Q: Gate behavior identification: which logic gate produces an output of 0 (logic low) whenever at least one of its inputs is 1 (logic high), and only outputs 1 when all inputs are 0?

Introduction / Context: Recognizing gate truth tables quickly is crucial when analyzing or simplifying digital circuits. The NOR gate is the logical complement of OR: it outputs 1 only when all inputs are 0 and outputs 0 if any input is 1. This distinctive behavior makes NOR a universal gate (alone it can realize any Boolean function). Given Data / Assumptions: Positive logic: high = 1, low = 0. We seek a gate whose output is forced low if any input is high. Multiple-input variants behave consistently with their 2-input counterparts. Concept / Approach: By definition, NOR = NOT(OR). The OR function goes high when any input is high; complementing that yields an output that goes low whenever any input is high, and remains high only when all inputs are low. Step-by-Step Solution: Recall OR truth: any 1 at inputs → OR = 1.Apply inversion: NOR = NOT(OR).Therefore, any 1 at inputs → NOR = 0; all 0s → NOR = 1.Select “NOR gate.” Verification / Alternative check: Truth table check for two inputs: inputs (0,0)→1; (0,1)→0; (1,0)→0; (1,1)→0 for NOR, matching the statement. Why Other Options Are Wrong: AND: outputs 1 only when all inputs are 1, not described behavior.OR: outputs 1 if any input is 1, opposite behavior.EX-OR: outputs 1 when inputs differ, not described behavior.None: incorrect; NOR fits exactly. Common Pitfalls: Confusing NOR with NAND; remembering that NOR is the complement of OR, while NAND is the complement of AND. Final Answer: NOR gate

Q: Ripple/parallel addition design: to build an m-bit binary adder that accepts two m-bit operands and a carry-in, how many full-adder cells are required in the standard ripple-carry configuration?

Introduction / Context: A full adder adds three one-bit values: A, B, and carry-in, producing a one-bit sum and a carry-out. By cascading full adders, we can construct an m-bit ripple-carry adder that computes the sum of two m-bit operands. The number of stages determines latency and hardware cost. Given Data / Assumptions: We are designing a basic ripple-carry (parallel) adder. Two m-bit inputs and an optional carry-in. Each stage handles one bit position and propagates carry. Concept / Approach: Each bit position requires a full adder because there is both bit addition and a potential carry from the previous less significant position. Therefore, for m bits, we need exactly m full-adder cells in series. A half adder would suffice only for the least significant bit if we guaranteed no carry-in, but standard parallel adders allow a carry-in, so all stages are full adders. Step-by-Step Solution: Define stage function: FA adds Ai, Bi, and Ci-1.Cascade from LSB to MSB with carry propagation.Count stages: one FA per bit → total = m FAs.Select “m.” Verification / Alternative check: Block diagrams in digital design show m 1-bit full adders for an m-bit ripple adder; only specialized designs (carry-lookahead, carry-skip) alter carry logic, not the count of sum cells. Why Other Options Are Wrong: m/2: too few; would cover only half the bits.m-1: would omit one necessary stage.m+1: unnecessarily adds hardware.None: a correct choice exists. Common Pitfalls: Confusing half adders with full adders at the LSB; overlooking the carry-in requirement that standardizes all stages as full adders. Final Answer: m

Q: Logic family transformation: what Boolean function results when you append a NOT (inverter) to the output of an AND gate, creating a single composite gate?

Introduction / Context: Combining basic gates yields new, widely used logic functions. By inverting the output of a gate, you form the complemented function. Understanding these compositions is important because NAND and NOR are universal gates, enabling the construction of any logic function using only one gate type. Given Data / Assumptions: Start with an AND gate producing Y = A * B (for two inputs; generalizes to more). Add a NOT gate to the output, so output becomes Y′ = NOT(Y). We must name the resulting composite function. Concept / Approach: By definition, NAND (Not-AND) is the logical complement of AND. If AND outputs 1 only when all inputs are 1, then NAND outputs 0 only when all inputs are 1 and 1 otherwise. This simple inversion transforms the behavior and yields a gate that is functionally complete for digital design. Step-by-Step Solution: Express AND: Y = A * B.Invert output: Z = NOT(Y) = NOT(A * B).Recognize Z as the NAND function.Select “NAND.” Verification / Alternative check: Truth table check confirms: AND truth table inverted equals NAND truth table. NAND is a universal gate used to implement NOT, AND, OR, XOR, and more through combinations. Why Other Options Are Wrong: OR: requires different input combination, not a simple inversion of AND.XOR: equality-based function; not obtained by inverting AND.NOR: inversion of OR, not AND.None: incorrect because NAND is correct. Common Pitfalls: Confusing NAND and NOR; forgetting that “complementing the output” means invert the entire function result, not each input. Final Answer: NAND

Q: Compute the binary division: 100011 (base 2) ÷ 101 (base 2). What is the quotient in binary?

Introduction / Context: Binary arithmetic mirrors decimal arithmetic, but it operates with base 2 digits (0 and 1). Dividing binary numbers is a fundamental digital design skill used in computer arithmetic, error checking, and algorithm analysis. Here, we divide 100011 (base 2) by 101 (base 2) and report the binary quotient. Given Data / Assumptions: Dividend = 100011 (base 2). Divisor = 101 (base 2). We want the exact quotient (no remainder reported, but we will check). Concept / Approach: The fastest way is to convert to decimal for validation, or to perform long division directly in binary. In decimal, 100011₂ = 32 + 2 + 1 = 35, and 101₂ = 4 + 1 = 5. Since 35 ÷ 5 = 7 and 7 in binary is 111₂, the expected quotient is 111₂. We can also verify by binary long division or by multiplying back: 111₂ * 101₂ should yield 100011₂. Step-by-Step Solution: Convert dividend: 100011₂ = 1*2^5 + 0*2^4 + 0*2^3 + 0*2^2 + 1*2^1 + 1*2^0 = 32 + 2 + 1 = 35. Convert divisor: 101₂ = 1*2^2 + 0*2^1 + 1*2^0 = 4 + 1 = 5. Compute decimal division: 35 / 5 = 7. Convert quotient back to binary: 7 = 4 + 2 + 1 = 2^2 + 2^1 + 2^0 → 111₂. Verification / Alternative check: Multiply to confirm: 111₂ * 101₂. Compute: 111₂ * 101₂ = 111₂*(1*2^2 + 0*2^1 + 1*2^0) = (111₂ << 2) + (111₂ << 0) = 11100₂ + 111₂ = 100011₂. This matches the original dividend exactly, confirming no remainder. Why Other Options Are Wrong: 100₂ (4) and 101₂ (5) are too small; their products with 101₂ would not reach 100011₂. 1010₂ (10) is larger than 111₂ (7) and would overshoot when multiplied by 101₂. “None” is invalid because a correct quotient exists. Common Pitfalls: Dropping leading zeros or misaligning shifts during binary long division; forgetting that subtraction in binary uses the same borrow rules as decimal but with base 2. Always verify by multiplying quotient and divisor to recover the dividend. Final Answer: 111 (base 2)

Q: In Boolean algebra and digital logic notation, the multiplication (× or ·) symbol corresponds to which logical operation?

Introduction / Context: Boolean algebra uses symbols that parallel arithmetic: addition corresponds to logical OR, and multiplication corresponds to logical AND. Understanding this mapping is essential when translating between algebraic expressions and logic gate circuits on schematics or HDL code. Given Data / Assumptions: We are using standard Boolean algebra conventions. “+” means OR and “·” (or implied multiplication) means AND. Complements are indicated by overbar or a prime (X′) to denote NOT. Concept / Approach: In Boolean algebra, X·Y (or simply XY) evaluates to 1 only when X=1 and Y=1, exactly matching the AND gate’s truth behavior. Addition “+” yields 1 if either input is 1, matching OR. This correspondence enables direct synthesis from algebraic formulas to gate-level implementations. Step-by-Step Solution: Recall truth table: AND outputs 1 only for (1,1). Map algebraic multiplication (·) to that behavior: X·Y = 1 iff X=1 and Y=1. Conclude that the times symbol denotes the AND operation. Verification / Alternative check: Standard logic texts and Karnaugh map derivations repeatedly use the product term (AND) nomenclature, confirming the mapping from multiplication to AND. Why Other Options Are Wrong: OR is denoted by “+”. NAND and NOR add inversion; they are not the basic multiplication symbol. “None” is invalid because a standard mapping exists. Common Pitfalls: Confusing algebraic “+” and “·” with arithmetic meanings; in Boolean algebra, they are logical OR and AND, not numeric addition/multiplication over real numbers. Final Answer: AND

Q: Compute the binary multiplication: 111 (base 2) × 101 (base 2). What is the product in binary?

Introduction / Context: Binary multiplication is essential for digital arithmetic units. The operation mirrors decimal long multiplication but uses shifts and additions in base 2. Here we multiply 111₂ by 101₂ and provide the binary product. Given Data / Assumptions: Multiplicand = 111 (base 2) = 7 decimal. Multiplier = 101 (base 2) = 5 decimal. We perform exact multiplication and present the result in binary. Concept / Approach: One approach is to convert to decimal (7×5=35) and convert back to binary (35 = 32+2+1 = 100011₂). Alternatively, perform binary long multiplication using partial products and left shifts corresponding to the multiplier’s 1-bits. Step-by-Step Solution: Partial products with multiplier 101₂ (bits from LSB to MSB): Bit 0 = 1 → add 111₂ shifted by 0 → 000111₂. Bit 1 = 0 → add 000₂ (no contribution) shifted by 1 → 000000₂. Bit 2 = 1 → add 111₂ shifted by 2 → 11100₂. Sum partials: 11100₂ + 000111₂ = 100011₂. Verification / Alternative check: Decimal cross-check: 111₂ = 7 and 101₂ = 5; 7×5=35. Convert 35 to binary: 35 = 32 + 2 + 1 → 100011₂, matching the computed result precisely. Why Other Options Are Wrong: 110011₂ (51 decimal) and 111100₂ (60 decimal) do not equal 35. 000101₂ equals 5 decimal, which is just the multiplier, not the product. “None” is invalid because 100011₂ is correct. Common Pitfalls: Misaligning shifts when forming partial products and forgetting that a 0 bit produces no partial product. Always add carefully and confirm by converting to decimal if in doubt. Final Answer: 100011 (base 2)

Q: Address space calculation: with 14 address lines, how many unique memory locations can be addressed?

Introduction / Context: Address lines determine how many distinct locations a processor can reference in memory-mapped systems. Each address bit doubles the addressable space because it represents a binary choice (0 or 1). Understanding this exponential relation is essential for estimating memory capacity and designing decoders. Given Data / Assumptions: Number of address bits, n = 14. Each unique address points to one memory location (often a byte or word depending on architecture). No banking or segmentation tricks—just pure binary addressing. Concept / Approach: The number of unique addresses is 2^n. With n=14, compute 2^14. This is a power-of-two calculation commonly encountered in digital design (e.g., 1K = 2^10 = 1024). Step-by-Step Solution: Recall powers: 2^10 = 1,024; 2^4 = 16. Compute 2^14 = 2^(10+4) = (2^10) * (2^4) = 1,024 * 16. Multiply: 1,024 * 16 = 16,384. Therefore, 14 address bits → 16,384 locations. Verification / Alternative check: Binary counting from 0 to 2^14 − 1 yields 2^14 distinct patterns; 2^14 − 1 = 16,383 is the maximum address value when starting from zero, confirming 16,384 total addresses. Why Other Options Are Wrong: 8,192 equals 2^13; too small. 4,096 equals 2^12; too small. 14 is the number of bits, not the number of locations. “None” is invalid because 16,384 is correct. Common Pitfalls: Mixing up address count with maximum address value; forgetting that counting starts at zero, which still yields 2^n distinct addresses. Final Answer: 16,384

Question 1

Convert the octal number 47₈ to its binary equivalent by replacing each octal digit with its 3-bit binary form and concatenating the results.

Accepted Answer

Introduction / Context: Because one octal digit represents exactly three binary bits, converting octal to binary is direct and avoids decimal intermediates. This skill is essential when reading or writing low-level values in documentation or device registers that use octal notation. Given Data / Assumptions: Octal input: 47₈ (digits 4 and 7). We use the fixed mapping from octal digits to 3-bit binaries. Leading zeros inside any digit’s 3-bit expansion must be preserved. Concept / Approach: Map octal 4→100 and 7→111. Then concatenate the groups to form the final binary string. No padding beyond each digit’s 3-bit mapping is required when converting octal to binary. Step-by-Step Solution: Write mappings: 4→100, 7→111.Concatenate: 100111.Append base: 100111₂.Select the matching option. Verification / Alternative check: Decimal check: 47₈ = 4*8 + 7 = 39. Binary 100111₂ = 32 + 4 + 2 + 1 = 39. Values agree. Why Other Options Are Wrong: 110011₂ or duplicate entries: incorrect mapping for digits 4 and 7.1111011₂: too many bits; does not equal 39 in decimal.None: incorrect because 100111₂ is correct. Common Pitfalls: Dropping leading zeros within a 3-bit group; reversing digit order; confusing hexadecimal (4-bit groups) with octal (3-bit groups). Final Answer: 100111₂

Question 2

Digital logic: an OR gate has two or more inputs. According to the standard truth table, if any one (or more) input is high (logic 1), what is the output state?

Accepted Answer

Introduction / Context: Basic gates (AND, OR, NOT) form the foundation of digital logic. The OR gate outputs a logic 1 whenever one or more of its inputs are at logic 1, capturing the inclusive “or” operation used throughout combinational circuits. Mastering this truth table is essential for designing and analyzing larger systems. Given Data / Assumptions: We consider conventional positive logic (high = 1, low = 0). OR gate has two or more inputs. We evaluate the output when any single input is high. Concept / Approach: The OR function is defined as output = 1 if at least one input = 1, else output = 0. This is inclusive OR (not XOR). Therefore, the correct state when any input is high is “high.” Options stating “low” or “0” contradict the gate’s truth table. Step-by-Step Solution: Recall truth table: for inputs A, B, …, output Y = A OR B OR …If any input = 1, then Y = 1.Therefore, the output is high.Select “high.” Verification / Alternative check: Construct a quick table for two inputs: (0,0)→0; (0,1)→1; (1,0)→1; (1,1)→1. This generalizes to more inputs. Why Other Options Are Wrong: low / 0: the OR function would output 1 when any input is 1, not 0.All of the above: impossible because “low” and “high” cannot both be correct simultaneously.None: incorrect because “high” is correct. Common Pitfalls: Confusing inclusive OR with exclusive OR; mixing active-low logic conventions; misreading multi-input truth tables. Final Answer: high

Question 3

Transistor–Transistor Logic (TTL) families differ in speed and power. Among the classic sub-families listed, which achieves the maximum speed due to Schottky clamping that prevents transistor saturation?

Accepted Answer

Introduction / Context: TTL logic evolved into several sub-families balancing speed and power. Speed is largely determined by how quickly transistors switch off. Schottky-clamped TTL uses Schottky diodes to prevent saturation, reducing storage delay and enabling faster transitions than earlier families. Given Data / Assumptions: We compare standard TTL, high-speed TTL, Schottky-clamped TTL, and low-power TTL. We consider classic families (e.g., 74, 74H, 74L, 74S) by relative switching speed. Advanced variants (e.g., 74AS/ALS) are not listed; we choose among provided options. Concept / Approach: Schottky clamping (TTL “S” series) keeps transistors out of deep saturation, drastically shortening turn-off time relative to standard or high-speed (H) TTL. Low-power TTL minimizes power at the cost of speed. Therefore, among the options, Schottky-clamped TTL is fastest. Step-by-Step Solution: Identify factors affecting speed: saturation depth and storage charge.Recall family traits: S > H > standard > L in speed.Select Schottky-clamped TTL as the maximum speed choice here. Verification / Alternative check: Datasheet families list typical propagation delays: 74S < 74H < 74 (standard) < 74L, confirming the ranking. Why Other Options Are Wrong: Standard TTL: slower than H and S.High-speed TTL: improved over standard but slower than Schottky.Low-power TTL: trades speed for power savings.None: incorrect because a fastest option is present. Common Pitfalls: Assuming “high-speed” is always the fastest; overlooking that “Schottky” specifically addresses saturation delay. Final Answer: Schottky-clamped TTL

Question 4

Gate behavior identification: which logic gate produces an output of 0 (logic low) whenever at least one of its inputs is 1 (logic high), and only outputs 1 when all inputs are 0?

Accepted Answer

Introduction / Context: Recognizing gate truth tables quickly is crucial when analyzing or simplifying digital circuits. The NOR gate is the logical complement of OR: it outputs 1 only when all inputs are 0 and outputs 0 if any input is 1. This distinctive behavior makes NOR a universal gate (alone it can realize any Boolean function). Given Data / Assumptions: Positive logic: high = 1, low = 0. We seek a gate whose output is forced low if any input is high. Multiple-input variants behave consistently with their 2-input counterparts. Concept / Approach: By definition, NOR = NOT(OR). The OR function goes high when any input is high; complementing that yields an output that goes low whenever any input is high, and remains high only when all inputs are low. Step-by-Step Solution: Recall OR truth: any 1 at inputs → OR = 1.Apply inversion: NOR = NOT(OR).Therefore, any 1 at inputs → NOR = 0; all 0s → NOR = 1.Select “NOR gate.” Verification / Alternative check: Truth table check for two inputs: inputs (0,0)→1; (0,1)→0; (1,0)→0; (1,1)→0 for NOR, matching the statement. Why Other Options Are Wrong: AND: outputs 1 only when all inputs are 1, not described behavior.OR: outputs 1 if any input is 1, opposite behavior.EX-OR: outputs 1 when inputs differ, not described behavior.None: incorrect; NOR fits exactly. Common Pitfalls: Confusing NOR with NAND; remembering that NOR is the complement of OR, while NAND is the complement of AND. Final Answer: NOR gate

Question 5

Ripple/parallel addition design: to build an m-bit binary adder that accepts two m-bit operands and a carry-in, how many full-adder cells are required in the standard ripple-carry configuration?

Accepted Answer

Introduction / Context: A full adder adds three one-bit values: A, B, and carry-in, producing a one-bit sum and a carry-out. By cascading full adders, we can construct an m-bit ripple-carry adder that computes the sum of two m-bit operands. The number of stages determines latency and hardware cost. Given Data / Assumptions: We are designing a basic ripple-carry (parallel) adder. Two m-bit inputs and an optional carry-in. Each stage handles one bit position and propagates carry. Concept / Approach: Each bit position requires a full adder because there is both bit addition and a potential carry from the previous less significant position. Therefore, for m bits, we need exactly m full-adder cells in series. A half adder would suffice only for the least significant bit if we guaranteed no carry-in, but standard parallel adders allow a carry-in, so all stages are full adders. Step-by-Step Solution: Define stage function: FA adds Ai, Bi, and Ci-1.Cascade from LSB to MSB with carry propagation.Count stages: one FA per bit → total = m FAs.Select “m.” Verification / Alternative check: Block diagrams in digital design show m 1-bit full adders for an m-bit ripple adder; only specialized designs (carry-lookahead, carry-skip) alter carry logic, not the count of sum cells. Why Other Options Are Wrong: m/2: too few; would cover only half the bits.m-1: would omit one necessary stage.m+1: unnecessarily adds hardware.None: a correct choice exists. Common Pitfalls: Confusing half adders with full adders at the LSB; overlooking the carry-in requirement that standardizes all stages as full adders. Final Answer: m

Question 6

Hazards and races: in which type(s) of digital circuits can dynamic race hazards occur due to unequal propagation delays along multiple paths to an output?

Accepted Answer

Introduction / Context: In real hardware, signals take time to propagate through gates and wires. When multiple paths with different delays drive the same output expression, temporary spurious transitions—called hazards—can appear. Understanding where dynamic race hazards arise helps engineers design glitch-free logic, especially for asynchronous interfaces. Given Data / Assumptions: Dynamic hazards are characterized by multiple unwanted transitions during a single input change. We distinguish hazards in combinational logic from race conditions in sequential elements. We assume standard gate-level realizations and non-zero propagation delays. Concept / Approach: Hazards are typically classified as static-0, static-1, and dynamic. Static hazards cause one spurious pulse; dynamic hazards cause multiple toggles when an output should make a single transition. These are properties of combinational networks implemented with real gates. Sequential circuits exhibit races (e.g., race-around in latches), but that phenomenon is distinct from the classical hazard types defined for combinational networks. Step-by-Step Solution: Model the Boolean function realized via multiple gate-level paths.Assign unequal delays to different paths.Observe that an input transition may cause an output to oscillate before settling (dynamic hazard).Note: while sequential circuits have “race” issues, the canonical hazard taxonomy refers to combinational logic. Verification / Alternative check: Karnaugh-map-based hazard analysis and consensus term insertion techniques are taught specifically for combinational circuits to remove dynamic and static hazards. Why Other Options Are Wrong: Sequential circuits only: conflates race conditions in storage elements with hazards in logic networks.Both combinational and sequential: overgeneralizes the dynamic hazard definition.None: hazards are well-documented in combinational implementations.None of the above: a correct option exists. Common Pitfalls: Mixing up hazard terminology with latch races; ignoring that proper synchronization or adding consensus terms can eliminate hazards in combinational paths. Final Answer: Combinational circuits only

Question 7

Logic family transformation: what Boolean function results when you append a NOT (inverter) to the output of an AND gate, creating a single composite gate?

Accepted Answer

Introduction / Context: Combining basic gates yields new, widely used logic functions. By inverting the output of a gate, you form the complemented function. Understanding these compositions is important because NAND and NOR are universal gates, enabling the construction of any logic function using only one gate type. Given Data / Assumptions: Start with an AND gate producing Y = A * B (for two inputs; generalizes to more). Add a NOT gate to the output, so output becomes Y′ = NOT(Y). We must name the resulting composite function. Concept / Approach: By definition, NAND (Not-AND) is the logical complement of AND. If AND outputs 1 only when all inputs are 1, then NAND outputs 0 only when all inputs are 1 and 1 otherwise. This simple inversion transforms the behavior and yields a gate that is functionally complete for digital design. Step-by-Step Solution: Express AND: Y = A * B.Invert output: Z = NOT(Y) = NOT(A * B).Recognize Z as the NAND function.Select “NAND.” Verification / Alternative check: Truth table check confirms: AND truth table inverted equals NAND truth table. NAND is a universal gate used to implement NOT, AND, OR, XOR, and more through combinations. Why Other Options Are Wrong: OR: requires different input combination, not a simple inversion of AND.XOR: equality-based function; not obtained by inverting AND.NOR: inversion of OR, not AND.None: incorrect because NAND is correct. Common Pitfalls: Confusing NAND and NOR; forgetting that “complementing the output” means invert the entire function result, not each input. Final Answer: NAND

Question 8

In computer architecture, which statement about semiconductor memory is generally accurate (considering typical RAM vs. historic magnetic core memory)?

Accepted Answer

Introduction / Context: Semiconductor memory underpins modern computing. Random access memory (RAM), particularly dynamic RAM (DRAM) and static RAM (SRAM), replaced magnetic core memory in mainstream systems because of speed, density, and cost advantages. This question asks you to identify the generally correct statement about semiconductor memory when contrasted with legacy core memory, while keeping in mind that we are talking about typical RAM behavior rather than every kind of semiconductor device (for example, ROM or flash). Given Data / Assumptions: Context focuses on main memory technologies historically: magnetic core memory vs. semiconductor RAM. Semiconductor RAM (DRAM/SRAM) is volatile; it requires power to retain data. Comparisons are about typical properties, not exceptional cases. Concept / Approach: Volatility is the most universal property of RAM: on power loss, the stored bits disappear. In contrast, magnetic core memory held data with persistent magnetic states and thus could retain information without power. Additionally, semiconductor RAM is generally much faster and far smaller in size per bit than core. Therefore, among the options presented, the statement identifying volatility is the consistent, domain-valid truth for RAM. Step-by-Step Solution: Identify the technology in scope: DRAM/SRAM as the typical “semiconductor memory.” Recall volatility: RAM needs power to preserve data; remove power and contents vanish. Contrast with core memory, which historically retained state without power. Select the option that correctly highlights volatility. Verification / Alternative check: Textbook summaries of memory technologies consistently classify RAM as volatile. Engineering handbooks list retention requirements and refresh cycles (for DRAM), reinforcing that power is essential for data retention. Why Other Options Are Wrong: “Somewhat slower than core” is false; semiconductor RAM is much faster. “Somewhat larger” is false; semiconductor memory is dramatically more compact than core memory. “All of the above” fails because the first and third statements are incorrect. Common Pitfalls: Confusing RAM (volatile) with semiconductor non-volatile memories like flash/ROM. The term “semiconductor memory” is broad, but in system design questions, unless specified otherwise, it typically refers to RAM used as main memory. Final Answer: A volatile memory (contents lost when power is removed)

Question 9

Compute the binary division: 100011 (base 2) ÷ 101 (base 2). What is the quotient in binary?

Accepted Answer

Introduction / Context: Binary arithmetic mirrors decimal arithmetic, but it operates with base 2 digits (0 and 1). Dividing binary numbers is a fundamental digital design skill used in computer arithmetic, error checking, and algorithm analysis. Here, we divide 100011 (base 2) by 101 (base 2) and report the binary quotient. Given Data / Assumptions: Dividend = 100011 (base 2). Divisor = 101 (base 2). We want the exact quotient (no remainder reported, but we will check). Concept / Approach: The fastest way is to convert to decimal for validation, or to perform long division directly in binary. In decimal, 100011₂ = 32 + 2 + 1 = 35, and 101₂ = 4 + 1 = 5. Since 35 ÷ 5 = 7 and 7 in binary is 111₂, the expected quotient is 111₂. We can also verify by binary long division or by multiplying back: 111₂ * 101₂ should yield 100011₂. Step-by-Step Solution: Convert dividend: 100011₂ = 1*2^5 + 0*2^4 + 0*2^3 + 0*2^2 + 1*2^1 + 1*2^0 = 32 + 2 + 1 = 35. Convert divisor: 101₂ = 1*2^2 + 0*2^1 + 1*2^0 = 4 + 1 = 5. Compute decimal division: 35 / 5 = 7. Convert quotient back to binary: 7 = 4 + 2 + 1 = 2^2 + 2^1 + 2^0 → 111₂. Verification / Alternative check: Multiply to confirm: 111₂ * 101₂. Compute: 111₂ * 101₂ = 111₂*(1*2^2 + 0*2^1 + 1*2^0) = (111₂ << 2) + (111₂ << 0) = 11100₂ + 111₂ = 100011₂. This matches the original dividend exactly, confirming no remainder. Why Other Options Are Wrong: 100₂ (4) and 101₂ (5) are too small; their products with 101₂ would not reach 100011₂. 1010₂ (10) is larger than 111₂ (7) and would overshoot when multiplied by 101₂. “None” is invalid because a correct quotient exists. Common Pitfalls: Dropping leading zeros or misaligning shifts during binary long division; forgetting that subtraction in binary uses the same borrow rules as decimal but with base 2. Always verify by multiplying quotient and divisor to recover the dividend. Final Answer: 111 (base 2)

Question 10

In Boolean algebra and digital logic notation, the multiplication (× or ·) symbol corresponds to which logical operation?

Accepted Answer

Introduction / Context: Boolean algebra uses symbols that parallel arithmetic: addition corresponds to logical OR, and multiplication corresponds to logical AND. Understanding this mapping is essential when translating between algebraic expressions and logic gate circuits on schematics or HDL code. Given Data / Assumptions: We are using standard Boolean algebra conventions. “+” means OR and “·” (or implied multiplication) means AND. Complements are indicated by overbar or a prime (X′) to denote NOT. Concept / Approach: In Boolean algebra, X·Y (or simply XY) evaluates to 1 only when X=1 and Y=1, exactly matching the AND gate’s truth behavior. Addition “+” yields 1 if either input is 1, matching OR. This correspondence enables direct synthesis from algebraic formulas to gate-level implementations. Step-by-Step Solution: Recall truth table: AND outputs 1 only for (1,1). Map algebraic multiplication (·) to that behavior: X·Y = 1 iff X=1 and Y=1. Conclude that the times symbol denotes the AND operation. Verification / Alternative check: Standard logic texts and Karnaugh map derivations repeatedly use the product term (AND) nomenclature, confirming the mapping from multiplication to AND. Why Other Options Are Wrong: OR is denoted by “+”. NAND and NOR add inversion; they are not the basic multiplication symbol. “None” is invalid because a standard mapping exists. Common Pitfalls: Confusing algebraic “+” and “·” with arithmetic meanings; in Boolean algebra, they are logical OR and AND, not numeric addition/multiplication over real numbers. Final Answer: AND

Question 11

Compute the binary multiplication: 111 (base 2) × 101 (base 2). What is the product in binary?

Accepted Answer

Introduction / Context: Binary multiplication is essential for digital arithmetic units. The operation mirrors decimal long multiplication but uses shifts and additions in base 2. Here we multiply 111₂ by 101₂ and provide the binary product. Given Data / Assumptions: Multiplicand = 111 (base 2) = 7 decimal. Multiplier = 101 (base 2) = 5 decimal. We perform exact multiplication and present the result in binary. Concept / Approach: One approach is to convert to decimal (7×5=35) and convert back to binary (35 = 32+2+1 = 100011₂). Alternatively, perform binary long multiplication using partial products and left shifts corresponding to the multiplier’s 1-bits. Step-by-Step Solution: Partial products with multiplier 101₂ (bits from LSB to MSB): Bit 0 = 1 → add 111₂ shifted by 0 → 000111₂. Bit 1 = 0 → add 000₂ (no contribution) shifted by 1 → 000000₂. Bit 2 = 1 → add 111₂ shifted by 2 → 11100₂. Sum partials: 11100₂ + 000111₂ = 100011₂. Verification / Alternative check: Decimal cross-check: 111₂ = 7 and 101₂ = 5; 7×5=35. Convert 35 to binary: 35 = 32 + 2 + 1 → 100011₂, matching the computed result precisely. Why Other Options Are Wrong: 110011₂ (51 decimal) and 111100₂ (60 decimal) do not equal 35. 000101₂ equals 5 decimal, which is just the multiplier, not the product. “None” is invalid because 100011₂ is correct. Common Pitfalls: Misaligning shifts when forming partial products and forgetting that a 0 bit produces no partial product. Always add carefully and confirm by converting to decimal if in doubt. Final Answer: 100011 (base 2)

Question 12

Address space calculation: with 14 address lines, how many unique memory locations can be addressed?

Accepted Answer

Introduction / Context: Address lines determine how many distinct locations a processor can reference in memory-mapped systems. Each address bit doubles the addressable space because it represents a binary choice (0 or 1). Understanding this exponential relation is essential for estimating memory capacity and designing decoders. Given Data / Assumptions: Number of address bits, n = 14. Each unique address points to one memory location (often a byte or word depending on architecture). No banking or segmentation tricks—just pure binary addressing. Concept / Approach: The number of unique addresses is 2^n. With n=14, compute 2^14. This is a power-of-two calculation commonly encountered in digital design (e.g., 1K = 2^10 = 1024). Step-by-Step Solution: Recall powers: 2^10 = 1,024; 2^4 = 16. Compute 2^14 = 2^(10+4) = (2^10) * (2^4) = 1,024 * 16. Multiply: 1,024 * 16 = 16,384. Therefore, 14 address bits → 16,384 locations. Verification / Alternative check: Binary counting from 0 to 2^14 − 1 yields 2^14 distinct patterns; 2^14 − 1 = 16,383 is the maximum address value when starting from zero, confirming 16,384 total addresses. Why Other Options Are Wrong: 8,192 equals 2^13; too small. 4,096 equals 2^12; too small. 14 is the number of bits, not the number of locations. “None” is invalid because 16,384 is correct. Common Pitfalls: Mixing up address count with maximum address value; forgetting that counting starts at zero, which still yields 2^n distinct addresses. Final Answer: 16,384

Question 13

Logic equivalence: a NOR gate is functionally equivalent to an OR gate followed by which single element?

Accepted Answer

Introduction / Context: Composite logic functions can be described by combining simpler gates. Recognizing these equivalences helps with circuit minimization and technology mapping. NOR, a universal gate, is defined as the logical negation of OR. Given Data / Assumptions: NOR outputs 1 only when all inputs are 0; otherwise outputs 0. We are asked to express NOR as an OR stage followed by one additional element. Standard logic gate behavior and naming are assumed. Concept / Approach: By definition, NOR(X, Y) = NOT(OR(X, Y)). Therefore, implement OR first, then invert its output using a single NOT (inverter). This identity is used widely when decomposing complex gates into primitive cells or CMOS pull-up/pull-down networks. Step-by-Step Solution: Form the intermediate signal S = OR(X, Y). Apply inversion: Z = NOT(S). Conclude Z equals NOR(X, Y). Verification / Alternative check: Truth table comparison confirms that for inputs (0,0) → OR=0 → NOT=1; for any case with a 1 input → OR=1 → NOT=0, matching NOR behavior exactly. Why Other Options Are Wrong: AND or XOR following OR does not invert the logic and thus cannot produce NOR. “XAND” is not a standard logic element. “None” is invalid because the inverter is correct. Common Pitfalls: Confusing NOR with NAND (which equals AND followed by NOT). Keep the pairings straight: NOR = OR then NOT; NAND = AND then NOT. Final Answer: Inverter (NOT)

Question 14

Convert the octal number 110 (base 8) to its decimal (base 10) equivalent.

Accepted Answer

Introduction / Context: Number base conversion is a foundational skill in digital systems. Octal (base 8) is convenient for grouping binary digits (3 bits per octal digit). Converting octal to decimal reinforces positional notation and prepares you to translate addresses, permissions, and encoded values across representations. Given Data / Assumptions: Octal number: 110₈. We need its base-10 (decimal) value. Octal digits range from 0 to 7. Concept / Approach: Use positional weights of powers of 8. For a three-digit octal number abc₈, value = a*8^2 + b*8^1 + c*8^0. Substitute digits from 110₈ to obtain the decimal sum. Step-by-Step Solution: Write weights: 8^2 = 64, 8^1 = 8, 8^0 = 1. Map digits: a=1, b=1, c=0. Compute: value = 1*64 + 1*8 + 0*1 = 64 + 8 + 0 = 72. Therefore, 110₈ equals 72 in decimal. Verification / Alternative check: Convert to binary first: each octal digit → 3 binary bits. 1→001, 1→001, 0→000. Thus 110₈ = 001 001 000₂ = 1001000₂. Convert binary to decimal: 64 + 8 = 72. The result agrees with the direct calculation. Why Other Options Are Wrong: 70 or 73 are close but result from arithmetic slips (e.g., mis-weighting digits). 110 (base 10) is a misinterpretation (reading symbols without considering base). “None” is invalid because 72 is correct. Common Pitfalls: Interpreting the same digits as decimal; forgetting that octal uses powers of 8; or mis-grouping when using the binary intermediary method (ensure exactly 3 bits per octal digit). Final Answer: 72 (base 10)

Question 15

In combinational logic terminology, which statement correctly describes a half-adder?

Accepted Answer

Introduction / Context: A half-adder is a fundamental building block for arithmetic in digital systems. It adds two one-bit inputs and produces two outputs: the sum and the carry. Recognizing its correct composition is essential before moving on to full adders and ripple/CLA adder architectures. Given Data / Assumptions: The half-adder has two inputs: A and B. Outputs are SUM and CARRY. We are identifying whether any single standard gate alone (AND, NAND, NOR, EX-OR) equals a half-adder. Concept / Approach: A half-adder is implemented with two distinct functions: SUM = A XOR B, and CARRY = A AND B. No single standard gate can provide both outputs simultaneously. Therefore, while XOR is part of the half-adder (for SUM), the half-adder is not “also known as” an XOR circuit; it requires both XOR and AND working together as a pair. Step-by-Step Solution: Write equations: SUM = A XOR B; CARRY = A AND B. Confirm that a single XOR gate yields only SUM, not CARRY. Confirm that a single AND/NAND/NOR gate cannot produce both SUM and CARRY. Conclude that none of the single-gate descriptions alone correctly define a half-adder. Verification / Alternative check: Truth table method: List all input combinations (00, 01, 10, 11). SUM column equals XOR of inputs; CARRY equals AND. This cannot be achieved by a single standard gate for both outputs, confirming “None of the above.” Why Other Options Are Wrong: AND/NAND/NOR: do not compute SUM and CARRY together. EX-OR circuit: computes SUM only, missing CARRY. Thus only “None of the above” fully reflects the correct definition. Common Pitfalls: Equating “half-adder” with “XOR” because the SUM expression is XOR; always remember the carry output requires an AND path. Final Answer: None of the above

Question 16

Basic data unit sizes: a byte consists of how many bits in standard digital systems?

Accepted Answer

Introduction / Context: Data sizes are fundamental to computing literacy. The byte is the most common addressable unit in modern architectures and forms the basis for kilobyte, megabyte, and higher-order storage measures. Understanding that a byte equals eight bits helps when interpreting memory sizes, network payloads, file formats, and character encodings. Given Data / Assumptions: We use the widely adopted standard where 1 byte = 8 bits. Historic variations existed, but 8-bit bytes are universal today. Bits are binary digits (0 or 1). Concept / Approach: An 8-bit byte supports 2^8 = 256 distinct values, which conveniently matches the range for extended ASCII/ISO-8859 encodings and fits neatly into word sizes used by processors. This convention underlies memory addressing and data path widths across contemporary digital systems. Step-by-Step Solution: Define a bit as the smallest unit of information: 0 or 1. Group 8 bits to form a byte, enabling 256 unique patterns. Recognize that most hardware, OS kernels, and languages define sizeof(char) = 1 byte = 8 bits (with rare exceptions). Therefore, select “Eight.” Verification / Alternative check: Standards (e.g., modern CPU and memory specs) and networking protocols operate on octets, explicitly defined as 8-bit bytes, reinforcing the answer. Why Other Options Are Wrong: Two/Four/Ten bits are nonstandard byte sizes in contemporary computing. “None” is invalid because eight is correct. Common Pitfalls: Confusing “byte” with “word,” which varies in width (16, 32, 64 bits). The term “octet” is sometimes used to avoid ambiguity and explicitly means 8 bits. Final Answer: Eight

Question 17

Power-of-two evaluation: what is the decimal value of 2^10 (two raised to the power 10)?

Accepted Answer

Introduction / Context: Powers of two are central to digital systems because binary representation scales by factors of 2. Memorizing key powers (2^10, 2^20, etc.) accelerates mental math for storage capacities, addressing, and algorithm complexity. Here we evaluate 2^10 in decimal. Given Data / Assumptions: Base = 2, exponent = 10. We want an exact decimal number. No approximations; exact integer evaluation. Concept / Approach: Use repeated doubling: each increment of the exponent multiplies the current value by 2. Alternatively, recall the common identity used in computing: 2^10 = 1024, which is close to 10^3 and often used for quick order-of-magnitude conversions between binary and decimal scales. Step-by-Step Solution: Start with 2^1 = 2; 2^2 = 4; 2^3 = 8. Continue doubling: 2^4 = 16; 2^5 = 32; 2^6 = 64; 2^7 = 128. Then 2^8 = 256; 2^9 = 512; 2^10 = 1024. Therefore, 2^10 equals 1024. Verification / Alternative check: Calculator or quick mental doubling confirms the result. Also, note that 1024 is widely used as a “binary thousand” in legacy contexts (though modern SI prefers 1000 while IEC uses 1024 = 1 kibibyte). Why Other Options Are Wrong: 4096 equals 2^12, not 2^10. 1000 is 10^3, not a power of two. 16 equals 2^4, far smaller than 2^10. “None” is invalid because 1024 is correct. Common Pitfalls: Mixing decimal and binary prefixes; remember that 2^10 = 1024 (1 Ki) while 10^3 = 1000 (1 k). In storage and networking, be attentive to whether SI (kilo) or IEC (kibi) units are used. Final Answer: 1024

Question 18

Digital electronics — ring counter concept: Instead of counting upward with binary numbers, a ring counter circulates “words” in which exactly one position (bit) is high at any time. In this context, that single high element is a:

Accepted Answer

Introduction / Context: A ring counter is a specialized sequential circuit used for timing, sequencing, and one-hot state encoding. Unlike a binary counter that increments through natural binary codes, a ring counter recirculates a pattern in which exactly one storage element is high (logic 1) at any given time. Understanding the “one-hot” nature of the pattern clarifies how ring counters simplify decoding and reduce hazards in digital systems. Given Data / Assumptions: The device discussed is a ring counter (e.g., using D or JK flip-flops). Its state is “one-hot,” meaning only one position contains a logic 1. The question asks what that single high element is called. Concept / Approach: A ring counter connects flip-flops in a loop so that a single logic 1 circulates. In an N-stage ring counter, the sequence length is N because the 1 advances one position per clock. Each flip-flop output represents one bit of the state word. Therefore, the “single high” element is a high bit. This one-hot scheme makes decoding trivial: each output line directly indicates a unique state without further combinational logic. Step-by-Step Solution: Identify the nature of the state word in a ring counter: one-hot.Map each flip-flop output to a bit in that word.Conclude that the single high item circulating is a high bit. Verification / Alternative check: Standard texts describe ring and Johnson counters as one-hot or quasi-one-hot encodings where each flip-flop corresponds to a bit position in the circulating word. Why Other Options Are Wrong: “Bytes” is 8 bits, not a single position. “Gate” and “chip” refer to hardware building blocks, not the position within the state word. “None of the above” is invalid because “bit” is correct. Common Pitfalls: Confusing ring counters with binary ripple/synchronous counters; assuming multiple bits may be high in a normal ring counter (they should not be). Final Answer: bit.

Question 19

Fundamental logic gates — complementation behavior: For which logic gate is the output always the Boolean complement (inverse) of the single input?

Accepted Answer

Introduction / Context: Digital systems are built from basic logic gates. The NOT gate (also called an inverter) performs the essential operation of complementation: converting logic 1 to logic 0 and vice versa. Recognizing which gate provides inversion is foundational for analyzing and designing combinational circuits. Given Data / Assumptions: Single-input gate behavior is under consideration. We seek the gate whose output is the complement of the input. Standard Boolean algebra conventions apply (0 ↔ 1 inversion). Concept / Approach: The NOT gate outputs Y = X’, where X’ denotes the logical complement of X. Multi-input gates like AND and OR compute different Boolean operations; XOR (exclusive-OR) compares two inputs to produce 1 when inputs differ, not the simple inversion of one input in general. Step-by-Step Solution: List gate behaviors: NOT inverts; AND and OR combine inputs; XOR is inequality of two inputs.Match “complement of the input” with inverter behavior.Select NOT (inverter) as the correct gate. Verification / Alternative check: Truth table for NOT: X=0 → Y=1; X=1 → Y=0. This directly matches the requirement. Why Other Options Are Wrong: AND/OR are aggregation functions; XOR requires two inputs and is not a simple complement of one input. Common Pitfalls: Misreading XOR as inversion; XOR with one input tied high behaves as NOT on the other, but that is a special case, not the definition. Final Answer: NOT.

Question 20

Linear integrated circuits — why “resistor ratio” design is preferred: In monolithic linear ICs, designers rely on resistor ratios rather than precise absolute values primarily because:

Accepted Answer

Introduction / Context: Monolithic linear integrated circuits (op-amps, references, instrumentation amplifiers) often implement functions using on-chip resistors. Semiconductor processes can produce resistors with good matching on the same die but relatively poor absolute tolerance. Consequently, designers specify performance using resistor ratios, which remain stable across process variations and temperature tracking. Given Data / Assumptions: On-chip passive components exhibit process-dependent absolute values. Layout techniques (common centroid, identical geometry) improve matching between resistors. The question seeks the primary reason for ratio-based design. Concept / Approach: Absolute resistance (e.g., target 10 kΩ) may deviate by ±20% or more due to sheet-resistance and dimensional variations. However, two resistors fabricated adjacently with identical shapes track tightly (e.g., mismatch in the few hundred ppm to several thousand ppm). If a circuit’s transfer function depends on R2/R1, the ratio remains accurate even when both R1 and R2 shift together. This enables precise gains, filters, and references without laser-trim or external components. Step-by-Step Solution: Identify process limitation: poor absolute tolerance, good matching.Exploit matching by designing functions dependent on ratios (e.g., gain = 1 + R2/R1).Conclude that the core reason is the impracticality of precise absolute values on chip. Verification / Alternative check: Datasheets often specify gain accuracy as a ratio-dependent spec and may use laser trimming only when absolute accuracy is required (e.g., precision references). Why Other Options Are Wrong: “Ratios increase input resistance” is not generally true; input resistance is largely set by topology/transistor devices. “Ratios increase amplifier gain” is misleading; ratios set gain accurately but do not universally “increase” it. Common Pitfalls: Assuming external resistor tolerances always fix accuracy; board-level tolerances and temperature coefficients must also be considered. Final Answer: Precise resistor values are not possible with IC processes.

Digital Computer Electronics Questions