Question 1

For a truth table with n binary inputs, the number of distinct input words (rows) is equal to which expression?

Accepted Answer

Introduction / Context: Truth tables enumerate all possible input combinations to fully define a combinational logic function. The size of the table grows exponentially with the number of inputs. Knowing the exact count of input words helps in planning testing, verification, and simplification steps like Karnaugh maps. Given Data / Assumptions: There are n independent binary inputs. Each input can be either 0 or 1. Inputs are considered simultaneously to form input words (vectors). Concept / Approach: Each binary input doubles the number of possible combinations. With one input, there are 2 possibilities. Adding a second input multiplies possibilities by 2 again, yielding 4. In general, with n independent binary inputs, the number of unique input words is 2^n. This exponential growth explains why exhaustive testing becomes infeasible for large n and why formal methods or random testing are used. Step-by-Step Solution: Start with 1 input: combinations = 2.Add inputs one by one: each new input multiplies combinations by 2.After n inputs: combinations = 2 * 2 * … (n times) = 2^n.Select option “2^n.” Verification / Alternative check: For n = 3, list inputs (A, B, C): there are 8 rows (000 through 111), which equals 2^3. For n = 4, there are 16 rows, consistent with 2^4. Why Other Options Are Wrong: 10^n corresponds to decimal digits, not binary choices; 4^n and 8^n would be correct only if each input had 4 or 8 states, which is not the case for binary inputs; “None” is false because 2^n is correct. Common Pitfalls: Misreading 2^n as 2*n; forgetting that binary inputs are independent; overlooking that outputs do not affect the count of input rows—only inputs do. Final Answer: 2^n

Question 2

In standard 7-bit ASCII, what is the binary code for the uppercase letter ‘‘T’’?

Accepted Answer

Introduction / Context: ASCII (American Standard Code for Information Interchange) defines 7-bit codes for letters, digits, punctuation, and control characters. Developers often need to translate between characters and their binary or hexadecimal codes for debugging, encoding, and protocol work. This item asks for the 7-bit code of uppercase “T.” Given Data / Assumptions: Use standard 7-bit ASCII, not extended 8-bit encodings. Uppercase letters ‘‘A’’–‘‘Z’’ occupy decimal 65–90. We want the binary code for ‘‘T.’’ Concept / Approach: Uppercase “T” is decimal 84 in ASCII. Converting 84 to binary: 84 = 64 + 16 + 4, corresponding to bits at positions 6, 4, and 2 set. Writing 84 in 7 bits yields 1010100. This representation is common in datasheets and protocol analyzers. In hexadecimal, the same code is 0x54, which many programmers recognize from C string dumps or UART traces. Step-by-Step Solution: Recall ASCII mapping: ‘‘A’’=65, ‘‘T’’ is the 20th letter, so 65 + 19 = 84.Convert 84 to binary: 84 = 64 + 16 + 4.Mark bits: 64 (2^6) → 1, 32 (2^5) → 0, 16 (2^4) → 1, 8 (2^3) → 0, 4 (2^2) → 1, 2 (2^1) → 0, 1 (2^0) → 0.Therefore the 7-bit pattern is 1 0 1 0 1 0 0 → 1010100. Verification / Alternative check: Hex check: 84 decimal = 0x54. Converting 0x54 to binary gives 0101 0100, which as 7-bit ASCII (drop the leading 0) is 1010100, confirming the result. Why Other Options Are Wrong: 1011010 (90) is ‘‘Z’’; 1011100 (92) and 1011111 (95) are punctuation/other codes in 7-bit context. “None” is incorrect because 1010100 is exactly “T.” Common Pitfalls: Mixing uppercase with lowercase codes (lowercase “t” is 116 decimal); forgetting to use 7-bit ASCII rather than 8-bit extended encodings; misplacing bit weights when converting decimal to binary. Final Answer: 1010100

Question 3

Convert the decimal number 61 (base 10) to its binary (base 2) representation. Show the place-value or division-by-2 reasoning.

Accepted Answer

Introduction / Context: Converting between decimal and binary is a routine skill in digital design and programming. It reinforces understanding of positional notation and is useful for interpreting register values, masks, and addresses. Here we convert 61 from base 10 to base 2. Given Data / Assumptions: Value: 61 decimal. Target: binary representation without leading zeros. Allowed method: place-value expansion or repeated division by 2. Concept / Approach: Use place-value expansion: represent 61 as a sum of powers of two. Alternatively, perform repeated division by 2 and read remainders bottom-up. Both methods should produce the same bit pattern. The highest power of two not exceeding 61 is 32 (2^5). Step-by-Step Solution: List powers: 2^5=32, 2^4=16, 2^3=8, 2^2=4, 2^1=2, 2^0=1.Select terms that sum to 61: 32 + 16 + 8 + 4 + 1 = 61.Mark bits: for 2^5,2^4,2^3,2^2,2^1,2^0 → 1,1,1,1,0,1.Write bits from 2^5 down to 2^0: 111101. Verification / Alternative check: Division method: 61/2=30 r1, 30/2=15 r0, 15/2=7 r1, 7/2=3 r1, 3/2=1 r1, 1/2=0 r1. Reading remainders upward yields 111101, which matches the place-value result. Why Other Options Are Wrong: 111100 equals 60; 110011 equals 51; 001101 equals 13; “None” is wrong because 111101 is correct. Common Pitfalls: Dropping a remainder in the division method; writing bits in reverse order; confusing the highest power needed and leaving a gap in the expansion. Final Answer: 111101

Question 4

Add the two binary numbers 110101 (base 2) and 101111 (base 2). Provide the final binary sum.

Accepted Answer

Introduction / Context: Binary addition underlies arithmetic logic units (ALUs) and software integer arithmetic. Practicing manual addition clarifies carry propagation and helps validate results when debugging or designing datapaths. We add 110101₂ and 101111₂ and express the result in binary. Given Data / Assumptions: Augend: 110101 (base 2). Addend: 101111 (base 2). Standard binary addition rules apply: 0+0=0, 0+1=1, 1+1=0 with carry 1, 1+1+carry=1 with carry 1, etc. Concept / Approach: Perform column-wise addition from least significant bit to most significant bit, tracking carries. Alternatively, convert to decimal, add, and reconvert to binary for a quick check. 110101₂ = 53₁₀ and 101111₂ = 47₁₀, so the sum should be 100₁₀, which hints that the binary result should be 1100100₂ (64+32+4). Step-by-Step Solution: Write the numbers aligned: 110101 + 101111.LSB column: 1+1=0 carry 1.Next: 0+1+carry1=0 carry 1.Next: 1+1+carry1=1 carry 1.Next: 0+1+carry1=0 carry 1.Next: 1+0+carry1=0 carry 1.MSB: 1+1(carry)=10 → write 10 at the left, final result 1100100. Verification / Alternative check: Decimal check: 53 + 47 = 100. Convert 1100100₂: 64 + 32 + 4 = 100. Both methods agree. Why Other Options Are Wrong: 1101000 equals 104; 1110111 equals 119; 110011 equals 51; “None” is wrong because 1100100 is correct. Common Pitfalls: Losing a carry in the middle columns; forgetting that 1+1=0 carry 1; misaligning bits. Always track carries systematically and verify with a decimal cross-check if time permits. Final Answer: 1100100

Question 5

Convert the decimal number 99 (base 10) to its octal (base 8) representation.

Accepted Answer

Introduction / Context: Octal (base 8) was historically used in computing because each octal digit maps neatly to three binary bits. Converting between decimal and octal is still useful for understanding legacy systems, permissions in Unix (e.g., 755), and compact binary representations. Here we convert 99₁₀ to base 8. Given Data / Assumptions: Value: 99 decimal. Target base: octal (digits 0–7). Use repeated division by 8 or binary grouping as methods. Concept / Approach: With repeated division by 8, we divide the decimal number by 8, record the remainder (octal digit), and continue with the quotient until it becomes 0. Reading remainders from last to first gives the octal number. Alternatively, convert 99 to binary (1100011₂) and group in 3-bit chunks to get octal 143 directly. Step-by-Step Solution: 99 / 8 = 12 remainder 3 → least significant octal digit = 3.12 / 8 = 1 remainder 4 → next octal digit = 4.1 / 8 = 0 remainder 1 → most significant octal digit = 1.Reading MSB to LSB: 1 4 3 → 143 (base 8). Verification / Alternative check: Binary grouping: 99₁₀ = 1100011₂ → 001 100 011 → 1 4 3 in octal. Also, compute back: 164 + 48 + 3 = 64 + 32 + 3 = 99, confirming correctness. Why Other Options Are Wrong: 121₈ = 81 decimal; 124₈ = 84 decimal; 119 is not a valid octal number (9 is invalid); “None” is wrong because 143 is correct. Common Pitfalls: Forgetting to read remainders in reverse; using base 16 groupings by mistake; including invalid digits 8 or 9 in octal results. Final Answer: 143

Question 6

In digital terminology, the standard abbreviation for “binary digit” is which term?

Accepted Answer

Introduction / Context: Digital systems represent information using two distinct states, commonly labeled 0 and 1. The smallest unit of information in this context is called a “binary digit.” Its widely accepted abbreviation is a core piece of vocabulary across computing, communications, and information theory. Given Data / Assumptions: We are dealing with binary (two-state) information units. The term sought is the standard, internationally recognized abbreviation. Options include distractors that sound similar or related but are not standard terms. Concept / Approach: “Binary digit” was shortened early in computing history to “bit,” a term credited to statisticians and information theorists. A bit can hold one of two values: 0 or 1. Collections of bits form larger units: 4 bits = nibble, 8 bits = byte (octet), 16 bits = word on some architectures, and so on. The bit underpins data sizes, bandwidth measures (bits per second), and storage metrics when used carefully (though storage vendors often use bytes). Step-by-Step Solution: Identify the full term: “binary digit.”Recall the accepted contraction used universally: “bit.”Select the option that exactly matches this standard abbreviation. Verification / Alternative check: Check common usage: Mbps (megabits per second), bitwise operations in programming (bit shifts, bit masks), and CPU instruction sets that manipulate bits—all use “bit” as the fundamental unit. Why Other Options Are Wrong: “0 and 1” describe possible values, not the unit name; “bingit” is not a standard term; “base” refers to a radix (e.g., base 2); “None” is wrong because “bit” is correct. Common Pitfalls: Confusing bits and bytes (a byte is typically 8 bits); mixing symbol cases in abbreviations (b vs B matters: b=bit, B=byte in many contexts). Final Answer: bit

Question 7

An inverter (NOT gate) has only how many input(s)? The output is always the logical complement of that input.

Accepted Answer

Introduction / Context: The inverter, also called a NOT gate, is one of the most basic logic elements. It flips the logic level: 0 becomes 1, and 1 becomes 0. Understanding its arity (number of inputs) and behavior is essential for building and analyzing combinational and sequential circuits. Given Data / Assumptions: The device discussed is an inverter/NOT gate. “Logical complement” means output = 1 when input = 0, and output = 0 when input = 1. We are referring to standard digital logic families (TTL, CMOS) where inverter symbols have a single input bubble output. Concept / Approach: An inverter is defined as a single-input gate. Formally, its Boolean function is Y = NOT(A), often written as Y = A′ or Y = ~A. Multi-input gates that produce complements exist (e.g., NAND, NOR), but those are not called inverters; they implement complements of multi-input functions. The minimal inverter requires only one input to produce the complement at its output. Step-by-Step Solution: Identify the gate: NOT (inverter).Recall the function: Y = NOT(A).Count inputs: exactly one input variable A.Therefore, choose “one.” Verification / Alternative check: Look at logic symbols: a triangle (buffer) with a small circle (inversion bubble) on the output represents an inverter with a single input pin. Datasheets list 6 inverters per IC (e.g., 74HC04) each with one input and one output. Why Other Options Are Wrong: Two or more inputs defines different gates (e.g., NAND, NOR, AND, OR). “All of the above” cannot be correct because the definition is specific. “None” is wrong since “one” is correct. Common Pitfalls: Confusing an inverter with a buffer (which does not invert); assuming a “universal” NAND/NOR gate is the same as an inverter—those require tying inputs together to mimic a single-input inversion, but they are not inverters by definition. Final Answer: one

Question 8

In digital electronics, Binary-Coded Decimal (BCD) numbers are most useful whenever human-friendly decimal information must be transferred into or out of a digital system (for example, keypad entry, numerical displays, and financial totals). Which n…

Accepted Answer

Introduction / Context: Binary-Coded Decimal (BCD) is a number representation scheme where each decimal digit (0 through 9) is encoded separately in a fixed number of bits, commonly 4 bits. BCD is widely used anywhere machines must interact closely with human-readable decimal values, such as cash registers, digital panel meters, time displays, and keypad input systems. This question tests your ability to recall why BCD exists and where it is preferred over pure binary or other codes. Given Data / Assumptions: We are discussing data transfer into or out of a digital system. Human operators frequently think and type in decimal digits. We must choose the numeric domain where BCD offers the clearest advantage. Concept / Approach: BCD preserves a one-to-one mapping between each decimal digit and its 4-bit code. That means converting the displayed or entered decimal digits does not require base conversion math; each digit is handled independently. This reduces coding effort for display drivers and simplifies debouncing and interpretation of decimal keypads, while avoiding rounding issues important in financial contexts. Step-by-Step Solution: Identify the use case: human-readable numbers at interfaces (keypad/display).Note that decimal digits must be mapped directly to an electronic format.BCD is designed specifically to carry decimal digits without base conversion.Therefore, BCD is useful whenever decimal information is transferred in or out. Verification / Alternative check: Compare with pure binary: while compact, binary needs conversion for each decimal digit on displays. ASCII encodes characters (including digits) but is less efficient for arithmetic on digits. Hexadecimal is a base-16 convenience for engineers, not end users. Why Other Options Are Wrong: Binary: efficient internally but requires conversion for decimal displays.ASCII: character-focused, not a numeric-digit code optimized for arithmetic on individual digits.Hexadecimal: not aligned with everyday decimal entry/reading.None of the above: incorrect because “decimal” is exactly right. Common Pitfalls: Assuming BCD is storage-efficient (it is not; it sacrifices compactness for easy decimal handling). Confusing ASCII digits with BCD nibbles. Forgetting that financial computations often use BCD or decimal floating formats to avoid binary rounding issues. Final Answer: decimal

Question 9

In basic logic design, which gate with two or more inputs produces a high (logic 1) output only when all of its inputs are high?

Accepted Answer

Introduction / Context: The fundamental Boolean gates—AND, OR, and NOT—form the foundation of digital logic. Knowing exactly how each gate responds to combinations of inputs is essential for building combinational circuits such as decoders, comparators, and arithmetic units. This item checks whether you can map a verbal specification (“high only when all inputs are high”) to the correct gate behavior. Given Data / Assumptions: The gate accepts two or more input signals. The output should be high only if every input is high. We use standard positive logic (high = 1, low = 0). Concept / Approach: By definition, the AND operation returns 1 if and only if all inputs are 1. Any input at 0 forces the output to 0. This matches exactly the phrasing “high only when all inputs are high.” OR, NAND, and NOR have different truth conditions that do not match the specification. Step-by-Step Solution: Recall truth condition: AND outputs 1 only for the all-ones input combination.Compare: OR outputs 1 if any input is 1; NOR and NAND involve inversion.Select the gate matching the described behavior: AND. Verification / Alternative check: Truth table check for two inputs A and B: AND gives 1 only for A=1 and B=1. For three or more inputs, the same rule generalizes—every input must be 1 to get a 1 at the output. Why Other Options Are Wrong: OR: outputs high if at least one input is high, not all.NAND: outputs the complement of AND (low only when all inputs are high).NOR: outputs 1 only when all inputs are low, the opposite extreme.None of the above: incorrect because AND fits perfectly. Common Pitfalls: Confusing “any” with “all,” and mixing up NAND with AND. Also, overlooking that multi-input versions follow the same rule as two-input gates. Final Answer: AND

Question 10

Number systems: Convert the decimal value 93₁₀ to its hexadecimal (base-16) representation.

Accepted Answer

Introduction / Context: Base conversion between decimal and hexadecimal is a routine skill in computing, especially for memory addresses, color codes, and low-level debugging. Hexadecimal compresses binary neatly because each hex digit corresponds to 4 binary bits. Given Data / Assumptions: Input value: 93 in base 10. Target base: 16 (hexadecimal). Standard digit mapping: 10→A, 11→B, 12→C, 13→D, 14→E, 15→F. Concept / Approach: Use repeated division by 16. The quotient and remainder sequence, read from last remainder to first, gives the hex digits. Alternatively, convert 93 to binary and group bits by 4; the division method is straightforward here. Step-by-Step Solution: Compute 93 / 16 → quotient = 5, remainder = 13.Map remainder 13 to hex digit D.Write hex as quotient-remainder pair: 5D16.Sanity check: 5 * 16 + 13 = 80 + 13 = 93. Verification / Alternative check: Binary route: 93₁₀ = 0101 1101₂ → split into nibbles 0101 (5) and 1101 (D) → 5D16. Why Other Options Are Wrong: 2D16: equals 45 in decimal (216 + 13), not 93.6216: equals 98 decimal (616 + 2), not 93.3116: equals 49 decimal (3*16 + 1), not 93.None of the above: incorrect because 5D16 is correct. Common Pitfalls: Mis-mapping remainders 10–15 to A–F and reversing the quotient/remainder order. Final Answer: 5D16

Question 11

Logic design terminology: Which gate is known as a “universal gate” because it can be combined to implement any Boolean function (including AND, OR, and NOT)?

Accepted Answer

Introduction / Context: Universal gates are building blocks from which any logic function can be constructed without needing other gate types. Recognizing them helps in cost-optimized designs, minimization, and standardizing on a small set of parts in hardware. Given Data / Assumptions: The question asks specifically for a gate that can implement any Boolean function. We consider standard families and functional completeness. Practical gate composition is allowed (multiple devices as needed). Concept / Approach: A functionally complete set of gates can realize all Boolean expressions. A single gate type is “universal” if repeated use of that one type can produce AND, OR, and NOT. NAND is universal because you can create NOT by tying its inputs together, AND by inverting the output of a NAND, and OR by applying De Morgan’s law with inverted inputs. Step-by-Step Solution: NOT from NAND: NAND(A, A) = NOT(A).AND from NAND: AND(A, B) = NOT(NAND(A, B)) = NAND(NAND(A, B), NAND(A, B)).OR from NAND via De Morgan: OR(A, B) = NOT(AND(NOT A, NOT B)) = NAND(NAND(A, A), NAND(B, B)).Hence, any expression composed of AND/OR/NOT can be built from NANDs. Verification / Alternative check: NOR is also universal, but the question asks for the gate “known as” a universal gate—NAND is the most commonly cited example in course material and component catalogs. Why Other Options Are Wrong: NOT: alone cannot compose both AND and OR.AND: cannot invert by itself.XOR: not functionally complete alone.None of the above: incorrect, because NAND is universal. Common Pitfalls: Thinking XOR is universal due to its usefulness in adders; it still needs inversion and other operations to be complete. Final Answer: NAND gate

Question 12

Conversion of an octal number 1428 to binary number is

Accepted Answer

11000102

Question 13

In synchronous digital systems, a single periodic coordination signal is sent to every flip-flop to synchronize state changes. What is this common signal called?

Accepted Answer

Introduction / Context: Synchronous digital design relies on a timing reference so that registers (flip-flops) capture data simultaneously at well-defined instants. This global coordination ensures predictable behavior across complex pipelines, state machines, and buses. The coordinating signal is central to the design methodology. Given Data / Assumptions: The system uses edge- or level-triggered flip-flops. All sequential elements receive a shared timing reference. We need the standard name of that reference signal. Concept / Approach: The “clock” is a periodic waveform that defines when state is sampled or transferred. On rising or falling edges (edge-triggered) or during active levels (level-sensitive latches), flip-flops capture inputs and update outputs. Distribution networks and clock trees are engineered to minimize skew and jitter so all parts of the system agree on timing. Step-by-Step Solution: Identify the element: flip-flops require a timing reference.Name the periodic, systemwide timing signal.Answer: the common signal is the clock. Verification / Alternative check: Observe any synchronous design: registers change on the same edge; asynchronous inputs must be synchronized to the clock domain to avoid metastability. Why Other Options Are Wrong: Latch: a level-sensitive storage element, not the system timing signal.Master/Slave: describe roles in certain flip-flop structures, not the global timing wave.None: incorrect because “clock” is the standard term. Common Pitfalls: Confusing clocks with enables or resets; overlooking clock domain crossing issues and the need for synchronizers. Final Answer: clock

Question 14

Two’s-complement practice: In a 16-bit microcomputer, what is the 2’s-complement representation of −24?

Accepted Answer

Introduction / Context: The two’s-complement system encodes signed integers so that addition and subtraction use the same binary adder. Negative values are formed by inverting all bits of the positive magnitude (one’s complement) and adding 1. This exercise reinforces that workflow in a 16-bit context. Given Data / Assumptions: Word size: 16 bits. Target value: −24. We assume standard two’s-complement arithmetic with sign extension. Concept / Approach: Procedure: write +24 in 16-bit binary, invert every bit to get the one’s complement, then add 1 to obtain the two’s complement. The high bit will be 1 to indicate a negative number in two’s complement. Step-by-Step Solution: Write +24: 24 = 16 + 8 → 0000 0000 0001 1000.Invert bits (one’s complement): 1111 1111 1110 0111.Add 1: 1111 1111 1110 0111 + 0000 0000 0000 0001 = 1111 1111 1110 1000.Thus, −24 = 1111 1111 1110 1000 (two’s complement, 16-bit). Verification / Alternative check: Add 1111 1111 1110 1000 to +24 and confirm wraparound to 0 with carry out discarded: result equals 0, validating the encoding. Why Other Options Are Wrong: 0000 0000 0001 1000: this is +24, not −24.1111 1111 1110 0111: one’s complement only; missing the +1 step.0001 0001 1111 0011: unrelated pattern; not derived from the two’s-complement steps.None: incorrect because option (c) is correct. Common Pitfalls: Forgetting to add 1 after inverting, or failing to maintain the full 16-bit width with proper leading ones for negative numbers. Final Answer: 1111 1111 1110 1000

Question 15

Memory unit prefixes (binary context): Express 8192 as K units, assuming K = 1024.

Accepted Answer

Introduction / Context: In computing, memory sizes are frequently expressed using powers of two. The shorthand “K” conventionally denotes 1024 bytes (2^10) in binary contexts, despite international discussions about kibibytes (KiB). Converting values like 8192 to K clarifies sizing for buffers, pages, or simple capacity planning. Given Data / Assumptions: K is taken to mean 1024 (2^10). Value to convert: 8192. We want a concise expression in K units. Concept / Approach: Compute the ratio 8192 / 1024. Because 8192 is a power-of-two multiple of 1024, the result should be an integer. This conversion is common when stating sizes like “8K cache” or “8K page.” Step-by-Step Solution: Use K = 1024.Compute 8192 / 1024 = 8.Thus, 8192 = 8K. Verification / Alternative check: Check by multiplication: 8K = 8 * 1024 = 8192, confirming the conversion. Why Other Options Are Wrong: 8 x 10^3 K: mixes decimal thousand with K; incorrect dimensionality.8.192 k: lower-case k typically means 1000 in SI; also the numeric value would be 8.192 * 1000, not 8192.All of the above: cannot be correct since other options are wrong. Common Pitfalls: Confusing SI decimal prefixes (k = 1000) with binary conventions (K = 1024) and mixing symbols inconsistently. Final Answer: 8K

Question 16

Number systems: Convert the hexadecimal value 1D7F16 to its decimal (base-10) equivalent. Show the positional weights used.

Accepted Answer

Introduction / Context: Hexadecimal-to-decimal conversion is common when interpreting addresses, error codes, and configuration values. Each hex digit corresponds to a power-of-16 weight, making manual conversion manageable for moderate-length values. Given Data / Assumptions: Input: 1D7F16, where D = 13 and F = 15. Weights from right to left: 16^0, 16^1, 16^2, 16^3, ... No sign or special encoding—just an unsigned value. Concept / Approach: Expand by positional weights and sum: value = d316^3 + d216^2 + d116^1 + d016^0. Substitute digits and compute each term, then sum to get the decimal number. Step-by-Step Solution: List digits: 1, D(13), 7, F(15).Compute weights: 16^3 = 4096; 16^2 = 256; 16^1 = 16; 16^0 = 1.Multiply: 14096 = 4096; 13256 = 3328; 716 = 112; 151 = 15.Sum: 4096 + 3328 + 112 + 15 = 7551. Verification / Alternative check: Convert to binary then to decimal: 1 = 0001, D = 1101, 7 = 0111, F = 1111 → 0001 1101 0111 1111₂; evaluating that binary confirms 7551. Why Other Options Are Wrong: 8771 / 5557 / 7781: sums do not match the correct positional expansion of 1D7F16.None of the above: incorrect because 7551 is correct. Common Pitfalls: Misreading hex digits above 9, forgetting digit weights, or adding in the wrong order. Final Answer: 755110

Question 17

Octal arithmetic: Compute the product of 47₈ and 52₈ and express the result in octal.

Accepted Answer

Introduction / Context: Working directly in octal reinforces base conversion fluency and place-value reasoning beyond decimal. While it is possible to multiply in octal directly, many learners find it reliable to convert to decimal (or binary), multiply, and then convert back to octal. Given Data / Assumptions: Multiplicands: 47₈ and 52₈. We will verify using a decimal intermediate. No signs or fractions are involved. Concept / Approach: Strategy: convert each octal number to decimal, multiply, then convert the product back to octal. Formula reminders: for abc₈ = a8^2 + b8 + c, and to convert decimal back to octal, repeatedly divide by 8 and collect remainders. Step-by-Step Solution: Convert 47₈ to decimal: 48 + 7 = 32 + 7 = 39.Convert 52₈ to decimal: 58 + 2 = 40 + 2 = 42.Multiply in decimal: 39 * 42 = 1638.Convert 1638 to octal via successive division: 1638 / 8 → remainders yield 3 1 4 6 (from last to first).Thus, 1638₁₀ = 3146₈, so 47₈ * 52₈ = (3146)₈. Verification / Alternative check: Binary route: convert both to binary, multiply or use shift-add, then group bits in threes to return to octal. This also yields 3146₈. Why Other Options Are Wrong: (3144)₈, (4147)₈, (3184)₈: none equal 1638 in decimal when converted back; they correspond to different decimal values.None of the above: incorrect because (3146)₈ is correct. Common Pitfalls: Mixing base-10 and base-8 multiplication tables, or making remainder/order mistakes during the base conversion back to octal. Final Answer: (3146)8

Question 18

Digital electronics fundamentals: which logic gate is considered a universal gate because any other logic function can be implemented using only this gate (possibly with multiple stages and interconnections)?

Accepted Answer

Introduction / Context: A universal gate is a building block from which any Boolean function can be constructed without needing any other gate types. In digital electronics, recognizing universal gates is essential for minimizing parts, simplifying procurement, and understanding functional completeness in logic design. Given Data / Assumptions: The term “universal” means functionally complete: able to realize NOT, AND, OR, and by extension any combinational network. We consider basic gate families: AND, OR, EX-OR (XOR), and NAND. Gate availability is unrestricted; multiple stages and interconnections are allowed. Concept / Approach: NAND is universal because you can build a NOT gate by shorting its inputs (NAND(A, A) = NOT A). Using NOT plus NAND, you can form AND (by NAND followed by NOT) and OR (via De Morgan transforms). Once AND, OR, and NOT are available, all Boolean expressions can be implemented. NOR is also universal, but it is not one of the provided correct choices here; among the listed options, NAND is the universal gate. Step-by-Step Solution: Construct NOT: NOT A = NAND(A, A). Construct AND: A AND B = NOT(NAND(A, B)). Construct OR: A OR B = NOT(NOT A AND NOT B) = NAND(NAND(A, A), NAND(B, B)). Since NOT, AND, OR are realizable, any Boolean function can be composed. Verification / Alternative check: De Morgan’s laws and truth tables confirm the correctness of these equivalences. Many standard logic libraries offer NAND-only realizations as cost or speed optimizations, reinforcing universality. Why Other Options Are Wrong: AND: not universal without NOT or OR; cannot directly create inversion. OR: similarly lacks inversion on its own. EX-OR (XOR): cannot on its own realize arbitrary inversion and is not functionally complete without additional gates. None of the above: incorrect because NAND is universal. Common Pitfalls: Confusing “universal” with “most commonly used.” Also, remembering NOR is universal does not negate NAND’s universality; both are universal, but you must select the one present among the options. Final Answer: NAND

Question 19

Binary-coded decimal (BCD): each decimal digit is represented using how many bits in standard BCD encoding?

Accepted Answer

Introduction / Context: Binary-coded decimal (BCD) is a numeric encoding in which each decimal digit (0 through 9) is encoded separately in binary. BCD is widely used in calculators, financial displays, and any system where exact decimal digit representation and easy decimal I/O are important. Given Data / Assumptions: We are discussing standard straight (8421) BCD. Valid digit codes range from 0000 (0) to 1001 (9). Upper binary combinations 1010–1111 are invalid in straight BCD. Concept / Approach: Standard BCD uses 4 bits per decimal digit. Each 4-bit group (a nibble) holds one digit. Multi-digit numbers are encoded as sequences of nibbles, not as a single pure-binary value. This facilitates easy conversion to human-readable decimal digits and prevents rounding errors that can occur with binary floating formats when decimal exactness is required. Step-by-Step Solution: Identify the unit of encoding in BCD: one decimal digit per group. Confirm bit-width: straight BCD maps 0–9 to 0000–1001, which fits in 4 bits. Therefore, each digit uses one nibble (4 bits). Verification / Alternative check: Check common examples: 59 in BCD is 0101 1001 (two nibbles), not 00111011 (which is binary 59). This reaffirms the 4-bit-per-digit rule. Why Other Options Are Wrong: One byte (8 bits): too large per digit; wastes space for straight BCD. One bit: insufficient, since a digit can be 0–9, needing at least 4 bits. All of the above: contradictory; only 4 bits is correct. None: invalid because the nibble answer is correct. Common Pitfalls: Confusing BCD with packed or unpacked representations: unpacked BCD may use a full byte per digit for convenience, but the encoding itself still uses 4 bits to carry the value 0–9. Final Answer: One nibble (4 bits)

Question 20

Integrated circuit density classification: what is the typical functional capacity range (in approximate gate count) used to classify large-scale integration (LSI) devices in classic terminology?

Accepted Answer

Introduction / Context: Historically, integrated circuits (ICs) have been classified by the approximate number of logic gates per chip: SSI, MSI, LSI, and VLSI. While modern chips far exceed these numbers, the classic definitions still appear in exams and foundational texts. Given Data / Assumptions: SSI (small-scale integration) denotes very few gates per chip. MSI (medium-scale integration) denotes tens of gates per chip. LSI (large-scale integration) denotes hundreds to thousands of gates per chip, before VLSI. Concept / Approach: The well-accepted classical ranges are: SSI ≈ 1–10 gates, MSI ≈ 10–100 gates, LSI ≈ 100–10,000 gates, VLSI > 10,000 gates. Exact boundaries vary by source, but these orders of magnitude are consistent across curricula. Step-by-Step Solution: Recall the conventional taxonomy: SSI → MSI → LSI → VLSI. Map values: LSI corresponds to hundreds to several thousands of gates. Select the option that matches: 100 to 10,000 gates. Verification / Alternative check: Cross-check against typical examples: early microprocessors and memory ICs were in the LSI/VLSI boundary zone around these counts, confirming the range. Why Other Options Are Wrong: 1 to 11 gates: SSI territory. 12 to 99 gates: MSI territory. More than 10,000 gates: VLSI, beyond LSI. None: invalid because the LSI range is specified. Common Pitfalls: Memorizing exact cutoffs rather than the order of magnitude; different textbooks may list slightly different thresholds, but the LSI band unquestionably spans roughly 10^2 to 10^4 gates. Final Answer: 100 to 10,000 gates

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