Question 1

For adding two 16-bit binary numbers using ripple-carry logic, how many half-adders and full-adders are required in the adder chain?

Accepted Answer

Introduction / Context:Binary adders are built from two primitives: the half-adder (adds two bits) and the full-adder (adds two bits plus a carry-in). Understanding how many of each are needed for an n-bit addition helps in datapath design and estimating delay through a ripple-carry chain.Given Data / Assumptions:We add two 16-bit operands.Ripple-carry architecture is used.The least significant stage has no incoming carry.Concept / Approach:The least significant bit (LSB) can be summed by a half-adder because there is no carry-in. Every other bit position requires a full-adder because it must include the carry generated by the previous stage. Therefore, for n bits: 1 half-adder and n−1 full-adders. For n = 16: 1 half-adder and 15 full-adders.Step-by-Step Solution:Determine the LSB stage requirement: no carry-in → half-adder.For the remaining 15 stages: carry-in present → full-adder per stage.Count devices: 1 half-adder, 15 full-adders.Choose the option that matches these counts.Verification / Alternative check:General formula: for n-bit addition, components = 1 HA + (n−1) FA. Substitute n=16 to verify: 1 + 15 = 16 stages total, which is correct for a 16-bit adder chain.Why Other Options Are Wrong:8/8 and 4/12 do not follow the n−1 full-adder rule; 16 half-adders cannot handle carry propagation; “None” is false since a correct option exists.Common Pitfalls:Using a half-adder at a non-LSB stage where a carry-in exists; forgetting that carry propagation dictates full-adders beyond the first stage.Final Answer:1 half-adders, 15 full-adders

Question 2

Convert the octal number 73 (base 8) to its binary (base 2) equivalent by expanding each octal digit into a 3-bit group.

Accepted Answer

Introduction / Context:Octal-to-binary conversion is straightforward because each octal digit corresponds to exactly three binary bits. This mapping is widely used in low-level work, especially where groups of three bits align nicely with permissions, opcodes, or compact numeric notation.Given Data / Assumptions:Octal value: 73 (base 8).Digit 7 and digit 3 must be mapped to 3-bit binary groups.No leading groups are omitted.Concept / Approach:Map each octal digit separately: 0→000, 1→001, 2→010, 3→011, 4→100, 5→101, 6→110, 7→111. Concatenate the resulting 3-bit groups in the same order to produce the binary value.Step-by-Step Solution:7 (octal) → 111 (binary).3 (octal) → 011 (binary).Concatenate: 111 and 011 → 111011.Select “111011.”Verification / Alternative check:Decimal cross-check: 73 (octal) = 7*8 + 3 = 56 + 3 = 59. Now binary 111011 = 32 + 16 + 8 + 2 + 1 = 59. Both methods agree.Why Other Options Are Wrong:110111 and 111100 represent different decimal values; 110010 is 50 decimal; “None” is wrong because 111011 is correct.Common Pitfalls:Dropping a leading 0 in the lower 3-bit group (e.g., writing 3 as 11 instead of 011) and thus producing a 5-bit result; reversing the order of digits.Final Answer:111011

Question 3

An odd-parity generator appends a parity bit so that the total number of 1s in the transmitted word is odd. At the receiver, an XOR parity check rejects words whose overall parity is even. Fill in the blanks.

Accepted Answer

Introduction / Context:Parity bits are a classic single-bit error detection method used in memory systems, serial links, and legacy storage. With odd parity, the transmitter ensures the total count of 1s (data bits + parity bit) is odd. The receiver recomputes parity to detect single-bit errors quickly.Given Data / Assumptions:Parity scheme: odd parity.Receiver uses XOR aggregation for parity checking.Rejection occurs when the check fails the expected parity.Concept / Approach:XOR of a set of bits equals 1 if and only if the set contains an odd number of 1s. Under an odd-parity scheme, a correct (error-free) received word should have XOR = 1. If noise flips a single bit, the parity flips and XOR = 0 (even number of 1s), which signals an error and the word is rejected (or flagged).Step-by-Step Solution:State the transmitted parity target: odd total count of 1s.Apply XOR property: XOR(all bits) = 1 for odd parity, 0 for even parity.Define the rejection criterion: reject even-parity words under an odd-parity scheme.Fill blanks: “odd, even.”Verification / Alternative check:Example: data 1011 has three 1s (odd). Odd-parity bit = 0 to keep total odd (still three). XOR over 10110 = 1, so accept. If one bit flips, say 10100 (two 1s), XOR = 0 (even) → reject.Why Other Options Are Wrong:“even, odd” mismatches odd-parity definition. “high/low” and “low/high” are not parity categories. “None” is false because “odd, even” is correct.Common Pitfalls:Confusing the parity of data alone versus data+parity; forgetting that XOR implements parity naturally—do not use AND/OR for parity checks.Final Answer:odd, even

Question 4

A half-adder is a combinational circuit that adds exactly how many input bits, producing a SUM and a CARRY output?

Accepted Answer

Introduction / Context:Adders are the core of arithmetic logic units (ALUs). The simplest adder is the half-adder, which demonstrates how binary addition creates both a sum and a carry. Knowing its exact input count is essential before graduating to full-adders and multi-bit adders.Given Data / Assumptions:Half-adder has no carry-in input.Outputs are SUM and CARRY.Inputs are equal-weight bits (no prior carry).Concept / Approach:A half-adder adds two single-bit inputs (A and B). The SUM output is A XOR B, and the CARRY output is A AND B. Because there is no carry-in, it cannot properly serve as an intermediate stage in a ripple-carry chain except at the least significant bit. For all higher bits where a carry-in exists, a full-adder is needed (adds three inputs: A, B, Cin).Step-by-Step Solution:Define half-adder inputs: A, B → exactly two inputs.Define outputs: SUM = A XOR B; CARRY = A AND B.Conclude the input count: 2 bits.Choose “2.”Verification / Alternative check:Truth table confirms: for inputs 00, 01, 10, 11, the SUM and CARRY match binary addition without a carry-in, validating the two-input definition.Why Other Options Are Wrong:Values like 8, 10, or 16 describe word sizes, not half-adder inputs; “None” is invalid because “2” is correct.Common Pitfalls:Assuming a half-adder can replace a full-adder at non-LSB positions; forgetting that CIN must be accommodated by a full-adder.Final Answer:2

Question 5

In Boolean algebra, a “sum-of-products” (SOP) expression is formed by taking several product terms (AND terms) and combining them with a sum (OR). Which description matches SOP?

Accepted Answer

Introduction / Context:Canonical logic forms simplify design and implementation. Two common normal forms are sum-of-products (SOP) and product-of-sums (POS). Recognizing SOP helps map logic to NAND/NOR realizations and guides minimization with Karnaugh maps or the Quine–McCluskey method.Given Data / Assumptions:An SOP uses product terms (ANDs) as building blocks.These product terms are ORed together to form the final expression.Variables may appear complemented or uncomplemented in each product term.Concept / Approach:Product term = AND of literals (e.g., A·B·C’). Sum = OR operation (+). Thus SOP = (product1) + (product2) + … . Examples: F = A’B + AC + BC’. This structure is convenient for two-level logic: first level AND gates, second level OR gate, possibly followed by inversion depending on technology mapping.Step-by-Step Solution:Identify that “products” means AND terms of literals.Combine these with OR to form the “sum.”Match the description: OR of several AND functions.Select option B.Verification / Alternative check:Draw a two-level gate diagram: multiple AND gates feeding a single OR gate → classic SOP topology, confirming the description.Why Other Options Are Wrong:Option A describes POS; C is OR of ORs (not SOP); D is AND of ANDs (a single larger product); “None” is false since B is correct.Common Pitfalls:Confusing SOP with POS; overlooking that complements in literals do not change the SOP structure, only the content of products.Final Answer:The OR function of several AND functions

Question 6

A typical microcomputer may address up to 65,536 locations in memory. Each individual addressable unit in this address space is usually called a memory location (i.e., one addressable register/cell in RAM/ROM).

Accepted Answer

Introduction / Context:When we say a system has, for example, 64K of memory, we mean it can address 65,536 discrete units. Understanding the terminology for each unit avoids confusion between physical chips, CPU internal registers, and the abstract addressing model used by programs.Given Data / Assumptions:The microcomputer has an address space with 65,536 addresses.Each address selects one storage unit (typically one byte).We need the term for that unit.Concept / Approach:A memory location is a single addressable storage cell in main memory. The word “address” is the label used to reference a memory location, not the location itself. “Register” usually refers to a CPU-internal fast storage element, not a cell in main memory. “Chip” is a physical device containing many locations.Step-by-Step Solution:Distinguish address (label) from the storage unit (location).Exclude CPU registers and physical chip terminology.Select the precise term for a single addressable unit: memory location.Verification / Alternative check:Programming languages and debuggers display memory as pairs of address:value, reinforcing that the stored cell is the memory location, while the hex code is the address used to access it.Why Other Options Are Wrong:Address: a reference label, not the storage entity. Registers: CPU internal, not main memory cells. Chip: hardware package, not a single cell. “None” is incorrect since “memory location” is standard.Common Pitfalls:Calling a memory cell a “register,” which can mislead when discussing CPU pipelines or register files; mixing logical addressing with physical packaging.Final Answer:memory location

Question 7

Convert the binary number 10000000000 (base 2) to its decimal (base 10) value. Show the power-of-two reasoning.

Accepted Answer

Introduction / Context:Recognizing powers of two in binary numbers is vital for quick conversions and for understanding address alignment, buffer sizes, and bit masks. A single 1 followed by n zeros in binary equals 2^n in decimal.Given Data / Assumptions:Binary value: 10000000000 (one 1 followed by ten zeros).Base is explicitly 2.We seek the base-10 value.Concept / Approach:If the only set bit is at position n (counting the rightmost bit as position 0), then the decimal value is 2^n. Here, there are ten zeros to the right of the leading 1, so n = 10. Therefore, the decimal value is 2^10 = 1024.Step-by-Step Solution:Count zeros to the right of the leftmost 1: there are 10.Compute 2^10 using powers of two: 2^10 = 1024.State decimal result: 1024.Choose option “1024.”Verification / Alternative check:Place-value expansion: 12^10 + 02^9 + … + 0*2^0 = 1024 + 0 + … + 0 = 1024, confirming the same result.Why Other Options Are Wrong:1034, 1124, and 1037 are unrelated to powers of two; they would correspond to different binary patterns. “None” is false because 1024 is correct.Common Pitfalls:Miscounting zeros (off-by-one errors) and forgetting that the rightmost position is 2^0. Always count positions carefully from the least significant bit.Final Answer:1024

Question 8

The 2-input XOR gate has a high output only when the input bits are _____

Accepted Answer

different

Question 9

A gate is a logic circuit with one or more input signals but _____ output signal.

Accepted Answer

one

Question 10

Conversion of decimal number 9910 to it's binary number equivalent is

Accepted Answer

11000112

Question 11

Conversion of an octal number 1258 to its decimal number is

Accepted Answer

8510

Question 12

Standard TTL has a multiple emitter input transistor and a _____ output

Accepted Answer

totem-pole

Question 13

The digits used in a binary number system are _____ and _____

Accepted Answer

0 and 1

Question 14

Conversion of binary number 10010102 to to its decimal number is

Accepted Answer

7410

Question 15

A 2's-complement adder-subtracter can add or subtract binary numbers. Sign-magnitude numbers represent _____ decimal numbers, and 2's complements stand for _____ decimal numbers.

Accepted Answer

positive, negative

Question 16

Conversion of an octal number 1068 to binary number is

Accepted Answer

10001102

Question 17

The 2764 is 65,536-bit EPROM organized as 8,192 words of 8 bits each. How many address lines does it have?

Accepted Answer

13

Question 18

In digital electronics, which specific logic circuit is used to convert a BCD (binary-coded decimal) input into an equivalent decimal (0–9) output for display or selection purposes?

Accepted Answer

Introduction / Context:A common task in digital systems is to take a 4-bit BCD input (0000 to 1001 representing 0–9) and produce a one-of-ten output that can directly drive indicators or select one of several circuits. The device dedicated to this mapping is a BCD-to-decimal decoder (often called a 1-of-10 decoder). Given Data / Assumptions: Input domain: valid BCD codes (0000–1001). Output domain: ten mutually exclusive lines (D0–D9), one active at a time. Invalid BCD codes (1010–1111) are either ignored or treated as don’t-cares depending on the design. Concept / Approach:A decoder converts coded inputs into a set of outputs in which exactly one line (or a defined pattern) represents each input code. BCD-to-decimal decoders, such as classic TTL/CMOS parts, assert exactly one of ten outputs corresponding to the decimal value of the BCD input. This is distinct from encoders, multiplexers, or generic “code converters.” Step-by-Step Solution: Identify the mapping need: 4-bit BCD → ten distinct outputs. Recognize the device that performs 1-of-N selection from binary inputs. Select the device class: decoder (specifically BCD-to-decimal). Verification / Alternative check:Standard logic catalogs list BCD-to-decimal decoders used to drive ten lamps, relays, or display segments (through drivers), validating the choice. Why Other Options Are Wrong: Encoder: performs the reverse (many inputs to a binary code). Multiplexer: selects one data input to pass to a single output. Code converter: too vague; many devices convert codes, but the precise, canonical device here is a decoder. None: incorrect because an exact device type exists. Common Pitfalls:Confusing decoders and demultiplexers (they are related but serve different immediate purposes). Also, forgetting to handle invalid BCD inputs can cause unintended outputs. Final Answer:decoder

Question 19

Integrated circuit (IC) scales: fill in the blanks. Small-scale integration (_____) typically refers to fewer than about 12 gates per chip; medium-scale integration (MSI) spans roughly 12 to 100 gates; large-scale integration (LSI) refers to more th…

Accepted Answer

Introduction / Context:Historically, integrated circuits were categorized by gate count per chip: SSI, MSI, LSI, and later VLSI/ULSI. While exact boundaries vary across textbooks, a widely taught convention places MSI up to about 100 gates and LSI above 100 gates (until VLSI crosses the thousand-gate mark). Given Data / Assumptions: SSI: fewer than ~10–12 gates per chip. MSI: approximately 12–100 gates per chip. LSI: more than ~100 gates per chip (up to ~1000 in many older references). Concept / Approach:Because the stem explicitly states MSI as 12 to 100 gates, the next tier (LSI) must start above 100. Therefore, the correct completion is SSI for the first blank and 100 for the second blank, aligning with the traditional banding. Step-by-Step Solution: Match “fewer than 12 gates” → SSI. Given MSI goes to 100 gates, LSI must be “more than 100.” Select the pair that reads “SSI, 100.” Verification / Alternative check:Classic digital design texts define SSI/MSI/LSI with these approximate thresholds; VLSI typically begins at >1000 gates historically, underscoring that LSI > 100 is the intended boundary here. Why Other Options Are Wrong: “75” contradicts the MSI upper bound of 100 in the stem. “1000” would skip LSI entirely and jump straight to VLSI-era metrics. “10000” is far beyond LSI in classic definitions. “None” is unnecessary because a consistent pair exists. Common Pitfalls:Assuming exact, universally fixed boundaries; remember these are historical categories used for teaching and vary slightly by source. Final Answer:SSI, 100

Question 20

Memory technologies: which device is ultraviolet-light erasable and electrically programmable, allowing repeated erase-and-reprogram cycles during development?

Accepted Answer

Introduction / Context:Developers historically used ultraviolet-erasable, electrically programmable memories to prototype firmware. By exposing the device window to UV light, stored charges are removed, enabling the chip to be reprogrammed many times before production. Given Data / Assumptions: We are comparing ROM, PROM, EPROM, and RAM. Key features: erasability and programmability mechanism. Use case: iterative program development and testing. Concept / Approach:EPROM (Erasable Programmable Read-Only Memory) is programmed electrically but erased using ultraviolet light through a quartz window. PROM can be programmed once only (fuses blown). ROM is factory-programmed and not field-alterable. RAM is volatile and not used for permanent firmware storage. Step-by-Step Solution: Identify device with UV erasure capability → EPROM. Confirm electrical programming method. Exclude irreversible (PROM) and non-erasable factory ROM. Verification / Alternative check:Device markings (e.g., 27xxx series) and quartz window packaging are characteristic of EPROMs, validating the answer. Why Other Options Are Wrong: ROM: not field erasable/programable. PROM: programmable once; not UV erasable. RAM: volatile storage; not for permanent firmware images. Common Pitfalls:Confusing EPROM with EEPROM/Flash (which are electrically erasable; no UV required). The question explicitly mentions ultraviolet light. Final Answer:EPROM

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