Question 1

Modulo-16 ripple counter — state after multiple clocks A MOD-16 (4-bit) ripple up-counter currently shows 1001₂ (decimal 9). After applying 31 additional clock pulses, what binary count will appear at its outputs?

Accepted Answer

Introduction / Context: A MOD-16 binary counter cycles through 16 distinct states (0–15) and then repeats. When asked for the state after many clock pulses, the key is to apply modular arithmetic rather than simulating every intermediate count. This is a staple technique for quickly determining counter outputs in design and troubleshooting. Given Data / Assumptions: The counter is a 4-bit ripple up-counter (natural binary sequence). Present state = 1001₂ = 9₁₀. Additional pulses = 31. Modulus = 16, so states wrap every 16 counts. Concept / Approach: Use modulo arithmetic: NewState = (Current + Pulses) mod 16. This yields the equivalent number of steps within a single 0–15 cycle. Convert the final decimal result back to binary for the required output format. Step-by-Step Solution: Compute sum: 9 + 31 = 40.Reduce modulo 16: 40 mod 16 = 40 − 2*16 = 40 − 32 = 8.Convert 8₁₀ to binary: 8 = 1000₂.Therefore, the counter reads 1000₂ after 31 additional pulses. Verification / Alternative check: Split 31 into 16 + 15. Advancing 16 steps leaves the counter unchanged (modulus property). From 9, advance 15 steps: 9 → 24, then wrap 24 − 16 = 8, which is 1000₂. Same result, confirming the calculation. Why Other Options Are Wrong: 1010₂ (10), 1011₂ (11), 1101₂ (13), 0111₂ (7): none equals 8, so they do not match the modulo arithmetic outcome. Common Pitfalls: Forgetting to apply modulo reduction and attempting to count linearly, which is time-consuming and error-prone. Misconverting between decimal and binary (e.g., confusing 1000₂ with 8 vs. 1001₂ with 9). Final Answer: 1000₂

Question 2

Display decoders — output polarity when the correct input is applied In most common LED/segment display decoders (e.g., BCD-to-7-segment with open-collector/sink outputs), when the correct input code is present the corresponding output line switches…

Accepted Answer

Introduction / Context: Many legacy and still-popular display decoders (such as the 74xx47 BCD-to-7-segment driver for common-anode displays) use active-LOW outputs that sink current to illuminate segments. Understanding the typical output polarity is essential for correct wiring and for avoiding inverted logic bugs during troubleshooting. Given Data / Assumptions: Typical decoder with sink-capable outputs driving LEDs (often common-anode displays). Outputs are active-LOW: a LOW asserts the segment/line. Appropriate current-limiting resistors are assumed. Concept / Approach: In an active-LOW driver, asserting a particular output means pulling that line to LOW so current can flow from Vcc through the LED segment into the decoder output transistor to ground. Therefore, when the correct input code is applied, the related output line goes LOW to turn the target segment on (or to select the intended line in a decoder tree). Step-by-Step Reasoning: Apply the valid input code to the decoder.Internal logic determines the required segment/line to activate.The corresponding output transistor turns on, pulling the output LOW (active state).Current sinks through the output, lighting the LED (for common-anode) or selecting the line. Verification / Alternative check: Datasheets typically mark outputs with overbars (e.g., a, b, c with bars) or specify “active-LOW.” Bench tests with a current-limited supply confirm that a LOW at the output lights the intended segment on a common-anode display module. Why Other Options Are Wrong: HIGH: would be active-HIGH behavior (common-cathode drivers), not typical of sink-type decoders. High impedance / open: tri-state behavior is not used for basic segment decode activation. Rapid toggling: not a normal decode response. Common Pitfalls: Reversing display polarity (common-anode vs common-cathode) and getting inverted results. Omitting current-limiting resistors, which can damage segments even if logic polarity is correct. Final Answer: LOW

Question 3

Synchronous (parallel) vs. asynchronous (ripple) counters — reason synchronous designs avoid ripple delay Why do synchronous counters eliminate the delay problems that occur in ripple counters?

Accepted Answer

Introduction / Context: Asynchronous (ripple) counters pass state changes from one flip-flop to the next, creating cumulative propagation delays and momentary false outputs. Synchronous counters are engineered to avoid this “ripple-through” behavior, which is critical for high-speed and glitch-sensitive applications such as address decoding or precise timing generation. Given Data / Assumptions: All flip-flops in a synchronous counter share a common clock. Combinational logic prepares the next state for each flip-flop before the clock edge. Proper setup/hold timing is satisfied. Concept / Approach: In synchronous counters, the clock edge arrives at all stages together. Therefore, every flip-flop samples its input at the same instant and updates simultaneously. This eliminates the sequential handoff of transitions that plagues ripple counters and prevents intermediate incorrect words at the multi-bit output. Step-by-Step Reasoning: Distribute the same clock to each stage.Ensure next-state logic is settled before the edge.At the clock edge, all Q outputs update together, avoiding cascaded delays. Verification / Alternative check: Oscilloscope traces show that in a synchronous counter all bits change on the same edge, whereas ripple counters show staggered edges from LSB to MSB separated by propagation delays (t_pd) of each stage. Why Other Options Are Wrong: First/last-stage-only or last-stage-only clocking preserves ripple behavior. “No clock” contradicts the definition of a counter. Using Gray code may reduce simultaneous bit changes but does not remove ripple delay; architecture does. Common Pitfalls: Assuming synchronous design removes all timing concerns; you must still meet setup/hold and account for combinational delays. Final Answer: input clock pulses are applied simultaneously to each stage.

Question 4

Divide-by-N by external gating — what action occurs at the predetermined terminal count? Any divide-by-N counter can be realized by adding external gating so that the counter will ________ when a chosen terminal count is reached.

Accepted Answer

Introduction / Context: When a required modulus is not a power of two, designers commonly “truncate” a binary counter by externally detecting a terminal count and forcing the counter back to a known state. This technique is foundational for creating arbitrary divide-by-N frequency dividers and sequence generators using standard binary counters. Given Data / Assumptions: Base counter is a binary up-counter. Combinational logic detects a particular output pattern (terminal count). On detection, a control input forces the counter to a known start state. Concept / Approach: To obtain an exact period of N clocks, detect the Nth state and assert the asynchronous clear (reset). The counter then immediately returns to 000… and begins counting the next cycle. Alternatively, for non-zero starting states, a synchronous/asynchronous load may be used, but the classic divide-by-N approach is to reset at the terminal count corresponding to N, yielding an output that repeats every N pulses. Step-by-Step Solution: Let the counter run freely from 0 upward.Logic monitors Q outputs and recognizes the terminal state.When recognized, assert RESET, forcing Q → 0.Release RESET so counting resumes from 0, forming an N-length cycle. Verification / Alternative check: Compare with a 6-to-10 decoder used to create a divide-by-10 from a 4-bit binary counter: detecting 1010₂ (decimal 10) and resetting yields MOD-10 behavior (a decade counter). Why Other Options Are Wrong: “HIGH” or “LOW”: logic levels, not actions. Preset: can implement a different start state but the textbook divide-by-N phrasing is to reset at N. Tri-state outputs: unrelated to modulus control. Common Pitfalls: Failing to consider reset pulse width and propagation delay; ensure reset cleanly clears one count only. Glitchy decode can cause premature resets; use synchronous reset or registered decode if necessary. Final Answer: reset

Question 5

Terminal count of a MOD-11 binary counter — identify the binary value For a modulus-11 (MOD-11) binary counter that counts 0 through 10 before resetting, what is the terminal count value in binary?

Accepted Answer

Introduction / Context: The terminal count of a truncated counter is the last valid state before the counter returns to its initial state. For MOD-N counters that begin at 0, the terminal count is N − 1. Converting this decimal value into binary yields the specific Q output pattern observed just prior to reset. Given Data / Assumptions: Counter starts at 0 and advances in natural binary. MOD-11 ⇒ states 0 to 10 (inclusive). Reset occurs immediately after the terminal count. Concept / Approach: Compute N − 1 and convert to binary. For N = 11, N − 1 = 10. Express 10 in 4-bit binary to match the minimal number of bits required to represent the state in a typical circuit using four flip-flops (truncated from the natural MOD-16). Step-by-Step Solution: Find terminal decimal value: 11 − 1 = 10.Convert 10₁₀ to binary: 10 = 8 + 2 = 1010₂.Therefore, the terminal count is 1010₂. Verification / Alternative check: Enumerate the sequence briefly: … 1000₂ (8), 1001₂ (9), 1010₂ (10), then reset to 0000₂. This matches the definition of terminal count for MOD-11 starting at 0. Why Other Options Are Wrong: 1000 (8) and 1001 (9): intermediate states, not terminal. 1100 (12) and 1011 (11): beyond the allowed range; a proper MOD-11 never reaches 11 before reset. Common Pitfalls: Confusing terminal count with modulus (10 vs 11), or assuming BCD behavior (MOD-10). Final Answer: 1010

Question 6

Truncated modulus — select the example that is not a power-of-two count Which of the following moduli is an example of a truncated counter (i.e., not equal to 2^n for any integer n)?

Accepted Answer

Introduction / Context: Binary counters naturally produce moduli that are powers of two because each new flip-flop doubles the number of states. When a design requires a modulus that is not a power of two, truncation via reset/load logic is used. Recognizing which moduli require truncation is important for choosing between a simple binary counter and additional decode/clear logic. Given Data / Assumptions: Binary counters without extra logic yield 2^n states (n = number of flip-flops). A truncated modulus requires additional gating. Concept / Approach: Identify whether the candidate modulus equals 2^n. Values like 8, 16, 32, and 64 are 2^3, 2^4, 2^5, 2^6 respectively. Any other integer (e.g., 13) is not a power of two and therefore must be implemented by truncation from the next higher power-of-two counter modulus (here 16) using decode/reset logic. Step-by-Step Reasoning: Check 8 = 2^3 ⇒ natural binary modulus.Check 16 = 2^4 ⇒ natural binary modulus.Check 32 = 2^5 and 64 = 2^6 ⇒ natural moduli.13 ≠ 2^n ⇒ must be truncated from 16 using decode/reset. Verification / Alternative check: Design sketch: use a 4-bit counter and decode decimal 13 (1101₂) to clear back to 0000; the counter then cycles through 0–12 (MOD-13). Why Other Options Are Wrong: 8, 16, 32, 64 are all natural binary moduli; no truncation required. Common Pitfalls: Assuming decimal-friendly moduli (10, 12) are natural; they are truncated unless implemented with decade counters. Final Answer: 13

Question 7

Checking a counter’s parallel load feature — choose a valid bench procedure Which procedure would correctly verify that a binary counter supports parallel loading of data (assume the LOAD input is active during the specified clock action)?

Accepted Answer

Introduction / Context: Many synchronous counters offer a parallel load capability: when the LOAD control is asserted, the next active clock transfers the pattern present on the DATA inputs directly into the internal registers, overriding the normal count sequence. Verifying this feature on the bench requires a simple, unambiguous procedure. Given Data / Assumptions: The counter has a synchronous LOAD input (active level known). CLR (asynchronous clear) is inactive unless explicitly noted. Outputs Q reflect either the counted value or the loaded value at the active clock edge. Concept / Approach: To test parallel load, hold a known pattern on DATA, assert LOAD, and apply one valid clock edge. If the feature works, Q will equal the DATA pattern immediately after the clock, independent of the previous count value. Using all LOWs (000…) or all HIGHs (111…) makes observation straightforward. Step-by-Step Procedure: Set DATA inputs to 000… (all LOWs).Assert LOAD to its active level.Provide one active clock edge.Observe Q outputs; they should become 000…, confirming the load operation. Verification / Alternative check: Repeat with DATA = 111… while keeping CLR inactive. After the clock with LOAD asserted, Q should read 111…. Remove LOAD and apply additional clocks; Q should now resume counting from the loaded value, demonstrating correct mode switching. Why Other Options Are Wrong: Preset LOAD then assert CLR: CLR forces Q to 0 regardless of DATA, so it does not test LOAD. Clock and CLR together: again verifies clear, not load. Driving Q terminals directly: outputs are not inputs; this is nonsensical. Toggling CLR with LOAD inactive: Q will clear but never mirror DATA. Common Pitfalls: Forgetting to keep CLR inactive when testing LOAD. Not asserting LOAD during the clock edge in a synchronous-load device. Final Answer: Apply LOWs to the parallel DATA inputs, pulse the CLK input with LOAD asserted, and check for LOWs on all the Q outputs.

Question 8

JK flip-flop chain troubleshooting (clocking issue) A sequential circuit shows no output activity at Q1 when observed on an oscilloscope. Conditions: all J–K inputs are held HIGH, the system CLK signal is present, Q0 is toggling, and the C (clock) i…

Accepted Answer

Introduction / Context: This question tests practical debugging of ripple or chained JK flip-flops. With J = K = 1, each JK device toggles on its clock edge. If the next stage's clock pin never sees transitions, that stage cannot toggle. Recognizing whether the issue is at the driving node (previous flip-flop output) or at the receiving node (next flip-flop clock input) is essential for rapid fault isolation. Given Data / Assumptions: All J–K inputs HIGH → each stage is configured to toggle on its clock. Global CLK present; Q0 is toggling, so FF0 is healthy internally. FF1's C (clock) input is observed constant LOW. Standard wiring: Q0 often drives the clock of FF1 in ripple chains. Concept / Approach: For FF1 to toggle, its clock input must receive transitions. If its clock node is stuck LOW while Q0 is known to toggle, there is likely a fault between Q0 and FF1.C. The two most probable hard faults are a short-to-ground at the driving node (FF0.Q0) or at the receiving node (FF1.C), either of which will clamp the signal line LOW and prevent edges from reaching FF1. Step-by-Step Solution: Confirm Q0 toggles at the pin of FF0 (not just in the schematic).Probe the net between FF0.Q0 and FF1.C; if the net is LOW at both ends, suspect a short-to-ground.Isolate by lifting (temporarily disconnecting) FF1.C or FF0.Q0 to see which device pin still reads LOW.If FF0.Q0 floats high/low correctly after isolation, the short is likely at FF1.C; otherwise, at FF0.Q0. Verification / Alternative check: An ohmmeter (with power off) often reveals a near-0 Ω path to ground on the affected node. Visual inspection for solder bridges or damaged vias also helps. Replacing the offending IC or repairing the short restores clock edges at FF1.C and Q1 toggling returns. Why Other Options Are Wrong: The Q0 to J connection (A): J controls toggle when J=K=1; the immediate symptom is a missing clock at FF1, not a data-path issue. Only FF0 output shorted (B) or only FF1 input shorted (C): Each is possible, but the best answer acknowledges either end of the same net could be shorted. Common Pitfalls: Assuming a logic design error when the probe already shows a hardware-level short; overlooking that a stuck clock pin will prevent any downstream activity even with correct J–K settings. Final Answer: Either the output of FF0 or the input of FF1 may be shorted to ground.

Question 9

Seven-segment display mapping For a standard seven-segment display (segments labeled a, b, c, d, e, f, g), which segments must be turned ON to display the decimal digit 6?

Accepted Answer

Introduction / Context: Seven-segment displays represent decimal digits by illuminating specific segments. Knowing which segments correspond to each digit is crucial when designing decoders (for example, BCD-to-7-segment) or when debugging display wiring and logic. Given Data / Assumptions: Segments are labeled a through g, with the usual arrangement: a is top, b is upper-right, c is lower-right, d is bottom, e is lower-left, f is upper-left, g is middle. We assume a common, non-stylized font used in textbooks and IC datasheets. Concept / Approach: The digit 6 typically lights every segment except b. It shows the top bar (a), both left bars (f, e), the middle bar (g), the bottom bar (d), and the lower-right bar (c). The upper-right bar (b) remains off to distinguish 6 from 8 and 0. Step-by-Step Solution: List the standard mappings: 0 → a b c d e f; 1 → b c; 2 → a b d e g; 3 → a b c d g; 4 → b c f g; 5 → a c d f g; 6 → a c d e f g; 7 → a b c; 8 → a b c d e f g; 9 → a b c d f g.From the mapping, digit 6 requires segments a, c, d, e, f, g.Verify visually: forms a closed loop on left side with both left segments lit and the top, middle, bottom, and lower-right segments lit.Confirm that b is OFF so the shape differs from 8 and 0. Verification / Alternative check: Consult any BCD-to-7-segment truth table (common-anode or common-cathode; the active level differs but the logical mapping is the same). Digit 6 has b = OFF in the standard glyph. Why Other Options Are Wrong: b, c, d, e, f, g: Missing a (top bar), so not a 6. a, b, c, d, and f: Missing g and e, shape becomes closer to 9 or malformed. b, c, d, e, f: Missing a and g, does not create a 6. Common Pitfalls: Confusing common-anode vs common-cathode wiring (active levels invert) and mislabeling segments, especially b and c. Final Answer: a, c, d, e, f, and g

Question 10

Truly synchronous multistage counter requirement For a multistage counter to be genuinely synchronous (all stages update together), the ________ of each stage must be tied to ________.

Accepted Answer

Introduction / Context: In synchronous counters, all flip-flops change state simultaneously on the same clock edge. This eliminates the ripple delays inherent in asynchronous counters and allows higher operating frequencies with predictable timing. The key design requirement is how the clock is distributed. Given Data / Assumptions: Each stage is a clocked flip-flop (e.g., D, JK, or T) with a clock input Cp. Combinational logic may generate next-state signals, but should not create multiple, skewed clock domains. We want every stage to respond to the same clock edge. Concept / Approach: A truly synchronous counter feeds a common clock to every flip-flop (Cp pins tied to the same line). Next-state logic determines which stages toggle/set/reset, but the actual state update occurs concurrently on the global clock transition. By contrast, ripple counters route a prior stage's output into the next stage's clock, creating accumulated delay and skew. Step-by-Step Solution: Distribute a buffered, low-skew clock to all flip-flop Cp pins.Use combinational logic (e.g., AND/OR/XOR) to compute each flip-flop's next state based on the present state.On each clock edge, all stages sample simultaneously and update together.Verify timing: single clock-to-Q path dominates; there is no ripple through clock chains. Verification / Alternative check: Timing simulation shows one uniform clock net and no multi-level clock chains. Hardware measurements reveal minimal skew between stage outputs at the active edge. Why Other Options Are Wrong: CE to same clock line: CE (clock enable) is not the system clock; tying enables together does not ensure synchronous sampling if clocks differ. Terminal count or “both clock input lines”: Misapplies ripple techniques or introduces ambiguous clocking, defeating synchronous behavior. Common Pitfalls: Deriving per-stage clocks from logic signals (gated clocks) which add skew and glitch sensitivity; proper synchronous design uses a common clock and enables within data paths, not clock gating wherever possible. Final Answer: Cp, the same clock input line

Question 11

Cascade division calculation A MOD-12 counter is cascaded with a MOD-10 counter. If the input clock is 60 MHz, what is the resulting output frequency from the cascade?

Accepted Answer

Introduction / Context: Cascaded counters divide an input clock by the product of their individual moduli. This is a common technique for generating precise lower-frequency time bases from a high-frequency oscillator in frequency synthesizers, digital clocks, and timing controllers. Given Data / Assumptions: First stage modulus: 12. Second stage modulus: 10. Input frequency: 60 MHz. Assume ideal counters with no missing counts and proper cascading. Concept / Approach: The overall division factor equals MOD_total = MOD1 * MOD2. Therefore, the output frequency equals f_out = f_in / MOD_total. We perform a straightforward numeric substitution and a unit conversion from MHz to kHz if needed for clarity. Step-by-Step Solution: Compute total modulus: MOD_total = 12 * 10 = 120.Compute output frequency: f_out = 60 MHz / 120.Calculate: 60 / 120 = 0.5 MHz.Convert: 0.5 MHz = 500 kHz. Verification / Alternative check: If each counter alone divided the clock, the first would produce 5 MHz (60/12), and the second would further divide by 10: 5 MHz / 10 = 0.5 MHz, consistent with 500 kHz. Why Other Options Are Wrong: 1,500 kHz: Corresponds to division by 40, not 120. 6 MHz: Division by 10 only, ignoring the MOD-12 stage. 5 MHz: Output after only the MOD-12 stage, not the cascade. Common Pitfalls: Forgetting to multiply moduli in series; confusing MHz and kHz; or mis-ordering the cascade (order does not affect the product result). Final Answer: 500 kHz

Question 12

Ripple counter speed limitation In an asynchronous (ripple) counter, the maximum speed is limited primarily by the propagation delay of which elements?

Accepted Answer

Introduction / Context: Ripple counters clock each successive flip-flop with the output of the previous stage. This creates a chain of sequential delays. Understanding which delays dominate sets expectations for maximum count frequency and informs when a synchronous counter is necessary. Given Data / Assumptions: Asynchronous topology: stage n+1 is clocked by stage n's output. Flip-flops have a finite clock-to-Q propagation delay. Minimal combinational gating lies in the clock path for simple ripple counters. Concept / Approach: The worst-case time for the most significant stage to settle equals the sum of the per-stage clock-to-Q delays of all preceding flip-flops. Thus, the flip-flop chain's propagation delays limit speed. External gates may appear in decoding, but they are not in the internal ripple clock path for basic counters. Step-by-Step Solution: Identify the number of stages N.Let t_pd(ff) be the per-flip-flop delay.Worst-case settle time to MSB ≈ N * t_pd(ff).Maximum count frequency must ensure period > worst-case settle time to avoid decoding hazards. Verification / Alternative check: Measure a ripple counter's outputs on a scope: higher-order bits lag the low-order bits by multiples of t_pd(ff), plainly showing the cumulative flip-flop delays. Why Other Options Are Wrong: All flip-flops and gates / the flip-flops only with gates / only gates: In basic ripple counters, the critical path is the flip-flop chain, not external gating. Common Pitfalls: Assuming synchronous timing or ignoring accumulated delay when decoding multiple bits immediately after a transition (can cause glitches). Final Answer: each flip-flop

Question 13

BCD to seven-segment interfacing requirement: In a seven-segment LED display system, a binary-coded decimal (BCD) value must be decoded into segment control signals suitable for driving the decimal digit display. Evaluate the statement.

Accepted Answer

Introduction / Context: Seven-segment LED displays render decimal digits by illuminating combinations of segments labeled a through g. Microcontrollers and counters often represent digits internally as BCD (binary-coded decimal). To light the correct segments for each digit, the BCD code must be converted to the required seven control lines, typically by a decoder/driver IC or equivalent logic. Given Data / Assumptions: Input digits are available in BCD format (0000 to 1001 represent 0–9). The visual output device is a seven-segment LED display (common-anode or common-cathode). Segment lines a–g require appropriate logic levels and current drive. Concept / Approach: A BCD-to-seven-segment decoder maps each valid BCD combination to a pattern that turns on the correct segments. Dedicated ICs (for example, classic decoder/drivers) simplify this mapping and may incorporate current limiting or transistor stages. Without decoding, the raw BCD bits do not directly correspond to segment lines; a translation is mandatory to obtain human-readable digits. Step-by-Step Solution: Accept a BCD input for a single digit (4 bits: D, C, B, A).Use a decoder truth table to produce seven outputs: a, b, c, d, e, f, g.Account for display type: common-anode requires active-LOW segment drive; common-cathode requires active-HIGH drive.Optionally buffer or limit current to protect LED segments. Verification / Alternative check: Wire a known decoder/driver to a seven-segment module and cycle BCD inputs through 0000…1001. Visually confirm that the displayed digits 0–9 are correct. For inputs 1010–1111, either blank or display a custom pattern per design. Why Other Options Are Wrong: Incorrect: BCD cannot directly drive segments without translation logic. Applies only to LCDs: LCDs also need decoding, but the statement is not restricted to LCDs. Only true for common-anode or only at 12 V: The need for decoding is independent of supply level and common configuration; only signal polarity and drivers change. Common Pitfalls: Mixing common-anode with common-cathode wiring; forgetting current-limiting resistors; not handling invalid BCD codes (10–15) which may cause unexpected patterns. Final Answer: Correct

Question 14

Synchronous binary counters — scope of use: The statement claims that synchronous binary counters can only be used for the timing of digital systems. Consider whether their applications are limited solely to timing, or if they also include frequency…

Accepted Answer

Introduction / Context: Synchronous binary counters are essential building blocks in digital design. The prompt suggests these counters can only be used for “timing” in digital systems. In practice, synchronous counters are far more versatile: they divide frequency, count events, generate addresses, control sequences, and participate in finite state machines. This question tests your understanding of the breadth of their applications beyond simple timing. Given Data / Assumptions: A synchronous counter changes all flip-flop stages simultaneously under a common clock. “Timing” typically refers to producing periodic intervals or delays tied to a reference clock. Typical synchronous counters support enable, clear, load, and sometimes up/down control. Concept / Approach: Because each stage toggles in a coordinated way, synchronous counters avoid the ripple delays of asynchronous designs. Their clean timing allows reliable high-speed counting and predictable decoding, enabling uses across system control, measurement, and communications. Therefore, restricting them to “only timing” is inaccurate. Step-by-Step Solution: Identify typical functions: frequency division, event counting, and code sequencing.Note system-level roles: address generation for memory scanning and pixel timing in displays.Consider control applications: step-through states in controllers and test pattern generators.Conclude that synchronous counters serve many roles beyond timing alone. Verification / Alternative check: Design examples commonly cascade synchronous counters to create modulo-N sequences, build baud-rate generators, or produce frame/line counters in video systems—none are solely “timing.” Why Other Options Are Wrong: Correct: Would imply exclusivity to timing, which is too narrow.Applies only to ripple counters / Valid only for divide-by-N clocks: These statements confuse counter type and application scope; synchronous counters are widely used beyond pure timing. Common Pitfalls: Equating “counter” with “timer” and overlooking state sequencing, decoding, and measurement; assuming that if a device can time, that is its only purpose. Final Answer: Incorrect

Question 15

Definition of cascading in digital design: The statement claims “To cascade is to connect in parallel.” Assess whether cascading refers to parallel wiring or to connecting stages in series so that the output of one feeds the input of the next.

Accepted Answer

Introduction / Context: “Cascading” is a common term in logic design, counters, decoders, multiplexers, and amplifiers. Understanding its meaning determines how you expand bit-width, range, or functionality. This question challenges the misconception that cascading is the same as connecting in parallel. Given Data / Assumptions: Counters, decoders, priority encoders, and multiplexers are frequently cascaded. Parallel connection means multiple devices share the same input or bus lines simultaneously. Series (cascade) connection means the output of one stage drives the input or control of the next stage. Concept / Approach: To cascade is to chain stages so that behavior accumulates across devices. For counters, the ripple or synchronous carry-out of one device connects to the enable/carry-in of the next to extend modulus. For decoders or multiplexers, cascade control pins are used similarly to increase address range or number of channels. Step-by-Step Solution: Recall the canonical use: two 4-bit counters cascaded to form an 8-bit counter.Observe that the first counter’s carry drives the second counter’s enable or clock input.This is a series connection that extends range, not a parallel connection.Therefore, the statement equating cascade to parallel is incorrect. Verification / Alternative check: Manufacturer datasheets show “cascading” diagrams labeled with carry/borrow links between packages to scale capacity—clearly a series relationship. Why Other Options Are Wrong: Correct: Would redefine “cascade” incorrectly.Only true for tri-state buses / Valid for analog multiplexing only: These distract from the basic definition; cascading is not inherently parallel in those contexts either. Common Pitfalls: Confusing bus width expansion (parallel) with range expansion (cascade), and miswiring carry/enable pins leading to nonfunctional designs. Final Answer: Incorrect

Question 16

Shift-register counters and reset logic: Evaluate the statement: “Shift register counters use logic functions to reset the registers when the desired count is reached,” as in ring and Johnson counters implemented with shift registers.

Accepted Answer

Introduction / Context: Shift-register-based counters, such as ring and Johnson counters, generate sequences by shifting bits through flip-flops. To maintain a specific sequence length (modulus), the circuit often uses feedback or decoding logic to initialize or recycle the pattern when a terminal condition is detected. This item asks whether logic functions are used to cause that reset or recycling action. Given Data / Assumptions: Shift registers can be wired into counters by feeding back selected taps. Ring counters circulate a single 1; Johnson counters circulate a complemented pattern. Decoding logic can detect illegal states or a terminal state to reinitialize the sequence. Concept / Approach: In practice, logic gates monitor the register contents. When a particular combination appears (desired terminal count or illegal state), a pulse is generated to clear/load the register. This ensures a stable, repeatable modulus and quick recovery if power-up places the register in an unintended state. Step-by-Step Solution: Consider a 4-stage ring counter requiring one-hot initialization.Use gates to detect “all zeros” or “multiple ones,” asserting a load or set line.In Johnson counters, detect the boundary pattern to wrap the sequence deterministically.Hence, logic functions do reset or recycle registers at the desired count. Verification / Alternative check: Example designs include a NAND of particular Q outputs to drive an asynchronous clear, or a synchronous load that inserts the seed pattern, confirming the role of logic functions in resetting. Why Other Options Are Wrong: Incorrect: Ignores widely used decoding and feedback techniques.Only true for binary ripple / only with async clear: Shift-register counters are not limited to those cases; synchronous load or gated feedback also work. Common Pitfalls: Failing to include a recovery path from illegal states; assuming power-up always sets a valid pattern without supervisory logic. Final Answer: Correct

Question 17

Up/down counters and direction control: The statement claims that once an up/down counter begins counting it cannot be reversed. Decide whether direction can be changed on-the-fly with proper control timing.

Accepted Answer

Introduction / Context: Up/down counters include a control input that selects the counting direction. Practical systems often need to reverse direction—for example, in position encoders, elevator controllers, or digital tuning. This question challenges the idea that once counting starts, reversal is impossible. Given Data / Assumptions: Counters have a direction or up/down input (often labeled U/D). Changing U/D synchronously with the clock prevents timing hazards. Asynchronous or synchronous clear/load inputs may also be present. Concept / Approach: Reversal is a core feature: the next state logic interprets the control line to add or subtract one. As long as setup/hold requirements for the U/D control relative to the active clock edge are met, the counter cleanly changes direction at the next count event without needing to stop or reset first. Step-by-Step Solution: Identify the U/D control input and its timing requirements.Ensure U/D is stable before the active clock edge.On the subsequent edge, the counter will step in the newly selected direction.Therefore, the claim that reversal is impossible is incorrect. Verification / Alternative check: Datasheets for standard up/down counters (for example, synchronous 4-bit devices) show truth tables where the state transition increments or decrements based on the U/D input each cycle. Why Other Options Are Wrong: Correct: Would deny a fundamental feature of up/down counters.Only true if reset first / Valid only for synchronous counters: Neither is generally required; both sync and ripple up/down counters can change direction with proper timing. Common Pitfalls: Switching U/D near the active edge and violating setup/hold, causing metastability or an incorrect step. Final Answer: Incorrect

Question 18

Creating time delay lines with shift registers: Assess the statement: “An effective time delay device can be constructed using the clocked propagation behavior of parallel shift registers to delay data by an exact number of clock cycles.”

Accepted Answer

Introduction / Context: Digital designers often need deterministic delays measured in clock cycles. While analog delays use components or specialized devices, digital systems commonly realize precise delays with shift registers that pass data along one stage per clock. The question asks whether shift registers can serve as effective time-delay elements. Given Data / Assumptions: A parallel shift register captures a multi-bit word simultaneously and shifts it one stage per clock. Clock period Tclk defines the discrete delay unit. N cascaded stages provide an overall delay of N * Tclk (ignoring minor t_pd effects). Concept / Approach: Using shift registers as digital delay lines is standard practice in serializers, de-serializers, video pipelines, and digital filters. Because each clock advances data by exactly one stage, the total delay is quantized and repeatable, making it highly effective for synchronization and alignment. Step-by-Step Solution: Choose the required delay D = N * Tclk.Instantiate N stages of flip-flops (or an N-stage shift register).Feed input at stage 0; tap output at stage N.The word emerges exactly N cycles later, achieving the desired delay. Verification / Alternative check: Simulation or timing diagrams easily show data propagating one stage per clock; hardware devices like 74HC595 chains serve as practical examples of controlled multi-cycle delays. Why Other Options Are Wrong: Incorrect: Overlooks widespread use of digital delay lines.True only with analog BBD devices / Valid only if no clock is used: These contradict the digital, clocked nature of shift-register delays. Common Pitfalls: Confusing asynchronous propagation delay with intentional clocked delay; neglecting that total delay is quantized in cycles, not continuous time. Final Answer: Correct

Question 19

Bidirectional shift registers — movement of data: Evaluate the statement: “Bidirectional shift registers can shift data either to the right or to the left under control.”

Accepted Answer

Introduction / Context: Shift registers move data through flip-flop stages on clock edges. Some applications require shifting in both directions—right or left—for operations like arithmetic shifts, data alignment, and bidirectional serial interfaces. This item asks whether bidirectional devices can indeed shift either way. Given Data / Assumptions: Bidirectional shift registers include a direction control input. On each clock, data moves one stage according to the direction selection. Devices may also support parallel load and serial I/O. Concept / Approach: With an explicit direction control, the internal multiplexing routes the proper neighbor stage to each flip-flop’s input. Thus, asserting “left” moves bits toward higher-order positions while “right” moves them toward lower-order positions. The capability is explicitly designed and documented for bidirectional parts. Step-by-Step Solution: Set direction = RIGHT and apply clocks; observe data shifts toward Q0.Set direction = LEFT and apply clocks; observe data shifts toward QN.Confirm that only the direction control changes; clocking behavior remains the same.Therefore, the statement is accurate for bidirectional devices. Verification / Alternative check: Datasheets for common bidirectional registers (for example, 74xx194/195 family) show mode tables with left/right shift options selected via control inputs. Why Other Options Are Wrong: Incorrect: Conflicts with the definition of “bidirectional.”Only true for serial-in/serial-out / Valid only when the clock is disabled: Directional shifting requires clocking and applies to multiple I/O modes. Common Pitfalls: Forgetting to provide valid serial data on the incoming side when shifting; not meeting setup/hold for the direction control relative to the clock. Final Answer: Correct

Question 20

Parallel in / parallel out (PIPO) register structure: Judge the statement: “A parallel in/parallel out register provides parallel input and parallel output buses for multi-bit words.”

Accepted Answer

Introduction / Context: Registers are classified by how data enters and exits: serial-in/serial-out (SISO), serial-in/parallel-out (SIPO), parallel-in/serial-out (PISO), and parallel-in/parallel-out (PIPO). PIPO registers are useful whenever an entire word must be captured and later presented simultaneously. This item asks if a PIPO device actually has parallel input and output buses. Given Data / Assumptions: “Parallel” implies multiple bits transferred at the same time. PIPO captures an N-bit word on load/clock and outputs all N bits together. Word width depends on the specific device (4, 8, 16, etc.). Concept / Approach: By design, PIPO devices expose N input pins (often D0..DN-1) and N output pins (Q0..QN-1). With a load or clock edge, the input word is sampled and then appears concurrently on the output bus, enabling quick data movement between subsystems without serializing. Step-by-Step Solution: Identify that “parallel in” requires N input lines.Identify that “parallel out” requires N output lines.Confirm that a load/clock edge transfers the word into the register and then to outputs.Therefore, the statement is correct by definition. Verification / Alternative check: Typical PIPO devices, including 74HC273 or similar, show N D-inputs and N Q-outputs with timing diagrams for parallel capture and update. Why Other Options Are Wrong: Incorrect: Contradicts standard terminology.True only for 8-bit / Valid only with serial I/O: Word size and presence of serial ports are independent of the PIPO definition. Common Pitfalls: Confusing PIPO with PISO/SIPO; forgetting that propagation and clock-to-Q delays still apply to “parallel” transfers. Final Answer: Correct

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