Q: Repair the subtraction rule: For 1-bit binary subtraction 0 − 1, the correct result is difference = 1 with a borrow-out of 1 (not 0). Evaluate the statement “difference = 1 borrow = 0.”

Introduction / Context: Single-bit subtraction cases underpin multi-bit subtractor design. Understanding the borrow behavior for each case prevents design and analysis mistakes in arithmetic circuits. Given Data / Assumptions: We consider 1-bit subtraction with minuend M and subtrahend S, no initial borrow-in. Case examined: M = 0, S = 1. Difference bit D and borrow-out Bout follow standard rules. Concept / Approach: If the minuend bit is smaller than the subtrahend bit, a borrow is required from the next higher bit position. For 0 − 1 at the LSB with no borrow-in, the difference is 1 (after borrowing 1 as 2 in binary) and the borrow-out is 1. Therefore, claiming borrow = 0 is incorrect. Step-by-Step Solution: Write 0 − 1 with no borrow-in.Borrow from the next higher position: 10₂ − 1₂ = 1₂.Set difference D = 1.Set borrow-out Bout = 1 to indicate the borrow occurred. Verification / Alternative check: Truth table for 1-bit subtraction confirms: M=0, S=1, Bin=0 ⇒ D=1, Bout=1. Hardware half-subtractor equations yield D = M ⊕ S, Bout = ~M * S (which evaluates to 1 for 0 − 1). Why Other Options Are Wrong: Correct: Contradicts the established subtractor behavior. Ambiguous / Cannot be determined: The case is explicit and standard. Common Pitfalls: Confusing borrow with carry, or forgetting that 0 − 1 requires borrowing even in the least significant position. Final Answer: Incorrect

Q: Adder performance: A carry look-ahead (CLA) adder uses extra logic to compute carries in parallel and is generally faster than a ripple-carry adder. Evaluate the claim that a look-ahead-carry adder is slower because it requires additional logic.

Introduction / Context: Adder architectures trade off complexity and speed. Ripple-carry adders propagate carry serially through each bit, while carry look-ahead (CLA) adders compute carry signals in parallel using generate/propagate logic to reduce delay. Given Data / Assumptions: Ripple adders have O(n) carry-chain delay for n bits. CLA adds extra logic (generate/propagate) to shorten the critical path. We compare speed, not just gate count. Concept / Approach: CLA adds logic to compute carries simultaneously across groups, transforming linear carry delay into a shorter, usually logarithmic or block-based delay. Although hardware cost rises, the net effect is faster addition for moderate and wide bit-widths compared to ripple schemes. Step-by-Step Solution: Define G = A * B (generate), P = A ⊕ B (propagate).Use CLA equations to compute C1, C2, ... in parallel from Cin and the G/P terms.Observe reduced dependency chain vs. bit-by-bit ripple.Conclude CLA is faster despite additional logic. Verification / Alternative check: ASIC/FPGA timing models consistently show CLA outperforming ripple for nontrivial widths; further enhancements (carry-select, carry-skip, parallel-prefix) push speed even higher. Why Other Options Are Wrong: Correct: Misinterprets added logic as a speed penalty; it is a speed optimization. Ambiguous / Cannot be determined: The architectural comparison is well-established. Common Pitfalls: Equating gate count with speed; in arithmetic circuits, structured parallelism often improves timing at the cost of area. Final Answer: Incorrect

Q: Single-bit subtraction coverage: There are four basic input combinations for subtracting two single-bit binary numbers (minuend, subtrahend) when considered case-by-case. Evaluate this statement.

Introduction / Context: Designing half- and full-subtractors starts with enumerating all 1-bit cases for the minuend (M) and subtrahend (S). Recognizing the complete set of cases avoids mistakes when deriving logic for the difference and borrow outputs. Given Data / Assumptions: M and S are single bits: each ∈ {0, 1}. No initial borrow-in considered (half-subtractor perspective). We classify cases by the (M, S) pair. Concept / Approach: With two binary inputs, there are 2^2 = 4 possible combinations. For subtraction, these are: 0−0, 0−1, 1−0, 1−1. Each case yields a specific difference bit and borrow-out pattern, forming the canonical truth table for the half-subtractor. Step-by-Step Solution: List cases: (M,S) = (0,0), (0,1), (1,0), (1,1).Compute differences and borrows for each case.Assemble the truth table to implement logic equations.Use equations in larger subtractor designs (with borrow-in/out). Verification / Alternative check: The complete table leads to the well-known relations: D = M ⊕ S (for half-subtractor) and Bout = ~M * S. Extending to full-subtractor adds a borrow-in term but preserves the base four cases per (M,S). Why Other Options Are Wrong: Incorrect: Understates or overstates the number of base cases for two inputs. Ambiguous / Cannot be determined: The enumeration is straightforward from combinatorics. Common Pitfalls: Forgetting to treat the borrow-in separately for full-subtractors; the base (M,S) cases remain four, with additional combinations when Bin is included. Final Answer: Correct

Q: Device fact check: The 74LS382 arithmetic/logic unit (ALU) is packaged as a 24-pin device. Evaluate this statement.

Introduction / Context: Members of the classic TTL ALU family (e.g., 74xx181/182 and related parts) are commonly found in 24-pin DIPs or equivalent SMT outlines. This question tests factual recall about the 74LS382's pin count rather than its function set. Given Data / Assumptions: 74LS382 is a 4-bit ALU-class device in the LS TTL family. Comparable ALUs (e.g., 74181) are 24-pin packages in DIP form. We assume standard packaging consistent with typical ALU devices. Concept / Approach: ALUs integrate multiple arithmetic and logic functions with carry/flag pins, requiring a relatively high pin count (operands, function selects, carries, outputs, power/ground). A 24-pin outline is therefore consistent with expectations for such devices and historically documented family members. Step-by-Step Solution: Identify typical ALU signals: A[3:0], B[3:0], F[3:0], function selects, carry in/out, mode, power/ground.Estimate needed pins; the total is consistent with ≈24 pins.Relate to known 74xx ALUs packaged in 24-pin DIP outlines.Conclude that a 24-pin package statement is consistent for 74LS382. Verification / Alternative check: Cross-reference with historical TTL catalogs shows ALU devices in 24-pin packages, matching the statement's characterization of the 74LS382. Why Other Options Are Wrong: Incorrect: Would contradict standard packaging conventions for ALUs of this era. Ambiguous / Cannot be determined: Package information is standard and widely reported. Common Pitfalls: Confusing the 74LS382 with other 74xx logic devices that use 14- or 16-pin packages; ALUs need more I/O and thus more pins. Final Answer: Correct

Q: ALU status flags: In a 4-bit ALU, overflow indicators (status flags) assert when addition or subtraction produces a result that does not fit in four bits according to the representation in use. Evaluate this statement.

Introduction / Context: Status flags (zero, carry/borrow, sign, overflow) summarize arithmetic outcomes. The overflow flag specifically signals that the computed result falls outside the representable range for the chosen width and signedness, which is essential for correct program flow or hardware state transitions. Given Data / Assumptions: ALU width is 4 bits for this discussion. Operations considered: addition and subtraction. Overflow semantics depend on signed interpretation; carry relates to unsigned arithmetic. Concept / Approach: For signed 2's complement, overflow occurs when adding two positives yields a negative, or two negatives yield a positive. In 4 bits, the representable signed range is −8 to +7; results outside this range set overflow. For unsigned, carry-out indicates out-of-range, while a separate overflow definition may be unused; the question's wording focuses on the general idea that results can exceed 4-bit capacity and that indicators exist for this condition. Step-by-Step Solution: Establish 4-bit ranges: unsigned 0..15; signed −8..+7.Check operation result against the relevant range.Set overflow (signed) or carry/borrow (unsigned) when range is exceeded.Use flags to trigger corrective handling (e.g., saturation, exceptions). Verification / Alternative check: Standard ALUs and CPU ISAs define these flags; 4-bit educational ALUs expose similar indicators for teaching purposes. Why Other Options Are Wrong: Incorrect: Ignores established flag semantics. Ambiguous / Cannot be determined: The behavior is well-specified in ALU designs. Common Pitfalls: Confusing carry with overflow; carry is an unsigned concept, while overflow is a signed-range violation. Both can inform about out-of-range conditions depending on interpretation. Final Answer: Correct

Q: Binary result format (fundamentals): In binary arithmetic, the written result (sum) is expressed using only the digits 0 and 1. Evaluate this statement.

Introduction / Context: This question checks the most fundamental property of binary notation: regardless of how addition is carried out internally (with carries and intermediate logic), the written result in base-2 uses only the symbols 0 and 1. Given Data / Assumptions: We are discussing standard base-2 positional representation. Arithmetic may generate carries between bit positions. We care about the final, recorded digits of the sum. Concept / Approach: Binary numbers are expressed with two symbols. During addition, each bit position computes a sum bit (0 or 1) and possibly a carry to the next position. No matter how many carries propagate, each output digit remains 0 or 1. The process may widen the word (e.g., an extra most-significant bit), but the alphabet of symbols does not change. Step-by-Step Solution: Add bitwise with carry: sum_bit = A xor B xor carry_in (0 or 1).Compute carry_out = majority(A, B, carry_in) (0 or 1).Concatenate all sum_bits (and a final carry if present) to form the base-2 result, which uses only 0 and 1. Verification / Alternative check: Example: 1011 + 0111 = 10010. Even with multiple carries, the digits in the result are still only 0s and 1s. Why Other Options Are Wrong: Incorrect: Contradicts the definition of base-2 representation.Only correct without carry: Carries do not introduce new symbols; they only extend width.Ambiguous for signed numbers: Signed encodings (two's complement) still use 0/1 digits.Insufficient context: The statement is precise and standard. Common Pitfalls: Confusing intermediate analog voltages or adder internals with the symbolic numeric result. Final Answer: Correct

Question 1

Adder notation convention: In adder schematics and equations, which Greek letter is commonly used to denote the summing (sum) output?

Accepted Answer

Introduction / Context: Symbols and notation are used in digital design schematics and academic texts for clarity. The sum output of adders is frequently labeled with a Greek character representing “sum.” Given Data / Assumptions: We are referring to the output node indicating the bit-wise sum of inputs and carry-in. Standard mathematical and engineering notation applies. Concept / Approach: “Sigma” (Σ) denotes summation in mathematics and is often adopted to label sum outputs in adders. While not mandatory, it is a conventional and widely recognized practice. Step-by-Step Solution: Identify the sum output symbol in common literature.Match it to the Greek letter conventionally associated with summation.Hence, the correct designation is sigma. Verification / Alternative check: Typical adder block diagrams show outputs labeled Σ (sum) and Cout (carry out), reinforcing the convention. Why Other Options Are Wrong: Omega, theta, lambda, and pi are not standard labels for the sum output in adder schematics. Common Pitfalls: Assuming Σ indicates the entire multi-bit arithmetic operation rather than the single-bit sum in a cell; context matters. Final Answer: sigma

Question 2

Ripple-carry adder limitation: What is a primary disadvantage of the ripple-carry adder architecture when compared with faster adder structures?

Accepted Answer

Introduction / Context: Adder design strongly influences CPU cycle time. The ripple-carry adder (RCA) is simple and area-efficient but suffers from worst-case delay proportional to bit-width. This question identifies the chief drawback. Given Data / Assumptions: Adder built by cascading full-adder stages. Each stage must wait for the carry from the previous stage. No special carry acceleration is present. Concept / Approach: Because each carry must ripple through all preceding stages, the overall delay grows linearly with the number of bits. This makes RCAs slow compared with carry-look-ahead or prefix adders. Step-by-Step Solution: Model delay: T_total ≈ n * t_carry + t_sum(msb).As n increases, T_total increases linearly.Hence, the principal disadvantage is speed (propagation delay), not interconnect complexity. Verification / Alternative check: Performance-optimized designs employ carry-look-ahead, carry-select, or parallel-prefix adders to reduce the carry chain depth. Why Other Options Are Wrong: Interconnections are simpler, not more complex, than advanced adders.The number of stages equals the bit-width; it is not “more stages to a full adder.”“All of the above” is false because only the speed disadvantage is correct.RCAs can be cascaded to arbitrary widths; they just become slower. Common Pitfalls: Confusing simplicity (an RCA is easy to wire) with performance advantages; they are often opposites. Final Answer: It is slow due to propagation time.

Question 3

Binary long division check: When the binary number 1100010₂ is divided by 0101₂ (which equals 5₁₀), what is the remainder in decimal?

Accepted Answer

Introduction / Context: Binary division appears in CRC computation, modular arithmetic, and hardware dividers. Translating between bases can simplify mental checks of remainders and quotients. Given Data / Assumptions: Dividend: 1100010₂. Divisor: 0101₂ (which is 5 in decimal). We need the remainder, reported in decimal. Concept / Approach: Either perform binary long division or convert both numbers to decimal, divide, and convert the remainder back if needed. Since only the remainder is asked and the divisor is small, decimal conversion is efficient. Step-by-Step Solution: Convert 1100010₂ to decimal: 164 + 132 + 016 + 08 + 04 + 12 + 01 = 98.Divisor 0101₂ = 5.Compute 98 / 5: quotient = 19, remainder = 3 (since 195 = 95).Therefore, the decimal remainder is 3. Verification / Alternative check: Binary check: 98 − 19*5 = 98 − 95 = 3; the remainder is less than the divisor, confirming correctness. Why Other Options Are Wrong: 2 or 4: plausible near values but incorrect; 98 mod 5 equals 3.6: remainder must be less than 5.0: would imply divisibility by 5, which 98 is not. Common Pitfalls: Arithmetic slips in base conversion (misvaluing bit weights) or forgetting the remainder must be less than the divisor. Final Answer: 3

Question 4

2’s-complement arithmetic: Before adding or subtracting signed binary numbers, operands of different original widths must be sign-extended so that the operation occurs at the same bit width. Evaluate the claim that it is not necessary to have the sa…

Accepted Answer

Introduction / Context: Signed arithmetic in digital systems is typically implemented with 2's complement representation. Correct results require consistent bit width so that sign bits align and carries/borrows propagate properly. Given Data / Assumptions: Operands may originate with different declared widths (e.g., 5-bit and 8-bit). 2's complement addition/subtraction rules apply. Overflow detection depends on consistent sign-bit alignment. Concept / Approach: When widths differ, you must sign-extend the smaller operand to the larger width, duplicating its sign bit into higher positions. Arithmetic is then performed at a single, common width. Saying “not necessary to have the same number of bits” is misleading; operations take place at one width after proper extension. Step-by-Step Solution: Identify operand widths n and m (assume n < m).Sign-extend the n-bit operand to m bits by repeating its MSB.Perform addition/subtraction at m bits.Evaluate overflow using sign-bit rules at width m. Verification / Alternative check: HDL synthesis and ISA specifications (e.g., sign-extension on load/extend instructions) formalize this requirement for correctness and overflow semantics. Why Other Options Are Wrong: Correct: Would imply mismatched widths can be directly added without extension. Ambiguous / Cannot be determined: The rule is well-defined in 2's complement arithmetic. Common Pitfalls: Zero-extending negative numbers (wrong) or ignoring width differences, causing sign errors and incorrect overflow flags. Final Answer: Incorrect

Question 5

Repair the subtraction rule: For 1-bit binary subtraction 0 − 1, the correct result is difference = 1 with a borrow-out of 1 (not 0). Evaluate the statement “difference = 1 borrow = 0.”

Accepted Answer

Introduction / Context: Single-bit subtraction cases underpin multi-bit subtractor design. Understanding the borrow behavior for each case prevents design and analysis mistakes in arithmetic circuits. Given Data / Assumptions: We consider 1-bit subtraction with minuend M and subtrahend S, no initial borrow-in. Case examined: M = 0, S = 1. Difference bit D and borrow-out Bout follow standard rules. Concept / Approach: If the minuend bit is smaller than the subtrahend bit, a borrow is required from the next higher bit position. For 0 − 1 at the LSB with no borrow-in, the difference is 1 (after borrowing 1 as 2 in binary) and the borrow-out is 1. Therefore, claiming borrow = 0 is incorrect. Step-by-Step Solution: Write 0 − 1 with no borrow-in.Borrow from the next higher position: 10₂ − 1₂ = 1₂.Set difference D = 1.Set borrow-out Bout = 1 to indicate the borrow occurred. Verification / Alternative check: Truth table for 1-bit subtraction confirms: M=0, S=1, Bin=0 ⇒ D=1, Bout=1. Hardware half-subtractor equations yield D = M ⊕ S, Bout = ~M * S (which evaluates to 1 for 0 − 1). Why Other Options Are Wrong: Correct: Contradicts the established subtractor behavior. Ambiguous / Cannot be determined: The case is explicit and standard. Common Pitfalls: Confusing borrow with carry, or forgetting that 0 − 1 requires borrowing even in the least significant position. Final Answer: Incorrect

Question 6

Adder performance: A carry look-ahead (CLA) adder uses extra logic to compute carries in parallel and is generally faster than a ripple-carry adder. Evaluate the claim that a look-ahead-carry adder is slower because it requires additional logic.

Accepted Answer

Introduction / Context: Adder architectures trade off complexity and speed. Ripple-carry adders propagate carry serially through each bit, while carry look-ahead (CLA) adders compute carry signals in parallel using generate/propagate logic to reduce delay. Given Data / Assumptions: Ripple adders have O(n) carry-chain delay for n bits. CLA adds extra logic (generate/propagate) to shorten the critical path. We compare speed, not just gate count. Concept / Approach: CLA adds logic to compute carries simultaneously across groups, transforming linear carry delay into a shorter, usually logarithmic or block-based delay. Although hardware cost rises, the net effect is faster addition for moderate and wide bit-widths compared to ripple schemes. Step-by-Step Solution: Define G = A * B (generate), P = A ⊕ B (propagate).Use CLA equations to compute C1, C2, ... in parallel from Cin and the G/P terms.Observe reduced dependency chain vs. bit-by-bit ripple.Conclude CLA is faster despite additional logic. Verification / Alternative check: ASIC/FPGA timing models consistently show CLA outperforming ripple for nontrivial widths; further enhancements (carry-select, carry-skip, parallel-prefix) push speed even higher. Why Other Options Are Wrong: Correct: Misinterprets added logic as a speed penalty; it is a speed optimization. Ambiguous / Cannot be determined: The architectural comparison is well-established. Common Pitfalls: Equating gate count with speed; in arithmetic circuits, structured parallelism often improves timing at the cost of area. Final Answer: Incorrect

Question 7

Single-bit subtraction coverage: There are four basic input combinations for subtracting two single-bit binary numbers (minuend, subtrahend) when considered case-by-case. Evaluate this statement.

Accepted Answer

Introduction / Context: Designing half- and full-subtractors starts with enumerating all 1-bit cases for the minuend (M) and subtrahend (S). Recognizing the complete set of cases avoids mistakes when deriving logic for the difference and borrow outputs. Given Data / Assumptions: M and S are single bits: each ∈ {0, 1}. No initial borrow-in considered (half-subtractor perspective). We classify cases by the (M, S) pair. Concept / Approach: With two binary inputs, there are 2^2 = 4 possible combinations. For subtraction, these are: 0−0, 0−1, 1−0, 1−1. Each case yields a specific difference bit and borrow-out pattern, forming the canonical truth table for the half-subtractor. Step-by-Step Solution: List cases: (M,S) = (0,0), (0,1), (1,0), (1,1).Compute differences and borrows for each case.Assemble the truth table to implement logic equations.Use equations in larger subtractor designs (with borrow-in/out). Verification / Alternative check: The complete table leads to the well-known relations: D = M ⊕ S (for half-subtractor) and Bout = ~M * S. Extending to full-subtractor adds a borrow-in term but preserves the base four cases per (M,S). Why Other Options Are Wrong: Incorrect: Understates or overstates the number of base cases for two inputs. Ambiguous / Cannot be determined: The enumeration is straightforward from combinatorics. Common Pitfalls: Forgetting to treat the borrow-in separately for full-subtractors; the base (M,S) cases remain four, with additional combinations when Bin is included. Final Answer: Correct

Question 8

Device fact check: The 74LS382 arithmetic/logic unit (ALU) is packaged as a 24-pin device. Evaluate this statement.

Accepted Answer

Introduction / Context: Members of the classic TTL ALU family (e.g., 74xx181/182 and related parts) are commonly found in 24-pin DIPs or equivalent SMT outlines. This question tests factual recall about the 74LS382's pin count rather than its function set. Given Data / Assumptions: 74LS382 is a 4-bit ALU-class device in the LS TTL family. Comparable ALUs (e.g., 74181) are 24-pin packages in DIP form. We assume standard packaging consistent with typical ALU devices. Concept / Approach: ALUs integrate multiple arithmetic and logic functions with carry/flag pins, requiring a relatively high pin count (operands, function selects, carries, outputs, power/ground). A 24-pin outline is therefore consistent with expectations for such devices and historically documented family members. Step-by-Step Solution: Identify typical ALU signals: A[3:0], B[3:0], F[3:0], function selects, carry in/out, mode, power/ground.Estimate needed pins; the total is consistent with ≈24 pins.Relate to known 74xx ALUs packaged in 24-pin DIP outlines.Conclude that a 24-pin package statement is consistent for 74LS382. Verification / Alternative check: Cross-reference with historical TTL catalogs shows ALU devices in 24-pin packages, matching the statement's characterization of the 74LS382. Why Other Options Are Wrong: Incorrect: Would contradict standard packaging conventions for ALUs of this era. Ambiguous / Cannot be determined: Package information is standard and widely reported. Common Pitfalls: Confusing the 74LS382 with other 74xx logic devices that use 14- or 16-pin packages; ALUs need more I/O and thus more pins. Final Answer: Correct

Question 9

ALU status flags: In a 4-bit ALU, overflow indicators (status flags) assert when addition or subtraction produces a result that does not fit in four bits according to the representation in use. Evaluate this statement.

Accepted Answer

Introduction / Context: Status flags (zero, carry/borrow, sign, overflow) summarize arithmetic outcomes. The overflow flag specifically signals that the computed result falls outside the representable range for the chosen width and signedness, which is essential for correct program flow or hardware state transitions. Given Data / Assumptions: ALU width is 4 bits for this discussion. Operations considered: addition and subtraction. Overflow semantics depend on signed interpretation; carry relates to unsigned arithmetic. Concept / Approach: For signed 2's complement, overflow occurs when adding two positives yields a negative, or two negatives yield a positive. In 4 bits, the representable signed range is −8 to +7; results outside this range set overflow. For unsigned, carry-out indicates out-of-range, while a separate overflow definition may be unused; the question's wording focuses on the general idea that results can exceed 4-bit capacity and that indicators exist for this condition. Step-by-Step Solution: Establish 4-bit ranges: unsigned 0..15; signed −8..+7.Check operation result against the relevant range.Set overflow (signed) or carry/borrow (unsigned) when range is exceeded.Use flags to trigger corrective handling (e.g., saturation, exceptions). Verification / Alternative check: Standard ALUs and CPU ISAs define these flags; 4-bit educational ALUs expose similar indicators for teaching purposes. Why Other Options Are Wrong: Incorrect: Ignores established flag semantics. Ambiguous / Cannot be determined: The behavior is well-specified in ALU designs. Common Pitfalls: Confusing carry with overflow; carry is an unsigned concept, while overflow is a signed-range violation. Both can inform about out-of-range conditions depending on interpretation. Final Answer: Correct

Question 10

Binary result format (fundamentals): In binary arithmetic, the written result (sum) is expressed using only the digits 0 and 1. Evaluate this statement.

Accepted Answer

Introduction / Context: This question checks the most fundamental property of binary notation: regardless of how addition is carried out internally (with carries and intermediate logic), the written result in base-2 uses only the symbols 0 and 1. Given Data / Assumptions: We are discussing standard base-2 positional representation. Arithmetic may generate carries between bit positions. We care about the final, recorded digits of the sum. Concept / Approach: Binary numbers are expressed with two symbols. During addition, each bit position computes a sum bit (0 or 1) and possibly a carry to the next position. No matter how many carries propagate, each output digit remains 0 or 1. The process may widen the word (e.g., an extra most-significant bit), but the alphabet of symbols does not change. Step-by-Step Solution: Add bitwise with carry: sum_bit = A xor B xor carry_in (0 or 1).Compute carry_out = majority(A, B, carry_in) (0 or 1).Concatenate all sum_bits (and a final carry if present) to form the base-2 result, which uses only 0 and 1. Verification / Alternative check: Example: 1011 + 0111 = 10010. Even with multiple carries, the digits in the result are still only 0s and 1s. Why Other Options Are Wrong: Incorrect: Contradicts the definition of base-2 representation.Only correct without carry: Carries do not introduce new symbols; they only extend width.Ambiguous for signed numbers: Signed encodings (two's complement) still use 0/1 digits.Insufficient context: The statement is precise and standard. Common Pitfalls: Confusing intermediate analog voltages or adder internals with the symbolic numeric result. Final Answer: Correct

Question 11

Full adder definition check: A full adder adds two data bits and, by definition, includes both a carry input and a carry output. Evaluate the claim that it “need not” have carry in/out.

Accepted Answer

Introduction / Context: This item distinguishes a full adder from a half adder. A full adder is the standard 1-bit building block used to construct multi-bit adders because it accepts a carry in (Cin) and produces a carry out (Cout). Given Data / Assumptions: Terminology: half adder vs full adder. Target: definition, not a specific IC. Concept / Approach: A half adder adds two bits (A, B) but has no Cin. A full adder adds A, B, and Cin, producing Sum and Cout. Multi-bit adders cascade full adders by wiring each stage’s Cout to the next stage’s Cin, enabling n-bit arithmetic. Step-by-Step Solution: Identify the device in question: “full adder.”Recall its interfaces: inputs A, B, Cin; outputs Sum, Cout.Compare with the claim: “need not have carry in/out” → contradicts definition → evaluation is “Incorrect.” Verification / Alternative check: Common logic equations: Sum = A xor B xor Cin; Cout = majority(A, B, Cin). These require Cin and produce Cout. Why Other Options Are Wrong: Correct: Would match a half adder, not a full adder.Only correct for ripple chains: Ripple needs Cin/Cout explicitly.Ambiguous/Implementation-dependent: The definition is standard, not optional. Common Pitfalls: Confusing half adders with full adders. Final Answer: Incorrect

Question 12

ALU scalability: Evaluate the statement: “Arithmetic logic unit (ALU) slices cannot be cascaded to operate on more than four bits.”

Accepted Answer

Introduction / Context: Classic CPUs and many TTL/MSI designs build wide datapaths by cascading smaller ALU slices (e.g., 4-bit units) with carry connections. The 74181 is a well-known historical example used to assemble 8-, 12-, or 16-bit ALUs. Given Data / Assumptions: ALU “slice” width is often 1, 2, or 4 bits. Cascading uses carry chains or look-ahead logic. Concept / Approach: To scale width, chain Cout of a lower slice into Cin of the next. Performance can be improved with carry look-ahead, carry-skip, or carry-select techniques. Therefore, not only is cascading possible, it is a standard approach in modular ALU design. Step-by-Step Solution: Identify claim: “cannot be cascaded.”Recall standard practice: 4-bit slices commonly chained.Conclude claim contradicts established design patterns → “Incorrect.” Verification / Alternative check: Datasheets and textbooks show ripple and look-ahead chains of 74181 slices forming wider ALUs. Why Other Options Are Wrong: Correct / serial-only: Both incorrect; parallel slice cascading is common.Ambiguous without timing: Timing affects speed, not possibility. Common Pitfalls: Equating four-bit demo circuits with inherent architectural limits. Final Answer: Incorrect

Question 13

HDL organization (VHDL context): “Constants must be included in a package.” Evaluate this statement about design organization.

Accepted Answer

Introduction / Context: Hardware description languages (e.g., VHDL) allow constants to be declared in various scopes. Packages are a recommended place to share constants across design units, but they are not the only legal place to define them. Given Data / Assumptions: Context is VHDL-style packaging of shared declarations. Claim uses the word “must.” Concept / Approach: In VHDL, constants can be declared inside an architecture, a package, or other declarative regions depending on desired visibility and reuse. Placing widely reused constants in a package improves modularity, but it is not mandatory. Therefore the statement, as written, is incorrect. Step-by-Step Solution: Identify that “must” implies a rule.Recall VHDL permits constants in multiple scopes.Conclude the requirement is not compulsory → “Incorrect.” Verification / Alternative check: Synthesis and simulation tools accept constants declared locally in architectures or entities. Packages simply expose them project-wide. Why Other Options Are Wrong: Correct / generics-only / toolchain-specific: Packages are good practice, not a must.Ambiguous across HDLs: Even in other HDLs, constants are not forced into packages. Common Pitfalls: Confusing best practice (“should”) with language requirement (“must”). Final Answer: Incorrect

Question 14

Adder notation: “The carry-out of a binary adder is denoted by the summation symbol (Σ).” Evaluate this statement.

Accepted Answer

Introduction / Context: In adder schematics and block diagrams, the symbol Σ commonly labels the sum function/output, not the carry. Carry signals are typically labeled Cin and Cout (or Cn and Cn+1) explicitly. Given Data / Assumptions: Standard adder notation: Σ for sum, C for carry. The claim conflates these roles. Concept / Approach: A one-bit full adder implements Sum = A xor B xor Cin and Cout = majority(A, B, Cin). Documentation and textbooks identify Σ with the Sum output; Cout is a distinct signal and is not denoted by Σ. Step-by-Step Solution: Check conventional labeling: Σ → sum output S.Carry uses C-labels (Cin/Cout) or explicit “carry.”Therefore, the statement is incorrect. Verification / Alternative check: Datasheets (e.g., 74xx/HC283 adders) show S outputs and a separate Cout pin; Σ is associated with summation, not carry. Why Other Options Are Wrong: Correct / parallel-only / ambiguous: All mischaracterize standard symbols.Insufficient context: The notation is broadly consistent. Common Pitfalls: Misreading Σ as “anything related to addition,” including carry. It designates the sum. Final Answer: Incorrect

Question 15

2-bit vector addition (A as MSB, B as LSB): Given [A] = 10 and [B] = 01 (both 2-bit unsigned), compute [A] + [B] as a 2-bit sum with possible carry.

Accepted Answer

Introduction / Context: This problem practices basic 2-bit binary addition using vector notation. We treat 10 as decimal 2 and 01 as decimal 1, both unsigned, and compute their sum. Given Data / Assumptions: 2-bit vectors: [A] = 10 (value 2), [B] = 01 (value 1). Unsigned addition; A is the most significant bit (MSB). Report the 2-bit sum (and note if a carry would appear). Concept / Approach: Add bitwise from LSB to MSB, propagating carry as needed. 2 + 1 = 3, which in binary is 11. Since 11 fits in 2 bits, no extra carry beyond the 2-bit field is needed. Step-by-Step Solution: LSB column: 0 + 1 = 1, carry = 0.MSB column: 1 + 0 + carry 0 = 1.Concatenate: result = 11 (decimal 3). Verification / Alternative check: Decimal check: 2 + 1 = 3 → binary 11. Why Other Options Are Wrong: [00]: Would imply 0, not 3.[10]: Equals 2, the augend, not the sum.[01]: Equals 1, the addend, not the sum.[11] with carry = 0: Redundant wording; the plain [11] already implies no extra carry. Common Pitfalls: Reversing bit order or misreading which bit is MSB. Final Answer: [11]

Question 16

Binary subtraction method used in computers: Evaluate the statement: “Digital computers use an easier subtraction method called one's complement.”

Accepted Answer

Introduction / Context: Modern binary arithmetic for signed integers overwhelmingly uses two's complement. Subtraction A − B is performed as A + (two's complement of B), which unifies add/sub hardware. Given Data / Assumptions: We consider mainstream digital computers. Signed integer encoding choice impacts subtraction method. Concept / Approach: Two's complement of B is formed by bitwise complement (one's complement) plus 1. While one's complement exists historically, it suffers from negative zero and complicates arithmetic. Two's complement removes negative zero and simplifies overflow rules, which is why it is standard. Step-by-Step Solution: Identify claim: “one's complement is the easier method.”Recall standard practice: two's complement for subtraction and signed representation.Conclude the statement is incorrect for modern computers. Verification / Alternative check: Instruction sets and ALUs implement add with an inverted operand plus one for subtraction, i.e., two's complement. Why Other Options Are Wrong: Correct/legacy/toolchain-dependent: Mainstream hardware uses two's complement; one's complement is rare and outdated.Ambiguous sign-magnitude: Not standard for arithmetic units. Common Pitfalls: Confusing the intermediate “invert bits” step (one's complement) with the complete two's complement operation. Final Answer: Incorrect

Question 17

Device recognition (TTL/CMOS): “74HC283” corresponds to which function in standard logic families?

Accepted Answer

Introduction / Context: Recognizing common 74xx/HC device numbers is useful in digital design. The 74HC283 (and 74LS283) are classic 4-bit binary adders providing both Cin and Cout pins for cascading. Given Data / Assumptions: Device code: 74HC283. Family: HC (High-speed CMOS), functionally similar to LS283 in TTL. Concept / Approach: The 74HC283 implements a 4-bit parallel adder. It adds two 4-bit operands plus a carry in to produce a 4-bit sum and a carry out, enabling simple ripple or look-ahead architectures for wider adders. Step-by-Step Solution: Match IC number to function from standard catalogs.Confirm presence of Cin and Cout pins for cascading.Identify it as a 4-bit full adder device. Verification / Alternative check: Datasheets list pinouts for A3..A0, B3..B0, S3..S0, Cin, and Cout, confirming adder functionality. Why Other Options Are Wrong: Decoder/demux, NAND, flip-flop, comparator: Different 74xx numbers correspond to those functions, not 74HC283. Common Pitfalls: Confusing 74HC283 (adder) with 74HC138 (3-to-8 decoder) or 74HC00 (quad NAND). Final Answer: 4-bit binary adder with carry in/out

Question 18

Binary addition check: Compute 0101 + 1111 and evaluate the proposed result 10100.

Accepted Answer

Introduction / Context: This verifies a straightforward binary addition. We add 0101 (5) to 1111 (15) and check whether the sum equals 10100 (20). Given Data / Assumptions: Operands are unsigned binary vectors. No fixed word-size truncation is imposed. Concept / Approach: Binary addition proceeds bitwise with carries. If the result exceeds the current word width, we extend the width to include the final carry, producing a wider sum. Step-by-Step Solution: LSB: 1 + 1 = 0, carry 1.Next: 0 + 1 + carry 1 = 0, carry 1.Next: 1 + 1 + carry 1 = 1, carry 1.MSB: 0 + 1 + carry 1 = 0, carry 1 → write final carry as new MSB.Result: 10100 (decimal 20). Verification / Alternative check: Decimal check: 5 + 15 = 20 → binary 10100. Why Other Options Are Wrong: Incorrect: Contradicts standard addition.Saturation/word-size ambiguity: Only relevant if a fixed 4-bit result is required; with natural-width arithmetic, 5-bit result is correct.BCD correction: Not applicable; operands are pure binary, not BCD. Common Pitfalls: Forgetting to emit the final carry as the new MSB. Final Answer: Correct

Question 19

BCD addition rule: Evaluate the BCD sum of 0101 (5) and 0100 (4). Is the result 0000 1001 (i.e., 00001001 BCD)?

Accepted Answer

Introduction / Context: Binary-coded decimal (BCD) encodes each decimal digit in a 4-bit nibble. We add 5 (0101) and 4 (0100) and verify the BCD correction rules and final encoding. Given Data / Assumptions: Operands are single-digit BCD values: 0101 and 0100. We use standard BCD correction (add 0110 when a nibble exceeds 1001). We present the result in 8-bit form as 0000 1001 for a one-digit sum. Concept / Approach: First perform binary addition of the low nibble. If the nibble sum is greater than 1001 (9) or a carry is generated, add 0110 to correct to valid BCD. Here, 0101 + 0100 = 1001 (9), which is already a valid BCD digit; no correction is needed. Step-by-Step Solution: Compute binary nibble sum: 0101 + 0100 = 1001.Check validity: 1001 ≤ 1001 and no nibble carry → valid BCD.Place result in 8-bit BCD: upper nibble 0000, lower nibble 1001 → 0000 1001. Verification / Alternative check: Decimal check: 5 + 4 = 9 → BCD 1001; extended to 8 bits as 0000 1001. Why Other Options Are Wrong: Incorrect / needs adjust: No adjust is needed because the nibble is ≤ 9.Correct only after adding 0110: Adding 0110 would corrupt a valid digit.Ambiguous without nibbles: Problem explicitly uses single-digit BCD. Common Pitfalls: Applying BCD correction even when the nibble is already valid. Final Answer: Correct

Question 20

Binary multiplication check – is the product of 1011 (base 2) and 0110 (base 2) equal to 01100110?

Accepted Answer

Introduction / Context: This item checks binary multiplication and result verification. Multiplying two unsigned binary numbers follows the same partial-products idea as decimal multiplication, but with base 2 digits (0/1). The claim proposes a specific product; we must compute and compare. Given Data / Assumptions: Multiplicand: 1011 (base 2) = 11 (decimal). Multiplier: 0110 (base 2) = 6 (decimal). Unsigned multiplication; no overflow truncation assumed. Exact equality to 01100110 is asserted. Concept / Approach: Convert to decimal (sanity check) or perform binary partial products. 11 * 6 = 66 (decimal). The correct 66 in binary is 1000010 (base 2). The proposed 01100110 equals 102 (decimal), which does not match 66, so the statement is wrong. Step-by-Step Solution: 1) 1011 * 0110 = 1011 * (2 + 4) = (1011 << 1) + (1011 << 2).2) (1011 << 1) = 10110; (1011 << 2) = 101100.3) Sum: 10110 + 101100 = 1000010 (binary) = 66 (decimal).4) Proposed result 01100110 = 102 (decimal) ≠ 66. Verification / Alternative check: Decimal first: 11 * 6 = 66. Convert 66 to binary by successive division: 66 = 64 + 2 → 1000010. Matches the computed binary sum, confirming correctness of our evaluation. Why Other Options Are Wrong: “Correct” is false because 01100110 ≠ 66. “Only if leading zeros are kept” is irrelevant; leading zeros change width, not value. “Only with signed operands” does not change 11 * 6. “Depends on bit-width wrapping” would matter only if truncation were specified; it was not, and even then, the given pattern still doesn’t match 66 exactly. Common Pitfalls: Misreading 01100110 as 66 due to hex association (0x66 = 102). Also, forgetting that shifting left by k equals multiplying by 2^k. Final Answer: Incorrect

Digital Arithmetic Operations and Circuits Questions