Question 1

Adder basics in digital electronics: A full adder takes _____ input bits and produces a SUM and a _____ output.

Accepted Answer

Introduction / Context: The full adder is a fundamental combinational block used to construct multi-bit adders. Unlike the half adder, which adds two bits, the full adder adds two operand bits and a carry-in, producing a sum and a carry-out. This behavior allows chaining many stages to create ripple-carry or more advanced adders. Given Data / Assumptions: Inputs: A, B (operand bits), and Cin (carry-in) → total 3 input bits. Outputs: SUM and Cout (carry-out). Standard Boolean implementation is assumed. Concept / Approach: Equations for a full adder are: SUM = A ⊕ B ⊕ Cin; CARRY = (A * B) + (Cin * (A ⊕ B)). This allows a per-bit computation that propagates carry to the next significant bit. Step-by-Step Solution: Recognize the distinction: half adder (2 inputs) vs. full adder (3 inputs including Cin).Recall outputs: SUM and CARRY.Therefore, select “3, CARRY.” Verification / Alternative check: Standard logic diagrams show the carry chain connecting Cout of bit i to Cin of bit i+1. Why Other Options Are Wrong: (a) and (c) are nonsensical pairings; (d) cannot be true because the others contradict full-adder behavior. Common Pitfalls: Misinterpreting carry-in as a separate operand; it is a chain input, but it makes the total input bit count three. Final Answer: 3, CARRY.

Question 2

Number system conversion — decimal to octal (with fraction): Find the octal equivalent of the decimal number 324.987 (rounded to at least three fractional octal digits).

Accepted Answer

Introduction / Context: Converting decimal numbers to octal involves separate handling of the integer and fractional parts. The integer part is repeatedly divided by 8, while the fractional part is repeatedly multiplied by 8. Octal is base-8, so each step cleanly extracts digits without resorting to binary intermediates. Given Data / Assumptions: Decimal value: 324.987. We need a correct octal equivalent to at least three fractional digits. Rounding or truncation is acceptable after three fractional octal digits, as typical in MCQs. Concept / Approach: For the integer part: use repeated division by 8 to obtain digits from least significant to most. For the fractional part: multiply by 8, record the integer part as the next octal digit, and iterate on the fractional remainder. Step-by-Step Solution: Integer part 324: 324 / 8 = 40 remainder 4 → LSB digit 4.40 / 8 = 5 remainder 0 → next digit 0.5 / 8 = 0 remainder 5 → MSB digit 5. Integer octal = 504.Fractional part 0.987: 0.987 * 8 = 7.896 → digit 7; remainder 0.896.0.896 * 8 = 7.168 → digit 7; remainder 0.168.0.168 * 8 = 1.344 → digit 1; remainder 0.344 (further digits would be 2, 6, 0, …).Combine: 324.987 ≈ 504.771 in octal (to three fractional octal digits). Verification / Alternative check: Back-convert: 504.771₈ ≈ 58^2 + 08 + 4 + 7/8 + 7/8^2 + 1/8^3 = 320 + 0 + 4 + 0.875 + 0.109375 + 0.001953125 ≈ 324.9863, closely matching 324.987 with small truncation error. Why Other Options Are Wrong: 640.781₈ and 815.234₈ decode to very different decimal values; 90.987 mixes bases incorrectly; they do not match 324.987. Common Pitfalls: Reversing division/multiplication steps, or rounding too early and losing accuracy. Keep integer and fractional conversions separate. Final Answer: 504.771.

Question 3

Octal to binary conversion: Convert the octal number 60₈ into binary and choose the correct binary representation.

Accepted Answer

Introduction / Context: Bases 8 and 2 are closely related because 8 = 2^3. Therefore, each octal digit maps to exactly three binary bits. This makes octal ↔ binary conversion especially quick and reliable without passing through decimal. Given Data / Assumptions: Octal input: 60₈. Binary output is required. We will map each octal digit to a 3-bit binary group. Concept / Approach: Use the mapping: 0 → 000, 1 → 001, 2 → 010, 3 → 011, 4 → 100, 5 → 101, 6 → 110, 7 → 111. Step-by-Step Solution: Write octal digits: 6 and 0.Map 6 → 110 and 0 → 000.Concatenate groups: 110000₂. Verification / Alternative check: Convert to decimal for a cross-check: 60₈ = 6*8 + 0 = 48; binary 110000₂ also equals 32 + 16 = 48. They match. Why Other Options Are Wrong: 10011₂ (19), 1100111₂ (103), and 1100110₂ (102) do not equal 48. Common Pitfalls: Forgetting to keep three bits per octal digit or dropping leading zeros for interior digits (though leading zeros on the far left can be omitted). Final Answer: 110000₂.

Question 4

Simplifying Boolean expressions: Besides Boolean algebra, another widely used visual method for simplifying sum-of-products (and product-of-sums) expressions is the _____ map.

Accepted Answer

Introduction / Context: The Karnaugh map (K-map) is a graphical tool that helps minimize Boolean expressions by grouping adjacent 1s (or 0s) to identify prime implicants. It provides intuition and a low-effort path to simplified SOP or POS forms, especially for 2–6 variables. Given Data / Assumptions: The question mentions sum-of-products; K-maps handle both SOP and POS simplifications. We assume Gray code arrangement on rows/columns to preserve adjacency. We consider basic digital logic curricula. Concept / Approach: By arranging truth-table outputs on a grid with Gray-coded indices, adjacent cells differ by one variable. Grouping 1s in powers of two (1, 2, 4, 8, …) yields simplified terms by eliminating changed variables in the group. Step-by-Step Solution: Map minterms onto a K-map grid.Form largest possible power-of-two groups (wrap-around allowed).Write simplified product terms for each group and sum them. Verification / Alternative check: Algebraic minimization (consensus theorem, absorption) can confirm the K-map result; both should match. Why Other Options Are Wrong: “De Morgan” is a law, not a map. “Standard” and “Schottky” are not simplification maps; Schottky refers to a device/family. Common Pitfalls: Missing wrap-around adjacency; making too many small groups instead of maximal ones; forgetting don’t-care utilization. Final Answer: Karnaugh.

Question 5

Binary arithmetic — multi-operand addition: Compute the sum of the five binary numbers 110101101₂, 111011101₂, 000111111₂, 100101101₂, and 111110111₂. Choose the correct final binary sum.

Accepted Answer

Introduction / Context: Adding multiple binary numbers follows the same column-wise procedure as decimal addition, with carries propagated when a column sum reaches 2 or more. Accuracy benefits from aligning bit widths and verifying with a decimal cross-check. Given Data / Assumptions: Numbers: 110101101₂, 111011101₂, 000111111₂, 100101101₂, 111110111₂. All are 9-bit values; pad if needed for alignment. We will compute the exact sum and validate against the options. Concept / Approach: Perform column addition from LSB to MSB, or convert to decimal to confirm. Carry out careful arithmetic to avoid bit slips. The final binary result must match one of the answer choices to be correct; otherwise “None of the above” applies. Step-by-Step Solution: Convert each to decimal for a reliable check: 110101101₂=429, 111011101₂=477, 000111111₂=63, 100101101₂=301, 111110111₂=503.Add: 429 + 477 = 906.906 + 63 = 969.969 + 301 = 1270.1270 + 503 = 1773 (decimal total).Convert 1773 back to binary: 1773₁₀ = 11011101101₂. Verification / Alternative check: Column-wise binary addition yields the same result 11011101101₂. Comparing with provided options shows none match 11011101101₂, hence “None of the above.” Why Other Options Are Wrong: Each listed binary choice corresponds to decimal values far from 1773; for example, 1011010111010₂=5818, 1111011000010₂=7874, etc., so they cannot be correct. Common Pitfalls: Mistyping bits, forgetting carries, or misaligning numbers. Always double-check with a decimal round-trip when possible. Final Answer: None of the above.

Question 6

Octal to decimal conversion: Convert the octal number 63₈ to its decimal value.

Accepted Answer

Introduction / Context: Octal (base-8) to decimal (base-10) conversion is straightforward using positional weights of powers of 8. This is a common skill in digital systems where octal and hexadecimal provide compact representations of binary values. Given Data / Assumptions: Octal input: 63₈. We seek the exact decimal value. Digits are valid octal digits (0–7). Concept / Approach: For a two-digit octal ab₈, the decimal value is a8 + b. Apply positional weights: the left digit has weight 8^1 = 8; the right digit has weight 8^0 = 1. Step-by-Step Solution: Compute: 68 + 3 = 48 + 3 = 51.Therefore, 63₈ = 51₁₀. Verification / Alternative check: Binary route: 6 → 110, 3 → 011; 110011₂ = 51₁₀ as 32 + 16 + 3 = 51. Matches. Why Other Options Are Wrong: 61₁₀, 57₁₀, and 53₁₀ correspond to other octal inputs; they do not match 63₈. Common Pitfalls: Mixing octal with decimal arithmetic or misreading the base subscript. Final Answer: 51₁₀.

Question 7

Sequential storage element definition: A flip-flop is a _____ element that stores a single binary digit (0 or 1) as a stable low or high voltage level.

Accepted Answer

Introduction / Context: Flip-flops are the fundamental 1-bit storage elements in synchronous digital systems. They hold state information across clock cycles and form registers, counters, and state machines. Understanding that a flip-flop is a memory element clarifies its role versus purely combinational logic. Given Data / Assumptions: Flip-flops are edge-triggered or level-sensitive bistable circuits. They store one bit per element. Binary values are represented by distinct low/high voltage levels. Concept / Approach: A flip-flop has two stable states and preserves its output until changed by a control event (e.g., clock edge or set/reset). Collections of flip-flops create registers; arrays of registers form larger memories or pipeline stages. Step-by-Step Solution: Identify device type: bistable storage.Relate function: holds one binary digit reliably.Select descriptor: it is a “memory” element. Verification / Alternative check: Textbook block diagrams of CPUs and controllers label register banks (arrays of flip-flops) as storage blocks, confirming the memory role. Why Other Options Are Wrong: “Chip” is a packaging/implementation term; many chips contain logic beyond flip-flops. “Bus” is an interconnect. “I/O” refers to interface functions, not intrinsic storage. Common Pitfalls: Confusing latches with flip-flops; both store bits, but timing characteristics differ. Regardless, they are memory elements. Final Answer: memory.

Question 8

Binary to decimal conversion in digital electronics: What is the decimal (base 10) value of the binary number 1100011₂?

Accepted Answer

Introduction / Context: Converting binary numbers to decimal is a foundational skill in digital electronics and computer engineering. The question asks for the base-10 value of the binary number 1100011₂. Understanding place values for powers of 2 ensures accurate and quick conversions needed for debugging, data interpretation, and low-level programming. Given Data / Assumptions: Binary input: 1100011₂. Target: decimal (base 10) representation. No leading zeros to consider; the bit-width remains 7 bits as written. Concept / Approach: Each bit in a binary number represents a power of 2, starting from the rightmost bit as 2^0, then 2^1, 2^2, and so on. To convert, multiply each bit by its corresponding power of 2 and sum the results. Only positions with bit value 1 contribute to the sum; positions with 0 contribute nothing. Step-by-Step Solution: Write positional weights for 7 bits: 2^6, 2^5, 2^4, 2^3, 2^2, 2^1, 2^0 → 64, 32, 16, 8, 4, 2, 1.Map bits: 1 1 0 0 0 1 1 (from MSB to LSB).Multiply and add only 1-bits: 64 + 32 + 2 + 1.Compute sum: 64 + 32 = 96; 96 + 2 = 98; 98 + 1 = 99.Therefore, 1100011₂ = 99₁₀. Verification / Alternative check: Group into convenient chunks: (110)_2 = 6 and (0011)_2 = 3 does not directly apply because the bit groups are unequal; however, a calculator or quick mental addition of powers (64 + 32 + 2 + 1) reconfirms the answer 99₁₀. Why Other Options Are Wrong: 97₁₀ or 93₁₀ omit one or more powers of 2 in the sum. 29₁₀ is far smaller than the MSB contribution (64) plus the next (32), so it cannot be correct for a 7-bit value starting with 11xxxxx. Common Pitfalls: Reading bits right-to-left incorrectly, forgetting a power (often 2^1 = 2), or adding a nonexistent 2^2 term. Always list the powers beneath the bits before summing. Final Answer: 99₁₀

Question 9

Decimal to hexadecimal conversion: Convert 99₁₀ to its hexadecimal (base 16) representation.

Accepted Answer

Introduction / Context: Hexadecimal is a compact way to express binary values, popular in computer architecture and debugging. Converting from decimal to hex trains you to think in powers of 16, which map neatly to 4-bit binary nibbles. Here we convert 99₁₀ to base 16. Given Data / Assumptions: Input: 99 in base 10. Output: base 16 (hexadecimal) notation. Digits in base 16 are 0–9 and A–F (10–15). Concept / Approach: Use repeated division by 16: the quotient is the high-order hex digit, and the remainder is the low-order hex digit. Since 16 = 2^4, each hex digit corresponds to four binary bits, making results easy to cross-check in binary if desired. Step-by-Step Solution: Divide: 99 ÷ 16 = 6 remainder 3.Map remainder 3 → hex digit 3; quotient 6 → hex digit 6.Assemble digits high-to-low: 6 then 3 → 63₁₆. Verification / Alternative check: Back-convert: 63₁₆ = 6*16 + 3 = 96 + 3 = 99, confirming correctness. Binary check: 63₁₆ → 0110 0011₂ → 99₁₀ (64 + 32 + 2 + 1). Both methods agree. Why Other Options Are Wrong: 08₁₆ = 8₁₀; 61₁₆ = 97₁₀; 50₁₆ = 80₁₀. None equals 99₁₀. Common Pitfalls: Writing digits in reverse order (putting remainder first as MSB). Always place the quotient digit to the left of the remainder digit when forming the final hex result. Final Answer: 63₁₆

Question 10

Binary to hexadecimal conversion: Convert the binary number 1101100111111000₂ to hexadecimal.

Accepted Answer

Introduction / Context: Because 16 = 2^4, converting from binary to hexadecimal is performed by grouping binary digits in sets of four (nibbles). Each group maps directly to a single hex digit. This is essential in digital design, firmware work, and reading machine code dumps. Given Data / Assumptions: Binary input: 1101100111111000₂. Output: hexadecimal representation. No leading zeros required because the bit-length is already a multiple of 4 (16 bits). Concept / Approach: Split the 16-bit string into 4-bit groups from left to right, translate each 4-bit group to its hex digit using the mapping 0000→0 to 1111→F, then concatenate the digits in order. Step-by-Step Solution: Group: 1101 1001 1111 1000.Map: 1101→D, 1001→9, 1111→F, 1000→8.Combine: D 9 F 8 → D9F8₁₆. Verification / Alternative check: Decimal spot-check of the first nibble: 1101₂ = 13 → D; last nibble: 1000₂ = 8 → 8. Intermediate nibbles match 9 and F, confirming D9F8₁₆ is consistent throughout. Why Other Options Are Wrong: D8F8₁₆ incorrectly maps the second nibble; A8B9₁₆ and 7891₁₆ are unrelated conversions. Only D9F8₁₆ matches the nibble mapping of the given binary. Common Pitfalls: Grouping bits from the right and misaligning nibbles; forgetting to preserve the original bit order when translating; misreading 1111 as E (14) instead of F (15). Final Answer: D9F8₁₆

Question 11

Hexadecimal to binary conversion: Convert the hexadecimal number 63₁₆ to its binary representation.

Accepted Answer

Introduction / Context: Converting hexadecimal to binary is straightforward because each hex digit corresponds to exactly four binary bits. This mapping is used constantly when reading memory dumps, register values, and microcontroller configurations. Given Data / Assumptions: Hex input: 63₁₆ (two hex digits). Target: binary representation. No need for leading zeros beyond those implied by each nibble. Concept / Approach: Translate each hex digit independently to a 4-bit pattern: 6 → 0110 and 3 → 0011. Concatenate the two nibbles to form the final binary number. Drop only superfluous leading zeros from the leftmost nibble if desired for compactness. Step-by-Step Solution: 6₁₆ → 0110₂.3₁₆ → 0011₂.Concatenate: 0110 0011 → 1100011₂ (dropping the leading zero). Verification / Alternative check: Back-convert to hex: 1100011₂ grouped as 0110 0011 → 6 and 3 → 63₁₆. Decimal cross-check: 1100011₂ = 64 + 32 + 2 + 1 = 99; 63₁₆ = 6*16 + 3 = 99, confirming consistency. Why Other Options Are Wrong: 1100111₂ corresponds to 67₁₆, not 63₁₆. 111011₂ corresponds to 3B₁₆ (with padding). Only 1100011₂ matches 63₁₆ precisely. Common Pitfalls: Misreading 3 as 0011 and then accidentally writing 0111; losing track of nibble boundaries; omitting necessary zero padding inside a nibble. Final Answer: 1100011₂

Question 12

Integration scale categories in digital ICs: What is the typical functional capacity of SSI (Small-Scale Integration) devices?

Accepted Answer

Introduction / Context: Digital integrated circuits are historically categorized by the number of logic gates (or equivalent complexity) per chip: SSI, MSI, LSI, and VLSI. Knowing these ranges helps engineers understand trade-offs in technology choices, cost, density, and performance while reading datasheets or legacy documentation. Given Data / Assumptions: The question asks specifically about SSI (Small-Scale Integration). Gate counts are approximate historical ranges used for classification. We assume standard textbook definitions for SSI/MSI/LSI/VLSI. Concept / Approach: Typical categorization: SSI roughly 1–10 gates, MSI roughly 10–100 gates, LSI roughly 100–10,000 gates, and VLSI beyond that. Some sources include 1–11 or 12–99 to make ranges non-overlapping. The essence remains: SSI is at the low end, handling a handful of basic logic functions per chip (e.g., gates, small flip-flop packages). Step-by-Step Solution: Identify SSI as the smallest integration scale.Recall standard range: about 1–10 gates (here presented as 1–11).Match to the provided options: 1 to 11 gates aligns with SSI.Therefore select: 1 to 11 gates. Verification / Alternative check: Common examples of SSI include chips like 7400 (quad 2-input NAND) where each package provides a small number of simple gates, reinforcing the low gate-count classification. Why Other Options Are Wrong: 12 to 99 gates corresponds to MSI. 100 to 10,000 gates is LSI. More than 10,000 gates moves into VLSI and beyond, not SSI. Common Pitfalls: Confusing modern transistor counts (millions to billions) with historical gate-count categories; mixing up MSI and SSI because some series (e.g., 74xx) have both small and medium scale parts. Final Answer: 1 to 11 gates

Question 13

Flip-flop storage capability: How many bits of information can a single flip-flop store?

Accepted Answer

Introduction / Context: A flip-flop is a basic bistable storage element used throughout sequential logic, registers, counters, and memory structures. Understanding its capacity clarifies how larger storage elements (like registers of width N) are built from multiple flip-flops working in parallel. Given Data / Assumptions: The device under discussion is a standard bistable flip-flop (e.g., SR, JK, D, T types). No special encoding or compression schemes are considered. We refer to fundamental, hardware-level storage. Concept / Approach: A flip-flop has two stable states representing logical 0 and logical 1. It can therefore store exactly one binary digit (one bit). Wider storage elements are built by combining multiple flip-flops—e.g., an 8-bit register contains 8 flip-flops clocked together to capture a byte of data. Step-by-Step Solution: Recognize that a flip-flop's output Q can be either 0 or 1 at any time.Two stable states → capacity is one bit.Therefore, a single flip-flop stores one bit of data. Verification / Alternative check: Practical designs: N-bit counters or registers are consistently implemented using N flip-flops, one per bit position, confirming the one-bit-per-flip-flop rule. Why Other Options Are Wrong: Two or three bits require 4 or 8 stable states, which a single flip-flop does not provide. “Any number of bits” is impossible for a single bistable element. Common Pitfalls: Confusing a latch/flip-flop with a memory cell array or register file that aggregates many flip-flops; conflating a flip-flop's control inputs (set, reset, clock) with stored information width. Final Answer: one bit of data

Question 14

Shift-register timing: For an input pulse train with clock period T, what total delay is produced by an n-stage shift register?

Accepted Answer

Introduction / Context: Shift registers move data one stage per clock, creating predictable time delays in serial data paths, serializers/deserializers, and digital timing circuits. Knowing the relationship between the number of stages and total delay is essential for timing closure and interface design. Given Data / Assumptions: Clock period: T (seconds per clock). Register length: n stages (each stage is typically a flip-flop). Data advances one stage per active clock edge. Concept / Approach: Each stage adds one clock-period delay because a bit placed at the input requires one clock to move into the first stage, another to move to the second, and so on. After n clocks, the bit emerges at the output of the nth stage, yielding a total delay proportional to the number of stages. Step-by-Step Solution: At time 0, inject the first input bit.After 1 clock (T), the bit occupies stage 1.After 2 clocks (2T), it occupies stage 2.…After n clocks (nT), it appears at the output of stage n. Verification / Alternative check: Simulate or sketch a timing diagram for a 3-stage register: the bit emerges after 3 clock cycles → total delay = 3T, confirming the general nT formula. Why Other Options Are Wrong: (n − 1)T and (n + 1)T underestimate or overestimate the progression by one cycle; 2nT doubles the actual time without basis. Common Pitfalls: Confusing pipeline latency with throughput; although latency is nT, once filled, a new bit exits each clock cycle. Final Answer: nT

Question 15

Positive vs negative logic equivalence: Which positive-logic function is equivalent to the OR function when interpreted in negative logic?

Accepted Answer

Introduction / Context: Logic families sometimes use negative logic, where logical 1 is represented by a low voltage and logical 0 by a high voltage. Understanding how operations translate between positive and negative logic is crucial for interpreting mixed-signal documentation and designing with active-low signals. Given Data / Assumptions: Positive logic: 1 = high, 0 = low. Negative logic: 1 = low, 0 = high (signal meanings are inverted). We seek the positive-logic function equivalent to OR under negative-logic interpretation. Concept / Approach: Using De Morgan’s transformations with signal complementation: an OR operation on active-low (negative logic) signals corresponds to an AND operation on their positive-logic complements. Intuitively, when low means asserted, combining two asserted-lows with OR (in the active-low sense) becomes an AND of their true (active-high) intents. Step-by-Step Solution: Let X̄ and Ȳ be active-low signals whose positive-logic meanings are X and Y.Negative-logic OR: X̄ OR Ȳ (low means true) corresponds to NOT( (NOT X) AND (NOT Y) ) in positive logic.Equivalently, the underlying positive-logic operation combining the true conditions is X AND Y.Thus, OR in negative logic ↔ AND in positive logic. Verification / Alternative check: Truth table mapping: when either X̄ or Ȳ is low (meaning X or Y is true), the combined condition holds exactly when X AND Y is the corresponding positive-logic combination for asserting active-low outputs. Formal De Morgan rules confirm this equivalence. Why Other Options Are Wrong: NOT and NOR do not capture the duality between OR under negative logic and positive-logic AND. Choosing OR ignores the change in signal polarity that flips the operator. Common Pitfalls: Forgetting that both signal levels and operators transform under negative logic; applying De Morgan to signals but not to the operator, or vice versa. Final Answer: AND

Question 16

Boolean algebra simplification: Reduce the expression (A + B)(A + C) to its simplest sum-of-terms form.

Accepted Answer

Introduction / Context: Simplifying Boolean expressions is central to minimizing logic circuits, reducing gate count, power, and delay. The expression (A + B)(A + C) appears frequently in control logic and can be simplified by applying distributive and absorption laws. Given Data / Assumptions: Variables A, B, C are Boolean (0 or 1). Operators: + is OR, juxtaposition (e.g., AB) is AND. Target: a minimal, logically equivalent expression. Concept / Approach: Use distributive and absorption properties: X + XY = X and X(X + Y) = X. By distributing A across the product and removing redundant terms, we reach a compact form with fewer gates when implemented in hardware. Step-by-Step Solution: Start: (A + B)(A + C).Distribute: = A(A + C) + B(A + C).Simplify first term: A(A + C) = A.Second term: B(A + C) = AB + BC.Combine: A + AB + BC.Apply absorption: A + AB = A.Result: A + BC. Verification / Alternative check: Construct a Karnaugh map for A, B, C and group minterms; the minimal cover yields A + BC. Alternatively, truth-table evaluation confirms equivalence across all 8 input combinations. Why Other Options Are Wrong: AB + AC is the result of partial distribution without final simplification. A + B + C overstates the function, turning don’t-care situations into always-on conditions. AC + B does not match the original for all inputs. Common Pitfalls: Stopping after distribution; forgetting the absorption rule; mixing operator precedence. Always look for X + XY = X patterns to reduce expressions. Final Answer: A + BC

Question 17

Bipolar logic families in practice: Which logic family has become the most widely used among classic bipolar families?

Accepted Answer

Introduction / Context: Several logic families have been developed over decades: DTL, TTL, ECL, and later MOS/CMOS. Among the bipolar options, TTL emerged as the most widely used in general-purpose digital circuits due to its balance of speed, noise margins, ease of use, and vast component availability in the 74xx series. Given Data / Assumptions: Scope is classic bipolar families (DTL, TTL, ECL), not modern CMOS dominance. We consider mainstream availability and widespread adoption in industry and education. Physical characteristics like speed, power, and fan-out matter for popularity. Concept / Approach: TTL offered practical switching speeds with reasonable power and excellent noise immunity compared to earlier DTL. Although ECL can be faster, it is power-hungry and used in niches demanding ultra-high speed. The 7400-series TTL parts became de facto standards, driving ecosystem adoption across boards, kits, and textbooks. Step-by-Step Solution: Compare DTL vs TTL: TTL integrates transistor-transistor stages for better performance and reliability.Compare TTL vs ECL: ECL is faster but consumes more static power and is less common in general designs.Assess market adoption: the breadth of the 74xx family and derivatives made TTL the most widely used bipolar logic family. Verification / Alternative check: Historical catalogs and educational kits predominantly feature TTL parts (e.g., 7400, 74138, 7474), reflecting its extensive use in laboratories and products prior to CMOS domination. Why Other Options Are Wrong: DTL is an earlier, slower predecessor supplanted by TTL. ECL, while very fast, is niche and power-intensive. MOS refers to a different device category (not strictly a bipolar family) and is typically associated with CMOS logic rather than bipolar. Common Pitfalls: Confusing present-day CMOS ubiquity with historical bipolar usage; assuming “fastest” (ECL) implies “most used” regardless of cost and power. Final Answer: TTL (Transistor–Transistor Logic)

Question 18

In number systems, the radix (base) of the binary representation is which value?

Accepted Answer

Introduction / Context: The concept of a radix, also called a base, is central to understanding how different number systems encode quantities. In positional systems, the radix determines the set of allowed digits and the place-value weights. This question asks for the radix of the binary system, which is the standard language of digital electronics and computing. Given Data / Assumptions: Binary numbers use only two symbols: 0 and 1. Each position to the left represents a higher power of the radix. We are dealing with unsigned positional notation for clarity. Concept / Approach: In any positional system with radix r, the allowed digit symbols run from 0 to r−1, and the positional weights are r^0, r^1, r^2, and so on. Because binary uses two symbols (0 and 1), its radix is 2. Every binary value can be interpreted as a weighted sum of powers of 2, which directly maps to how digital circuits represent and manipulate information using low (0) and high (1) voltage levels. Step-by-Step Solution: Identify the symbol set of binary: {0, 1} → two symbols.By definition, the count of distinct digit symbols equals the radix.Therefore, radix = 2.Select option “2.” Verification / Alternative check: Compare with decimal (radix 10, digits 0–9) and hexadecimal (radix 16, digits 0–9 and A–F). Binary’s two symbols make its base unambiguously 2. Why Other Options Are Wrong: 3 would imply digits 0–2; 1 is not a valid radix for positional systems; 10 is the decimal base, not binary; “None of the above” is incorrect because 2 is correct. Common Pitfalls: Confusing the number 10 written in binary (which equals 2 decimal) with “base 10”; the string “10” in any base represents the radix value itself, but that does not make the radix 10. Always separate numeral meaning from base identity. Final Answer: 2

Question 19

In a buffer register with a LOAD control, when LOAD is active the input word is captured on the next positive clock edge.

Accepted Answer

Introduction / Context: Edge-triggered registers are the fundamental storage elements in synchronous digital systems. A buffer register typically has a LOAD (enable) input that gates whether new data is latched. Understanding exactly when data is captured helps avoid setup/hold violations and ensures correct timing in pipelines and interfaces. Given Data / Assumptions: The register is clocked and edge-triggered. The LOAD input is active (asserted). “Positive edge” means a transition from logic 0 to logic 1 on the clock signal. Concept / Approach: For a positive-edge-triggered register, data present at the inputs is sampled and stored only at the instant of the rising clock edge, provided the LOAD (enable) is asserted. Between edges, the output remains stable. Designers must meet setup and hold times: data must be stable for a minimum time before and after the rising edge to guarantee reliable capture. Step-by-Step Solution: Identify the trigger mechanism: positive-edge-triggered register.Check control: LOAD active → register is enabled to accept new data.At the next rising transition of the clock, the input word is sampled and stored.Therefore, the correct word to fill the blank is “clock.” Verification / Alternative check: Consult typical timing diagrams for D-type flip-flops with enable: they show sampling on the rising clock edge when enable is high. This behavior underpins synchronous design methodologies and static timing analysis tools. Why Other Options Are Wrong: “Pulse” is vague and not the synchronous reference edge; “register” and “transistor” are components, not timing references; “None of the above” fails because “clock” is correct and precise. Common Pitfalls: Confusing level-sensitive latches with edge-triggered flip-flops; forgetting to meet setup/hold windows; assuming LOAD changes can occur exactly at the edge—controls must also meet timing requirements. Final Answer: clock

Question 20

Identify the logic gate whose output is 1 if and only if at least one input is 0 (i.e., it outputs 0 only when all inputs are 1).

Accepted Answer

Introduction / Context: Understanding the behavior of basic logic gates is essential for designing and simplifying digital circuits. This question asks for the gate that produces a high output in every case except when all inputs are high. Recognizing this pattern allows quick identification without building full truth tables every time. Given Data / Assumptions: The gate is multi-input (2 or more) but the rule must hold for any number of inputs. Output is 0 only when all inputs are 1. In every other input combination, the output is 1. Concept / Approach: The NAND function is the logical NOT of AND. Because AND outputs 1 only when all inputs are 1, NAND outputs 0 only in that same case and outputs 1 otherwise. This matches the statement “output is 1 if at least one input is 0.” NAND is functionally complete, meaning any Boolean function can be built using only NAND gates, which is why it is so common in IC design. Step-by-Step Solution: Recall AND truth: all 1s → 1, otherwise → 0.Invert (NAND): all 1s → 0, otherwise → 1.Compare with the condition: “output 1 if at least one input is 0.”Therefore, the gate is NAND. Verification / Alternative check: For two inputs, list cases: 00→1, 01→1, 10→1, 11→0 for NAND. This exactly matches the described behavior. Why Other Options Are Wrong: AND outputs 0 whenever any input is 0; OR outputs 1 if any input is 1 (different rule); NOR outputs 1 only when all inputs are 0; “None” is false since NAND is correct. Common Pitfalls: Confusing “at least one input is 0” (NAND) with “at least one input is 1” (OR). Also mixing up NOR (only 1 when all 0s) with NAND (only 0 when all 1s). Final Answer: NAND gate

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