Question 1

Gray Code — Primary Practical Use What is the primary real-world use of Gray code in digital systems and instrumentation?

Accepted Answer

Introduction / Context: Gray code is a binary numbering scheme where adjacent codes differ by only one bit. This single-bit-change property drastically reduces the chance of ambiguous readings when multiple bits do not change simultaneously—an issue common with mechanical sensors and encoders. Given Data / Assumptions: Mechanical or optical encoders produce discrete position codes as a shaft rotates. Signal transitions are not perfectly simultaneous due to mechanical tolerances and electrical skew. A robust position readout should avoid misinterpretation between adjacent positions. Concept / Approach: Because only one bit changes between successive Gray codes, momentary overlap or timing skew cannot create a codeword that corresponds to a distant position. Thus Gray code is ideal for shaft encoders and other position transducers, ensuring reliable, monotonic readings as the shaft moves. Step-by-Step Solution: Consider an encoder disk with tracks producing a multi-bit code.During a boundary transition between positions, only one Gray bit toggles.Therefore, any transient read error is limited to one-bit ambiguity (adjacent position), not a multi-bit error.Hence, Gray code is primarily used to represent mechanical shaft position. Verification / Alternative check: Industrial encoders and robotics sensors commonly specify Gray output; controller firmware often converts Gray→binary after sampling. Why Other Options Are Wrong: turning on/off software switches: Too generic; not a coding scheme’s primary use. ASCII/hexadecimal for angular position: ASCII and hexadecimal are unrelated encodings for this purpose. Common Pitfalls: Confusing Gray code with error-correcting codes; it reduces transition ambiguity but does not add redundancy for correction. Final Answer: coded representation of a shaft's mechanical position

Question 2

1-of-10 BCD Decoder — Required Number of Inputs A 1-of-10 BCD decoder activates exactly one of ten outputs (0–9). How many input lines are required to represent the BCD value applied to the decoder?

Accepted Answer

Introduction / Context: Binary-Coded Decimal (BCD) represents decimal digits 0–9 using a 4-bit binary pattern. A 1-of-10 decoder uses this 4-bit code to select one of ten output lines. Understanding the required number of inputs is key when interfacing counters and displays. Given Data / Assumptions: Target device: 1-of-10 decoder (e.g., 74xx42 family). Valid inputs: decimal 0 through 9 only. Input coding: standard 8421 BCD. Concept / Approach: BCD uses 4 bits per decimal digit. Therefore, a BCD decoder that uniquely identifies digits 0–9 requires four input lines (A, B, C, D). Illegal codes 1010–1111 are typically treated as don’t-care or invalid. Step-by-Step Solution: Recognize 8421 BCD → 4 bits.Map 4-bit combinations 0000–1001 to outputs 0–9.Ignore 1010–1111 (not used) or handle per datasheet.Hence, the decoder needs 4 input lines. Verification / Alternative check: Examine datasheets for BCD decoders; pin listings show four data inputs plus enable pins. Why Other Options Are Wrong: 8: Would be used for two BCD digits (8 bits). 10: Misinterprets outputs as inputs. 1: Cannot represent 10 distinct values. Common Pitfalls: Confusing the number of outputs (10) with the number of input lines (4 for BCD). Final Answer: 4

Question 3

Multiplexer Enable — Global Control Function On a multiplexer (MUX) chip, what is the function of the enable (EN) input with respect to the device outputs?

Accepted Answer

Introduction / Context: Most multiplexers include an enable input that globally turns the device’s data path on or off. This allows hierarchical selection, bus sharing, and power savings. When disabled, the MUX typically forces a defined output state or high-impedance (depending on the family), regardless of select inputs. Given Data / Assumptions: Device: generic MUX with select lines and an enable pin. Enable polarity can be active-HIGH or active-LOW, but the functional role is the same. When disabled, internal selection logic is effectively gated off. Concept / Approach: The enable input acts as a global gate. With EN asserted, the MUX responds to select lines and routes the chosen input to the output. With EN deasserted, the output is disabled or forced to a fixed state, independent of the select code. Step-by-Step Solution: Identify EN's role: global permission for data selection.EN asserted → output follows selected input.EN deasserted → output disabled or forced, selection ignored.Therefore, EN activates the entire MUX for normal operation. Verification / Alternative check: Datasheets (e.g., 74HC157/158, 74HC151) show enable pins that disable outputs or tri-state them when inactive. Why Other Options Are Wrong: to apply Vcc / connect ground: Power and ground are separate pins, not functions of EN. to activate one half of the chip: Some dual MUX packages have separate enables per section, but the generic single MUX chip uses EN for the whole device. Common Pitfalls: Assuming EN is optional; if left floating, it may disable the device unintentionally. Always drive EN to a defined logic level. Final Answer: to activate (enable) the entire MUX so that selection and data routing occur

Question 4

Demultiplexers — Why They Are Called Data Distributors Why is a demultiplexer (DEMUX) commonly referred to as a “data distributor” in digital systems?

Accepted Answer

Introduction / Context: A demultiplexer performs the inverse function of a multiplexer. While a MUX selects one of many inputs to drive a single output, a DEMUX routes a single input to exactly one of many outputs based on select lines. This behavior motivates the nickname “data distributor.” Given Data / Assumptions: One data input line is present. Several output lines are available (e.g., 1-to-4, 1-to-8, 1-to-16). Select lines choose which output receives the input. Concept / Approach: For each select-code value, precisely one output is enabled to forward the input signal; the others are disabled (logic 0 or high-impedance depending on the part). Hence, the single input is “distributed” spatially across outputs over time as the select code changes, enabling time-division or address-based routing. Step-by-Step Solution: Recognize DEMUX structure: 1 input, n outputs, select lines.Operation: given a select code, connect input to the corresponding output line.Name rationale: this routing distributes a single data stream to different destinations → “data distributor.”Therefore, the correct description is that the input is distributed to one of the outputs. Verification / Alternative check: Review standard parts like 74HC138 (decoder used as DEMUX) or dedicated DEMUX devices; truth tables show one active output per select value reflecting the input. Why Other Options Are Wrong: One of the inputs will be selected for the output: That describes a MUX, not a DEMUX. The output will be distributed to one of the inputs: Direction is reversed. All outputs active simultaneously: Incorrect; exactly one output is selected for normal operation. Common Pitfalls: Confusing encoder/decoder/MUX/DEMUX roles; always track the number of inputs vs outputs and the direction of data flow. Final Answer: The input will be distributed to one of the outputs.

Question 5

Gray code identification in digital systems Which binary coding scheme progresses so that only one bit changes between any two successive code words (minimizing transition errors at boundaries)?

Accepted Answer

Introduction / Context: Digital encoding schemes are used to represent numbers, characters, and states. In applications like position encoders, asynchronous state transitions, and analog-to-digital interfaces, it is desirable to ensure that only one bit changes between successive values to avoid transient misreads. The code specifically designed for this purpose is the reflected Gray code. Given Data / Assumptions: We compare common codes: 8421 (pure binary/BCD), Excess-3, nine's-complement, and Gray code. The target property: only one bit changes between successive code words. Successive means numerically adjacent values (e.g., 3 to 4). Concept / Approach: Gray code (reflected binary code) is constructed so adjacent integers differ by a Hamming distance of 1. This minimizes decoding ambiguity during transitions, a crucial advantage for mechanical and high-noise environments. Other codes do not guarantee single-bit adjacency across the entire counting sequence. Step-by-Step Solution: List property: need Hamming distance = 1 between adjacent numbers.Check Gray code: by definition, adjacent code words differ in one bit.Check 8421 (binary): multiple bits often change (e.g., 0111 to 1000 flips four bits).Check Excess-3 and nine's-complement: neither ensures single-bit transitions.Therefore, the required code is Gray code. Verification / Alternative check: Enumerate 3-bit Gray: 000, 001, 011, 010, 110, 111, 101, 100 → each step flips one bit. Compare with pure binary sequence and note multi-bit changes at boundaries such as 3→4. Why Other Options Are Wrong: Nine's-complement: used for decimal subtraction, not single-bit adjacency. 8421 code: standard binary/BCD; many transitions flip multiple bits. Excess-3: decimal code offset by +3; does not enforce single-bit transitions. Common Pitfalls: Confusing Gray code with BCD variants; assuming any “special” decimal code has single-bit transitions—it does not. Final Answer: Gray code

Question 6

Decoder size from truth-table labeling A truth table whose output columns are labeled 0 through 15 most likely corresponds to which decoder integrated circuit (IC) type?

Accepted Answer

Introduction / Context: Decoders assert exactly one of many outputs based on a binary input code. When a truth table shows output lines numbered 0–15, it strongly suggests a 1-of-16 decode, commonly described as a “hexadecimal” (base-16) decoder because there are 16 distinct outputs for the 16 possible 4-bit input combinations. Given Data / Assumptions: Outputs explicitly labeled 0,1,2,…,15. Binary input width implied: 4 bits (2^4 = 16). Standard active-high decoding assumed (one-hot output). Concept / Approach: A 1-of-n decoder maps an n-ary selection from log2(n) binary inputs to n outputs. For 16 outputs, log2(16) = 4 inputs. Therefore a 1-of-16 decoder (often called hexadecimal) fits the 0–15 labeling perfectly. Step-by-Step Solution: Note: 0–15 labels → 16 distinct outputs.Determine inputs: need 4 inputs to select 1 of 16.Match IC type: hexadecimal 1-of-16 decoder. Verification / Alternative check: Typical parts (e.g., 74HC154) implement 4-to-16 decoding with outputs Y0..Y15, aligning with the 0–15 truth-table columns. Why Other Options Are Wrong: Dual octal outputs: would indicate two separate 1-of-8 sections, not a single 0–15 set. Binary-to-hexadecimal: a converter/encoder term, not a one-hot decoder. Hexadecimal-to-binary: describes an encoder, not a decoder with 16 distinct output lines. Common Pitfalls: Confusing decoders (one-hot outputs) with encoders/code converters (multi-bit outputs). Labeling 0–15 almost always implies 4-to-16 decoding. Final Answer: hexadecimal 1-of-16

Question 7

Gray code vs. Excess-3 — conceptual check Evaluate the statement: “In Gray code, each number is 3 greater than the binary representation of that number.”

Accepted Answer

Introduction / Context: Gray code and Excess-3 are distinct numbering schemes used for different purposes. Gray code minimizes bit changes between adjacent values, while Excess-3 is a decimal code that offsets each BCD digit by +3 to simplify certain arithmetic processes and display logic. Mixing these definitions leads to common misconceptions on exams. Given Data / Assumptions: Statement claims: Gray code equals binary + 3. Gray code property: single-bit change between successive values. Excess-3 property: each BCD digit = binary of (digit + 3). Concept / Approach: Check the defining properties. Gray code is not a numeric offset of binary; it is a different mapping designed for adjacency with Hamming distance 1. Excess-3, not Gray, uses the “+3” rule per decimal digit. Therefore the statement is false. Step-by-Step Solution: Recall Gray code: reflected mapping; adjacency differs by 1 bit.Recall Excess-3: for decimal digit d, the code equals binary(d + 3).Compare to claim: “+3” applies to Excess-3, not Gray.Conclude: the statement is false. Verification / Alternative check: Example for decimal 5: Excess-3 gives 1000 (since 5+3=8), while Gray(5) in 3-bit Gray is 111; clearly unrelated to a +3 offset model. Why Other Options Are Wrong: Choosing True would conflate two different coding schemes and ignore the adjacency design of Gray code. Common Pitfalls: Memorizing “+3” and misattributing it to the wrong code; forgetting that Gray code emphasizes transition reliability, not arithmetic offset. Final Answer: False

Question 8

Decimal-to-BCD encoder inputs How many separate input lines must a decimal-to-BCD encoder provide to uniquely encode decimal digits 0 through 9?

Accepted Answer

Introduction / Context: An encoder maps many input lines into a smaller set of output bits. A decimal-to-BCD encoder accepts one-of-ten decimal inputs (0–9) and outputs a 4-bit binary-coded-decimal representation. Understanding the number of input lines clarifies pin counts and truth-table sizing for practical IC selection and schematic design. Given Data / Assumptions: Digits 0–9 must be represented. Only one decimal input is active at a time (one-hot input model). Outputs are 4 BCD lines representing 0000–1001. Concept / Approach: For N distinct symbols in a one-hot input scheme, the encoder needs N inputs. Decimal digits are ten symbols, so 10 input lines are required, each mapped to a unique 4-bit BCD output. Step-by-Step Solution: List decimal symbols: {0,1,2,3,4,5,6,7,8,9} → 10 items.Provide one input per symbol → 10 inputs.Encode to 4-bit BCD (0000..1001). Verification / Alternative check: Examine common encoder ICs or truth tables (e.g., 10-to-4 encoders): there are 10 input pins, often with enable/priority features. Why Other Options Are Wrong: 4 and 8 are fewer than the number of symbols; 16 would imply hexadecimal. Common Pitfalls: Confusing encoder inputs with decoder outputs; a decoder would have 10 outputs for BCD-to-decimal. Final Answer: 10

Question 9

Input width of a 1-of-16 decoder How many binary select inputs are required to address exactly one of sixteen outputs (a 1-of-16 decoder)?

Accepted Answer

Introduction / Context: A 1-of-n decoder asserts exactly one output line out of n possibilities, based on a binary input code. The number of input bits needed equals log2(n). This relationship is central when selecting decoder sizes and building memory address decoders or chip-select trees. Given Data / Assumptions: n = 16 desired outputs. Binary selection, one-hot output. No additional enable lines considered for the bit count. Concept / Approach: Compute k where 2^k = 16 → k = 4. Thus, four input bits uniquely select one of sixteen outputs. Step-by-Step Solution: Apply formula: inputs = log2(16) = 4.Confirm uniqueness: 4 bits represent 16 combinations (0000..1111).Therefore, 4 inputs suffice for a 1-of-16 decoder. Verification / Alternative check: Check standard ICs like 74HC154 (4-to-16 line decoder): four select pins A,B,C,D with 16 outputs Y0..Y15. Why Other Options Are Wrong: 2 bits → 4 outputs; 8 bits → 256 outputs; 16 bits → 65,536 outputs. Common Pitfalls: Counting enable pins as “inputs” for selection width; enable does not increase address space, it gates operation. Final Answer: 4

Question 10

XNOR count for equality comparison — 8-bit comparator In a typical equality comparator built from basic gates, how many exclusive-NOR (XNOR) gates are required to compare two 8-bit words for equality?

Accepted Answer

Introduction / Context: An equality comparator outputs HIGH when two multi-bit inputs are identical. A common logic-level implementation uses one XNOR per bit to test equality, followed by an AND that combines all bitwise equal signals. Knowing the gate count supports quick resource estimation in discrete logic designs and FPGA LUT budgeting. Given Data / Assumptions: Two 8-bit inputs: A7..A0 and B7..B0. Use XNOR per bit: A_i XNOR B_i = 1 when bits match. Combine via multi-input AND to form overall EQ. Concept / Approach: For n-bit equality, use n XNOR gates to produce n equality flags, then an n-input AND (or a tree of AND gates). Thus, for 8 bits, exactly 8 XNORs are needed for the bitwise comparisons. Step-by-Step Solution: Per bit i: E_i = XNOR(A_i, B_i).Overall equality: EQ = AND(E_7, E_6, …, E_0).Count XNORs: one per bit → 8 total. Verification / Alternative check: Check standard 74xx series comparators: internal structures mirror bitwise equality checks aggregated by AND/OR logic; discrete implementations follow the same pattern. Why Other Options Are Wrong: 4 or 6 XNORs leave some bits untested; 10 exceeds the requirement. Common Pitfalls: Using XOR instead of XNOR; XOR must be inverted for equality, which implicitly recreates XNOR behavior. Final Answer: 8

Question 11

Signal direction in analog multiplexers/demultiplexers For analog MUX/DEMUX devices, are the channel and common terminals inherently uni-directional or bi-directional for analog signals?

Accepted Answer

Introduction / Context: Analog multiplexers/demultiplexers (e.g., 74HC4051/4052/4053, CD4051 family) are built from CMOS transmission gates that act as electronically controlled analog switches. Understanding whether signals can flow in both directions is crucial when routing audio, sensor voltages, or DAC/ADC channels through a single IC. Given Data / Assumptions: Device type: analog MUX/DEMUX using transmission gates. Channels and common terminal connect through low-resistance CMOS switches. Operation within voltage ranges specified by the datasheet. Concept / Approach: Transmission gates are symmetrical pass elements; when enabled, they pass analog voltages/currents in either direction with approximately the same on-resistance and bandwidth (within limits). Therefore, analog multiplexer channels are bidirectional; a device used as a mux can often be used as a demux and vice versa. Step-by-Step Solution: Identify structure: complementary MOSFET pair forms a bilateral switch.Enabled switch conducts regardless of signal direction.Thus, channel–common path is bidirectional for analog signals. Verification / Alternative check: Datasheets specify I/O as bidirectional with an equivalent on-resistance R_on and leakage in both directions; functional diagrams show bilateral gates. Why Other Options Are Wrong: Unidirectional applies to many digital buffers, not these analog pass elements. Parity and BCD are unrelated to analog routing. Common Pitfalls: Driving signals outside supply rails or violating current limits; forgetting that on-resistance and charge injection can distort high-precision signals though direction itself is bidirectional. Final Answer: bidirectional

Question 12

Hardware method to convert BCD to binary Which approach is commonly used to implement a BCD-to-binary conversion purely in hardware (without firmware/software)?

Accepted Answer

Introduction / Context: BCD-to-binary conversion can be implemented in hardware for speed or simplicity, especially in older or minimalist embedded systems. Medium-Scale Integration (MSI) devices—such as adders, multiplexers, small ROMs/PLAs, or dedicated converter ICs—offer practical gate-level solutions for translating BCD digits into a binary number without using a CPU or microcode. Given Data / Assumptions: We need a hardware-only conversion path from BCD to straight binary. MSI building blocks (adders, counters, ROMs) are available. No reliance on serial peripherals or software routines. Concept / Approach: One method is to use adders and shift/weighting logic: binary = 10 * tens + ones (and extend for more digits). Another is to employ a small ROM/PLA that maps BCD inputs to binary outputs. Both are classic MSI techniques. Devices like keyboard encoders or UARTs do not perform numeric base conversion; an ALU could, but typically belongs inside a CPU, not as a simple MSI “hardware approach.” Step-by-Step Solution: Use MSI adders: compute binary_out = 8digit + 2digit (for tens weighting) + ones.Alternatively, use a ROM lookup: address = BCD_in, data = binary_out.Scale to multiple digits by cascading weighting stages (10, 100, etc.). Verification / Alternative check: Historical designs and application notes show 74xx adders and multiplexers used for code conversions; small PROMs were also popular as fixed translators. Why Other Options Are Wrong: Keyboard encoder: converts key matrix positions to codes, not BCD-to-binary arithmetic. ALU: generally part of a processor; while possible, it is not the simple hardware-only MSI pathway implied. UART: handles serial-to-parallel and framing, not numeric base conversion. Common Pitfalls: Ignoring invalid BCD inputs (1010–1111); proper designs must detect or map them safely. Final Answer: with MSI IC circuits

Question 13

Priority encoder behavior with multiple active inputs If two input lines are active simultaneously on a priority encoder, which input value is represented on the binary output?

Accepted Answer

Introduction / Context: A priority encoder resolves conflicts when more than one input is asserted. It outputs the code for the highest-priority (commonly the highest-index) active input and typically provides additional flags (e.g., valid output). Understanding priority behavior prevents ambiguity in interrupt controllers, keyboard encoders with rollover, and bus request arbiters. Given Data / Assumptions: Standard priority convention: higher-numbered input has higher priority. Two inputs become active at the same time. Device asserts only one output code and may raise a “valid” flag. Concept / Approach: By definition, a priority encoder selects the highest-priority asserted line. Therefore, if inputs i and j are active and i > j, the binary output corresponds to i. Lower-priority inputs are effectively masked. Step-by-Step Solution: Identify active lines: e.g., I6 and I2 are 1.Apply priority rule: I6 overrides I2.Output code = binary of the higher-index input; assert valid. Verification / Alternative check: Datasheets for devices like 74HC147 (10-line to 4-line priority encoder) show truth tables where higher inputs dominate when multiple are active. Why Other Options Are Wrong: Lower value: contradicts the “priority” specification. Neither/both: real priority encoders are designed to deliver a single, unambiguous code when multiple inputs are active. Common Pitfalls: Forgetting to synchronize inputs to avoid metastability; misunderstanding active-low conventions (some encoders use active-low inputs/outputs, but priority still favors the higher-numbered line). Final Answer: the higher value

Question 14

Truth-table size of a BCD decoder A BCD (Binary-Coded Decimal) decoder accepts valid 4-bit BCD inputs for digits 0–9. How many rows (valid input cases) appear in its functional truth table for the decimal digits?

Accepted Answer

Introduction / Context: BCD represents decimal digits using 4-bit codes 0000 through 1001. A BCD decoder activates a unique output for each valid digit (e.g., for seven-segment displays or one-of-ten lines). Counting the number of valid input rows in the truth table clarifies device behavior and helps when designing with invalid-code handling for 1010–1111. Given Data / Assumptions: Valid BCD inputs: 0000–1001 (ten patterns). Invalid inputs: 1010–1111, typically ignored or treated as don’t-care by the decoder. Decoder maps each valid input to a specific output pattern/line. Concept / Approach: Count the valid BCD codes. There are ten decimal digits, hence ten valid code words and ten corresponding rows in the functional truth table that map directly to decimal outputs. Some datasheets show additional rows for invalid inputs, but the canonical “BCD decoder truth table for digits” has 10 main rows. Step-by-Step Solution: Enumerate digits: 0..9 → 10 inputs.Each input corresponds to one output pattern/line.Therefore, the truth table includes 10 valid rows for the digits. Verification / Alternative check: Review 7447/7448 (BCD-to-seven-segment) documentation: ten entries illustrate segment patterns for digits 0–9; entries for 10–15 are either undefined or treated as blank/reset. Why Other Options Are Wrong: 9 and 8 undercount the valid decimal symbols; 3 is unrelated. Common Pitfalls: Confusing a complete 4-bit truth table (16 rows including invalid codes) with the functional, digit-focused mapping (10 valid rows). Final Answer: 10

Question 15

IC decoders — typical output polarity when a valid input is applied In most integrated-circuit (IC) decoders used for display driving or address decoding (for example, 74xx138 or BCD-to-7-segment drivers), when the correct input code is present the …

Accepted Answer

Introduction / Context: IC decoders translate binary codes into one-of-N outputs and are common in address decoding and seven-segment display driving. Many standard TTL/CMOS decoder families use outputs that are active-LOW, especially when their outputs are intended to sink current from LEDs or enable lines with negative-true logic. Recognizing the usual polarity prevents wiring mistakes and simplifies troubleshooting. Given Data / Assumptions: Generic decoder devices as used in displays and address selection. Outputs intended to select one line or illuminate LED segments. Typical TTL open-collector or CMOS sink-capable outputs. Concept / Approach: In an active-LOW decoder, the asserted state is logic 0 at the output. This is advantageous for driving common-anode displays and for wire-OR configurations with open-collector outputs. When the correct input code is applied, the decoder activates the matching output by pulling it LOW, while all others remain HIGH (or inactive). Step-by-Step Solution: Identify intended load: LEDs or enable pins often expect a sinking driver.Decoder design: provide open-collector/sink outputs that pull lines LOW when selected.Therefore, correct input ⇒ corresponding output becomes active-LOW. Verification / Alternative check: Examine typical devices such as 74xx138 (3-to-8 decoder) and 74xx47 (BCD-to-7-segment for common-anode displays). Datasheets label outputs with overbars or specify “active-LOW”; bench tests confirm that outputs go LOW when selected, sinking current through the load. Why Other Options Are Wrong: active-HIGH: common for some drivers (e.g., common-cathode displays) but not the default for many decoder ICs aimed at sinking loads. high impedance / open: pertains to tri-state or open outputs when inactive, not the asserted state. toggle: decoders do not toggle; they select a single line deterministically. Common Pitfalls: Mismatching display polarity (common-anode vs common-cathode) with decoder output polarity. Omitting current-limiting resistors when driving LEDs. Final Answer: active-LOW

Question 16

Functional replacement by a multiplexer (MUX) In digital design, a single multiplexer can effectively take the place of which of the following elements when used to implement an arbitrary truth table or selection logic?

Accepted Answer

Introduction / Context: Multiplexers (MUXes) are universal combinational building blocks. By wiring data inputs to logic 0/1 (or signals) and steering with select lines, a MUX can implement many logic functions that would otherwise require multiple small-scale integration (SSI) gates. Understanding this equivalence helps minimize chip count and PCB area. Given Data / Assumptions: A standard n-to-1 multiplexer is available (e.g., 4:1, 8:1, 16:1). We can tie MUX inputs to constants or real signals. Select lines represent address bits of a truth table. Concept / Approach: Any truth table with up to n selectable minterms can be realized by mapping select lines to the table’s variables and data inputs to the desired output values for each address. This directly replaces networks of AND/OR/NOT gates used in sum-of-products or other combinational forms. Step-by-Step Solution: Choose MUX size so that 2^k data inputs ≥ number of minterms to implement.Connect select lines to k input variables.Wire each data input to 0, 1, or a signal such that the MUX output equals the truth table for that address.Result: the MUX replaces many SSI gates or an entire combinational subcircuit. Verification / Alternative check: Compare gate counts: implementing a 4-variable logic function via canonical SOP may require numerous gates, whereas a single 16:1 MUX plus inverters on select lines can implement the same function compactly and predictably. Why Other Options Are Wrong: “Several Ex-NOR gates”: overly specific; a MUX can emulate them but also far more. Only “SSI gates” or only “combinational circuits” is incomplete. The best description includes both. Counters are sequential devices; a MUX cannot replace their state storage. Common Pitfalls: Ignoring the need for input inversion; sometimes inverters on select lines are required. Overlooking propagation delay and fan-out when replacing large gate networks. Final Answer: several SSI logic gates or combinational logic circuits

Question 17

7485 magnitude comparator — determine outputs from given inputs and cascade (expansion) signals Given a 7485 4-bit magnitude comparator with the following values: A0 = 0, B0 = 1, (IA < B) = 0; A1 = 1, B1 = 0, (IA = B) = 1; A2 = 1, B2 = 0, (IA > B) =…

Accepted Answer

Introduction / Context: The 7485 is a 4-bit magnitude comparator providing three primary outputs: A > B, A = B, and A < B. It also includes cascade (expansion) inputs to chain devices for wider word lengths. Correctly interpreting bit significance and the expansion inputs is essential to determine the final comparison result. Given Data / Assumptions: A3..A0 and B3..B0 represent 4-bit words with A3/B3 as the most significant bits. Expansion inputs are: (IA < B), (IA = B), (IA > B) = 0, 1, 0 respectively, meaning higher-order sections are equal. Bit values provided: A3 = 0, B3 = 0; A2 = 1, B2 = 0; A1 = 1, B1 = 0; A0 = 0, B0 = 1. Concept / Approach: When the expansion input IA = B = 1, the present 4-bit section decides the result. Evaluate from MSB to LSB: the first bit position where A and B differ determines the outcome. If they are equal through all bits, output A = B = 1; otherwise assert A > B or A < B at the first differing position with A or B larger. Step-by-Step Solution: Compare MSB (A3 vs B3): 0 vs 0 → equal so continue.Next bit (A2 vs B2): 1 vs 0 → A is greater here; decision made.Lower bits do not change the outcome once a higher bit decides the comparison.Therefore: A > B = 1, A < B = 0, A = B = 0. Verification / Alternative check: Interpreting A and B as unsigned: A = 0b0110x (given lower bits) is larger than B since A2 = 1 and B2 = 0 at the first differing position. This aligns with the standard priority of higher-order bits in magnitude comparison. Why Other Options Are Wrong: Any option asserting A < B contradicts the MSB-first comparison (A2 > B2). Asserting A = B is impossible because multiple bits differ. All zeros or all ones outputs cannot occur for valid comparisons with IA = B = 1. Common Pitfalls: Reading bits LSB-to-MSB; must evaluate MSB-first. Ignoring expansion inputs; IA = B must be 1 to let this stage decide. Final Answer: A = B = 0, A < B = 0, A > B = 1

Question 18

Comparator expansion inputs — purpose and target width In a standard 4-bit magnitude comparator (such as the 7485), the expansion (cascading) inputs are provided to allow easy expansion to what wider system size?

Accepted Answer

Introduction / Context: Magnitude comparators include cascade or expansion inputs that let multiple devices be chained to compare wider words. Understanding the purpose of these pins helps designers scale from 4-bit building blocks to larger bit widths without redesigning the internal comparison logic. Given Data / Assumptions: A single IC compares 4 bits (A3..A0 vs B3..B0). Expansion inputs accept results from a more significant comparator stage. Outputs from a less significant stage feed into a more significant stage when cascading multiple devices. Concept / Approach: To compare two n-bit numbers larger than 4 bits, chain devices so the most significant stage decides the overall relation unless equality holds there, in which case the next stage decides, and so on. The simplest and most common cascade is two 4-bit comparators forming an 8-bit comparator. Step-by-Step Solution: Use one 4-bit comparator for bits 7..4 and another for bits 3..0.Connect the outputs of the MSB comparator to system outputs.Tie the expansion outputs of the LSB stage into the expansion inputs of the MSB stage to propagate “less than, equal, greater than” status upward.Result: an 8-bit comparison using two 4-bit ICs. Verification / Alternative check: Datasheets typically show example circuits where two 7485 devices are cascaded to achieve 8-bit comparison; more devices can be chained for 12 or 16 bits, but the canonical next step from 4 bits is an 8-bit system using two ICs. Why Other Options Are Wrong: 4-bit: no expansion. BCD system and counter system: unrelated to comparator width expansion. 12-bit “without additional logic”: possible with three devices, but the immediate and most direct expansion target is 8 bits. Common Pitfalls: Wiring expansion pins incorrectly (IA=B must be driven by the lower stage). Forgetting proper priority from MSB stage downward. Final Answer: 8-bit system

Question 19

Gray code advantage for rotary position encoding Why is Gray code preferred for encoding the angular position of a rotating shaft in encoders and position sensors?

Accepted Answer

Introduction / Context: Rotary encoders and absolute position sensors often use Gray code rather than straight binary to minimize reading errors during transitions. Mechanical contacts, optical sensors, and sampling circuits rarely change all bits at exactly the same instant. Gray code reduces ambiguity when the shaft moves between adjacent positions while being read. Given Data / Assumptions: We compare properties of Gray code vs straight binary. Adjacent codewords in Gray code differ by a Hamming distance of 1. Sampling may occur during a code transition due to mechanical or timing skew. Concept / Approach: The key Gray code property is that consecutive counts change only one bit. If a sensor or reader samples during the change, at most a single bit may be ambiguous, limiting the interpreted error to at most one adjacent code rather than a large jump that could occur with multi-bit changes in binary code. Step-by-Step Reasoning: Consider binary ...0111 → 1000, where four bits change simultaneously; any skew can momentarily produce invalid patterns far from the true value.In Gray code, the equivalent transition changes only one bit, so intermediate readings differ by at most 1 from the true position.Therefore, Gray code greatly reduces transition ambiguity and misreads. Verification / Alternative check: Plot Hamming distance between successive Gray codewords: it is uniformly 1. In contrast, binary’s worst-case transition changes many bits, which is problematic for asynchronous sampling and mechanical bounce. Why Other Options Are Wrong: All digits or two digits change: contradicts Gray code definition. Weighted code with parity: Gray is non-weighted and does not inherently include parity. Fewer bits: both Gray and binary use the same number of bits for the same range. Common Pitfalls: Assuming Gray code simplifies arithmetic; it is mainly for transition robustness, not computation. Final Answer: Only one digit changes between counts.

Question 20

Gray-to-binary (and binary-to-Gray) conversion — which gate type is fundamental to the conversion network? Choose the logic function most commonly used to implement these code converters.

Accepted Answer

Introduction / Context: Gray code differs from binary by ensuring only one bit changes between consecutive values. Converters between Gray and binary are standard textbook circuits, and their structure hinges on a particular logic function applied cumulatively across bits. Given Data / Assumptions: MSB in Gray equals MSB in binary. Each lower binary bit is formed by XORing the next-higher binary bit with the corresponding Gray bit. Symmetry exists for binary-to-Gray conversion: Gray bit = XOR of adjacent binary bits. Concept / Approach: The defining operation of these conversions is XOR (exclusive-OR). For Gray→binary: b[n] = g[n], and for each i from n-1 down to 0, b[i] = b[i+1] XOR g[i]. The cumulative XOR chain ensures that each binary bit correctly reflects the parity of the Gray prefix as required. Step-by-Step Solution: Start with MSB: b[n] = g[n].For next bit: b[n-1] = b[n] XOR g[n-1].Repeat down to LSB using XOR chaining.Binary-to-Gray uses XOR of adjacent binary bits: g[i] = b[i+1] XOR b[i]. Verification / Alternative check: Build a 4-bit converter and truth-table-check conversions both directions; XOR mapping will match expected sequences. Why Other Options Are Wrong: XNOR gates: Would invert the parity relationship and produce incorrect mapping without extra inversions.AND / OR gates: Do not implement the required parity (mod-2 sum) relationship. Common Pitfalls: Miswiring the XOR chain direction (using Gray instead of the previous binary bit).Forgetting that the MSB is copied directly. Final Answer: XOR gates

Code Converters and Multiplexers Questions