Question 1

Definition of transparency in latches When is a latch or flip-flop said to be transparent in practical digital design terminology?

Accepted Answer

Introduction / Context: Transparency refers to whether the stored element passes input changes directly to the output during a certain control interval. This concept distinguishes level-sensitive latches from edge-triggered flip-flops and is fundamental for timing closure in pipelined and gated data paths. Given Data / Assumptions: Device: level-sensitive latch (for example, D latch) controlled by an enable or gate signal. Transparency applies during the active level of the enable. Edge-triggered flip-flops are not transparent; they sample only at transitions. Concept / Approach: A latch is called transparent when, while enabled, its output Q follows (tracks) the input D essentially in real time, limited only by propagation delay. When the enable is inactive, the latch becomes opaque, holding the last sampled value and blocking further input changes from affecting Q until re-enabled. Step-by-Step Solution: Enable asserted: Q ≈ D → transparent behavior.Enable deasserted: Q holds last value → opaque behavior.Edge-triggered FF: not transparent; samples only on clock edge → distinct behavior.Design implication: control the enable window to prevent hazards from propagating. Verification / Alternative check: Timing diagrams show Q tracking D during the enable window and flat otherwise. Simulation confirms that glitches on D may propagate while transparent, underscoring the need for clean combinational inputs during that interval. Why Other Options Are Wrong: Opposite the input: That describes inversion, not transparency. See-through packaging: Humorous distractor; unrelated to logic behavior. Ignores the input regardless of control: That would be an inoperative or constantly opaque device. Common Pitfalls: Assuming flip-flops behave transparently (they do not) and overlooking that transparency can allow hazards or race-through if enable timing is poorly constrained. Final Answer: when the Q output follows the input

Question 2

VHDL component placement In VHDL, in which declaration region is a COMPONENT typically declared before it is instantiated and port-mapped?

Accepted Answer

Introduction / Context: Structural VHDL allows designers to build larger systems by wiring together component instances. A component declaration introduces the interface of a reusable block so that it can be instantiated and connected within an architecture. Knowing where to place this declaration is critical for correct coding style and tool compatibility. Given Data / Assumptions: We are using component instantiation (as opposed to direct entity instantiation). The design unit includes an entity and an architecture. Libraries and packages are already made visible via library/use clauses. Concept / Approach: A COMPONENT is declared in the declarative region of an architecture (or in a package for broader reuse). The declaration specifies the component's generics and ports so that later, within the same architecture, you can create labeled instances with port map associations. The LIBRARY clause merely makes libraries visible; it does not host component declarations. The ENTITY describes the external interface of the current design unit, not third-party component declarations. A port map is part of an instantiation statement, not a declaration region. Step-by-Step Solution: Place the COMPONENT declaration at the top of the architecture declarative section.Later in the architecture's statement region, instantiate it: U1 : component_name port map (...).Alternatively, declare the component in a package to share across multiple architectures and use the package in each architecture.Direct entity instantiation can bypass component declarations but follows a different syntax. Verification / Alternative check: Tool templates and vendor examples show component declarations inside architectures or packages, followed by instantiations referencing those declarations. Why Other Options Are Wrong: Library: Used for visibility and referencing, not for component declarations. Entity: Declares the interface of the current module, not external components for instantiation. Port map: Part of the instantiation statement, not a declaration section. Common Pitfalls: Confusing component instantiation with direct entity instantiation; placing the declaration after use; or omitting needed packages that define subtypes used in the component ports. Final Answer: Architecture

Question 3

Effect of asynchronous inputs Do asynchronous inputs (such as preset or clear) cause a flip-flop to respond immediately, independent of the clock input?

Accepted Answer

Introduction / Context: Flip-flops generally have two classes of inputs: synchronous inputs that are evaluated with respect to a clock (for example, D, J, K, T) and asynchronous inputs that act immediately (for example, PRE and CLR). Understanding this distinction is crucial for safe reset design, power-on initialization, and robust timing closure. Given Data / Assumptions: Asynchronous inputs are typically labeled preset (PRE) and clear (CLR) and are often active-LOW. When asserted, they override normal synchronous operation. Clock presence or level is irrelevant to their immediate effect. Concept / Approach: Asynchronous controls bypass the clocked input path and directly force the internal storage node to a defined state. For example, PRE forces Q = 1 and CLR forces Q = 0 (polarity depending on the device). Because these signals do not wait for an active clock edge, they are described as causing an immediate response with respect to the clock. Step-by-Step Solution: Assert PRE (active level): Q is driven to 1 immediately, even if the clock is inactive or static.Assert CLR (active level): Q is driven to 0 immediately, independent of the clock.Release asynchronous input: device returns to normal synchronous operation at subsequent clock edges.Design reset networks to avoid violating removal/recovery times that can cause metastability upon deassertion. Verification / Alternative check: Datasheet timing sections specify asynchronous set/reset timing (recovery/removal) separately from synchronous setup/hold, confirming independence from the clock for assertion, but requiring care when deasserting around clock edges. Why Other Options Are Wrong: False: Would imply asynchronous controls wait for the clock, which contradicts their definition and practical behavior. Common Pitfalls: Mixing synchronous resets (implemented through clocked logic) with asynchronous resets (through PRE/CLR pins), and failing to respect recovery/removal constraints when releasing asynchronous signals. Final Answer: True

Question 4

Edge-triggered flip-flops — identifying the internal circuit In synchronous digital design, which internal circuit is primarily responsible for certain flip-flops being designated as edge-triggered (i.e., responding only at the clock transition rath…

Accepted Answer

Introduction / Context: An edge-triggered flip-flop changes state only during a brief instant when the clock transitions (rising or falling edge). This behavior contrasts with level-sensitive latches, which remain transparent while their enable level is asserted. Understanding what internal circuit enforces this edge-only sensitivity is crucial for reliable synchronous system design. Given Data / Assumptions: The device should respond at the clock edge (LOW→HIGH or HIGH→LOW), not during the entire HIGH or LOW level. We are comparing possible internal building blocks: edge-detection circuit, NOR/NAND latches, and pulse-steering. Setup and hold time requirements must still be met around the sampling instant. Concept / Approach: Edge-triggered flip-flops typically synthesize a very narrow sampling pulse from the transition of the external clock. This internal, short-duration pulse gates the data path only momentarily, thereby preventing the output from tracking input changes during the steady HIGH or LOW clock levels. The functional element that creates this pulse is the edge-detection circuit (also called a pulse transition detector). Step-by-Step Solution: 1) Desired behavior: sample inputs only around the clock transition.2) Mechanism: generate a narrow internal pulse synchronized to the clock edge.3) Effect: enable storage path briefly, then lock the state for the remainder of the cycle.4) Conclusion: the essential internal circuit is the edge-detection circuit. Verification / Alternative check: Inspect logic diagrams of common edge-triggered D/JK flip-flops; you will find differentiating networks or logic that create a spike (narrow pulse) from the clock transition to gate data transfer briefly. Why Other Options Are Wrong: NOR latch / NAND latch: these are storage primitives; by themselves they are level-sensitive, not edge-triggered. Pulse-steering circuit: used in master–slave or gating arrangements but does not inherently create the edge-only sampling pulse. Common Pitfalls: Assuming that fast response alone makes a device edge-triggered; without a pulse transition detector, the device would still be sensitive across a level, not just the edge. Final Answer: edge-detection circuit.

Question 5

VHDL fundamentals — primitive JK flip-flop interface In synthesizable VHDL describing a primitive JK flip-flop, how many primary inputs does the device have (considering J, K, and clock as the essential inputs; ignore optional asynchronous set/clear…

Accepted Answer

Introduction / Context: Hardware description languages such as VHDL capture the behavior of fundamental storage elements. A JK flip-flop is a classic device with two data-type inputs (J and K) and a clock input that governs when the stored state may change. Clarifying the count of the essential inputs avoids confusion when writing entity/architecture declarations and testbenches. Given Data / Assumptions: We focus on the basic, clocked JK flip-flop behavior. Asynchronous preset/clear lines, if present, are not counted for this question. We consider the inputs required to define the synchronous operation. Concept / Approach: At minimum, a JK flip-flop needs the J input, the K input, and the clock that determines the exact instant of sampling. Additional lines (like asynchronous set or clear) modify state outside the clocked path but are not part of the primitive synchronous interface under discussion here. Step-by-Step Solution: 1) Identify synchronous data/control: J and K (two inputs).2) Identify timing control: clock (one input) to trigger state change.3) Sum essential inputs: 2 (J,K) + 1 (CLK) = 3.4) Exclude optional async signals from this count per the problem statement. Verification / Alternative check: Examine a typical VHDL entity for a synchronous JK FF: entity ports usually include J, K, and CLK as required; PRE/CLR lines are optional extras. Why Other Options Are Wrong: 2: omits the clock, which is mandatory for a flip-flop. 4 or 5: would imply inclusion of extra asynchronous controls that the prompt asked us to ignore. Common Pitfalls: Confusing a JK latch (level-sensitive enable) with a clocked flip-flop; the clock is essential in the flip-flop case. Final Answer: 3

Question 6

S–R flip-flop — typical application Which of the following is a representative and practical use of an S–R (Set–Reset) flip-flop in digital electronics?

Accepted Answer

Introduction / Context: The S–R flip-flop is the simplest form of a bistable memory element built from cross-coupled gates. It stores one bit of information until inputs command a change. Knowing where an S–R flip-flop fits among common digital building blocks helps in recognizing appropriate use cases and in avoiding misuse where edge-triggered devices are required. Given Data / Assumptions: S–R flip-flop can hold a logic 0 or logic 1 stably. Inputs S and R control set and reset actions (active-HIGH in NOR version, active-LOW in NAND version). No clocking is required in the basic latch configuration. Concept / Approach: Because the S–R flip-flop maintains a stable state until commanded otherwise, it naturally serves as a 1-bit storage cell. By combining several S–R (or derived) flip-flops, designers form registers that store multi-bit values. Oscillators and pulse generators require specific feedback timing and are not direct applications of a stand-alone S–R latch. Step-by-Step Solution: 1) Identify the core capability: bistability and memory.2) Map capability to function: long-term holding of a single bit.3) Recognize multi-bit extension: multiple flip-flops form a binary storage register.4) Conclude that “binary storage register” captures a primary application. Verification / Alternative check: Look at register files and state machines where set/reset actions initialize or store state; the S–R mechanism underlies many synchronous elements when properly gated or clocked. Why Other Options Are Wrong: Racer: not a standard digital component; likely refers to race conditions, which are undesirable. Astable oscillator: requires timing networks and feedback not provided by a bare S–R latch. Transition pulse generator: typically uses differentiators/monostables, not a static S–R latch alone. Common Pitfalls: Using an S–R latch where a clocked D or JK flip-flop is required, leading to asynchronous hazards and metastability. Final Answer: binary storage register

Question 7

Frequency division using cascaded flip-flops A frequency divider consists of 12 cascaded toggle flip-flops (each divides by 2). If the input clock is 20.48 MHz, determine the output frequency after the twelfth stage.

Accepted Answer

Introduction / Context: Cascaded flip-flops are widely used to divide down a high-frequency clock to lower rates for timing, baud-rate generation, and scanning. Each flip-flop in toggle mode halves the frequency at its output, so N stages divide by 2^N overall. This problem checks your ability to apply that rule and handle units correctly. Given Data / Assumptions: Number of stages N = 12. Each stage divides by 2 in toggle mode. Input frequency fin = 20.48 MHz. Concept / Approach: The overall division factor is 2^N. Compute 2^12 and divide the input frequency accordingly. Ensure conversion from MHz to kHz if required when comparing to options. Step-by-Step Solution: 1) Compute division: 2^12 = 4096.2) Apply to fin: fout = 20.48 MHz / 4096.3) Perform the division: 20.48e6 / 4096 = 5000 Hz.4) Convert to kHz: 5000 Hz = 5 kHz → matches the choice. Verification / Alternative check: Stage-by-stage view: 20.48→10.24→5.12→2.56→1.28→0.64→0.32→0.16→0.08→0.04→0.02→0.01→0.005 MHz = 5 kHz; the twelfth stage indeed yields 5 kHz. Why Other Options Are Wrong: 10.24 kHz: corresponds to one stage fewer (divide by 2048). 30.24 kHz and 15 kHz: do not correspond to any power-of-two divisions from 20.48 MHz. Common Pitfalls: Miscounting the number of stages or forgetting that each stage divides by exactly 2, not by 10 or another factor. Final Answer: 5 kHz

Question 8

Triggering terminology — recognizing edge activation In digital logic, if an input action occurs specifically on a signal transition (rising or falling), that input is said to be ________.

Accepted Answer

Introduction / Context: Precise terminology matters when describing how a circuit responds to its control signals. When the response occurs only at the instant a control signal changes state (rather than while it is held at a level), we use the term edge-triggered. This distinction is key for timing analysis and for writing accurate specifications and HDL models. Given Data / Assumptions: The action is tied to a transition, not a sustained level. Both rising-edge and falling-edge cases qualify as edge-triggered. We are discussing general terminology independent of device family. Concept / Approach: Edge-triggering restricts sensitivity to a narrow time window around the signal edge. This minimizes the risk of unintended state changes caused by signal variations during the remainder of the cycle and enables stable synchronous design with defined setup/hold windows. Step-by-Step Solution: 1) Identify the nature of activation: action happens on transition only.2) Map the definition: such activation is called “edge-triggered.”3) Note examples: positive-edge D flip-flops, negative-edge counters.4) Conclude the correct term is edge-triggered. Verification / Alternative check: Consult typical datasheets: trigger types are specified as rising-edge, falling-edge, or level-sensitive; “edge-triggered” is the umbrella term for the first two. Why Other Options Are Wrong: Toggle triggered: informal; typically refers to JK FF behavior when J=K=1, not a general trigger type. Clock triggered: ambiguous; many devices use a clock but can be level-sensitive. Noise triggered: undesirable behavior, not a design goal or term of art. Common Pitfalls: Confusing level-sensitive latches with edge-triggered flip-flops; they have very different timing implications. Final Answer: edge-triggered

Question 9

Gated D flip-flop (level-enabled) — correct behavior Which statement correctly describes a gated (enable-controlled) D flip-flop or D latch when the enable input is asserted?

Accepted Answer

Introduction / Context: Gated D devices (often called D latches when level-controlled) are transparent while enabled and hold data when disabled. Being able to state their behavior succinctly is essential when building registers, FIFOs, or two-phase latch pipelines. Given Data / Assumptions: There is a data input D, an enable input (EN), and outputs Q and possibly Q'. Enable is active-HIGH. We are focusing on functional behavior, not electrical details. Concept / Approach: When EN is HIGH, the internal gating opens, allowing D to propagate to Q (subject to propagation delay). When EN goes LOW, the path is blocked and Q retains the last value sampled just before disabling. This is the hallmark of a level-sensitive D storage element. Step-by-Step Solution: 1) EN = HIGH → device transparent → Q ≈ D.2) EN = LOW → device opaque → Q holds its previous value.3) Therefore the correct statement is: Q follows D when EN is HIGH.4) Complement output, if present, remains Q' = NOT(Q). Verification / Alternative check: Timing diagrams show Q changing only during intervals where EN is asserted; otherwise it remains constant despite D activity. Why Other Options Are Wrong: Toggling with one input HIGH (a) and mutual exclusion of inputs (b) describe JK behavior, not D devices. Output complement follows input (c) is incorrect; Q follows D, not Q'. Common Pitfalls: Confusing a gated D latch with an edge-triggered D flip-flop, which updates only at a clock edge rather than during the entire enable interval. Final Answer: Q output follows the input D when the enable is HIGH.

Question 10

S–R flip-flop limitation — identifying the classic disadvantage What is one widely recognized disadvantage of an S–R (Set–Reset) flip-flop implementation?

Accepted Answer

Introduction / Context: The basic S–R flip-flop (or latch) is instructive but has a well-known drawback. Recognizing this limitation guides designers toward D or JK flip-flops when deterministic behavior is required, especially in synchronous systems where simultaneous assertion of set and reset must be handled predictably. Given Data / Assumptions: S–R built from cross-coupled NOR (active-HIGH) or NAND (active-LOW) gates. Two outputs: Q and Q' intended to be complementary. We evaluate functional limitations, not specific vendor implementations. Concept / Approach: With a NOR-based S–R latch, the input combination S=1 and R=1 drives both outputs LOW, violating complementarity and creating an invalid/forbidden state. Similarly, for NAND-based active-LOW inputs, S=0 and R=0 is forbidden. The aftermath when the inputs return to the idle state can be unpredictable and dependent on device delays, which undermines reliable operation. Step-by-Step Solution: 1) Identify forbidden input combination for the chosen implementation.2) Observe that outputs conflict or become indeterminate under that condition.3) Note potential metastability or race when exiting the forbidden state.4) Conclude that the S–R flip-flop has an inherent invalid state. Verification / Alternative check: Simulate with a gate-level model: apply the forbidden combination and then release simultaneously—final Q may depend on slight delay differences, confirming the hazard. Why Other Options Are Wrong: No enable input: not inherent; enables can be added externally. No clock input: true for a latch, but clocked S–R flip-flops exist. Single output: incorrect; Q and Q' are both available. Common Pitfalls: Forgetting which input polarity is active for NOR vs. NAND versions when identifying the forbidden combination. Final Answer: It has an invalid state.

Question 11

Switch bounce and erratic register outputs — diagnosing the cause A push-button switch is used to input data to a digital register, but the register's output appears erratic. What is the most likely cause of the problem?

Accepted Answer

Introduction / Context: Mechanical push-buttons and toggle switches rarely make a single, clean transition. Instead, their contacts briefly chatter between open and closed states for a few milliseconds. When feeding an edge-sensitive digital input, this “bounce” can be interpreted as multiple presses, leading to unpredictable register contents or multiple counts in counters. Given Data / Assumptions: Human-operated push-button connected to a register input (directly or via minimal conditioning). Observed symptom: erratic or non-reproducible register output. No evidence yet of power or IC failure. Concept / Approach: Contact bounce produces a burst of rapid transitions that digital logic may interpret as many presses. The standard remedy is debouncing, either with hardware (RC + Schmitt trigger, dedicated debounce IC) or with firmware filtering (if sampling via a microcontroller). Without debouncing, even a healthy power supply and ICs will exhibit erroneous behavior when driven by a raw mechanical contact. Step-by-Step Solution: 1) Recognize symptom: multiple unintended transitions per press.2) Identify cause: physical contact bounce intrinsic to mechanical switches.3) Apply mitigation: RC low-pass with Schmitt trigger, SR latch debouncer, or software time filtering.4) Verify fix: observe clean single transition per actuation after debouncing. Verification / Alternative check: Scope the switch node during a press; you will see a burst of transitions over 1–10 ms instead of a single clean edge. Why Other Options Are Wrong: Noisy power supply: would affect many nodes, not just the buttoned input. Corroded socket contacts: possible but typically produce static faults, not consistent per-press chattering. Intermittent IC: far less common; always eliminate bounce first. Common Pitfalls: Driving direct, non-Schmitt inputs from switches; neglecting pull-ups/downs that set a defined idle level. Final Answer: The switch contacts are bouncing.

Question 12

Master–slave J–K flip-flop — input stability requirements A correct and predictable output is achieved from a master–slave J–K flip-flop only if its J and K inputs remain stable while the ________.

Accepted Answer

Introduction / Context: Master–slave flip-flops use two level-sensitive latches in sequence. The master latch is typically enabled when the clock is HIGH (for a positive-logic design), and the slave latch is enabled when the clock is LOW, resulting in an effective edge-controlled transfer of data. Understanding when inputs must be stable avoids unintended toggling and races. Given Data / Assumptions: Master enabled on clock HIGH; slave enabled on clock LOW (typical convention). Inputs J and K can command set, reset, toggle, or hold depending on their levels. We seek the period requiring input stability for correct operation. Concept / Approach: Because the master latch is transparent when the clock is HIGH, any change on J or K during that interval can alter what the master captures, leading to incorrect or unpredictable outcomes. Therefore, J and K must be stable throughout the HIGH phase so that the master captures a single, well-defined state to transfer to the slave on the subsequent LOW phase (or edge-equivalent moment). Step-by-Step Solution: 1) Identify transparency window: master is transparent while CLK = HIGH.2) Requirement: keep J and K steady during this interval.3) Slave transfers when CLK = LOW; by then, the master state is fixed.4) Result: predictable output transition without double toggling. Verification / Alternative check: Refer to timing diagrams; when J and K vary during clock HIGH, the master can respond multiple times, producing errors once the slave samples. Why Other Options Are Wrong: Clock LOW: during this time the master is closed; J and K changes do not affect the captured value. Slave transferring: phrasing is vague and does not define the critical input window. Flip-flop is reset: unrelated to the timing requirement for normal operation. Common Pitfalls: Forgetting that level-sensitive windows exist within master–slave designs; assuming they behave like ideal edge-triggered devices. Final Answer: clock is HIGH

Question 13

Gated S–R flip-flop — condition for output change For a gated S–R flip-flop (latch with an enable/control input), the Q output changes state only if which condition about the control input and data is true?

Accepted Answer

Introduction / Context: A gated S–R flip-flop includes an enable or control input that determines when the S and R inputs are allowed to affect the stored state. Understanding this gating is essential for preventing unintended state changes and for correct interfacing with asynchronous sources, such as mechanical switches or combinational logic that may glitch. Given Data / Assumptions: The device has S and R inputs plus a control (enable) input. When the control input is inactive, the latch is opaque and holds state. When the control input is active, the latch is transparent to S/R commands. Concept / Approach: The output changes only when the control input allows the internal gating to pass the effects of S and/or R to the storage nodes. In many texts, this is phrased as “the output changes only when the control input changes from inactive to active (or is asserted).” In other words, a meaningful change on the control/enable that admits S/R information is required; otherwise, the latch holds its current state regardless of S and R line chatter. Step-by-Step Solution: 1) With control inactive → no change at Q even if S or R toggles.2) Assert/activate control input → the latch responds to S or R levels.3) If S=1,R=0 under enable → set; if S=0,R=1 under enable → reset; if S=R=0 → hold.4) Therefore, a change/activation of the control input permitting data through is the prerequisite for output change. Verification / Alternative check: Examine a gated S–R symbol and truth table: the enable (or C) input qualifies S and R; without enable, the next-state function equals present state. Why Other Options Are Wrong: “Flip-flop is set” or “flip-flop is reset” are results, not conditions that permit change. “Input data has no change” directly contradicts the need for a qualifying control action. Common Pitfalls: Confusing edge-triggered control with level-enabled control; for gated latches, sustained enable keeps transparency, so care must be taken to avoid hazards while enabled. Final Answer: control input data has changed

Question 14

Master–Slave J–K Flip-Flop — Toggle Condition and Timing In a master–slave J–K flip-flop, the toggle condition (J = 1 and K = 1) implies that Q and Q' will switch to their ________ state(s) at the ________. Choose the option that correctly completes…

Accepted Answer

Introduction / Context: The J–K flip-flop is a versatile bistable device widely used in counters and state machines. A particularly important operating case is the toggle condition, which occurs when both inputs J and K are set to logic 1. Understanding exactly when Q and its complement Q' change state, and how that relates to the clock edge in a master–slave arrangement, is essential for robust synchronous design. Given Data / Assumptions: Device: master–slave J–K flip-flop, functionally equivalent to an edge-triggered J–K FF. Toggle condition: J = 1 and K = 1. Edge sensitivity: the device responds at the active clock edge (rising or falling, per symbol). Concept / Approach: In the toggle condition, the next state is the logical complement of the present state. In a master–slave architecture, the master latch samples inputs on one clock level; the slave transfers the master’s state to the output on the opposite level, yielding an effective edge-triggered response. Therefore, the externally observed change of Q and Q' happens at the active clock edge and results in both outputs switching to their opposite states. Step-by-Step Solution: Assume present state Q(n). With J = K = 1, the next state is Q(n+1) = Q(n)'.In a master–slave implementation, the master captures the toggle intent during its transparent phase.At the active clock edge, the slave updates the external output to the captured next state.Thus, Q and Q' switch to their opposite states at the active clock edge. Verification / Alternative check: Examine the truth/characteristic table for the J–K FF: with J = K = 1, Q(next) = Q(prev)'. Timing diagrams confirm that the visible change aligns with the specified triggering edge, not with the entire clock level. Why Other Options Are Wrong: inverted, positive clock edge: “Inverted” is ambiguous and redundant; the precise phrase is “opposite” state, and the edge may be rising or falling depending on the device. quiescent, negative clock edge: “Quiescent” suggests no change, contradicting toggling. reset, synchronous clock edge: Toggle does not force reset; it complements the state. Common Pitfalls: Confusing internal latch levels with the observed edge-triggered behavior, or assuming toggling happens throughout a clock level rather than at the edge. Final Answer: opposite, active clock edge

Question 15

HDL/IP Conventions — Identify the Non-Altera Primitive Port Name Among common Intel (Altera) primitive/IP port identifiers, which of the following is not a standard Altera-style name?

Accepted Answer

Introduction / Context: When instantiating vendor primitives or parameterized IP blocks in FPGA designs, using the correct port names is crucial for synthesis tools to bind signals properly. Intel (Altera) flows have recognizable conventions such as clk for clock and ena for enable. This question checks familiarity with naming that appears across megafunctions and primitive libraries. Given Data / Assumptions: Typical Altera-style port names include clk, ena, aclr/sclr, aset/sset. Legacy TTL-style names like prn (preset) are common in 74xx datasheets, not in Altera IP naming. Context is HDL-based FPGA design, not discrete logic ICs. Concept / Approach: Map each option to vendor conventions. clk and ena are ubiquitous in Altera examples. While clr may sometimes appear, the canonical asynchronous clear is aclr and synchronous clear is sclr. The identifier prn for preset aligns with TTL flip-flops (e.g., 74LS devices) rather than Altera IP, making it the odd one out. Step-by-Step Solution: Recognize clk: standard clock input label in Altera IP.Recognize ena: standard enable input label.Consider clr: while aclr/sclr are preferred, clr is closer to the convention than TTL-style prn.Identify prn as a TTL-era preset label not typical for Altera primitives. Verification / Alternative check: Review Altera example instantiations (counters, RAMs, PLL wrappers). You will consistently find clk, ena, and clear/set ports with the a/s prefix; prn does not appear in these IP templates. Why Other Options Are Wrong: clk: Correct and common. ena: Correct and common. clr: While not the most canonical, it aligns better with vendor style than prn. Common Pitfalls: Assuming TTL datasheet naming directly transfers to FPGA IP; vendors standardize their own HDL-friendly conventions to improve clarity and automation. Final Answer: prn

Question 16

Connecting components together using HDL is not difficult.

Accepted Answer

True

Question 17

VHDL was created as a very flexible language and it allows us to define the operation of clocked devices in the code without relying on logic primitives.

Accepted Answer

True

Question 18

A gated S-R flip-flop goes into the SET condition when S is HIGH, R is LOW, and EN is HIGH.

Accepted Answer

True

Question 19

All multivibrators require feedback.

Accepted Answer

True

Question 20

Most basic latches and flip-flops are available in IC packages of eight latches or flip-flops with a common clock.

Accepted Answer

True

Flip-Flops Questions