Question 1

Concept check: Because a register stores a multi-bit word using flip-flops, its storage capability makes it an important (albeit small and fast) form of memory in digital systems. Evaluate this statement.

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Introduction / Context:Digital systems distinguish between combinational logic (no memory) and sequential logic (with memory). A register is a set of clocked flip-flops that holds data across time, providing the fastest, smallest-capacity memory close to arithmetic units and control logic.Given Data / Assumptions:Registers are built from edge-triggered flip-flops or latches.They hold bits until updated by a clocked event or control signal.Architectures include multiple register files and pipeline registers.Concept / Approach:Memory is any element that retains state. Registers retain state deterministically under clock and control, thus constituting memory. They differ from large arrays (SRAM, DRAM) in capacity and use, not in the fundamental property of storing bits. This is why pipeline stages, CPU registers, microarchitecture scoreboards, and configuration registers are all categorized as memory elements.Step-by-Step Solution:1) Identify storage: flip-flops store Q values across cycles.2) Recognize control: enables, resets, and loads manage when values change.3) Observe use: registers buffer data between combinational stages and serve as programmer-visible storage in CPUs.4) Conclude that register storage capacity is meaningful memory within a system hierarchy.Verification / Alternative check:Textbooks classify registers under sequential circuits; CPU block diagrams list 'register file' as on-chip memory.Why Other Options Are Wrong:“Incorrect” and 'purely combinational' misunderstand registers; they are sequential and stateful. Requiring SRAM would conflate a specific memory array type with the broader concept of memory.Common Pitfalls:Confusing memory hierarchy levels; assuming 'memory' only means large RAM arrays.Final Answer:Correct

Question 2

Motion profile: A stepper motor rotates in discrete step increments determined by its construction and drive sequence. Does it inherently produce smooth continuous motion?

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Introduction / Context:Stepper motors are chosen for precise position control because they move in fixed angular steps. The visual appearance of motion can be smoothed with control techniques, but fundamentally the motor advances in discrete increments rather than a continuous rotation like a typical DC motor without commutation steps.Given Data / Assumptions:Stepper step angle is set by stator/rotor geometry (e.g., 1.8°, 7.5°).Drive sequences energize phases to advance one step at a time.Microstepping modulates currents to interpolate intermediate positions.Concept / Approach:Without microstepping, torque ripple and stepwise motion are evident. Microstepping applies sinusoidal or quasi-sinusoidal currents to create smaller effective steps; however, even then the motion is realized as a series of controlled increments. Calling the motion 'smooth continuous' mischaracterizes the technology, although proper microstepping can make it feel smooth for many applications.Step-by-Step Solution:1) Energize phase sequence → rotor aligns to the nearest stable step position.2) Advance the sequence → rotor jumps to the next stable position.3) With microstepping → intermediate current ratios yield intermediate equilibrium positions.4) Motion remains discrete by control, not inherently continuous physics.Verification / Alternative check:Torque vs position plots show detent and holding torque at discrete positions; microstepping reduces ripple but does not change the underlying incremental nature.Why Other Options Are Wrong:“Correct” ignores discrete stepping; limiting correctness to certain stepper types or microstepping still does not create truly continuous, brushless-DC-like rotation.Common Pitfalls:Assuming microstepping eliminates all resonance and ripple; misinterpreting smoothness at low load as continuous rotation.Final Answer:Incorrect

Question 3

Johnson counter fact check: An n-bit Johnson (twisted-ring) counter produces 2n unique states before repeating. For an 8-bit Johnson counter, is the sequence length eight states?

Accepted Answer

Introduction / Context:Johnson counters are widely used for sequence generation and timing because they visit a set of unique states with simple feedback. Knowing the exact sequence length is essential for designing decoders and detecting end-of-sequence conditions.Given Data / Assumptions:In a Johnson counter, the inverted output of the last flip-flop feeds the input of the first.There are n flip-flops in the chain.All flip-flops shift simultaneously on the clock edge.Concept / Approach:A Johnson counter cycles through 2n states, not n. For n = 8, the sequence length is 16 states. This differs from a simple ring counter, which has exactly n states (one-hot patterns). Understanding the distinction prevents incorrect assumptions about decoding and resource allocation.Step-by-Step Solution:1) Start from a valid seed state.2) On each clock, shift and feed back the inverted last bit to the first.3) Count the number of distinct states before repetition; algebra and examples show 2n unique states.4) For n = 8, obtain 16 unique states.Verification / Alternative check:Enumerate states for a 4-bit Johnson counter (2n = 8) to build intuition, then generalize to 8-bit for 16 states.Why Other Options Are Wrong:“Correct” confuses Johnson with ring counters. Other qualifiers do not alter the fundamental 2n property.Common Pitfalls:Misidentifying counter type; assuming one-hot behavior for Johnson sequences.Final Answer:Incorrect

Question 4

Recovery and clarification: Assume a shift register with an active-LOW shift enable input (SE_L). When SE_L is LOW, the data path advances one bit on each active clock pulse. Under this common convention, does SE_L = 0 cause one-bit shifting per clo…

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Introduction / Context:The original stem omitted the control signal name. Applying the recovery-first policy, we interpret the intent as describing an active-LOW shift-enable input (often labeled SE_L or /SE). Many shift-register ICs use an enable that, when asserted, allows shifting to occur on each qualifying clock event.Given Data / Assumptions:Assume a registered, clocked shift register.Assume a dedicated active-LOW shift enable (SE_L).Shifting occurs on the specified clock edge when enabled.Concept / Approach:Enable pins gate the action of the internal flip-flops. With SE_L asserted (LOW), each active clock edge transfers data from one stage to the next, achieving a one-bit-per-clock shift. When SE_L is deasserted (HIGH), the contents hold (no shift). Note that some parts invert this convention (enable HIGH); always check the function table.Step-by-Step Solution:1) Tie SE_L = 0 to assert shift enable.2) Apply a sequence of clock edges; on each edge, stages advance by one position.3) When SE_L = 1, hold state; no shifting occurs.4) Confirm behavior against the device timing diagram.Verification / Alternative check:Examine example datasheets where /SE or /CE gates shifting; simulation or bench tests show one-bit advancement per clock when the enable is active.Why Other Options Are Wrong:“Incorrect” would imply enable LOW prevents shifting, which contradicts the assumed active-LOW definition. Options that bind behavior to level clocks or asynchronous operation misstate typical edge-triggered, synchronous shifting.Common Pitfalls:Forgetting enable polarity; assuming shifting occurs on clock level rather than edges; ignoring hold behavior when enable is inactive.Final Answer:Correct

Question 5

Part-specific statement: The 74194 4-bit bidirectional universal shift register is versatile and finds use in data conversion, sequence generation, and interface tasks. Evaluate the claim that it has a wide range of applications.

Accepted Answer

Introduction / Context:The 74194 class of devices implements a bidirectional universal shift register with serial and parallel paths plus directional control. Such capability allows designers to bridge serial and parallel domains, manipulate data order, and implement compact state machines or counters with minimal glue logic.Given Data / Assumptions:Device supports parallel load and serial shift left/right.Exposes both serial and parallel I/O.Provides mode and direction control pins.Concept / Approach:With parallel load, data from a bus can be captured and then shifted out serially; conversely, serial data can be shifted in and then presented in parallel. Bidirectional shifting enables MSB-first or LSB-first sequencing as needed. These features underpin wide application: format conversion (PISO/SIPO), time-division serialization, sequence generation, simple counters, and programmable delays.Step-by-Step Solution:1) Use parallel load to capture a 4-bit word.2) Select shift direction to serialize MSB-first or LSB-first.3) Chain multiple devices for wider words.4) Employ mode controls to alternate between load and shift phases.Verification / Alternative check:Typical application notes show the device in serial/parallel conversion, temporary storage, and small sequence generators.Why Other Options Are Wrong:“Incorrect” and claims of 'only ring counter' or 'only serial output' contradict the documented capabilities.Common Pitfalls:Ignoring timing (synchronous vs asynchronous load); misinterpreting direction controls; not providing proper chaining for wider widths.Final Answer:Correct

Question 6

Shift Registers — In parallel load mode, data is applied to all bit positions and all flip-flops capture their respective inputs at the same clock event (i.e., loaded simultaneously). Evaluate the statement.

Accepted Answer

Introduction: Parallel loading is a foundational capability of many shift registers and register files. The idea is that each flip-flop in the register receives its own data input line (a parallel bus), and when the correct control and clock conditions occur, all flip-flops capture their assigned bits at the same instant. This item asks you to confirm that “parallel load means to load all flip-flops at the same time.”Given Data / Assumptions:A register contains multiple flip-flops, typically one per bit.A control input (often labeled LOAD or SHIFT/LOAD) selects between parallel load and shift modes.Loading occurs on a specified active clock edge if the control selects parallel load.Concept / Approach: In parallel load mode, each stage’s D input is driven by the corresponding bit of the parallel data word. When the clock edge arrives (synchronous design), the entire word is sampled and stored simultaneously, eliminating the per-bit skew that would appear if bits were shifted serially over multiple cycles.Step-by-Step Solution:Identify the control: when LOAD=active, the input multiplexers route the external data bus to each flip-flop.On the active clock edge, every flip-flop samples its D input.Because the same edge triggers all stages, capture is simultaneous for the whole word.Verification / Alternative check:Datasheets for universal shift registers (e.g., 74195 family) describe “parallel load” as synchronous, single-edge capture across all stages.Why Other Options Are Wrong:Incorrect: Contradicts standard register operation.Only true for ripple registers: Ripple refers to asynchronous carry or clocking in counters, not parallel load behavior.Depends on duty cycle: Duty cycle does not alter the fact of simultaneous sampling at an edge.Common Pitfalls:Confusing parallel load with serial shift (which takes multiple clocks).Assuming “simultaneous” means “asynchronous”; it is synchronous to a single edge.Final Answer:Correct

Question 7

74164 Behavior — The SN74164 is a serial-in/parallel-out shift register with no synchronous parallel load feature. Evaluate the claim: “To synchronously load parallel data on the negative clock edge, set the parallel enable LOW.”

Accepted Answer

Introduction: Not all shift registers support both serial shifting and parallel loading. The SN74164 device is a classic serial-in/parallel-out (SIPO) register: it shifts data in serially and presents the accumulated bits at parallel outputs; it does not provide a “parallel load” control that captures an entire word at once. This question asks you to judge a statement that incorrectly attributes a parallel-load capability to the 74164.Given Data / Assumptions:Device: 74164 (or HC/HCT164 variants).Inputs: serial A/B (gated), clock, asynchronous clear.Outputs: parallel Q pins, updated by shifting.Concept / Approach: A true parallel load requires per-bit data inputs and a load control that selects those inputs at a clock edge. The 74164 exposes no per-bit data bus for parallel capture; it accepts serial data only. Therefore, any wording about “parallel enable LOW” for synchronous parallel loading is incompatible with the device’s feature set and is false.Step-by-Step Solution:Check the pinout: only serial inputs (A and B) feed the internal chain.Check controls: there is no LOAD/SHIFT select; only clock and asynchronous clear exist.Conclude: parallel loading is not supported; the claim is invalid regardless of clock edge polarity.Verification / Alternative check:Compare with a universal shift register (e.g., 74194/74195) which does provide a SHIFT/LOAD control and parallel inputs; 74164 does not.Why Other Options Are Wrong:Correct: Would require a feature the 74164 lacks.True only with external gating: External logic can preload lines but still enters serially; not parallel load.Depends on asynchronous clear: Clear resets the chain; it is unrelated to parallel capture.Common Pitfalls:Confusing SIPO devices (74164) with universal shift registers (74195).Thinking “parallel outputs” implies “parallel load.” These are different features.Final Answer:Incorrect

Question 8

Register Terminology — In a multi-bit register, a “stage” refers to a single flip-flop (one storage element) that holds one bit. Evaluate the statement: “A stage is two storage elements in a register.”

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Introduction: The word “stage” is used consistently in the context of registers and shift registers to denote one bit of storage, typically implemented by a single flip-flop (FF). This question challenges a wrong definition that doubles the storage per stage.Given Data / Assumptions:Registers are built from cascaded flip-flops.Each flip-flop holds 1 bit.Shift registers connect stages in series for data movement.Concept / Approach: A 4-bit register has four stages; each stage is one flip-flop. While an edge-triggered FF can be implemented internally as a master-slave pair of latches, the architectural unit of storage seen by the designer is still one bit per stage. Therefore the statement “A stage is two storage elements” conflates internal FF construction with top-level register staging.Step-by-Step Solution:Define “stage” = one FF = one bit.Master-slave implementation detail does not double the register’s bit count.Thus, a 4-stage register stores 4 bits, not 8.Verification / Alternative check:Datasheets and textbooks enumerate “n-stage shift register” to mean “n bits capacity,” confirming one bit per stage.Why Other Options Are Wrong:Correct: Misstates the standard definition.True only for master-slave FFs: Internal latches do not define the stage count.Depends on bit width: Stage definition is independent of total width.Common Pitfalls:Equating internal latching with external capacity.Assuming two latches equal two bits; in a master-slave FF they represent one bit.Final Answer:Incorrect

Question 9

Ring & Johnson Counters — These are widely used shift-register-based counters (sequence generators), not “uncommon” circuits. Evaluate the statement.

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Introduction: Ring counters and Johnson (twisted-ring) counters are classic sequence generators built from shift registers. They appear in timing generators, state machines with simple decoding, LED chasers, and clock-division applications. The claim that they are “uncommon” is inaccurate in both education and industry contexts.Given Data / Assumptions:Ring counter: a single 1 circulates through a ring of flip-flops.Johnson counter: the inverted output of the last stage is fed back to the first, producing 2N distinct states with N flip-flops.Synchronous clocking is typical; decoding is straightforward.Concept / Approach: Because these counters provide easily decodable one-hot or two-hot patterns with low logic overhead, they are favored where simple, glitch-resistant state outputs are needed. Their usage is common in CPLD/FPGA designs and discrete TTL/CMOS projects alike.Step-by-Step Solution:Recognize ring/Johnson as shift-register feedback topologies.Note applications: timing sequences, divide-by-N chains, LED scanning.Conclude that describing them as “uncommon” is a mischaracterization.Verification / Alternative check:Survey introductory digital design textbooks and vendor app notes; both cover these counters prominently.Why Other Options Are Wrong:Correct: Would imply rarity; contradicted by widespread usage.True only in ASICs / FPGAs: Both discrete and programmable logic use them.Common Pitfalls:Confusing these with binary ripple or synchronous up/down counters; they serve different needs.Overlooking the ease of decoding compared to binary counters.Final Answer:Incorrect

Question 10

SIPO Recognition — Which shift register type inherently requires access to all Q outputs of each stage to use the data?

Accepted Answer

Introduction: Different shift register configurations expose data at different interfaces. Some deliver bits serially, while others present a word on parallel output pins. Identifying which type needs access to all Q outputs helps clarify how data is retrieved from the structure.Given Data / Assumptions:Shift register stages each have an output (Q).We are comparing SISO, SIPO, PISO, and related variants.“Requires access to all Q outputs” means you must read each stage’s Q pin to obtain the word.Concept / Approach: A serial-in/parallel-out (SIPO) register accepts data serially but makes the accumulated contents available as a parallel word. To read that word, the system connects to each stage’s Q pin (one per bit). In contrast, SISO streams data out on one serial line; PISO loads a word in parallel but shifts it out serially; neither requires external access to all Q pins for normal operation.Step-by-Step Solution:Define SIPO: serial input, Q outputs for every bit.Reading the word ⇒ sample all Q pins simultaneously.Therefore, SIPO matches the requirement in the prompt.Verification / Alternative check:Check a 74HC595 style device: it provides parallel outputs (Q0..Q7) to drive loads directly.Why Other Options Are Wrong:PISO: Uses parallel inputs, not parallel outputs; normal readout is serial.SISO / Bidirectional SISO: Only a single serial output line is used; no need to tap every Q.Common Pitfalls:Confusing “has Q pins internally” with “requires external access” to all Q lines.Assuming any parallel path implies both directions (in and out); SIPO is output-parallel only.Final Answer:Serial in / parallel out (SIPO)

Question 11

Johnson Counter States — For a Johnson (twisted-ring) counter constructed with N flip-flops, how many unique states are produced per full sequence?

Accepted Answer

Introduction: A Johnson counter (also called a twisted-ring counter) is formed by feeding the inverted output of the last flip-flop back to the input of the first. Unlike a normal ring counter, which has N states for N flip-flops, the Johnson counter produces more unique patterns for the same number of stages.Given Data / Assumptions:N flip-flops clocked synchronously.Feedback: inverted last stage to first stage.Sequence repeats after a fixed number of clocks.Concept / Approach: The inverted feedback creates a sequence where runs of 1s and 0s expand and contract through the register. The resulting cycle length is 2N states, offering easier decoding (many “two-hot” combinations) and glitch-resistant outputs for one-of-K control signals.Step-by-Step Solution:Start with all zeros; each clock injects a 1 at one end while previous bits shift.After N clocks, the register is all ones; subsequent clocks inject zeros and shrink the run of ones.Total distinct patterns until returning to the start = 2N.Verification / Alternative check:Simulate a small case (N=4): states = 0000 → 1000 → 1100 → 1110 → 1111 → 0111 → 0011 → 0001 → 0000 (8 = 2N states).Why Other Options Are Wrong:N: That is the count for a simple ring counter, not a Johnson counter.2^N: That is the maximum for a generic N-bit register, not this constrained sequence.N + 1: Underestimates the Johnson sequence length.Common Pitfalls:Confusing ring counters (N states) with Johnson counters (2N states).Counting “transitions” instead of “distinct states.”Final Answer:2N

Question 12

74195 Operation — With SHIFT/LOAD held LOW on a 74195 universal shift register (selecting parallel load), when do the outputs update?

Accepted Answer

Introduction: The 74195 is a universal shift register supporting both serial shifting and parallel loading via a SHIFT/LOAD control input. Understanding whether the parallel load is synchronous or asynchronous is essential for correct timing design.Given Data / Assumptions:Device: 74195 (or HC/HCT195 family).Control: SHIFT/LOAD = LOW selects parallel load mode.Clocked (synchronous) operation; asynchronous clear may also exist but is separate.Concept / Approach: In parallel load mode, each stage takes its input from the parallel data bus. However, loading occurs on the defined active clock edge (typically the rising edge) so that all bits are captured simultaneously and deterministically. Therefore, outputs do not change “immediately”; they update synchronously on the next clock edge while SHIFT/LOAD remains LOW.Step-by-Step Solution:Set SHIFT/LOAD = LOW to select parallel inputs.Apply the desired parallel data on the input pins.Toggle the clock to its active edge; the register outputs update together.Verification / Alternative check:Consult timing diagrams: output transitions are edge-aligned; there is no level-sensitive, asynchronous load in normal operation.Why Other Options Are Wrong:Immediately: Would imply asynchronous load, which is not the case.Only if CLOCK is LOW: Level of clock is irrelevant; the edge triggers the load.Depends on J and K: The 74195 does not use J/K controls; that wording confuses with JK flip-flops.Common Pitfalls:Assuming “parallel load” equals “asynchronous load.”Forgetting to hold SHIFT/LOAD LOW through the active edge.Final Answer:On the next active clock edge

Question 13

4-Bit Ring Counter — A ring counter is loaded with a single 1 (one-hot). What is the output frequency at any given stage relative to the clock?

Accepted Answer

Introduction: In a ring counter with a single circulating 1 (one-hot encoding), only one stage is HIGH at a time, and that HIGH token advances by one position each clock. Understanding the per-stage output frequency helps with timing, decoding, and LED scanning applications.Given Data / Assumptions:4 stages; synchronous shifting on each clock edge.Exactly one stage is HIGH in any state.Outputs are observed at the Q pins of the stages.Concept / Approach: In four stages, the HIGH appears at a given stage once every 4 clocks. Therefore that stage’s output is HIGH for one clock out of four and LOW for the other three, yielding a frequency that is 1/4 of the clock frequency (and a 25% duty cycle, assuming one-cycle-wide HIGH).Step-by-Step Solution:Label stages S0..S3; initialize 1000.Sequence per clock: 1000 → 0100 → 0010 → 0001 → 1000 ...S0 is HIGH once every 4 clocks ⇒ f_out(S0) = f_clk / 4.Verification / Alternative check:Count rising edges between consecutive HIGH occurrences at one stage; the interval is 4 clocks.Why Other Options Are Wrong:Same as the clock: Would require the bit to toggle every clock, which it does not.Twice the clock / Half the clock: Not consistent with one-hot circulation among four positions.Common Pitfalls:Confusing per-stage frequency with the shift frequency of the token.Assuming multiple 1s (two-hot) as in Johnson counters; this is one-hot.Final Answer:One-fourth the clock frequency

Question 14

4-Bit Johnson Counter — If the counter is initially cleared (all zeros) and after the first clock pulse the output is 0001, what is the output after the next clock pulse (standard Johnson feedback)?

Accepted Answer

Introduction: A Johnson counter (twisted-ring) feeds back the inverted last stage into the first stage, producing a sequence of two-hot patterns before returning to all zeros. The exact bit ordering in questions varies by textbook convention; here the second state is given as 0001, and you must identify the next state under standard Johnson feedback with right-shifted insertion of the inverted last bit.Given Data / Assumptions:4 flip-flops, synchronously clocked.Initial: 0000; after first clock: 0001 (given).Feedback: insert NOT(Q_last) into the vacated end on each shift; shift direction consistent with the given transition.Concept / Approach: From 0001 to the next state under Johnson operation, the register shifts the pattern and inserts the inverted last bit. With the given convention (moving the “1” toward the MSB while inserting NOT of the dropping bit), the two-hot run grows: 0001 → 0011, then 0111, then 1111, etc., before shrinking back toward 0000.Step-by-Step Solution:State S1 = 0001 (given).Invert last bit (1 → 0) and shift according to the stated convention.Result S2 = 0011, extending the run of ones.Verification / Alternative check:Enumerate the classic 4-bit Johnson sequence under this bit order: 0000 → 0001 → 0011 → 0111 → 1111 → 1110 → 1100 → 1000 → 0000.Why Other Options Are Wrong:0010: Would shrink, not extend, the run; not Johnson behavior here.1000: Belongs to the later part of the cycle in this ordering.0110: Does not follow directly from 0001 under standard Johnson feedback.Common Pitfalls:Mixing left vs right shift conventions; follow the one implied by the given step.Confusing Johnson with simple ring sequences.Final Answer:0011

Question 15

PIPO Recognition — Which shift register configuration does not connect a stage’s Q (or Q̅) output into the next stage’s input (i.e., no serial chain between stages)?

Accepted Answer

Introduction: In many registers, stages are daisy-chained so a bit can shift from one flip-flop to the next. But in a purely parallel register, each stage is loaded and read independently with no inter-stage serial path. Recognizing this architecture helps you choose the right component for buffering or synchronous capture tasks.Given Data / Assumptions:We compare PISO, SIPO, SISO, and PIPO structures.“Q to next-stage input” refers to the serial chain typical in shift registers.We are looking for the configuration without that chain.Concept / Approach: A PIPO register provides parallel inputs and parallel outputs. Each flip-flop captures its input on the clock edge and presents it at its own Q pin. There is no requirement that Q of stage i drives the D of stage i+1; hence there is no serial shifting. This makes PIPO devices ideal for temporary data latching and bus interfacing where word-wide simultaneous capture is required.Step-by-Step Solution:Identify the serial chain requirement: excluded by definition in PIPO.Confirm PISO/SISO/SIPO all involve serial movement of data.Conclude PIPO is the only option with no Q-to-next interconnect.Verification / Alternative check:Check typical logic families: “octal D-type registers” (PIPO) show no serial connections between stages on the datasheet block diagram.Why Other Options Are Wrong:PISO: Has a serial output chain from stage to stage.SIPO: Has a serial input chain linking stages.SISO: Entire operation depends on a single serial chain.Common Pitfalls:Assuming “parallel outputs” implies no chain; SIPO still has a serial input chain internally.Conflating latches/registers with shift registers; PIPO is a register without shifting.Final Answer:Parallel in / parallel out (PIPO)

Question 16

Shift-register delay selection: An 8-bit serial-in/parallel-out (SIPO) shift register is clocked at 4 MHz (Tclk = 0.25 µs) and is used to delay a serial digital signal by exactly 1.25 µs. Which parallel tap (QA through QH) provides the output with t…

Accepted Answer

Introduction / Context:Serial-in/parallel-out (SIPO) shift registers are frequently used to create precise, clock-synchronous delays. Each stage holds the input bit for one clock period, so selecting the correct tap provides a deterministic delay useful in timing alignment, serialization, de-serialization, and pulse stretching.Given Data / Assumptions:Register type: 8-bit SIPO with taps QA (stage 1) through QH (stage 8).Clock frequency fclk = 4 MHz, so Tclk = 1 / 4 MHz = 0.25 µs.Required total delay = 1.25 µs relative to the serial input bit entering the register.Standard stage labeling: QA is the first stage, QH is the eighth.Concept / Approach:In a SIPO, a bit appears at stage n after n clock edges. Thus, the delay at tap Qn equals n * Tclk. Choose n so that n * 0.25 µs = 1.25 µs. The integer n that satisfies this is n = 5, corresponding to the fifth stage (QE).Step-by-Step Solution:Compute Tclk = 1 / 4,000,000 = 0.25 µs.Set n * 0.25 µs = 1.25 µs.Solve for n: n = 1.25 / 0.25 = 5.Tap for stage 5 is QE (QA=1, QB=2, QC=3, QD=4, QE=5).Verification / Alternative check:List delays per tap: QA=0.25 µs, QB=0.50 µs, QC=0.75 µs, QD=1.00 µs, QE=1.25 µs, QF=1.50 µs, QG=1.75 µs, QH=2.00 µs. The required delay matches QE exactly.Why Other Options Are Wrong:QF: 6 * 0.25 µs = 1.50 µs (too large).QG: 7 * 0.25 µs = 1.75 µs (too large).QH: 8 * 0.25 µs = 2.00 µs (too large).Common Pitfalls:Off-by-one mistakes in stage numbering; confusing SIPO stage index with bit significance; ignoring that each stage adds exactly one Tclk of latency. Always confirm the device's pinout for QA–QH mapping.Final Answer:QE

Question 17

Parallel-in/serial-out timing: A 4-bit parallel-in/serial-out (PISO) shift register is loaded with a binary number using the parallel-load control. After the parallel load occurs, how many clock pulses are required before the first bit of the serial…

Accepted Answer

Introduction / Context:PISO (parallel-in/serial-out) registers convert a parallel word into a serial bitstream. Understanding when the first bit becomes available is essential for designing serializers, time-division multiplexers, and interface adapters where precise cycle-by-cycle timing is critical.Given Data / Assumptions:Device: 4-bit PISO shift register.A valid parallel load cycle places the word into the internal stages.Serial output taps the register stage designated for the first transmitted bit (often MSB).Standard synchronous design: load and shift controls are distinct.Concept / Approach:On a parallel load, each internal stage immediately holds its corresponding bit. Because the serial output is wired to the stage that is transmitted first, that bit is already present at the output right after the load completes. No additional shift clock is required to make the first bit appear; the next clocks shift the remaining bits out in order.Step-by-Step Solution:Assert parallel-load to capture D3..D0 into stages Q3..Q0.Observe serial output pin connected (commonly) to Q3 (MSB first).Immediately after load, serial output = first bit (e.g., D3).Subsequent shift clocks move D2, D1, D0 to the output in turn.Verification / Alternative check:Consult generic PISO timing: the first bit is valid post-load; tCO (clock-to-output) or load-to-output timing applies, not an extra shift. Only if the design specifies LSB-first with a different tap would the same rule apply to that designated stage.Why Other Options Are Wrong:1, 2, 3 clocks: These assume a shift is needed to expose the first bit, which contradicts the PISO's parallel load behavior.Common Pitfalls:Confusing PISO with SISO behavior; overlooking that the output is hardwired to one stage that already contains a valid bit immediately after load; misreading the data sheet's sequence diagram.Final Answer:0

Question 18

Bit shifting equivalence: In binary arithmetic, shifting a number left by one bit position is equivalent to which arithmetic operation on an unsigned value (ignoring overflow)?

Accepted Answer

Introduction / Context:Shifts are among the most efficient arithmetic operations in digital systems. Understanding their equivalence to multiply/divide by powers of two helps optimize algorithms, reduce resource usage in hardware, and reason about fixed-point scaling in DSP pipelines.Given Data / Assumptions:Binary, unsigned interpretation (or signed two’s complement without overflow).One-bit left shift (logical or arithmetic behaves identically on positive magnitudes).Overflow and bit growth are ignored for the equivalence statement.Concept / Approach:In base-2, each left shift moves every bit to a position with double the weight. Therefore, a left shift by k positions multiplies the numeric value by 2^k. For k = 1, the factor is 2. The lowest bit becomes 0, and the most significant bit may be lost if the width is fixed, which is the typical source of overflow concerns.Step-by-Step Solution:Let X be a binary word with value V.After a left shift by 1: value becomes V * 2.Bitwise view: each 1 moves to a position of double weight.Hence, left shift 1 ↔ multiply by two, provided no overflow occurs.Verification / Alternative check:Example: 0011₂ (3) Why Other Options Are Wrong:Multiplying by four: That corresponds to left shift by 2.Dividing by two/four: These correspond to right shifts by 1 or 2, not left shifts.Common Pitfalls:Forgetting fixed-width overflow; assuming the rule applies to signed negative values with arithmetic edge cases; confusing logical vs arithmetic shift (relevant for right shifts, not left).Final Answer:multiplying by two

Question 19

Clearing without reset: An 8-bit serial-in/parallel-out (SIPO) shift register has no asynchronous clear. You apply a logic 0 at the serial input to flush the register. After how many clock cycles will that injected 0 reach the QH (8th) output tap?

Accepted Answer

Introduction / Context:When a shift register lacks a dedicated clear pin, designers sometimes “clear” it by clocking in zeros until all stages are overwritten. Knowing exactly how many cycles are needed ensures predictable recovery from unknown states and clean start-up sequences.Given Data / Assumptions:Register: 8-bit SIPO with taps QA..QH (QA is the first stage, QH the eighth).No asynchronous/synchronous clear available.A logic 0 is applied to the serial data input.We want the cycle count until that specific 0 reaches QH.Concept / Approach:Each clock advances the serial input bit one stage deeper into the register. After k clocks, the injected bit resides at stage k. Therefore, to reach stage 8 (QH), the bit must traverse 8 stages and thus needs 8 clock cycles from the moment of injection.Step-by-Step Solution:Clock 1: 0 moves to QA (stage 1).Clock 2: 0 moves to QB (stage 2)....Clock 8: 0 moves to QH (stage 8), appearing at the QH output.Verification / Alternative check:General rule: to push a known bit from the input to stage n, apply exactly n clocks. Simulation waveforms of standard 8-bit SIPO devices validate this propagation behavior.Why Other Options Are Wrong:1 or 7 clocks: Insufficient to reach the 8th stage.9 clocks: One extra clock would push the original 0 out while a new input bit enters.Common Pitfalls:Counting from zero instead of one; confusing stage index with tap label; overlooking pipeline latency when back-to-back zeros are clocked in.Final Answer:8

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