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

It is not necessary to have the same number of bits when adding or subtracting signed binary numbers in the 2's-complement system.

Accepted Answer

False

Question 5

The binary subtraction 0 – 1 = is difference = 1 borrow = 0

Accepted Answer

False

Question 6

The look-ahead-carry adder is slower than the ripple-carry adder because it requires additional logic circuits.

Accepted Answer

False

Question 7

There are four possible combinations for subtracting two binary numbers.

Accepted Answer

True

Question 8

The 74LS382 ALU is a 24-pin arithmetic/logic unit.

Accepted Answer

False

Question 9

Overflow indicators in ALU circuits indicate when add or subtract operations produce results that are too large to fit into four bits.

Accepted Answer

True

Question 10

A binary sum is made up of only 1s and 0s.

Accepted Answer

True

Question 11

Full adders can add two numbers and need not have a carry input or a carry output.

Accepted Answer

False

Question 12

ALU circuits cannot be cascaded to perform functions on more than four bits.

Accepted Answer

False

Question 13

Constants must be included in a package.

Accepted Answer

True

Question 14

The carry-out of a binary adder is identified using the summation symbol, sigma.

Accepted Answer

False

Question 15

If [A] = 10 and [B] = 01, then [A] + [B] = [ ].

Accepted Answer

False

Question 16

Digital computers use an easier method to subtract binary numbers, called one's complement.

Accepted Answer

False

Question 17

A 74HC283 can be used to implement a 4-bit full adder.

Accepted Answer

True

Question 18

The solution to the binary problem 0101 + 1111 is 10100.

Accepted Answer

True

Question 19

The solution to the BCD problem 0101 + 0100 is 00001001BCD.

Accepted Answer

True

Question 20

The solution to the binary problem 1011 × 0110 is 01100110.

Accepted Answer

False

Digital Arithmetic Operations and Circuits Questions