Question 1

Exclusive-NOR as an equality detector Why is an exclusive-NOR (XNOR) gate also referred to as an equality gate?

Accepted Answer

Introduction / Context: In digital logic, comparators detect when two signals match. The exclusive-NOR (XNOR) gate implements the simplest 1-bit equality function, making it a building block for multi-bit equality comparators. Given Data / Assumptions: XNOR truth table yields 1 when inputs are the same (00 or 11). Opposite inputs (01 or 10) produce 0. Inputs are single-bit logical values. Concept / Approach: Equality means A equals B. The Boolean function for equality is X = A XNOR B. This can also be written as X = (A AND B) OR (NOT A AND NOT B). Step-by-Step Solution: List pairs: (0,0) → 1, (0,1) → 0, (1,0) → 0, (1,1) → 1.Observe that output is 1 exactly when inputs match.Therefore, the device directly tests for equality of two bits.This property generalizes: cascading bitwise XNORs and ANDing the results yields n-bit equality. Verification / Alternative check: Algebraic form X = A*B + A'*B' matches the equality condition because each product term covers one of the equal-input cases. Why Other Options Are Wrong: The output is false if the inputs are equal: This describes XOR, not XNOR. The output is true if the inputs are opposite: Also describes XOR, not XNOR. Common Pitfalls: Confusing XOR and XNOR symbols or names, and forgetting that the bar (or bubble) on XOR indicates inversion of the XOR result. Final Answer: The output is true if the inputs are equal.

Question 2

Odd parity assignment for data words For each 7- or 8-bit data word, determine the required parity bit P to achieve odd parity: • 1011101 • 11110111 • 1001101

Accepted Answer

Introduction / Context: Parity bits provide simple error detection over a set of data bits. For odd parity, the total number of 1s—including the parity bit—must be odd. This question asks you to compute the parity bit P for three separate data words. Given Data / Assumptions: Words: 1011101, 11110111, 1001101. Odd parity required: total count of 1s (data + P) must be odd. We append one bit P to each word independently. Concept / Approach: Count the number of 1s in each data word. If the count is already odd, set P = 0. If the count is even, set P = 1 to make the total odd. This follows directly from parity definition and requires only a popcount of each word. Step-by-Step Solution: Word 1: 1011101 has 1s at positions 1,3,4,5,7 → 5 ones (odd) → P = 0.Word 2: 11110111 has ones count 7 (odd) → P = 0.Word 3: 1001101 has ones count 4 (even) → P = 1.Therefore results: P = 0, P = 0, P = 1. Verification / Alternative check: Add P and recount: Word 1 total ones 5, Word 2 total ones 7, Word 3 total ones 5 → all odd, confirming correctness. Why Other Options Are Wrong: P = 1, 1, 0 and P = 1, 1, 1: Do not satisfy odd parity for the words with already-odd one counts. P = 0, 0, 0: Fails for the third word which has an even number of ones and requires P = 1. Common Pitfalls: Miscounting ones, assuming the same parity bit for all words, or forgetting that each word’s parity is computed independently. Final Answer: P = 0, P = 0, P = 1

Question 3

Identify the equality gate Which logic gate outputs HIGH when both inputs are HIGH or both inputs are LOW (i.e., when the inputs are equal)?

Accepted Answer

Introduction / Context: Digital systems frequently need to compare two signals. The gate that outputs HIGH when the inputs are the same implements equality at the 1-bit level and is commonly used in comparators and parity circuits. Given Data / Assumptions: Two binary inputs A and B. We seek a gate that asserts when A equals B (00 or 11). Concept / Approach: The exclusive-NOR (XNOR) function is 1 exactly when inputs match. Boolean form: X = A*B + A'B'. This is the complement of XOR, which is 1 when inputs differ. Step-by-Step Solution: Evaluate truth table pairs: 00 → 1, 01 → 0, 10 → 0, 11 → 1.This matches the equality condition.Therefore, the equality gate is XNOR.Other common gates (OR, NAND) do not meet this pattern. Verification / Alternative check: Construct XNOR from simpler gates: two ANDs (AB and A'*B') feeding an OR implements equality, confirming the behavior. Why Other Options Are Wrong: OR gate: Produces 0 only for 00; does not assert for 11 exclusively. Ex-OR gate: Asserts when inputs differ (01 or 10). NAND gate: Produces 0 only for 11; not an equality detector. Common Pitfalls: Confusing XOR with XNOR because of similar names; remember XNOR is the complemented XOR and corresponds to equality. Final Answer: Ex-NOR gate

Question 4

Parity system basics Parity systems are defined as either ________ or ________, and they add an extra ________ to the digital information being transmitted.

Accepted Answer

Introduction / Context: Parity is a simple and widely used error-detection technique. Depending on whether the system uses odd or even parity, one extra bit is appended to ensure the total number of 1s in a word is odd or even, respectively. Given Data / Assumptions: Digital data words are sequences of bits. We add exactly one parity bit per word. Two common schemes: odd parity and even parity. Concept / Approach: In even parity, the parity bit is chosen so that data+parity contains an even count of 1s. In odd parity, it is chosen so the count is odd. This single-bit redundancy allows detection of all single-bit errors within a word (and some multi-bit patterns depending on the error model). Step-by-Step Solution: Compute the number of 1s (popcount) in the data word.If even parity: set P = 0 for even counts; P = 1 for odd counts.If odd parity: set P = 1 for even counts; P = 0 for odd counts.Append this parity bit to the data and transmit/stored the composite word. Verification / Alternative check: After adding parity, recount the ones. For even parity, the total must be even; for odd parity, total must be odd. Any single-bit flip will toggle the total's parity and thus be detected. Why Other Options Are Wrong: positive, negative, byte: Parity is not described by polarities and does not add a whole byte. upper, lower, digit: Not relevant terminology for parity. on, off, decimal: Mixed and incorrect concepts unrelated to parity bits. Common Pitfalls: Confusing parity with checksums or CRCs, or believing parity can correct (it only detects) single-bit errors. Final Answer: odd, even, bit

Question 5

Generating odd vs even parity How is an odd-parity bit generated differently from an even-parity bit for the same data word?

Accepted Answer

Introduction / Context: Odd and even parity differ only in the chosen value of the parity bit for a given data word. Understanding the simple relationship between them helps when designing parity-generation hardware using XOR networks. Given Data / Assumptions: Parity bit is a single bit appended to the data word. Even parity requires an even total count of 1s; odd parity requires an odd total. Even-parity bit can be formed as the XOR (mod-2 sum) of all data bits. Concept / Approach: Because the even-parity bit equals XOR of the data bits, the odd-parity bit is simply the inversion of that result. Hardware-wise, you can generate even parity with a tree of XOR gates and then pass it through a NOT gate to obtain odd parity. Step-by-Step Solution: Let Peven = XOR of all data bits.For odd parity, set Podd = NOT(Peven).Append the chosen parity bit to the data word as required by the protocol.Verify by counting ones: data + Peven → even total; data + Podd → odd total. Verification / Alternative check: Try a sample word with an odd number of 1s. Even parity produces 1 to make the total even; inverting yields 0 for odd parity, preserving the odd total. The complement relationship holds for all words. Why Other Options Are Wrong: The first output is inverted: There is no notion of a “first” output; parity is a single bit. Use AND in place of XOR: AND does not produce a parity (mod-2) sum. Double the parity field: Parity uses one bit, not two. Common Pitfalls: Confusing parity generation with majority or checksum logic, or forgetting that XOR implements addition modulo 2 across bits. Final Answer: The parity bit is the logical NOT of the even-parity bit (invert the even-parity output).

Question 6

Using XOR as a controlled inverter Show from the truth table how an exclusive-OR (XOR) gate can invert the data on one input when the other input acts as a control.

Accepted Answer

Introduction / Context: The XOR function is invaluable for conditional inversion. Treating one input as a control line turns XOR into a 1-bit multiplexer between “pass-through” and “invert” behaviors, used in arithmetic units and data encoding. Given Data / Assumptions: XOR definition: X = A XOR B = 1 when inputs differ, 0 when equal. Let A be the control input; B is the data input. We want X to equal B when A = 0, and X to equal NOT B when A = 1. Concept / Approach: From the XOR truth table: X = A XOR B = (NOT A AND B) OR (A AND NOT B). This directly encodes pass-through for A = 0 and inversion for A = 1. Step-by-Step Solution: If A = 0: X = 0 XOR B = B → pass-through.If A = 1: X = 1 XOR B = NOT B → inversion.Truth-table rows: (A,B) = (0,0)→0, (0,1)→1 (same as B); (1,0)→1, (1,1)→0 (inverted).Therefore XOR acts as a controlled inverter with A as the control. Verification / Alternative check: Algebraic rewrite: X = B XOR A = B when A = 0; X = NOT B when A = 1. Hardware implementations of adders use this to conditionally complement operands. Why Other Options Are Wrong: Same as B for both A values: Ignores XOR behavior; that would be a buffer. Inverse when A = 0 and same when A = 1: Reversed from XOR’s property. Inverse for both A values: That would be a pure inverter, not controlled. Common Pitfalls: Swapping the role of control and data, or confusing XOR with XNOR (which is equality, not inversion control). Final Answer: Using A as the control, when A = 0, X is the same as B. When A = 1, X is the inverse of B.

Question 7

Identify “one-or-the-other” logic Which logic circuit outputs HIGH if one input or the other input, but not both, is HIGH?

Accepted Answer

Introduction / Context: Sometimes we require a signal that indicates exclusivity: exactly one of two conditions is true. The exclusive-OR (XOR) gate implements this function at the 1-bit level and is widely used in arithmetic and parity generation. Given Data / Assumptions: Two inputs A and B. Desired behavior: X = 1 when A ≠ B; X = 0 when A = B. Concept / Approach: The XOR truth table is 1 for input pairs 01 and 10, and 0 for 00 and 11. This matches “one or the other, but not both.” Step-by-Step Solution: List cases: A=0,B=0 → X=0; A=0,B=1 → X=1.A=1,B=0 → X=1; A=1,B=1 → X=0.Therefore XOR provides the exclusive condition.Other common gates (OR, NAND, XNOR) do not satisfy exclusivity. Verification / Alternative check: Boolean identity: X = A XOR B = (A + B) * (NOT(A*B)). It is OR gated with the complement of AND, ensuring not both. Why Other Options Are Wrong: Ex-NOR: True for equality (both same), opposite of exclusivity. OR: True for 11 as well, violating “not both.” NAND: False only when both are 1; not the exclusive function. Common Pitfalls: Assuming OR implies exclusivity; OR includes the both-true case, whereas XOR excludes it. Final Answer: Ex-OR gate

Question 8

Error detection in digital communication — understanding parity checking Which of the following statements best describes the parity method of error detection used during code transmission between two locations?

Accepted Answer

Introduction / Context: Parity checking is a classic, low-cost error-detection technique widely used in memory systems and serial links. It appends one extra bit (parity bit) to a data word so that the total number of 1s becomes either even (even parity) or odd (odd parity). This question tests whether you know what kind of errors parity can reliably detect and what it cannot. Given Data / Assumptions: The channel might flip bits during transmission. Only a single parity bit is added to each data word. No additional redundancy (like CRC or ECC) is present. Concept / Approach: Parity adds one bit to enforce a parity rule. If exactly one bit in the data word flips, the total number of 1s no longer satisfies the rule and the receiver flags an error. However, if two bits flip, the parity may still appear correct (an even number of errors can cancel out), so parity cannot reliably detect double-bit errors, nor can it locate or correct any error by itself. Step-by-Step Solution: 1) Define parity: add 1 bit so that total count of 1s is even (even parity) or odd (odd parity).2) Consider a single-bit error: parity changes from valid to invalid → detection succeeds.3) Consider a double-bit error: two flips can keep parity valid → detection may fail.4) Correction requires locating the flipped bit(s); a single parity bit cannot do this. Verification / Alternative check: Try a concrete example: with even parity, data 1011 (three 1s) gets parity 1 to make total 4 (even). If one bit flips, total becomes odd → error detected. If two bits flip, total returns to even → error can slip through. Why Other Options Are Wrong: Double-bit detection (option a): parity cannot guarantee detection of two errors. Not suitable for single-bit errors (option b): the opposite is true; it is ideal for single-bit error detection. Detecting and correcting (option d): correction needs codes like Hamming; parity only detects. Common Pitfalls: Confusing detection with correction; assuming parity works for all multi-bit errors; forgetting that parity gives zero information about which bit flipped. Final Answer: Parity checking is best suited for detecting single-bit errors in transmitted codes.

Question 9

Naming the EX-NOR logic function The Exclusive-NOR (EX-NOR) gate is sometimes also referred to by which descriptive names in digital logic?

Accepted Answer

Introduction / Context: Logic functions often have descriptive aliases that highlight their behavior. The EX-NOR gate outputs HIGH when its inputs are the same, making it valuable in comparators and parity checking. Recognizing these common names helps when reading datasheets and design notes from different vendors. Given Data / Assumptions: We are discussing the EX-NOR (Exclusive-NOR) operation. Two-input case is the most common reference for naming. Terminology should reflect the function's behavior. Concept / Approach: For a two-input EX-NOR, the output is 1 if inputs are equal (00 or 11) and 0 if they differ (01 or 10). Because it reports equality, it is often called an equality gate. In parity circuits, an EX-NOR outputs 1 when an even number of inputs are 1s (for two inputs, 0 or 2 ones), matching an even-parity detector—hence the name parity gate. Step-by-Step Solution: 1) Write truth table pairs: 00 → 1, 01 → 0, 10 → 0, 11 → 1.2) Observe that output is HIGH when inputs are the same → equality behavior.3) Note that the count of ones is even at HIGH outputs → parity relation.4) Therefore, common aliases are equality gate and parity gate. Verification / Alternative check: Compare to XOR (EX-OR), which outputs 1 when inputs differ; EX-NOR is exactly the complement, reinforcing the equality interpretation. Why Other Options Are Wrong: Inverted OR: that is a NOR gate, not EX-NOR. Singular names (only parity or only equality) are partially correct, but the most complete name includes both common aliases. Common Pitfalls: Confusing EX-NOR with NOR because of the “NOR” syllable; remember EX-NOR is the complement of XOR, not of OR. Final Answer: parity gate or the equality gate

Question 10

Adding two single bits using logic Which type of logic gate directly produces the sum output when adding two one-bit operands (ignoring the carry)?

Accepted Answer

Introduction / Context: Binary addition of two single bits (A and B) produces a sum and a carry. In half adders and full adders, distinct logic functions generate these outputs. Knowing which gate corresponds to the sum output is a foundational digital design skill. Given Data / Assumptions: Operands: A and B are 1-bit values. We are asking only about the sum output, not the carry. Standard gate-level realization is assumed. Concept / Approach: The truth table for the binary sum without carry is: A B → Sum: 00→0, 01→1, 10→1, 11→0. This exactly matches the Exclusive-OR (XOR, written Ex-OR) function. The carry-out for a half adder is A * B (AND), but that is not requested here. Step-by-Step Solution: 1) Write sum truth table: 00→0, 01→1, 10→1, 11→0.2) Compare with XOR truth table: identical behavior.3) Therefore, the gate that outputs the bitwise sum is Ex-OR.4) Recognize carry is separate (AND), confirming XOR is specific to sum. Verification / Alternative check: Construct a half adder: Sum = A XOR B, Carry = A AND B. Simulation or logic-equation derivation yields the same mapping. Why Other Options Are Wrong: Ex-NOR: outputs 1 when inputs are equal (00 or 11) → not the sum. Ex-NAND: not a standard primitive function for addition. NOR: outputs 1 only when both inputs are 0. Common Pitfalls: Confusing the roles of XOR and AND in a half adder; remember XOR is for sum, AND for carry. Final Answer: Ex-OR

Question 11

The Ex-NOR is sometimes called the equality gate.

Accepted Answer

True

Question 12

The odd/even parity system would require a sixth bit to be added to a 4-bit system.

Accepted Answer

False

Question 13

In an exclusive-OR, both inputs cannot be HIGH to provide a HIGH output.

Accepted Answer

True

Question 14

Using the CPLD design environment, we can simulate any combinations of inputs and observe the resulting output to check for proper circuit operation.

Accepted Answer

True

Question 15

In a parity generator circuit, an error is signaled on an error indicator.

Accepted Answer

False

Question 16

Parity generator and checker circuits are available in single IC packages.

Accepted Answer

True

Question 17

Electrical noise does not affect the transmission of binary information.

Accepted Answer

False

Question 18

The exclusive-OR provides a LOW input if one input or the other input is HIGH.

Accepted Answer

False

Question 19

A parity checker is constructed in the same way as a parity generator, except that in a 4-bit system there must be five inputs, and the output is used as the error indicator.

Accepted Answer

True

Question 20

The exclusive-OR is written in a Boolean equation as a plus sign with a circle around it.

Accepted Answer

True

Ex-OR and Ex-NOR Gates Questions