selectors.” This device is central to datapaths, bus sharing, and signal routing. Given Data / Assumptions: Multiplexer inputs: multiple data lines (e.g., D0..Dn-1). Control inputs: select lines that choose which data appears at the output. Output: a single line carrying the selected input. Concept / Approach: The alternate name “data selector” emphasizes selection rather than routing. Select lines encode the index of the chosen input: for N inputs, log2(N) select bits are required. By cascading multiplexers, designers build wide selectors, arithmetic units, or implement conditional moves in hardware. Step-by-Step Solution: Identify the function: choose one input among many.Relate the function to the common alias: “data selector.”Select “data selector.” Verification / Alternative check: Datasheets (e.g., 74HC157, 74HC151) explicitly describe multiplexers as data selectors, and many schematic symbols label them as such. Why Other Options Are Wrong: Coder/decoder refer to encoders/decoders, not multiplexers. Multivibrator is an oscillator/bistable/monostable circuit category, unrelated to input selection. Common Pitfalls: Confusing multiplexers (many-to-one) with demultiplexers (one-to-many) due to similar names. Final Answer: data selector.

Question 18

Cyclic codes in digital communication: In which application domain are cyclic codes (e.g., CRC) primarily used?

Accepted Answer

Introduction / Context: Cyclic codes, especially cyclic redundancy checks (CRCs), are a class of linear block codes widely used to detect errors in digital communication and storage. They operate on blocks of bits using polynomial division over a finite field, appending a check sequence that receivers can verify to detect corruption. Given Data / Assumptions: The question asks for the primary, practical use of cyclic codes. We consider mainstream engineering practice in networks, serial links, and storage devices. Options include unrelated domains like analog (continuously varying) signals and CPU arithmetic/logic internals. Concept / Approach: Cyclic codes are optimized for error detection in bit- or byte-oriented data streams. They are computationally efficient, implementable in hardware shift-register logic, and provide strong burst-error detection. While computation and analog representation are important fields, they are not where cyclic codes are primarily applied. Step-by-Step Solution: Recognize that cyclic codes compute a remainder (check value) over a data block.Transmit data + check; receiver recomputes and compares for integrity.This mechanism is integral to data links, packets, and storage sectors.Therefore, the correct application domain is data transfer (and storage). Verification / Alternative check: Examples include Ethernet CRC-32, USB CRC-5/16, CAN CRC-15, and disk sector CRCs—each used to detect transmission or media errors, confirming the domain is data transfer/storage integrity rather than analog signal representation or arithmetic logic operations. Why Other Options Are Wrong: Continuously varying signal representation is an analog task, not a coding-theory application. Arithmetic/logic computation occurs inside processors and ALUs; while error-detecting codes exist in some CPUs, cyclic codes are not primarily used for general ALU operations. Common Pitfalls: Confusing error detection (CRC) with error correction (e.g., Hamming, Reed–Solomon). CRCs detect with high probability but do not correct by themselves. Final Answer: data transfer

Question 19

Binary to hexadecimal conversion: What is the hexadecimal equivalent of the binary number 110000₂?

Accepted Answer

Introduction / Context: Hexadecimal (base 16) is convenient for expressing binary values compactly because 16 = 2^4, so each hex digit maps to exactly four binary bits. Converting binary to hex is a matter of grouping bits into 4-bit chunks and translating each group. Given Data / Assumptions: Binary input: 110000₂. Target representation: hexadecimal (base 16). No change in numeric value—only the base changes. Concept / Approach: Group the binary number into 4-bit nibbles from the right. If the most significant group has fewer than 4 bits, pad with leading zeros. Then use the binary-to-hex table: 0000→0, 0001→1, …, 1111→F. Step-by-Step Solution: Write 110000₂ as 0011 0000₂ to form two 4-bit groups.Translate 0011₂ → 3₁₆.Translate 0000₂ → 0₁₆.Combine: 0x30 → written as 3016. Verification / Alternative check: Decimal cross-check: 110000₂ = 48₁₀; 48 / 16 = 3 remainder 0, so 0x30 (3016). Both methods agree. Why Other Options Are Wrong: 4016, 4316, and 2716 correspond to different binary values. Only 3016 matches 110000₂ exactly. Common Pitfalls: Forgetting to pad the most significant bits to multiples of four; misreading 0011 as 4 instead of 3. Final Answer: 3016

Question 20

Logic synthesis using only NAND gates: How many two-input NAND gates are required to implement a two-input OR function?

Accepted Answer

Introduction / Context: NAND is a universal gate, meaning any Boolean function can be built using only NAND gates. A classic exercise is to realize OR using only NANDs via De Morgan's law and input inversion with NAND-as-inverter constructs. Given Data / Assumptions: Available gate: 2-input NAND only. Target: 2-input OR, i.e., Y = A OR B. Allowed technique: use NAND gates as inverters by tying inputs together. Concept / Approach: By De Morgan's law, A OR B = NOT( NOT(A) AND NOT(B) ). With NANDs: NOT(x) = x NAND x. Therefore, implement NOT(A) with one NAND, NOT(B) with one NAND, and then NAND their results to realize the outer NOT of AND. Step-by-Step Solution: Form ~A using NAND: N1 = A NAND A.Form ~B using NAND: N2 = B NAND B.Compute OR: Y = N1 NAND N2 = (~A) NAND (~B) = A OR B.Total NAND gates used: 3. Verification / Alternative check: Build the truth table for A,B ∈ {0,1} and confirm that Y equals A OR B. The construction matches for all four combinations. Why Other Options Are Wrong: 1 or 2 NANDs cannot both invert two inputs and implement the final combination. 4 is possible with non-minimal designs, but 3 is the standard minimal count. Common Pitfalls: Forgetting you can invert with a NAND by shorting inputs; misapplying De Morgan's law signs. Final Answer: 3

Digital Computer Electronics Questions