Question 1

K-map verification of a Boolean simplification: Consider the Boolean expression X = A*C*D + A*B*(C*D + B*C). By Boolean algebra (distributive law and idempotence), this reduces to X = A*C*(B + D). A Karnaugh-map (K-map) drawn for variables A, B, C, …

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Introduction / Context: Karnaugh maps (K-maps) are a visual technique used to verify and minimize Boolean expressions. When an expression is algebraically simplified, a consistent K-map grouping must reproduce the same minimized function. Here we check whether the algebraic reduction of X = A*C*D + A*B*(C*D + B*C) to X = A*C*(B + D) is reflected correctly on a K-map implementation. Given Data / Assumptions: Variables: A, B, C, D with conventional binary interpretation. Boolean laws apply: distributive, absorption, idempotence, and commutativity. K-map cell adjacency uses Gray code ordering to allow grouping by powers of two. Concept / Approach: Start by expanding and simplifying algebraically, then map the resulting minterms on a 4-variable K-map. Consistency requires that the groups of 1-cells correspond to the product terms found in the simplified sum-of-products form. Step-by-Step Solution: Expand the bracket: X = A*C*D + A*B*C*D + A*B*B*C.Use idempotence B*B = B to get: X = A*C*D + A*B*C*D + A*B*C.Apply absorption: A*C*D absorbs A*B*C*D (the latter is more restrictive), yielding X = A*C*D + A*B*C.Factor A*C: X = A*C*(D + B) = A*C*(B + D).Thus the minimal SOP contains two implicants: A*C*B and A*C*D. Verification / Alternative check: On a 4-variable K-map, mark 1s for minterms where A=1 and C=1 and (B=1 or D=1). These cells form two groups (or one combined arrangement depending on adjacency) that correspond to A*C*B and A*C*D. The K-map therefore matches the simplified expression. Why Other Options Are Wrong: “Incorrect” contradicts the algebraic derivation. “Correct only for POS form” is irrelevant since we are discussing SOP. “Depends on Gray code ordering” is a red herring; Gray code affects cell placement, not the underlying function. Common Pitfalls: Forgetting absorption (dropping ABCD when ACD is present), confusing grouping sizes, or mislabeling K-map axes leading to wrong implicants. Final Answer: Correct

Question 2

Combinational versus memory behavior: A “combinational logic circuit” is claimed to have memory that remembers prior inputs after they are removed. Evaluate this statement in the context of digital design.

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Introduction / Context: Digital circuits fall broadly into two classes: combinational logic and sequential logic. Knowing which category a circuit belongs to determines how its outputs are interpreted and how it is verified or synthesized in hardware description languages and gate-level implementations. Given Data / Assumptions: Combinational logic: outputs depend only on present inputs (function of current values). Sequential logic: outputs depend on present inputs and past history (state), usually via feedback or storage (latches/flip-flops). No implicit storage elements are included unless specifically designed. Concept / Approach: The defining property of a combinational circuit is the absence of memory. Its output can be written as Y = f(X) for current inputs X. If memory were present, we would need state variables S and next-state equations to describe behavior (Y = g(X, S)). Therefore, stating that combinational logic “remembers” past inputs is contrary to definition. Step-by-Step Solution: Identify circuit class: “combinational logic.”Recall definition: Y depends only on current inputs, not previous values.Conclude: the claim that it remembers inputs is false.Any apparent memory arises only if external feedback or storage is added, converting it into sequential logic. Verification / Alternative check: Truth tables for logic gates (AND, OR, XOR, etc.) and for combinational blocks (adders, encoders, multiplexers) map present inputs directly to outputs; no timing history or clock is needed. In contrast, latches and flip-flops require a clock or feedback path to hold information. Why Other Options Are Wrong: “Correct” contradicts the standard definition. Hazards or high-impedance outputs do not provide memory in the Boolean sense; they are implementation details or bus states, not logic storage. Common Pitfalls: Confusing propagation delay or glitching with memory; assuming that because combinational circuits can be embedded in systems with memory, they themselves must remember data. Final Answer: Incorrect

Question 3

Full-adder sum implementation using XOR gates: In a standard 1-bit full adder, the SUM output (S) can be realized using two exclusive-OR (XOR) gates arranged so that S = A XOR B XOR Cin. Assess this claim.

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Introduction / Context: The full adder is a fundamental combinational block that sums two bits and an incoming carry. Its outputs are SUM (S) and CARRY OUT (Cout). Understanding minimal gate implementations of S helps in efficient hardware design and HDL coding. Given Data / Assumptions: Inputs: A, B, Cin. Outputs: S = A XOR B XOR Cin, and Cout = majority(A, B, Cin) = A*B + B*Cin + A*Cin. Exclusive-OR gate property: X XOR Y = (X and not Y) + (not X and Y). Concept / Approach: The XOR function is the natural expression of bitwise addition without carry. For two inputs, S2 = A XOR B. Adding Cin requires XORing that intermediate sum with Cin: S = (A XOR B) XOR Cin. This can be implemented with exactly two XOR gates in cascade. Step-by-Step Solution: Compute partial sum: P = A XOR B using one XOR gate.Compute final sum: S = P XOR Cin using a second XOR gate.No additional gates are required for S; Cout is implemented separately.Therefore, two XORs suffice for the SUM path in a full adder. Verification / Alternative check: Build the truth table for A, B, Cin and compute S via XOR chaining. The resulting S matches binary addition modulo 2, confirming the implementation. Why Other Options Are Wrong: “Incorrect” ignores the standard identity. “Only correct for carry-lookahead adders” is irrelevant; the SUM path is the same across architectures. “Requires three XOR gates” overcounts the needed logic. Common Pitfalls: Mixing SUM and CARRY logic; assuming Cout also needs XOR chaining rather than majority logic; forgetting XOR associativity (A XOR B XOR Cin is well defined). Final Answer: Correct

Question 4

Ripple-adder carry behavior: In an n-bit ripple adder, the carry output of each stage is claimed to provide an additional SUM output bit of that same stage. Judge this statement carefully.

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Introduction / Context: Ripple adders chain 1-bit full adders so that each stage’s carry out feeds the next stage’s carry in. Understanding the role of carry versus sum bits prevents architectural misunderstandings and timing misinterpretations in digital arithmetic. Given Data / Assumptions: Each 1-bit full adder produces SUM (Si) and CARRY OUT (Ci+1). Inputs at stage i: Ai, Bi, and Ci (from previous stage). Overall n-bit addition yields an n-bit sum plus a final carry (n+1th bit). Concept / Approach: The SUM bit at a stage is Si = Ai XOR Bi XOR Ci. The CARRY OUT is Ci+1 = Ai*Bi + Bi*Ci + Ai*Ci. The carry is not an additional SUM output of that stage; instead, it becomes the carry input of the next higher-order stage. Only the final carry out (from the most significant stage) can be viewed as an extra high-order bit of the overall result. Step-by-Step Solution: Write per-stage relations: Si depends on present bits and incoming carry.Carry out propagates forward; it is not reported as another SUM at the same position.Overall number of SUM bits = n; optional extra MSB arises from the final carry only.Hence, the statement that each stage’s carry provides an additional SUM bit is false. Verification / Alternative check: Add 1111 + 0001: the four SUM bits are 0000 with a final carry 1. The single extra bit appears only at the most significant end, not at every stage. Why Other Options Are Wrong: “Correct” confuses carry propagation with sum generation. References to “least-significant stage” or “end-around carry” do not change the fundamental role of stage carries in a ripple adder. Common Pitfalls: Treating carry chains as additional data outputs per stage; miscounting result width; ignoring that carry is positional and feeds only the next stage. Final Answer: Incorrect

Question 5

K-map grouping rules: “Single looping in groups of three is a common Karnaugh-map simplification technique.” Evaluate this claim about allowed group sizes in K-map minimization.

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Introduction / Context: Karnaugh maps (K-maps) are used to group adjacent 1-cells (for SOP) or 0-cells (for POS) to reduce Boolean expressions. The rules for legal group sizes are central to obtaining correct simplified results. Given Data / Assumptions: We are considering standard K-maps for 2–6 variables. Groups must be rectangular and contain 1, 2, 4, 8, 16, ... cells. Wrap-around adjacency applies across map edges due to Gray code ordering. Concept / Approach: Legal groups always have sizes that are powers of two. Grouping in threes is not permitted because it does not eliminate a consistent set of variables, and it cannot represent a valid implicant in minimized SOP/POS. Allowed group sizes correspond to removing one, two, three, or more variables per group, each time halving the number of fixed literals. Step-by-Step Solution: Identify 1-cells to be grouped (for SOP).Form the largest possible rectangular groups of size 1, 2, 4, 8, ... with edge wrap-around as needed.Each grouping produces a product term where changing variables are eliminated.Note that a “group of three” cannot be rectangular and cannot preserve the power-of-two rule; thus it is invalid. Verification / Alternative check: Try to write a product term for a supposed 3-cell group; one variable will flip within the group, defeating elimination and proving it is not a valid implicant. Why Other Options Are Wrong: “Correct” contradicts K-map fundamentals. References to don’t-care or POS do not legalize 3-sized groups; even with don’t-care cells, final chosen groups must still be powers of two. Common Pitfalls: Attempting irregular shapes, ignoring wrap-around adjacency, or choosing many small groups instead of fewer larger groups. Final Answer: Incorrect

Question 6

Decoder selection with active-LOW outputs: A 4-line-to-16-line decoder (inputs weighted 1, 2, 4, 8) has active-LOW outputs. If the input is 1110, the claim is that output line 7 is driven LOW. Decide if this is correct.

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Introduction / Context: Decoders convert binary inputs to one-of-N outputs. With active-LOW outputs, exactly one output line is asserted LOW for any valid input. Correctly interpreting bit weights and active-level conventions is essential to determine which line is selected. Given Data / Assumptions: 4-to-16 decoder with inputs weighted 1, 2, 4, 8 (LSB to MSB). Active-LOW outputs: selected line goes LOW; others remain HIGH. Input presented as 1110 (bit order 8 4 2 1 → 1 1 1 0). Concept / Approach: Translate the input code to its decimal equivalent using the given weights. The selected output index equals the binary input value. Active-LOW simply flips the asserted level, not the selected index. Step-by-Step Solution: Interpret bits (8,4,2,1) = (1,1,1,0) → value = 8 + 4 + 2 + 0 = 14.Therefore, output line 14 is the selected line.Because outputs are active-LOW, line 14 goes LOW while all others stay HIGH.The claim that “line 7 goes LOW” is incorrect; line 7 would correspond to input 0111. Verification / Alternative check: Cross-check with truth tables for 74xx154-style decoders (dual-enable aside). The index equals the binary input value regardless of output polarity. Why Other Options Are Wrong: “Correct” wrongly maps 1110 to 7. Output polarity does not change selected index, only logic level. Enable pins, while necessary, do not remap bit weights in normal operation. Common Pitfalls: Reversing bit order, misreading LSB/MSB, or confusing active-LOW assertion with index inversion. Final Answer: Incorrect

Question 7

XOR gate characteristic: An exclusive-OR (XOR) gate asserts HIGH only when exactly one— but not both— of its two inputs is HIGH. Evaluate this behavior statement.

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Introduction / Context: The XOR gate is ubiquitous in arithmetic (adders), parity generation/checking, and data comparison. It outputs true when the inputs differ and false when they are the same, making it the digital equivalent of “either/or but not both.” Given Data / Assumptions: Two-input XOR gate with ideal Boolean behavior. Input symbols: A and B. Output: Y. Truth table basis rather than analog timing effects. Concept / Approach: The XOR function can be written as Y = A XOR B = A*~B + ~A*B. This indicates Y is 1 when A ≠ B and 0 when A = B. Therefore, Y is HIGH if exactly one input is HIGH, and LOW if both are LOW or both are HIGH. Step-by-Step Solution: Evaluate cases: (A,B) = (0,0) → Y=0; (0,1) → Y=1; (1,0) → Y=1; (1,1) → Y=0.Identify the “one but not both” condition corresponds to (0,1) and (1,0).Confirm the descriptive statement matches the truth table.Conclude the behavior statement is correct. Verification / Alternative check: In parity logic, XOR of a set of bits returns 1 when the number of 1s is odd. For two inputs, an odd count occurs only when exactly one input is 1, reinforcing the result. Why Other Options Are Wrong: “Incorrect” contradicts the truth table. Hardware styles (open-collector, CMOS, TTL) and rise-time details do not change the Boolean mapping. Common Pitfalls: Confusing XOR with OR (which is true when any input is 1, including both); forgetting that XOR is addition modulo 2. Final Answer: Correct

Question 8

K-map yields a minimized Boolean equation: A correctly completed Karnaugh map (K-map) always allows derivation of a simplified Boolean equation by grouping adjacent cells in powers of two and eliminating changing variables. Judge this general statem…

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Introduction / Context: K-maps provide a graphical method for simplifying Boolean expressions. They help designers reduce logic by combining adjacent minterms (for SOP) or maxterms (for POS) into larger implicants, directly producing a minimized or near-minimized Boolean equation. Given Data / Assumptions: Proper K-map layout using Gray coding for variable ordering. Valid entries: 1s (minterms), 0s, and optional don’t-care (X) cells. Grouping rules: rectangles of size 1, 2, 4, 8, ... with wrap-around adjacency. Concept / Approach: By grouping adjacent 1s for SOP (or 0s for POS), variables that change within the group are eliminated from the product (or sum) term, yielding simpler expressions. Don’t-care cells can be used to enlarge groups, further simplifying the final equation. Step-by-Step Solution: Map the truth table onto the K-map cells according to Gray code.Form the largest valid groups that cover all 1s (SOP) or all 0s (POS).Translate each group into a term by keeping only variables that do not change across the group.Sum (OR) the product terms (for SOP) or multiply (AND) the sum terms (for POS) to form the simplified equation. Verification / Alternative check: Compare the derived equation against algebraic simplification or Quine–McCluskey results for the same function. The K-map approach should match or closely approximate the minimum form. Why Other Options Are Wrong: It is not limited to three variables; 2–6 variable maps are standard. The presence or absence of don’t-cares does not prevent simplification; it often helps. Common Pitfalls: Making non-rectangular groups, ignoring wrap-around adjacency, or using non-power-of-two group sizes. Final Answer: Correct

Question 9

Meaning of TTL: The acronym “TTL” is expanded as “transistor-technology-logic.” Decide whether this expansion is accurate in digital electronics.

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Introduction / Context: TTL is one of the classic digital logic families widely used in 74xx integrated circuits. Knowing the correct expansion of the acronym helps avoid confusion in documentation, procurement, and study materials. Given Data / Assumptions: “TTL” is a standard term in digital logic. It refers to how the gates are constructed internally. The expansion appears in manufacturer datasheets and textbooks. Concept / Approach: TTL stands for “transistor-transistor logic,” not “transistor-technology logic.” The name reflects the use of bipolar junction transistors for both the input logic and the output stage (as opposed to older resistor-transistor logic, RTL, where resistors provided input combining). Step-by-Step Solution: Identify the correct expansion: transistor-transistor logic.Note the historical contrast with RTL (resistor-transistor logic) and DTL (diode-transistor logic).Recognize the importance of the transistor pair usage in both logic and amplification within the gate structure.Conclude the provided expansion is incorrect. Verification / Alternative check: Check any 74xx family datasheet or a standard digital design text; the name is uniformly given as “transistor-transistor logic.” Why Other Options Are Wrong: “Correct” repeats the wrong expansion. Packaging or specific subfamilies (LS, ALS, S, F) do not alter the acronym’s meaning. Common Pitfalls: Assuming “technology” appears in the term; confusing logic family naming with semiconductor processes or packaging. Final Answer: Incorrect

Question 10

Even parity evaluation: In an even-parity system, determine whether the data word 1010011 should generate a parity bit of 1.

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Introduction / Context: Parity bits are simple error-detection mechanisms. In an even-parity scheme, the parity bit is chosen so that the total number of 1s (data bits + parity bit) is even. Correctly determining the parity bit is essential for reliable data transmission and memory integrity checks. Given Data / Assumptions: Parity type: even parity. Data word: 1010011 (7 bits). No special ordering or markers; count literal 1s in the data. Concept / Approach: Count the number of 1s in the data. If the count is already even, the parity bit must be 0 to keep the total even. If the count is odd, the parity bit must be 1 to make the total even. This rule applies regardless of bit positions or endianness. Step-by-Step Solution: Count 1s in 1010011: positions with 1s are 1, 3, 6, 7 → total = 4.The count (4) is even.Even parity requires the total number of 1s (data + parity) to be even.Therefore, the parity bit should be 0, not 1. Verification / Alternative check: Append parity 0 to obtain 10100110: number of 1s remains 4 (even). If we appended 1, the number of 1s would be 5 (odd), violating even parity. Why Other Options Are Wrong: “Correct” would imply an odd total of 1s is acceptable for even parity. References to odd parity or MSB position do not apply, because parity concerns only the count of 1s, not their positions. Common Pitfalls: Miscounting bits; confusing even and odd parity rules; assuming parity depends on which bit is the most significant. Final Answer: Incorrect

Question 11

Digital design practice — meaning and purpose of a pull-up resistor In digital electronics, a pull-up resistor is used so that a signal node does not float. State whether the following is correct or incorrect: “A pull-up resistor is a resistor used …

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Introduction / Context: A pull-up resistor is one of the most common passive components in digital circuits. It prevents undefined or “floating” logic levels on inputs or on wired-OR/open-drain style outputs whenever no active device is driving the line LOW. This question checks whether you understand the role of a pull-up and why it results in a stable logic HIGH under idle conditions. Given Data / Assumptions: The node can be tri-stated, open-collector/open-drain, or simply an input with no strong source. Logic HIGH/LOW thresholds follow the family in use; the details of exact voltages are not needed. “Active state” here means “when a driver is actively pulling the line to a definite level.” Concept / Approach: Connect a resistor from the node to the positive supply (Vcc). When no device asserts a LOW, a small current through the resistor biases the node to Vcc, interpreting as logic 1. When an active device pulls the line LOW, the device sinks current through the pull-up, forcing a logic 0 within safe current limits. Value selection balances noise immunity, rise time, and current consumption. Step-by-Step Solution: Model the node as an input with very high impedance.Attach R to Vcc; node rises toward Vcc → interpreted as HIGH.Add an active pull-down driver; it sinks current I = (Vcc - VOL) / R, driving a valid LOW.When driver releases the line, the node returns HIGH via the resistor. Verification / Alternative check: Scope measurements show the node sitting near Vcc when idle and transitioning to a valid LOW when a driver engages. Data sheets show recommended pull-up values (for example, 1 kΩ–100 kΩ depending on speed and family). Why Other Options Are Wrong: “Incorrect” ignores the standard definition. Technology-specific choices (TTL vs. CMOS or open-collector only) are too narrow; pull-ups are used widely wherever a passive HIGH bias is needed. Common Pitfalls: Selecting too large a resistor causes slow rising edges and susceptibility to noise; too small wastes current and can overstress drivers. Forgetting that inputs should never be left floating is another frequent error. Final Answer: Correct

Question 12

Karnaugh map capabilities — what it does best in logic minimization Choose the most accurate description of what a Karnaugh map (K-map) provides for Boolean simplification and implementation planning.

Accepted Answer

Introduction / Context: Karnaugh maps (K-maps) are a visual method for simplifying Boolean functions, especially up to 4 or 5 variables. They help designers move from raw truth tables to minimal gate-level forms quickly. This question asks what a K-map most directly achieves in practice. Given Data / Assumptions: We are minimizing a Boolean function defined by minterms/don’t-cares. Standard grouping rules (1, 2, 4, 8, … cells) apply. We are focusing on the K-map’s primary output: a minimal expression. Concept / Approach: K-maps identify adjacent 1-cells (or 0-cells for POS) to form implicants, then essential prime implicants, yielding a minimal expression. The canonical result most directly associated with K-maps is a minimal sum-of-products (SOP); alternatively, by mapping 0s, one may obtain minimal product-of-sums (POS). Step-by-Step Solution: Place truth-table 1s (and X don’t-cares) on the K-map.Group adjacent cells in powers of two to cover all 1s with as few groups as possible.Translate each group into a product term (for SOP).Sum all product terms to form the minimal SOP expression. Verification / Alternative check: Compare with algebraic minimization (Quine–McCluskey or computer-aided tools). The K-map solution’s gate count should be minimal or tie for minimal with other methods. Why Other Options Are Wrong: Eliminating the need for simplification is misleading; K-maps are a simplification tool. “Any circuit with just AND/OR” is not a unique K-map promise. “Signal flow picture” describes block diagrams, not K-maps’ tabular adjacency logic. Common Pitfalls: Incorrect wrapping adjacency and overlooking don’t-cares often lead to nonminimal results. Mixing SOP and POS rules mid-process causes errors. Final Answer: produce the simplest sum-of-products expression

Question 13

VHDL control flow — which statement evaluates a variable’s status value In VHDL coding of combinational decisions, identify the statement that evaluates the current status/value of a variable or signal to select among labeled choices.

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Introduction / Context: Hardware description languages provide multiple control constructs to describe logic behavior. In VHDL, selection among several discrete alternatives is often written more clearly with a construct tailored to evaluate a status value and choose one of many branches. Given Data / Assumptions: We need to choose a construct that matches a single expression against enumerated choices. We are describing combinational selection (e.g., decoders, controllers). Readability and synthesizability are desired. Concept / Approach: In VHDL, a CASE statement evaluates one expression (e.g., a std_logic_vector or enumerated type) and selects exactly one when branch whose choice list matches. It is ideal when the set of options is mutually exclusive and collectively exhaustive (others => … can be included). IF/THEN and IF/THEN/ELSE handle boolean conditions or ranges but are less compact for many discrete choices. Step-by-Step Solution: Identify the driving expression (e.g., op_code or state).Write CASE expression IS … WHEN value1 => … WHEN value2 => …Ensure coverage of all legal values; include WHEN OTHERS => for safety.Synthesize to multiplexers/decoders as appropriate. Verification / Alternative check: Comparing a multi-branch IF cascade to CASE shows the CASE is clearer when matching exact codes; synthesis results are equivalent. Why Other Options Are Wrong: IF/THEN and IF/THEN/ELSE are condition-based rather than value-selection constructs. ELSIF is part of IF chains, not a stand-alone statement. Common Pitfalls: Forgetting WHEN OTHERS or overlapping choice lists leads to unintended latches or mismatches. Using CASE with unresolved types without proper casting can cause tool errors. Final Answer: CASE

Question 14

BCD-to-7-segment 7447A — purpose of ripple blanking input/output (RBI/RBO) For the 7447A BCD-to-7-segment decoder/driver, what is the practical purpose of the ripple blanking input and output lines in multi-digit displays?

Accepted Answer

Introduction / Context: Driving multi-digit 7-segment displays from BCD often requires suppressing nonsignificant zeros to improve readability. The 7447A includes ripple blanking lines (RBI/RBO) designed to cascade blanking across multiple digits in a numeric field. Given Data / Assumptions: Digits are arranged in a multi-decade display (e.g., thousands, hundreds, tens, ones). Leading zeros are typically undesired; sometimes trailing zeros are also suppressed, depending on application. RBI is an input; RBO is an output for blanking propagation. Concept / Approach: When a more significant digit is blank (e.g., zero that is nonsignificant), its RBO signal can be used to blank the next digit via RBI. This creates a chain so that a sequence of zeros at the high end does not light the corresponding segments. Systems can also be configured to suppress trailing zeros in certain formats. Step-by-Step Solution: Feed BCD=0000 to a high-order digit and assert RBI to check for blanking conditions.If blanking applies, the 7447A turns off segment outputs and asserts RBO.Cascade RBO → RBI of the next lower digit to continue blanking across digits.Stop blanking when a nonzero BCD value is encountered. Verification / Alternative check: Observe the display: numbers like “0047” render as “47” with ripple blanking properly wired. Oscilloscope or logic-analyzer traces verify RBO/RBI transitions during updates. Why Other Options Are Wrong: “Any zero” would incorrectly suppress significant zeros (e.g., in 1010). “Nonsignificant digit” is vague and not specifically tied to leading/trailing suppression. “Test all segments” describes lamp/test functions, not RBI/RBO. Common Pitfalls: Miswiring RBO/RBI prevents correct blanking propagation. Forgetting current-limiting resistors or common-anode/cathode compatibility causes dim or damaged displays. Final Answer: turn off the display for leading or trailing zeros

Question 15

Standard XOR behavior — choose the correct complete truth-table output Select the correct 2-input XOR truth table description for all input combinations (A,B → Y).

Accepted Answer

Introduction / Context: The exclusive-OR (XOR) gate outputs logic 1 when its inputs differ and outputs logic 0 when they are the same. This fundamental pattern drives applications such as parity generation, binary addition without carry, and conditional inversion. Given Data / Assumptions: Two binary inputs A and B. XOR is defined as Y = A ⊕ B. “Differ” means exactly one of A or B equals 1. Concept / Approach: Use the definition “1 when inputs are unequal.” That is equivalent to the Boolean expression Y = A · B̄ + Ā · B, or to Y = (A + B) · (Ā + B̄). This ensures Y=0 for 00 and 11, and Y=1 for 01 and 10. Step-by-Step Solution: A=0, B=0 → equal → Y=0.A=0, B=1 → different → Y=1.A=1, B=0 → different → Y=1.A=1, B=1 → equal → Y=0. Verification / Alternative check: Implement XOR with two ANDs, two inverters, and one OR as Y = A·B̄ + Ā·B; simulate or hand-evaluate to see the same outputs. Another check: think of binary addition without carry; sum bit is XOR of inputs. Why Other Options Are Wrong: Each incorrect option either asserts 1 when inputs are equal or assigns 0 to one of the unequal cases, contradicting the XOR definition. Common Pitfalls: Confusing XOR with OR (which is 1 if any input is 1) or with XNOR (which is 1 when inputs are equal). Final Answer: 00→0, 01→1, 10→1, 11→0

Question 16

Keyboard to BCD — identify the appropriate MSI function block A circuit must encode one of ten decimal key presses (0–9) into a 4-bit BCD output. Which MSI function best fits this requirement?

Accepted Answer

Introduction / Context: Human interfaces like numeric keypads produce one-of-ten active signals. Digital systems commonly require a compact 4-bit binary-coded decimal (BCD) representation. Recognizing the MSI block that compresses multiple inputs into a coded output is essential in front-end design. Given Data / Assumptions: Ten inputs (0–9), only one active at a time under normal operation. Desired output is 4-bit BCD (0000 to 1001). Occasional multiple activations must be resolved predictably. Concept / Approach: An encoder performs input compression: a one-hot input line is converted into a binary code. To handle possible simultaneous presses (two keys), a priority encoder assigns precedence (e.g., highest-numbered input) to guarantee a valid, deterministic code. Typical MSI devices include the 74147/74148 families for decimal/octal priority encoding. Step-by-Step Solution: Map each key line to the corresponding code (e.g., key 7 → 0111).Use a priority encoder variant to define which key wins on overlap.Buffer and debounce as needed; optionally latch the BCD code.Drive downstream display decoders or arithmetic blocks with the BCD output. Verification / Alternative check: Simulate multiple key-press cases; confirm priority behavior and verify exactly one valid BCD code appears at the outputs at a time. Why Other Options Are Wrong: A decoder expands code to many outputs (reverse function). A multiplexer selects one of many data inputs; it does not produce a code. A demultiplexer routes one input to many outputs; again, not a coder. Common Pitfalls: Omitting diodes or logic to prevent ghosting; ignoring key bounce and failing to latch the stable code leads to flicker or misreads. Final Answer: priority encoder

Question 17

VHDL base types — identify the item that is not a valid standard type In common VHDL practice with IEEE libraries, which of the following is not a recognized standard data type name?

Accepted Answer

Introduction / Context: VHDL type correctness is critical for synthesizable code and clean simulations. IEEE packages (such as std_logic_1164) define widely used resolved types and arrays. Knowing the exact type names avoids compilation errors and unintended implicit conversions. Given Data / Assumptions: We refer to standard types defined in the IEEE library and the language core. Case is not significant in VHDL, but exact identifiers matter. Array type names follow specific conventions. Concept / Approach: Core types include BIT and BIT_VECTOR. With IEEE std_logic_1164, the resolved scalar is STD_LOGIC and the array is STD_LOGIC_VECTOR. There is no standard type named STD_VECTOR; using it would produce a compile error unless a user-defined type shadows that name. Step-by-Step Solution: List known scalar types: BIT, STD_LOGIC.List matching vector types: BIT_VECTOR, STD_LOGIC_VECTOR.Compare options; detect that STD_VECTOR is not one of the standard names. Verification / Alternative check: Examine IEEE package declarations or compile a small VHDL snippet declaring signals of each listed type to see which names pass type checking. Why Other Options Are Wrong: BIT, BIT_VECTOR, and STD_LOGIC are legitimate standard types and compile under typical toolchains with the right library use clauses. Common Pitfalls: Mixing BIT/STD_LOGIC without proper conversions; forgetting to include IEEE std_logic_1164 leads to unknown identifiers for STD_LOGIC-based types. Final Answer: STD_VECTOR

Question 18

4-bit ripple adder — compute the binary sum from given A, B, and initial carry A 4-bit adder has inputs C0 = 0, A1 = 0, A2 = 1, A3 = 0, A4 = 1 and B1 = 0, B2 = 1, B3 = 1, B4 = 1 (index 1 = LSB). Determine the 5-bit output (carry out concatenated wit…

Accepted Answer

Introduction / Context: Understanding bit ordering and initial carry is essential when interpreting multi-bit adder problems. Here, you are asked to compute the result of adding two 4-bit numbers with an initial carry-in of zero and report the 5-bit result (carry out plus sum). Given Data / Assumptions: Bit indices: A1/B1 are least significant bits (LSB), A4/B4 are most significant bits (MSB). C0 (carry-in) is 0. Output format is Cout S4 S3 S2 S1 as a 5-bit string. Concept / Approach: Form the two 4-bit numbers from the given bits: A = (A4 A3 A2 A1) and B = (B4 B3 B2 B1). Add them as unsigned integers with the carry-in. Convert the decimal sum back to binary; include final carry as the leftmost bit. Step-by-Step Solution: Construct A: A4=1, A3=0, A2=1, A1=0 → A = 1010₂ = 10₁₀.Construct B: B4=1, B3=1, B2=1, B1=0 → B = 1110₂ = 14₁₀.Add with C0=0: 10 + 14 = 24.Convert 24 to binary: 24 = 11000₂ (5 bits already show carry position).Thus the reported 5-bit output is 11000. Verification / Alternative check: Perform column-wise binary addition with carries: 0+0=0 (carry0), 1+1=0 carry1, 0+1+carry1=0 carry1, 1+1+carry1=1 carry1, final carry=1 → result 11000. Why Other Options Are Wrong: Other options correspond to arithmetic errors or misinterpreting bit order and carry concatenation (e.g., missing the final carry bit or reversing bit significance). Common Pitfalls: Confusing index order (A1 as MSB), omitting carry-out from the reported result, or misreading one of the provided bits. Final Answer: 11000

Question 19

Odd parity generation — which data words require a parity bit of 1? In an odd-parity system, the parity bit is chosen so that the total number of 1s (data bits + parity) is odd. For which of the following data words will the parity bit equal 1?

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Introduction / Context: Parity bits provide simple error detection on serial links and memory. For odd parity, we set the parity bit so that the overall count of 1s is odd. The question asks you to determine, for several example data words, whether the parity bit should be 1 or 0. Given Data / Assumptions: Odd-parity convention: total number of 1s (including the parity bit) must be odd. We consider only single additional parity bit (not multi-bit ECC). No endianness concerns affect the count of 1s. Concept / Approach: Let ones(data) be the number of 1s in the data. If ones(data) is even, choose parity=1 to make even+1 = odd. If ones(data) is odd, choose parity=0 to keep the total odd. Therefore, parity bit equals 1 exactly when ones(data) is even. Step-by-Step Solution: Count ones in 1010011: 1+0+1+0+0+1+1 = 4 (even) → parity = 1.Count ones in 1111000: 1+1+1+1+0+0+0 = 4 (even) → parity = 1.Count ones in 1100000: 1+1+0+0+0+0+0 = 2 (even) → parity = 1.Thus all three require parity bit = 1 in an odd-parity system. Verification / Alternative check: Append the parity bit and recount: totals become 5, 5, and 3 respectively — each odd, satisfying the rule. Why Other Options Are Wrong: Each individual option (a–c) is true, so selecting only one would exclude the other correct cases. Therefore “All of the above” is the only fully correct choice. Common Pitfalls: Confusing odd and even parity conventions; assuming parity depends on bit order (it does not); forgetting to recount after appending the parity bit. Final Answer: All of the above

Question 20

In digital communications and error detection circuits, parity generators and parity checkers are commonly implemented using specific logic gates. Which type(s) of logic gates are typically used to generate a parity bit and to check parity (even or …

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Introduction / Context: Parity is one of the simplest and most widely used error-detection techniques in digital systems, serial links, memory subsystems, and communication protocols. A parity generator produces a single parity bit that makes the total number of 1s either even (even parity) or odd (odd parity). A parity checker then verifies that property on reception. Understanding which logic primitives naturally realize these functions is essential in combinational logic design and exam preparation. Given Data / Assumptions: Task: identify the gate type(s) used for parity generation and checking. Even parity requires an even count of 1s; odd parity requires an odd count. We assume standard two-level logic without encoded tricks. Concept / Approach: The XOR operation outputs 1 if and only if the number of 1s at its inputs is odd. Cascading XORs across multiple bits therefore produces an output that represents the odd-parity function. Conversely, XNOR outputs 1 if and only if the number of 1s is even. Thus, XOR naturally generates odd parity and can be inverted for even parity; XNOR naturally generates even parity and can be inverted for odd parity. Parity checkers compare the received data bits and parity bit using XOR or XNOR to verify integrity. Step-by-Step Solution: For N data bits d0, d1, …, dN-1, compute P = d0 XOR d1 XOR … XOR dN-1.If odd parity is required, transmit parity_bit = P; for even parity, use parity_bit = P’ or use XNOR across all bits.At the receiver, XOR all bits including the parity bit; a 0 result indicates even parity satisfied (or choose XNOR with the corresponding rule). Verification / Alternative check: Truth-table inspection for two inputs shows XOR counts odd ones; XNOR counts even ones. Cascading preserves this count property for more bits. Why Other Options Are Wrong: Exclusive-OR (XOR) gates only: Parity checking often uses XNOR directly for even-parity validation.Exclusive-NAND gates only / Exclusive-AND gates: These are not standard parity primitives and do not map directly to odd/even count logic. Common Pitfalls: Confusing XOR with OR; XOR is parity-sensitive whereas OR is not.Forgetting that XNOR is the complement parity function (even count detector). Final Answer: Exclusive-OR (XOR) and Exclusive-NOR (XNOR) gates

Combinational Logic Circuits Questions