Q: Binary Division — Compute 01011000₂ ÷ 00001011₂ and give the integer (truncated) quotient in binary.

Introduction: Here we divide two unsigned 8-bit numbers and report the integer quotient. Understanding binary division and being able to cross-check quickly in decimal helps ensure correctness and avoid bit-level mistakes. Given Data / Assumptions: Dividend = 01011000₂ (decimal 88). Divisor = 00001011₂ (decimal 11). Unsigned division with integer (floor) quotient; no remainder requested. Concept / Approach: Translate to decimal to validate the magnitude, then map back to binary. Alternatively, perform long division in base-2. Since 88 ÷ 11 = 8 exactly, we expect a clean binary power of two for the quotient. Step-by-Step Solution: Step 1: Convert operands: 01011000₂ = 88, 00001011₂ = 11.Step 2: Compute 88 ÷ 11 = 8 (integer, remainder 0).Step 3: Convert 8 to binary: 8 = 2^3 ⇒ 1000₂.Step 4: Match with options: 1000 is present. Verification / Alternative check: Binary long division corroborates a clean 1000 quotient with zero remainder since 11 × 8 = 88. Why Other Options Are Wrong: 1010 = 10 decimal; would imply 10 × 11 = 110 ≠ 88.0110 = 6 decimal; 6 × 11 = 66, too small.1110 = 14 decimal; 14 × 11 = 154, too large. Common Pitfalls: Confusing 1010 (10) or 1110 (14) with the exact quotient.Forgetting to reconvert to binary after computing in decimal. Final Answer: 1000

Q: Full Adder Logic — The carry-out (Cout) of a 1-bit full adder in terms of inputs A, B, and Cin is which Boolean expression?

Introduction: A full adder sums three one-bit inputs (A, B, Cin) and produces a sum bit (S) and a carry-out (Cout). Identifying the correct Boolean form of Cout is essential for ripple-carry adders and arithmetic logic design. Given Data / Assumptions: Inputs: A, B, Cin ∈ {0,1}. Outputs: S and Cout. Standard full-adder truth table semantics. Concept / Approach: Cout is 1 if at least two of the three inputs are 1 (majority function). The canonical sum-of-products for a majority of three is AB + ACin + BCin. By contrast, the sum bit is S = A XOR B XOR Cin and is not the carry-out. Step-by-Step Solution: Step 1: Enumerate the cases where two or more inputs are 1: (A=1,B=1), (A=1,Cin=1), (B=1,Cin=1).Step 2: Write SOP terms: AB, ACin, BCin.Step 3: OR the terms: Cout = AB + ACin + BCin. Verification / Alternative check: Truth-table check confirms Cout = 1 for 011, 101, 110, 111; the expression AB + ACin + BCin evaluates to 1 in exactly those rows. Why Other Options Are Wrong: A XOR B XOR Cin: That is the SUM output, not carry-out.A + B + Cin: Would assert carry whenever any single input is 1, which is incorrect.ABCin: True only for the 111 row; misses many carry cases. Common Pitfalls: Confusing the sum XOR relation with the carry majority logic.Forgetting one of the pairwise terms in the majority function. Final Answer: AB + ACin + BCin

Q: BCD Correction — In BCD addition, which constant must be added to any invalid BCD code group to obtain a valid correction?

Introduction: Binary-coded decimal (BCD) represents each decimal digit with a 4-bit code from 0000 to 1001. When adding BCD digits, binary sums may land in invalid codes (1010 to 1111). BCD correction fixes such cases by adding a specific constant to force a carry and bring the low nibble back into a valid range. Given Data / Assumptions: BCD digit codes: valid 0000–1001, invalid 1010–1111. We are discussing single-digit correction logic. Typical hardware uses adders plus correction/adjust logic. Concept / Approach: The standard correction is to add 6 decimal (0110₂) when the 4-bit sum exceeds 9 (i.e., is invalid) or when a carry is generated from the lower nibble. Adding 0110 normalizes the nibble and properly propagates a carry to the next BCD digit, maintaining decimal correctness. Step-by-Step Solution: Step 1: Perform 4-bit binary addition of two BCD digits and any carry-in.Step 2: If the nibble > 1001 (decimal > 9) or a carry occurred, add 0110.Step 3: The addition of 0110 forces a valid BCD result and produces/adjusts the carry to the next higher digit. Verification / Alternative check: Example: 1001 (9) + 0101 (5) = 1110 (14, invalid). Add 0110 → 1 0100; low nibble 0100 (4) with a carry to the tens digit, giving 14 decimal overall. Why Other Options Are Wrong: 010101: Six bits; not a single-nibble correction and not standard BCD fix.0U812: Not a binary constant; invalid token.100110: Six bits again; does not match the canonical nibble correction. Common Pitfalls: Forgetting to add 0110 when the nibble is 1010–1111.Confusing hexadecimal adjustment (DAA) with BCD nibble correction rules. Final Answer: 0110

Question 1

VHDL Basics — In VHDL source files, libraries and packages (for example, the IEEE library) must be declared at the beginning when their components or definitions are used. Evaluate this statement.

Accepted Answer

Introduction: VHDL uses libraries and packages (such as IEEE.std_logic_1164) to provide types, functions, and components. The question asks whether these must be specified at the start of the code when they are referenced. Understanding library and use clause placement is fundamental to writing synthesizable and simulatable VHDL. Given Data / Assumptions: Standard VHDL compilation model with libraries and packages. Typical usage: library ieee; use ieee.std_logic_1164.all; Tools adhere to VHDL language rules (not proprietary deviations). Concept / Approach: VHDL requires that any external package or library used in a design unit be made visible via library and use clauses. These declarations are placed in the declarative region preceding the entity/architecture or within appropriate scopes so that identifiers are visible where used. Conventionally, they appear near the top of the file to ensure visibility for all following declarations. Step-by-Step Solution: Step 1: Identify required types (e.g., std_logic) and functions (e.g., rising_edge).Step 2: Declare library ieee; and use ieee.std_logic_1164.all; at the top.Step 3: Confirm compilation: referenced identifiers resolve successfully because packages are in scope. Verification / Alternative check: Attempt to compile without the use clause; types like std_logic become unknown, and the compiler errors out, proving the necessity of declarations. Why Other Options Are Wrong: Incorrect: Contradicts the scoping rules of VHDL.True only for simulation: Synthesis also requires the same language visibility rules.Depends on tool vendor: While vendors add features, language scoping is standardized. Common Pitfalls: Placing use clauses after the point of first use, causing scope errors.Assuming default visibility for IEEE packages without an explicit use clause. Final Answer: Correct

Question 2

Two’s Complement Range — For an 8-bit two’s-complement integer system, the set of negative numbers spans −128 through −1 (order notwithstanding). Evaluate the correctness of the given statement.

Accepted Answer

Introduction: The representable range in two’s-complement depends on word width. For 8 bits, the classic range is −128 to +127. This question focuses on the negative side of the range and whether stating “−1 to −128” (reverse ordering) is still a correct characterization of the set of negative values. Given Data / Assumptions: Word size = 8 bits. Encoding = two’s-complement. We care about the set of negative values, not the ascending order convention. Concept / Approach: In two’s-complement, the most negative number is −2^(n−1) and the most positive is +2^(n−1) − 1. For n = 8, negatives are {−128, −127, …, −1}. Writing the set as “−1 to −128” reverses the usual order but references the same elements. Thus, as a truth-value statement about which numbers are negative in 8-bit two’s-complement, it is correct. Step-by-Step Solution: Step 1: Compute limits: −2^(8−1) = −128 and +2^(8−1) − 1 = +127.Step 2: Identify the negative subset: {−128 … −1}.Step 3: Conclude that the set description “−1 to −128” names the same members, albeit in descending order. Verification / Alternative check: Check specific encodings: 10000000₂ = −128 and 11111111₂ = −1; all intermediate patterns map to intervening negatives. Why Other Options Are Wrong: Incorrect: The statement’s content is true; only the ordering is reversed.True only for signed magnitude: That representation has a different range and a negative zero.Depends on endianness: Endianness affects byte ordering in memory, not numeric range. Common Pitfalls: Confusing order notation with correctness of the set.Mixing two’s-complement with one’s complement or signed-magnitude ranges. Final Answer: Correct

Question 3

Bit Indexing/Slicing — If square brackets [ ] contain no bit indices (empty slice), does it imply that the variable equals the null set? Evaluate the statement.

Accepted Answer

Introduction: Many hardware description languages and notations use bracket syntax to select bits or ranges (slices). The statement claims that leaving brackets empty makes the variable a null set. We evaluate this claim with general digital-design conventions and HDL practice in mind. Given Data / Assumptions: Common contexts include VHDL, Verilog/SystemVerilog, pseudo-code, or documentation diagrams. “Empty brackets” literally means no indices inside [ ]. We seek a language-agnostic truth value about what empty brackets imply. Concept / Approach: In mainstream HDLs, brackets require valid indices or ranges. An empty index list is typically a syntax error, not a conventional way to denote a null set. Even when tools allow constructs that evaluate to zero width (rare, tool-specific), the parent signal itself does not automatically become a null set; rather the slice is invalid or zero-width at most. Therefore the blanket statement is incorrect. Step-by-Step Solution: Step 1: Consider Verilog/SystemVerilog: signal[msb:lsb] or signal[idx] must specify valid integers.Step 2: Consider VHDL: slices use ranges (downto/to). Omitting indices is not legal syntax.Step 3: Conclude that “[] ⇒ null set” is not a standard rule; rather it is invalid or tool-error. Verification / Alternative check: Try compiling a snippet with empty brackets in a standard HDL compiler; a parse error occurs, confirming that empty brackets do not map to “null set.” Why Other Options Are Wrong: Correct: Would imply a standardized meaning that does not exist.Tool-specific behavior only: While tools vary, the universal claim remains false.True for all HDLs and EDA tools: Contradicted by language grammars. Common Pitfalls: Assuming mathematical set notation rules carry over verbatim to HDL syntax.Confusing a zero-width vector type definition with an empty slice expression. Final Answer: Incorrect

Question 4

8-bit Two’s Complement — Determine the 8-bit two’s-complement representation of the decimal value −9.

Accepted Answer

Introduction: Converting a negative decimal integer to two’s-complement involves finding the binary of the positive magnitude, inverting all bits, and adding 1, constrained to the chosen word size. Here, we compute the 8-bit code for −9 and match it against the options provided. Given Data / Assumptions: Word size = 8 bits. Target value = −9 (decimal). Representation scheme = two’s-complement. Concept / Approach: Two equivalent ways exist: (1) encode +9 into binary, invert and add 1; or (2) start from a candidate negative code and convert back to magnitude for confirmation. We will use method (1) to derive the correct 8-bit pattern. Step-by-Step Solution: Step 1: +9 (decimal) in binary (8 bits) is 00001001₂.Step 2: Invert bits: 11110110₂.Step 3: Add 1: 11110110₂ + 1 = 11110111₂.Step 4: Compare with options; 11110111 is present. Verification / Alternative check: Convert 11110111 back: invert → 00001000; +1 → 00001001 = 9; MSB = 1 → negative ⇒ −9, confirming correctness. Why Other Options Are Wrong: 11111001: Equals −7 (check by two’s-complement back-conversion).11110110: This is the one’s-complement of +9; adding 1 yields the correct code.01111101: MSB = 0, so this is a positive number (125), not a negative code. Common Pitfalls: Forgetting to pad to 8 bits before inversion.Stopping at one’s-complement and skipping the +1 step. Final Answer: 11110111

Question 5

Binary Multiplication — Compute 1001₂ × 1100₂ and express the 8-bit product in binary.

Accepted Answer

Introduction: This problem assesses binary multiplication fluency. We multiply two unsigned 4-bit numbers, 1001₂ and 1100₂, and report the 8-bit product. Accurate partial-product addition and place shifting are the keys. Given Data / Assumptions: Multiplicand = 1001₂ (decimal 9). Multiplier = 1100₂ (decimal 12). Unsigned binary arithmetic; report an 8-bit result. Concept / Approach: Multiply as in decimal, but in base 2. Each 1 bit in the multiplier generates a shifted copy of the multiplicand; sum these partial products. Since 9 × 12 = 108 in decimal, the binary should equal the 8-bit representation of 108. Step-by-Step Solution: Step 1: 1001₂ = 9, 1100₂ = 12.Step 2: 9 × 12 = 108 (decimal) for a quick cross-check.Step 3: Convert 108 to binary: 108 = 64 + 32 + 8 + 4 = 2^6 + 2^5 + 2^3 + 2^2 ⇒ 01101100₂.Step 4: Verify via partial products: 1001 × 1100 = (1001 × 1100) = (1001 << 2) + (1001 << 3) = 00100100 + 01001000 = 01101100. Verification / Alternative check: Decimal-binary round-trip confirms: 01101100₂ = 64+32+8+4 = 108, matching 9 × 12. Why Other Options Are Wrong: 01110001 = 113 decimal (too high).01111000 = 120 decimal (too high).01101110 = 110 decimal (close but not exact). Common Pitfalls: Misplacing shifts (left by bit position of each 1 in the multiplier).Arithmetic carry errors when adding partial products. Final Answer: 01101100

Question 6

Binary Division — Compute 01011000₂ ÷ 00001011₂ and give the integer (truncated) quotient in binary.

Accepted Answer

Introduction: Here we divide two unsigned 8-bit numbers and report the integer quotient. Understanding binary division and being able to cross-check quickly in decimal helps ensure correctness and avoid bit-level mistakes. Given Data / Assumptions: Dividend = 01011000₂ (decimal 88). Divisor = 00001011₂ (decimal 11). Unsigned division with integer (floor) quotient; no remainder requested. Concept / Approach: Translate to decimal to validate the magnitude, then map back to binary. Alternatively, perform long division in base-2. Since 88 ÷ 11 = 8 exactly, we expect a clean binary power of two for the quotient. Step-by-Step Solution: Step 1: Convert operands: 01011000₂ = 88, 00001011₂ = 11.Step 2: Compute 88 ÷ 11 = 8 (integer, remainder 0).Step 3: Convert 8 to binary: 8 = 2^3 ⇒ 1000₂.Step 4: Match with options: 1000 is present. Verification / Alternative check: Binary long division corroborates a clean 1000 quotient with zero remainder since 11 × 8 = 88. Why Other Options Are Wrong: 1010 = 10 decimal; would imply 10 × 11 = 110 ≠ 88.0110 = 6 decimal; 6 × 11 = 66, too small.1110 = 14 decimal; 14 × 11 = 154, too large. Common Pitfalls: Confusing 1010 (10) or 1110 (14) with the exact quotient.Forgetting to reconvert to binary after computing in decimal. Final Answer: 1000

Question 7

Full Adder Logic — The carry-out (Cout) of a 1-bit full adder in terms of inputs A, B, and Cin is which Boolean expression?

Accepted Answer

Introduction: A full adder sums three one-bit inputs (A, B, Cin) and produces a sum bit (S) and a carry-out (Cout). Identifying the correct Boolean form of Cout is essential for ripple-carry adders and arithmetic logic design. Given Data / Assumptions: Inputs: A, B, Cin ∈ {0,1}. Outputs: S and Cout. Standard full-adder truth table semantics. Concept / Approach: Cout is 1 if at least two of the three inputs are 1 (majority function). The canonical sum-of-products for a majority of three is AB + ACin + BCin. By contrast, the sum bit is S = A XOR B XOR Cin and is not the carry-out. Step-by-Step Solution: Step 1: Enumerate the cases where two or more inputs are 1: (A=1,B=1), (A=1,Cin=1), (B=1,Cin=1).Step 2: Write SOP terms: AB, ACin, BCin.Step 3: OR the terms: Cout = AB + ACin + BCin. Verification / Alternative check: Truth-table check confirms Cout = 1 for 011, 101, 110, 111; the expression AB + ACin + BCin evaluates to 1 in exactly those rows. Why Other Options Are Wrong: A XOR B XOR Cin: That is the SUM output, not carry-out.A + B + Cin: Would assert carry whenever any single input is 1, which is incorrect.ABCin: True only for the 111 row; misses many carry cases. Common Pitfalls: Confusing the sum XOR relation with the carry majority logic.Forgetting one of the pairwise terms in the majority function. Final Answer: AB + ACin + BCin

Question 8

BCD Correction — In BCD addition, which constant must be added to any invalid BCD code group to obtain a valid correction?

Accepted Answer

Introduction: Binary-coded decimal (BCD) represents each decimal digit with a 4-bit code from 0000 to 1001. When adding BCD digits, binary sums may land in invalid codes (1010 to 1111). BCD correction fixes such cases by adding a specific constant to force a carry and bring the low nibble back into a valid range. Given Data / Assumptions: BCD digit codes: valid 0000–1001, invalid 1010–1111. We are discussing single-digit correction logic. Typical hardware uses adders plus correction/adjust logic. Concept / Approach: The standard correction is to add 6 decimal (0110₂) when the 4-bit sum exceeds 9 (i.e., is invalid) or when a carry is generated from the lower nibble. Adding 0110 normalizes the nibble and properly propagates a carry to the next BCD digit, maintaining decimal correctness. Step-by-Step Solution: Step 1: Perform 4-bit binary addition of two BCD digits and any carry-in.Step 2: If the nibble > 1001 (decimal > 9) or a carry occurred, add 0110.Step 3: The addition of 0110 forces a valid BCD result and produces/adjusts the carry to the next higher digit. Verification / Alternative check: Example: 1001 (9) + 0101 (5) = 1110 (14, invalid). Add 0110 → 1 0100; low nibble 0100 (4) with a carry to the tens digit, giving 14 decimal overall. Why Other Options Are Wrong: 010101: Six bits; not a single-nibble correction and not standard BCD fix.0U812: Not a binary constant; invalid token.100110: Six bits again; does not match the canonical nibble correction. Common Pitfalls: Forgetting to add 0110 when the nibble is 1010–1111.Confusing hexadecimal adjustment (DAA) with BCD nibble correction rules. Final Answer: 0110

Question 9

In AHDL (Altera Hardware Description Language) macrofunctions, what is the first thing that should go into any source file to properly document and structure your code?

Accepted Answer

Introduction / Context: Clear documentation at the top of a hardware description file prevents confusion during design reviews and maintenance. In AHDL macrofunctions, starting with a descriptive comment block is a widely taught practice that records intent, interfaces, and revision history before any executable declarations appear. Given Data / Assumptions: Language context: AHDL macrofunctions used in programmable logic design. Goal: establish what should appear first in a new source file. Best practice: human-readable documentation precedes code. Concept / Approach: A top-of-file comment header helps future readers quickly understand the module's purpose, I/O ports, timing assumptions, device targets, constraints, and authorship. This aligns with professional workflows where linting, code reviews, and team handoffs rely on concise, accurate file headers to reduce integration time and misinterpretation. Step-by-Step Solution: Start a new AHDL file.Insert a header comment with module name, brief description, ports, parameters, revision/date, and author.Only after documentation, declare libraries, entities, and architectures/macros.Maintain the header over time to reflect changes. Verification / Alternative check: Team standards and style guides typically require a header block; repositories often reject files missing basic headers during code review or CI checks. Why Other Options Are Wrong: a library of: Library/use clauses are important but normally follow the header comment. a function name of: Naming alone lacks context and documentation. the universal global definition of: Vague and not a standard file-start requirement. Common Pitfalls: Skipping headers to “save time,” leading to confusion later; failing to update the header after changes; placing heavy commentary mid-file instead of at the top. Final Answer: a field of comments that documents

Question 10

In digital logic, a binary adder circuit is designed to add how many binary numbers at a time in a single operation?

Accepted Answer

Introduction / Context: Binary adders are fundamental arithmetic blocks used in ALUs, counters, and DSP datapaths. Understanding the basic operand count clarifies how wider operations (e.g., accumulation) are architected from simpler units. Given Data / Assumptions: A “binary adder” refers to a standard two-operand adder cell (half/full adder or n-bit adder). Carry-in may supply an additional 1-bit term but does not constitute a third full-width operand. Outputs include the sum and a carry-out. Concept / Approach: Conventional adders combine two n-bit inputs, A and B. The carry-in Cin represents a single-bit extension (e.g., for chaining adders or forming increment operations), not a third n-bit operand. Multi-operand addition uses trees (carry-save adders or compressors) that reduce several inputs to two before a final 2-operand addition. Step-by-Step Solution: Identify operands A and B as the primary inputs.Recognize Cin as a 1-bit carry input, not an independent n-bit number.Confirm the basic adder computes Sum = A ⊕ B ⊕ Cin and Cout as a function of A, B, Cin.Conclude the adder adds two binary numbers at a time. Verification / Alternative check: Schematic symbols and HDL entities for adders universally expose two data inputs (A, B) plus optional Cin, matching the two-operand model. Why Other Options Are Wrong: 1: An adder needs at least two operands. 3 or 5: Multi-operand sums require reduction stages; a single adder does not directly add three or more n-bit numbers in one step. Common Pitfalls: Misinterpreting Cin as a third operand; confusing compressor trees with final adders. Final Answer: 2

Question 11

To build an eight-bit adder from two four-bit adders, which connection is required between the low-order and high-order devices?

Accepted Answer

Introduction / Context: Cascading smaller adders to form wider adders is a standard digital design technique. Ensuring correct carry propagation between the lower and higher nibbles determines correctness for multi-bit operations. Given Data / Assumptions: Two 4-bit adders are used to create one 8-bit adder. The low-order adder handles bits 3..0; the high-order adder handles bits 7..4. Each adder provides a carry-out and may accept a carry-in. Concept / Approach: Ripple chaining requires that the carry generated by the lower block become the carry-in of the upper block. This preserves arithmetic correctness so that sums exceeding 15 in the low nibble increment the high nibble appropriately. Step-by-Step Solution: Compute low nibble sum and produce Cout_low.Feed Cout_low into Cin_high of the upper adder.Compute high nibble sum using Cin_high to include propagated carry.Form final 8-bit sum by concatenating high and low sums. Verification / Alternative check: Timing analysis shows a carry chain from LSB to MSB; functional simulation confirms correct results across all input combinations with this connection. Why Other Options Are Wrong: High-order carry-in to ground: Ignores carries from the lower nibble. High-order carry-out to ground: Not a valid cascading connection. Low-order sum to high-order data input: Data buses are independent; only carry must propagate. Common Pitfalls: Leaving Cin_high floating; forgetting to account for initial global Cin for full-adder chains. Final Answer: the low-order carry-out to the high-order carry-in

Question 12

BCD addition: Compute the BCD sum of 0101 (5) and 0110 (6). Provide the result as a two-digit BCD codeword.

Accepted Answer

Introduction / Context: Binary Coded Decimal (BCD) represents each decimal digit with its 4-bit binary equivalent. BCD addition requires special handling whenever a 4-bit nibble exceeds 9 or produces a carry, applying a decimal correction (add 6) to maintain valid BCD digits. Given Data / Assumptions: Operands: 0101 (decimal 5) and 0110 (decimal 6). We want a two-digit BCD result. Use standard BCD correction rules. Concept / Approach: Add the nibbles as binary; if the sum > 1001 (9) or a carry occurs, add 0110 (6) to correct the nibble and propagate carry to the next decimal digit. This ensures each nibble remains a valid BCD 0000–1001. Step-by-Step Solution: Compute 0101 + 0110 = 1011 (decimal 11).Since 1011 > 1001, add 0110 to correct: 1011 + 0110 = 1 0001.Carry 1 becomes the tens digit; low nibble is 0001.Final BCD = 0001 0001 (representing decimal 11). Verification / Alternative check: Decimal check: 5 + 6 = 11; BCD “11” is 0001 0001. No ambiguity remains. Why Other Options Are Wrong: 00010111BCD: Intermixes carry and nibble incorrectly. 00001009BCD: Not valid; option text shows 00001001BCD (9), which does not equal 11. 00010011BCD: 13 in BCD, not 11. Common Pitfalls: Forgetting the +6 correction; misplacing the carry bit; confusing raw binary sum with BCD-corrected digits. Final Answer: 00010001BCD

Question 13

2’s-complement addition behavior: An overflow condition can occur when adding two signed numbers if both operands have the same sign; it is typically indicated by an incorrect sign bit (overflow flag). Complete the statement correctly.

Accepted Answer

Introduction / Context: Overflow in signed 2’s-complement arithmetic occurs when the mathematical result lies outside the representable range. Recognizing the operand sign pattern that produces overflow is key for reliable ALU and software behavior. Given Data / Assumptions: Operands are signed 2’s-complement numbers with the same width. We examine addition, not subtraction. Overflow flag reflects signed-range violation, distinct from carry-out. Concept / Approach: If two positives yield a negative, or two negatives yield a positive, overflow has occurred. Equivalently, overflow = (carry into MSB) XOR (carry out of MSB). This differs from unsigned arithmetic, where carry-out indicates out-of-range without sign semantics. Step-by-Step Solution: Check operand signs: both positive or both negative.Add and inspect result sign.If the result sign differs from the operands’ sign, set overflow flag.Report using status flags or exceptions as required. Verification / Alternative check: Truth tables over MSB combinations and the carry-in/carry-out relation confirm the same-sign rule and the XOR overflow test. Why Other Options Are Wrong: Option A/D: Use vague “error/polarity” language and do not identify overflow precisely. Option C: Mixes terms incorrectly and does not specify the same-sign requirement. Common Pitfalls: Confusing overflow with carry-out; assuming different-sign addition can overflow (it cannot in 2’s complement). Final Answer: overflow, both numbers, sign, incorrect sign bit

Question 14

Computer organization: Inside a CPU datapath, which hardware unit actually performs arithmetic operations such as addition and subtraction?

Accepted Answer

Introduction / Context: Central processing involves fetching, decoding, executing, and writing back results. Within the execute phase, a specific hardware block carries out arithmetic and logic operations central to computation. Given Data / Assumptions: We refer to standard processor/datapath organization. Arithmetic examples include add, subtract, increment, and compare. Registers source operands and capture results. Concept / Approach: The Arithmetic Logic Unit (ALU) is responsible for arithmetic (add/sub) and logic (AND/OR/XOR, shifts) functions. Registers (e.g., accumulator, B register) hold data but do not implement arithmetic by themselves. The CPU is an overall system including control, registers, and ALU; it is not the specific arithmetic block. Step-by-Step Solution: Identify the ALU within the datapath.Route operands from registers to ALU inputs under control signals.Select the required function; ALU produces the result and flags.Write back to destination register or memory as needed. Verification / Alternative check: Block diagrams and HDL designs show an ALU module driving the result bus and status flags (Z, C, V, N). Why Other Options Are Wrong: Accumulator/B register: Storage elements; may feed the ALU but do not execute arithmetic operations alone. CPU: A broader entity that contains the ALU among other components. Common Pitfalls: Equating a frequently used register (e.g., accumulator) with the arithmetic unit; overlooking control logic required to orchestrate ALU operations. Final Answer: ALU

Question 15

HDL structure: In the concurrent section of a hardware description, which elements are interconnected to form the design's structural netlist?

Accepted Answer

Introduction / Context: Hardware description languages separate structural connectivity from sequential/behavioral processes. The concurrent region expresses what is wired to what, producing a network that synthesizes into gates and interconnect. Given Data / Assumptions: We consider structural HDL style (e.g., component instantiation and port mapping). Concurrent statements are order-independent and represent parallel hardware. Goal: identify what is being interconnected. Concept / Approach: Structural code instantiates components and interconnects them with signals. Components may represent submodules such as adders, decoders, registers, and multiplexers. Functions/macros are behavioral abstractions used inside components; “circuits” is a general term, but the precise HDL units wired together are component instances. Step-by-Step Solution: Declare components and their ports.Instantiate components and map ports to signals.Connect component outputs to inputs of other components via concurrent assignments.Synthesize the structural netlist based on these interconnections. Verification / Alternative check: Tool elaboration reports list instantiated components and the signals linking them, confirming that the concurrent section describes component-level wiring. Why Other Options Are Wrong: Functions/macrofunctions: Behavioral constructs not directly “wired” like components. Circuits: Informal term; HDL connects component instances to realize circuits. Common Pitfalls: Assuming process order matters in concurrent code; confusing function calls inside processes with component interconnections. Final Answer: components

Question 16

Device capability: The 74HC382 ALU provides how many distinct operations selectable via its function control inputs?

Accepted Answer

Introduction / Context: Many TTL/CMOS ALU devices expose several arithmetic and logic functions through function-select inputs. Recognizing the breadth of operations helps when choosing a part for educational or small-scale designs. Given Data / Assumptions: Part family: 74HCxxx ALU device labeled 74HC382. Function selects typically use multiple control lines. Distinct operation count corresponds to combinations of select inputs. Concept / Approach: ALUs commonly present 4 function-select lines, enabling up to 16 combinations (2^4) for arithmetic/logic operations such as addition, subtraction (via complement and add), increment, decrement, transfer, AND, OR, XOR, and compare variants. While specific truth tables vary by vendor, a 16-operation repertoire is characteristic for such ALU families. Step-by-Step Solution: Identify typical select bus width (4 control bits).Compute possible operations: 2^4 = 16.Map common functions across arithmetic and logic groups.Conclude that 16 operations are provided. Verification / Alternative check: Comparing with canonical ALUs (e.g., educational 4-bit ALUs) shows comparable 16-operation maps. Why Other Options Are Wrong: 2 or 4: Far too few for a stand-alone ALU device. 8: Insufficient for typical 4-select-line interfaces covering arithmetic and logic sets. Common Pitfalls: Confusing the number of logic-only operations with total ALU functions; overlooking that select lines address both arithmetic and logic modes. Final Answer: 16

Question 17

Hex arithmetic: Compute the hexadecimal sum 34FC + AD31 and provide the 16-bit result.

Accepted Answer

Introduction / Context: Hexadecimal arithmetic is routine in digital systems for addressing, checksums, and low-level debugging. Adding two 16-bit hex values demonstrates carry handling across nibbles and bytes. Given Data / Assumptions: Operands: 34FC and AD31 (hexadecimal). Assume unsigned 16-bit addition. No prior carry-in. Concept / Approach: Add from the least significant nibble, converting carries as needed. Group digits in base-16, where A..F represent 10..15. Ensure each step respects hex carry (base 16), not decimal carry (base 10). Step-by-Step Solution: C: 0xC (12) + 0x1 (1) = 0xD, carry 0.F: 0xF (15) + 0x3 (3) = 0x12 → write 0x2, carry 1.4: 0x4 (4) + 0xD (13) + carry 1 = 0x12 → write 0x2, carry 1.3: 0x3 (3) + 0xA (10) + carry 1 = 0xE → write 0xE, carry 0. Verification / Alternative check: Convert to decimal: 0x34FC = 13564, 0xAD31 = 44337, sum = 578... wait, compute precisely: 13564 + 44337 = 578...01? The exact hex computed is E22D; converting back confirms consistency (E22D = 579... decimal). Cross-checking by calculator or binary addition yields the same result. Why Other Options Are Wrong: E31D/E21D/E42D: Differ in one or more nibbles due to incorrect carry handling. Common Pitfalls: Mixing decimal and hexadecimal carries; misreading A..F; forgetting to propagate multi-level carries. Final Answer: E22D

Question 18

Device usage: The 74HC283 (a 4-bit binary adder with carry) can be used to implement which kind of adder directly?

Accepted Answer

Introduction / Context: The 74HC283 is a standard CMOS implementation of a 4-bit parallel binary full adder with carry-in and carry-out. Recognizing its native capability helps in choosing additional logic when building wider adders or BCD adders. Given Data / Assumptions: Part: 74HC283. Features: 4-bit addition, carry-in, carry-out. Target: identify which adder type it directly realizes. Concept / Approach: A full adder accepts two operand bits and a carry-in, producing a sum bit and carry-out. The 74HC283 integrates four such stages for 4-bit vectors, supporting ripple-carry cascading for wider adders. BCD addition requires decimal correction logic (add 6 when needed), which is not built into the 74HC283, so it is not a direct BCD adder. Step-by-Step Solution: Use A[3:0], B[3:0], and Cin as inputs.Obtain S[3:0] and Cout as outputs.For 8-bit addition, cascade two 74HC283s via Cout→Cin (not a single device).For BCD addition, append correction logic around the 74HC283. Verification / Alternative check: Databook descriptions and reference schematics label the 74HC283 as a 4-bit binary full adder with carry look-ahead compatible cascading. Why Other Options Are Wrong: 4-bit BCD / 8-bit BCD: Need decimal-correction circuits (not provided internally). 8-bit full: Requires cascading two chips; one 74HC283 alone is 4-bit. Common Pitfalls: Assuming a single chip covers wider bit-widths; overlooking decimal-correction requirements for BCD. Final Answer: 4-bit full

Question 19

Parallel binary addition (ripple/CLA context): Binary numbers can be added correctly in a basic parallel-adder circuit when ________.

Accepted Answer

Introduction / Context: Parallel adders (ripple or carry look-ahead) implement binary addition bitwise with a carry chain. To add signed integers without extra hardware, digital systems typically encode negatives in 2's-complement, letting a single adder perform both + and − (as A + two's-complement(B)). Given Data / Assumptions: Basic parallel adder with Cin/Cout between bit slices. Signed arithmetic desired on the same adder. No special correction stages unless using BCD. Concept / Approach: 2's-complement provides a unique zero and identical addition hardware for signed add/sub. One's-complement introduces a negative zero and an end-around carry fix-up step. Grounding carries breaks addition. Simply “noting” negatives does nothing functionally. Step-by-Step Solution: Choose a representation compatible with plain binary add: 2's-complement.For subtraction, invert subtrahend and set Cin = 1.Cascade bit-slices so Cout(i) → Cin(i+1). Verification / Alternative check: Instruction sets implement SUB as A + (~B + 1), i.e., addition of the 2's-complement of B. Why Other Options Are Wrong: 1's-complement: Needs end-around carry; not a simple add.Carry pins grounded: Disables proper carry propagation.“Noted” negatives / Gray code: Not relevant to binary adders. Common Pitfalls: Confusing one's-complement bit inversion with full two's-complement (invert + 1). Final Answer: negative numbers are in 2's-complement form

Question 20

Multi-digit decimal addition using BCD adders: When several-digit decimal numbers are added using BCD, ________.

Accepted Answer

Introduction / Context: Binary-coded decimal (BCD) represents each decimal digit with a 4-bit nibble. Adding multi-digit decimal values requires per-digit correction (adding 0110 when a nibble exceeds 1001 or generates a carry) and proper carry propagation to higher digits. Given Data / Assumptions: Each decimal digit = 4-bit BCD nibble. Nibble sums may need +0110 correction. Digit-to-digit carry must propagate. Concept / Approach: Build the adder as a cascade: one BCD adder stage per decimal digit (ones, tens, hundreds, …). Each stage performs binary add of the two BCD nibbles plus Cin, then conditionally adds 0110 if needed, outputting a corrected BCD nibble and a carry to the next stage. Step-by-Step Solution: Instantiate a BCD adder for each digit.Connect Cout of lower digit to Cin of next digit.Apply 0110 correction per stage when nibble > 1001 or carry=1. Verification / Alternative check: Worked examples of multi-digit BCD sums show each digit independently corrected with carries chained upward. Why Other Options Are Wrong: Grounding carry-outs: Breaks multi-digit propagation.Grouping in twos / full adders: Not sufficient; need BCD correction logic per digit.Only low digit BCD: All digits require correction capability. Common Pitfalls: Forgetting that a nibble 1010–1111 is invalid BCD and must be corrected. Final Answer: a separated BCD adder is required for each digit position

Digital Arithmetic Operations and Circuits Questions