Question 1

Serial loading time of a shift register: With a clock frequency of 50 kHz, how long does it take to serially enter six bits into a shift register (assume 1 bit per clock)?

Accepted Answer

Introduction / Context:Shift registers load or transmit one bit per clock edge in simple serial schemes. Timing estimates rely on the reciprocal relationship between frequency and period, then multiplying by the number of bits. Given Data / Assumptions: Clock frequency f_clk = 50 kHz. Bits to shift N = 6. Assume one bit shifts per clock cycle and no extra overhead. Concept / Approach:Compute the clock period T = 1 / f_clk. Total time T_total = N * T. Convert seconds to microseconds for the final answer. Step-by-Step Solution:Find T: T = 1 / 50,000 s = 0.00002 s = 20 µs.Multiply by bits: T_total = 6 * 20 µs = 120 µs.Therefore, six serial bits require 120 µs at 50 kHz. Verification / Alternative check:Reasonableness: At 100 kHz, 6 bits would take 60 µs; halving the frequency doubles the time, yielding 120 µs at 50 kHz—consistent. Why Other Options Are Wrong:12 µs and 3 µs imply much higher clock rates.12 ms and 120 ms are off by factors of 100–1000 (milliseconds instead of microseconds). Common Pitfalls:Mixing units (kHz vs. Hz) or forgetting that total serial time scales linearly with the number of bits. Final Answer:120 µs

Question 2

Ring counters vs. Johnson (twisted-ring) counters In digital electronics, ring shift counters and Johnson (twisted-ring) counters are best classified as which type of counters?

Accepted Answer

Introduction / Context:Ring counters and Johnson (twisted-ring) counters are special shift-register-based counting circuits used for sequence generation, state machines, timing, and divide-by-N operations. Unlike ripple (asynchronous) counters, they use a common clock to advance through well-defined state sequences. This question checks whether you can correctly classify their timing and nature relative to synchronous/asynchronous and true binary counters. Given Data / Assumptions: A ring counter circulates a single 1 or 0 through a shift register. A Johnson counter feeds the inverted output of the last flip-flop back to the input of the first flip-flop. Both are clocked shift-register structures. “True binary counter” implies natural binary sequences (… 000, 001, 010, 011, …). Concept / Approach:Because every flip-flop in these counters receives the same clock edge, their state transitions occur simultaneously. That is the hallmark of synchronous operation. However, their sequences are not natural binary counts; they are coded sequences (one-hot for ring; two-hot patterns for Johnson), so they are not “true binary counters.” Step-by-Step Solution:Identify the clocking method: common clock → synchronous.Identify the sequence type: one-hot or twisted ring → not natural binary.Therefore, classification: synchronous counters (but not true binary). Verification / Alternative check:Datasheets and textbooks show that a 4-bit ring counter has 4 valid states (one-hot), and a 4-bit Johnson counter has 8 valid states. Both are advanced by one common clock; no ripple path is used, confirming synchronous behavior. Why Other Options Are Wrong: Asynchronous counters: use ripple timing; not applicable here. True binary counters: produce natural binary sequences; ring/Johnson do not. Synchronous and true binary: incorrect because they are not natural binary. Common Pitfalls: Confusing synchronous clocking with binary coding. Assuming any “counter” must be binary; many are coded or one-hot. Final Answer:Synchronous counters

Question 3

Serially loading a shift register — role of the strobe (enable) signal In serial data loading, how is a strobe (enable) signal typically used with a shift register?

Accepted Answer

Introduction / Context:Shift registers load and move data on clock edges. In practical systems, an extra control called a strobe (or enable) is often used to precisely gate when the register should accept data or shift. Understanding the strobe's purpose prevents timing errors and unintended shifts. Given Data / Assumptions: Serial loading means data enters bit-by-bit at the serial input. Clock pulses advance the bits through the flip-flops. An enable (strobe) input is commonly provided on many shift-register ICs. Concept / Approach:The strobe/enable acts as a gating control. When asserted, the register responds to the clock (loading or shifting). When deasserted, the register ignores clocking for the operation in question, thereby preventing unintended data movement or capture. This is often used to align sampling moments with valid data on the serial line. Step-by-Step Solution:Recognize that the clock alone would shift data on every edge.Use the strobe to permit shifting only during valid windows.Therefore, the strobe's role is to enable/disable the register’s action, effectively “turning acceptance on or off.” Verification / Alternative check:Datasheets for common shift registers (e.g., 74HC595, 74HC164 families) show enable or latch signals that gate shift or update behavior. Timing diagrams illustrate strobe windows during which serial data is latched or ignored. Why Other Options Are Wrong: Controlling the number of clocks: The strobe does not generate or count clocks; it gates behavior. Determining which Q outputs are used: That is a wiring/selection issue, not strobe function. Choosing internal flip-flops: The device's internal structure is fixed; strobe doesn’t reconfigure flip-flops. Common Pitfalls: Confusing strobe/enable with a latch clock or storage register clock; some devices separate shift and storage clocks. Assuming the strobe changes the data path rather than gating action. Final Answer:To enable or disable the register (turn data acceptance on or off) at the proper moment

Question 4

Universal shift register What do we call a shift register that supports parallel loading, bidirectional serial loading, and internal shifting features?

Accepted Answer

Introduction / Context:Shift registers range from simple serial-in/serial-out devices to fully flexible units that can load in parallel, shift left or right, and accept serial data from either side. The industry uses a specific name for the most flexible category, frequently found in timing, communication, and data format conversion tasks. Given Data / Assumptions: Device supports parallel input (loads multiple bits simultaneously). Supports bidirectional serial loading (left or right). Supports internal shifting (controlled by clock and mode signals). Concept / Approach:A “universal shift register” is designed to perform parallel load, shift-left, shift-right, and hold operations under mode control. It can ingest serial data from either end, making it useful for interfacing between serial and parallel buses and for bit manipulation tasks. Step-by-Step Solution:Recognize feature set: parallel load + bidirectional serial + shift = universal capability.Match to standard terminology: “universal shift register.”Therefore, the correct name is Universal shift register. Verification / Alternative check:Common parts such as the 74HC194/74HC195 are expressly identified as universal shift registers in datasheets, supporting load and bidirectional shifting via mode inputs S1/S0 and a clock. Why Other Options Are Wrong: End-around: Usually refers to connecting serial output back to serial input to form ring/Johnson sequences, not parallel load capability. Tristate: Describes output driving capability, not shift functionality. Conversion register: Not a standard term for this feature set. Common Pitfalls: Confusing “bidirectional” with “tristate” outputs. Assuming “universal” means any IC with both serial and parallel paths; true universal devices provide left/right shifting under control. Final Answer:Universal shift register

Question 5

Predicting the next state in a 4-bit Johnson (twisted-ring) counter Given on the 5th clock pulse: Q0 = 0, Q1 = 1, Q2 = 1, Q3 = 1. What is the state on the 6th clock pulse?

Accepted Answer

Introduction / Context:A 4-bit Johnson counter (twisted-ring) is a shift register whose input is the inversion of its last stage. It generates a 2n-state sequence (for n flip-flops), useful for decoded timing signals and sequence control. Predicting the next state requires understanding the feedback and shift direction. Given Data / Assumptions: 4-bit Johnson counter with stages Q0 (first) through Q3 (last). On the 5th clock: Q0=0, Q1=1, Q2=1, Q3=1. Feedback to the first stage is NOT(Q3). On each rising clock edge, bits shift toward higher-index stages. Concept / Approach:For a typical Johnson implementation, the new Q0 becomes the inverted previous Q3. Each stage Qi (i>0) takes the previous value of Qi-1. Thus, compute the inverted feedback for Q0, then shift prior values forward by one position. Step-by-Step Solution:Previous Q3 = 1 → new Q0 = NOT(1) = 0.New Q1 = old Q0 = 0.New Q2 = old Q1 = 1.New Q3 = old Q2 = 1.Therefore, next state: Q0=0, Q1=0, Q2=1, Q3=1. Verification / Alternative check:Listing the 8 states of a 4-bit Johnson sequence confirms the transition from 0 1 1 1 to 0 0 1 1 under the assumed shift direction and feedback polarity. Why Other Options Are Wrong: Q0=1,1,1,0: Would imply Q0=NOT(Q3)=0 is violated. Q0=0,0,0,1 and Q0=1,0,0,0: Do not match the correct shift of old bits. Common Pitfalls: Mixing up shift direction (left vs. right) and which stage feeds which. Forgetting the inversion in Johnson feedback. Final Answer:Q0 = 0, Q1 = 0, Q2 = 1, Q3 = 1

Question 6

Interfacing parallel data to serial equipment Which device is commonly used to convert between a system’s parallel data format and an external serial format?

Accepted Answer

Introduction / Context:Many microprocessors and microcontrollers natively handle parallel data on internal buses, but external links (RS-232/TTL serial, UART-over-USB bridges, etc.) are serial. A specialized interface converts between these formats reliably and with timing control. Given Data / Assumptions: Parallel data inside the host system (e.g., CPU bus or memory-mapped I/O). Serial link to external equipment (asynchronous serial). Need framing, start/stop bits, optional parity, and baud-rate generation. Concept / Approach:A UART is designed to serialize parallel bytes into asynchronous serial frames and to deserialize incoming serial frames into parallel bytes. It handles timing via baud-rate generators and often provides buffers/FIFOs for reliable data transfer. Step-by-Step Solution:Identify requirement: parallel-to-serial and serial-to-parallel conversion for asynchronous links.Select device class providing framing and baud control: UART.Therefore, the correct interface choice is a UART. Verification / Alternative check:Microcontroller datasheets list on-chip UARTs explicitly for serial communication. External bridge ICs (e.g., USB-UART) use the same concept, confirming the role of UARTs in parallel/serial interfacing. Why Other Options Are Wrong: Key matrix: Input scanning hardware, not a serial interface. Memory chip: Storage, not a parallel/serial conversion interface. Series-in, parallel-out register: Converts serial to parallel but lacks framing/baud control and reverse direction. Common Pitfalls: Confusing simple shift registers with full serial communication interfaces; shift registers do not manage asynchronous framing or baud rates. Overlooking the need for parity, start, and stop bits in many serial protocols. Final Answer:UART (Universal Asynchronous Receiver/Transmitter)

Question 7

Parallel-loaded 4-bit shift register behavior A 4-bit shift register that initially receives 4 bits of parallel data will, on each clock pulse, shift in which direction and by how many positions?

Accepted Answer

Introduction / Context:Shift registers can be configured for serial-in/serial-out, parallel-in/serial-out, and other modes. After a parallel load, each subsequent clock typically shifts the stored bits uniformly in a predefined direction by one stage. This question tests your grasp of the basic shift behavior per clock edge. Given Data / Assumptions: A 4-bit register can load 4 bits simultaneously (parallel load). After loading, the register shifts with each clock. No special “shift by N>1” hardware is implied. Concept / Approach:Standard shift registers move data one stage per clock to the left or right, depending on the design (e.g., right-shift for SIPO like 74HC164, or left-shift variants). Unless explicitly stated otherwise, “shift” means one position per active clock edge in the specified direction. Many textbook examples assume right shift for serial-out from Q0 or Q3, but direction is device-dependent; here the common interpretation is right by one. Step-by-Step Solution:Parallel load places bits simultaneously into all four stages.On each clock, all bits step to the next stage.Thus, the register shifts right by one position per clock in the typical configuration. Verification / Alternative check:Examine timing diagrams for common ICs (e.g., 74HC194/195 universal shift registers): they show one-position shifts per clock edge when in shift mode. Why Other Options Are Wrong: Left by one: Possible for left-shift devices, but not assumed here; question implies the standard right-shift convention. Right by two / Left by three: Multi-position shifting per clock is nonstandard without special logic. Common Pitfalls: Assuming multi-bit jumps per clock; normal devices shift one stage per edge. Ignoring that actual direction depends on IC; always check the datasheet mode pins. Final Answer:Right by one position per clock

Question 8

Adding recirculation to a 4-bit PISO register By adding recirculating lines to a 4-bit parallel-in, serial-out register, it becomes which kind of register (in/out capabilities)?

Accepted Answer

Introduction / Context:A 4-bit PISO (parallel-in, serial-out) shift register outputs bits serially after a parallel load. If you add recirculation (feeding the serial output back to a serial input), you extend its capability to also accept serial data and to re-use or loop data, creating more flexible in/out modalities. This question asks you to identify the resulting I/O capabilities in standard shorthand. Given Data / Assumptions: Base device: 4-bit parallel-in, serial-out register. Added feature: recirculation path (serial feedback), possibly with gating. Parallel outputs are inherently available at the register stages. Concept / Approach:With recirculation, the device can accept serial input (via the feedback/gated serial path) in addition to its original parallel load. The stored contents can be observed in parallel (tap Q outputs) and shifted out serially. A concise descriptor is “serial-in, parallel, serial” (i.e., serial in, parallel out available, serial out available). Step-by-Step Solution:Original: parallel-in → serial-out (PISO).Add recirculation/gated serial path → now supports serial-in as well.Stage taps provide parallel observations → “parallel.”Thus, capabilities: serial-in, parallel (out), serial (out). Verification / Alternative check:Universal shift registers (e.g., 74HC194) illustrate how mode control enables serial-in left/right, parallel load, and serial/parallel outputs. A PISO with added feedback approaches this flexibility. Why Other Options Are Wrong: Parallel-in, serial, parallel: Omits serial-in; not reflecting the added capability. Bidirectional in, parallel, series / Series-parallel-in, series, parallel: Nonstandard/ambiguous phrasing and not the common shorthand. Common Pitfalls: Assuming recirculation only loops the same data but does not enable serial-in; with gating, it can accept new serial data. Forgetting that parallel taps are inherently available on many shift registers. Final Answer:Serial-in, parallel, serial

Question 9

Purpose of a buffer circuit in digital logic Which statement best describes the function of a buffer?

Accepted Answer

Introduction / Context:Buffers are ubiquitous digital building blocks. They sit between a logic source and a load, offering higher current drive, improved fan-out, and isolation without changing logic polarity. Recognizing the buffer’s role helps with signal integrity and proper interfacing. Given Data / Assumptions: Logic-level input must drive multiple loads or a heavier single load. We want no inversion of logic level. We prefer isolating the driving gate from load-induced effects. Concept / Approach:A buffer outputs the same logic level it sees at its input. It neither inverts (like a NOT gate) nor necessarily “cleans up” noise in the sense of a Schmitt trigger (though some buffers include Schmitt inputs). Its key advantage is stronger output drive and isolation, maintaining logic level integrity at the load. Step-by-Step Solution:Input logic level is presented to the buffer.Buffer re-drives the signal with higher current capability and lower output impedance.Output logic level matches input polarity (no inversion). Verification / Alternative check:Examine datasheets for noninverting buffers (e.g., 74HC125/126 family) that specify higher IOH/IOL ratings than basic logic gates and no inversion symbol on the schematic. Why Other Options Are Wrong: “Clean up input/output”: That suggests conditioning like hysteresis; standard buffers may not add this unless they are Schmitt-trigger buffers. Inverted replica: That is an inverter, not a buffer. Common Pitfalls: Assuming every buffer is a line driver with Schmitt inputs; some are plain CMOS/TTL buffers. Confusing “buffer” with “repeater + equalizer” used in high-speed links. Final Answer:To provide an output equal to its input (no inversion) with increased drive capability

Question 10

Serial loading time for a 5-bit shift register How many clock pulses are required to load all 5 bits serially into a 5-bit shift register?

Accepted Answer

Introduction / Context:Serial-in shift registers accept one bit per clock edge. To completely fill an n-bit register starting empty (or regardless of previous contents, if we define loading as bringing in n new bits), you must shift in n bits. This question checks that basic but crucial timing relationship. Given Data / Assumptions: Register length n = 5 bits. One new input bit is captured per qualifying clock edge. No parallel load operation is being used. Concept / Approach:Each clock captures the current serial input into the first stage and shifts existing contents one position. After k pulses, k of the most recent input bits occupy the first k stages. Therefore, to fully load 5 fresh bits, we need 5 clock pulses. Step-by-Step Solution:Pulse 1 → 1st bit enters stage 0.Pulse 2 → 2nd bit enters stage 0; 1st bit shifts to stage 1.…Pulse 5 → 5th bit enters; previous 4 bits have shifted, filling all 5 stages. Verification / Alternative check:Timing diagrams for SISO/SIPO registers demonstrate that n pulses are required to move n new bits fully into an n-bit structure. Why Other Options Are Wrong: 2, 3, 4 pulses: Insufficient; fewer than n pulses yield fewer than n newly-loaded bits. Common Pitfalls: Mistaking the time to output the first bit at the far end for the time to fully load the register. Assuming double-data-rate or multi-bit-per-clock behavior; standard shift registers are one bit per qualifying edge. Final Answer:5

Question 11

Distinguishing a ring shift counter from a Johnson (twisted-ring) counter What is the key difference between a ring counter and a Johnson counter?

Accepted Answer

Introduction / Context:Both ring and Johnson counters are built from shift registers with feedback, but their feedback polarity changes the sequence of states, the number of states, and common use cases. Correctly identifying the feedback difference is essential to predicting sequences and designing decoders. Given Data / Assumptions: Ring counter: last stage output fed directly (non-inverted) back to the first stage. Johnson counter: last stage output is inverted before being fed back to the first stage. Both are clocked synchronously. Concept / Approach:With non-inverted feedback, a ring counter simply circulates a single 1 (or 0), producing N states from N flip-flops (one-hot). With inverted feedback, a Johnson counter produces 2N states by toggling runs of 1s and 0s, which are convenient for evenly distributed decoded outputs. Step-by-Step Solution:Identify feedback path: ring → direct; Johnson → inverted.Relate to number of states: ring → N; Johnson → 2N.Therefore, the distinguishing feature is inverted feedback in the Johnson counter. Verification / Alternative check:Enumerating states for a 4-bit example: ring → 4 states; Johnson → 8 states. Textbook sequences confirm the feedback polarity difference. Why Other Options Are Wrong: “No difference” is incorrect; the feedback polarity is fundamental. Speed claims (always faster) are context-dependent and not a defining difference. Common Pitfalls: Confusing Johnson with ring due to similar shift-register structures. Forgetting that Johnson counters are sometimes called twisted-ring counters precisely because of the inverted feedback. Final Answer:The feedback is inverted in the Johnson counter, non-inverted in the ring counter

Question 12

Shift registers — serially loading a nibble (4 bits) To shift a nibble (four bits) of data serially into a shift register, how many clock pulses are required?

Accepted Answer

Introduction / Context:Serial shift registers accept data one bit per clock edge. Knowing how many clock pulses are required to load a specified number of bits is fundamental to timing analysis in digital systems, serial interfaces, and logic simulation. Given Data / Assumptions: The word length to be shifted in is one nibble, i.e., 4 bits. Data is applied to the serial input and clocked in bit by bit. Each active clock edge shifts the register by exactly one bit position. Concept / Approach: A serial-in register advances its contents by one bit for each clock. Therefore, loading N bits requires N clock pulses, independent of the bit values (0 or 1). The number of 1s in the data does not affect the number of clocks; 0s shift just like 1s. Step-by-Step Solution: Let N = number of bits to load = 4.Each clock edge inserts one new serial bit and shifts existing bits by one stage.Total required clock pulses = N = 4. Verification / Alternative check: Simulate a 4-stage register (Q3..Q0) with any 4-bit pattern. After the first clock, one bit resides in Q0; after the second, two bits are present; after the fourth, all four bits occupy the register. This confirms the one-clock-per-bit rule. Why Other Options Are Wrong: One clock pulse: loads only one bit, not four. Eight clock pulses: would be needed for a byte, not a nibble. One pulse for each 1 in the data: incorrect; clocks are independent of bit values. Common Pitfalls: Confusing serial loading with parallel loading (which would take a single strobe, not multiple clocks). Assuming data value affects clock count. Final Answer: four clock pulses

Question 13

SIPO timing — keeping parallel output data accurate To keep output data accurate, 4-bit serial-in, parallel-out (SIPO) shift registers typically employ which control line?

Accepted Answer

Introduction / Context:Serial-in, parallel-out (SIPO) shift registers receive bits serially but often need to present a stable 4-bit word to downstream logic. Designers use a separate output latch with a strobe (latch enable) to avoid glitches while shifting. Given Data / Assumptions: A 4-bit SIPO is shifting new bits on each clock. We want a clean, accurate parallel output word only when loading is complete. Common solution uses an internal or external latch controlled by a strobe line. Concept / Approach: During shifting, internal register stages change every clock. A strobe (latch enable) captures the register contents into an output latch at a defined instant, isolating downstream circuitry from intermediate transitions. This improves data integrity and timing closure. Step-by-Step Solution: Shift in 4 bits over 4 clocks while outputs remain latched.Assert the strobe once to transfer the stable 4-bit word to the parallel outputs.Downstream logic now reads a coherent nibble until the next strobe. Verification / Alternative check: Datasheets for classic parts (e.g., 74HC595) show a shift register feeding a storage register; the storage register is updated by a RCLK/LATCH signal (strobe), ensuring accurate outputs. Why Other Options Are Wrong: Divide-by-4 clock: changes frequency, not output coherency. Sequence generator: unrelated; generates patterns, not latching. Multiplexer: selects signals, does not freeze outputs. Common Pitfalls: Driving loads directly from shifting stages without a latch, causing visible ripple on displays. Final Answer: strobe line

Question 14

Getting parallel data from a shift register How can parallel data be taken simultaneously from a shift register built from flip-flops?

Accepted Answer

Introduction / Context:A shift register stores one bit per stage. To expose the entire word in parallel, each stage's Q output is treated as one bit of the bus. This is the basis of SIPO devices and parallel reads from serial streams. Given Data / Assumptions: Register consists of cascaded flip-flops. Each flip-flop Q holds one bit. We need a simultaneous parallel representation of the stored word. Concept / Approach: Parallel readout is accomplished by wiring each flip-flop's Q to a dedicated output pin or bus line. Optionally, an output latch can isolate the internal shifting from the bus, but the fundamental method is one Q per bit line. Step-by-Step Solution: Identify number of stages n → n parallel outputs required.Connect Q0..Q(n−1) to the parallel bus lines D0..D(n−1).Read all bits simultaneously on the bus or latch them with a strobe if needed. Verification / Alternative check: Examine typical SIPO IC pinouts (e.g., 74HC164, 74HC595). Pin names Q0..Q7 expose each flip-flop's output for parallel access. Why Other Options Are Wrong: First/last Q only: yields one bit, not a word. Tying all Qs together: creates bus contention/shorts and corrupts data. Common Pitfalls: Forgetting to buffer or latch outputs when long loads or timing integrity is required. Final Answer: Use the Q output of each flip-flop.

Question 15

Why stepper motors are popular in digital automation Stepper motors have become popular in digital automation systems because ________.

Accepted Answer

Introduction / Context:Stepper motors bridge digital control and mechanical motion. They move in discrete steps when coils are energized in a sequence, making them ideal for printers, CNC axes, and robotics where repeatable positioning is required. Given Data / Assumptions: Digital controllers can output timed sequences easily. Applications need repeatable, incremental motion. Feedbackless (open-loop) operation is often sufficient due to step quantization. Concept / Approach: Sequential digital drive (full-step, half-step, microstepping) energizes phases to produce controlled rotor increments. Because each pulse advances the motor by a known angle, repetitive movements are straightforward to generate and repeat. Step-by-Step Solution: Identify key benefit 1: digital sequencing directly drives motion.Identify key benefit 2: inherent stepwise motion yields repeatability.Combine both for the most complete explanation of popularity. Verification / Alternative check: Common drive ICs and microcontroller libraries implement phase sequences; applications like 3D printers rely on precise repetitive movements without encoders at moderate loads. Why Other Options Are Wrong: Low cost alone is not the main reason. Only sequential drive or only repetitive motion is incomplete; popularity stems from both features together. Common Pitfalls: Assuming stepper speed/torque equals servo performance; steppers excel mainly in controlled, repetitive positioning. Final Answer: they are driven by sequential digital signals and can be used to provide repetitive mechanical movement

Question 16

Identifying the register type from its behavior Which register accepts a complete binary number serially (one bit at a time) and also shifts stored bits out one at a time?

Accepted Answer

Introduction / Context:Register classifications describe how data enters and leaves: serial or parallel, in and out. Matching behavioral descriptions to register types is basic digital logic literacy. Given Data / Assumptions: Input arrives as a bitstream (serial). Output also departs as a bitstream (serial). No requirement for simultaneous multi-bit output. Concept / Approach: By definition, a serial-in, serial-out (SISO) register uses a single data line for input and typically exposes the last stage as the serial output. On each clock, data shifts one stage forward; after N clocks, an N-bit word has been fully entered or exited. Step-by-Step Solution: Input mode: serial → must be “serial-in”.Output mode: serial → must be “serial-out”.Therefore, the correct type is serial-in, serial-out. Verification / Alternative check: Contrast with SIPO (serial-in, parallel-out), which would provide all bits at once; PISO (parallel-in, serial-out), which would load in parallel but shift out serially; PIPO, which does both in parallel. Why Other Options Are Wrong: PIPO and PISO include parallel operation, contradicting the description. SIPO outputs in parallel, not one bit at a time. Common Pitfalls: Confusing direction of data movement with width of the I/O interface. Final Answer: serial-in, serial-out

Question 17

SIPO register contents after loading a nibble A serial-in/parallel-out (SIPO) 4-bit shift register initially contains 1111. The incoming data nibble is 0111. After four clock pulses (one per bit), what contents appear in the register?

Accepted Answer

Introduction / Context:Predicting a shift register's state after a defined number of clock pulses is a staple timing exercise. Here, a 4-bit SIPO starts at 1111 and then receives exactly four serial bits: 0, 1, 1, 1 (a nibble 0111). Given Data / Assumptions: Register width = 4 bits. Initial contents = 1111. Incoming data nibble = 0 then 1 then 1 then 1 (serial order). One bit is shifted in per clock; after 4 clocks the initial bits have been fully overwritten. Concept / Approach: In a SIPO, each clock pushes the incoming bit into the first stage and shifts prior bits toward the last stage. After N = 4 clocks, the register contains exactly the 4 bits that arrived, in the same order they were clocked (assuming conventional left/right mapping consistent with the device). Step-by-Step Solution: Clock 1 → 0xxxClock 2 → 01xxClock 3 → 011xClock 4 → 0111 Verification / Alternative check: Regardless of the initial 1111, after four fresh bits have been introduced, all original bits have shifted out. The final state must equal the incoming nibble 0111. Why Other Options Are Wrong: 0000: would require four zeros. 1111: unchanged; impossible after injecting new bits. 1000: would occur if the nibble were 1000. Common Pitfalls: Reversing bit order; ensure you use the order presented by the incoming serial stream. Final Answer: 0111

Question 18

Preset pattern for a ring counter What preset condition is required for a ring shift counter before clocking begins?

Accepted Answer

Introduction / Context:A ring counter is a shift register with its serial output fed back to the serial input. It cycles a single active bit through its stages, generating one-hot codes widely used in sequencing and timing applications. Given Data / Assumptions: Ring counter built from n flip-flops. Feedback is direct (not inverted), so the active state is 1. We want proper operation immediately after preset. Concept / Approach: For a direct ring counter, the valid circulating pattern is one-hot: exactly one stage = 1 and all others = 0. Each clock shifts the 1 into the next flip-flop. Presetting all 1s or all 0s would stall the counter; a single 0 with the rest 1 would circulate zeros, not the intended one-hot 1. Step-by-Step Solution: Preset Q0..Q(n−1) such that one chosen stage is 1 and all others are 0.On each clock, the 1 advances by one stage.After n clocks, the pattern repeats. Verification / Alternative check: Waveforms of ring counters show the single 1 stepping through outputs; alternative preset patterns fail to produce the canonical sequence. Why Other Options Are Wrong: All 1s or all 0s: no single active bit to circulate. Single 0, rest 1: describes an inverted (0-hot) behavior, not the standard ring counter. Common Pitfalls: Confusing ring counters with Johnson counters, which use inverted feedback and produce 2n states. Final Answer: a single 1, the rest 0

Question 19

State count of a Johnson (twisted-ring) counter In a 6-bit Johnson counter, how many distinct states (bit patterns) are produced in the complete sequence?

Accepted Answer

Introduction / Context:A Johnson counter (twisted-ring) is a shift register with complemented feedback. Its key property is that an n-stage Johnson counter cycles through 2n unique states, making it efficient for sequence generation and timing signals. Given Data / Assumptions: Number of stages n = 6. Inverted feedback from last stage to input. Clocking advances the pattern by one bit each cycle. Concept / Approach: Unlike a one-hot ring counter (n states), the Johnson counter gradually fills with 1s and then empties, creating two distinct sequences of length n each. Therefore, the total number of unique states is 2n. Step-by-Step Solution: Compute total states = 2 * n.For n = 6, total = 2 * 6 = 12.These 12 states repeat cyclically with no duplicates within one cycle. Verification / Alternative check: Enumerating states for a smaller n (e.g., n = 3) yields 6 unique patterns, reinforcing the 2n rule; scaling to n = 6 yields 12. Why Other Options Are Wrong: 2 and 6: too few; confuse with ring or stage count. 24: would correspond to 4n or 2^(n+1), which is not a Johnson property. Common Pitfalls: Mixing Johnson (2n states) with regular ring (n states) or binary counters (2^n states). Final Answer: 12

Question 20

Storage capacity per stage in a shift register In a shift register, how much storage capacity does each stage (flip-flop) represent?

Accepted Answer

Introduction / Context:Shift registers are arrays of flip-flops used for temporary storage and serial/parallel conversion. Each flip-flop is a 1-bit memory element, so capacity scales linearly with stage count. Given Data / Assumptions: A stage equals one edge-triggered flip-flop. Binary storage is being considered (not multi-level memory). Concept / Approach: A standard flip-flop stores one of two states (0 or 1). Therefore, a k-stage shift register stores k bits. Larger units such as nibbles and bytes are formed by grouping 4 or 8 such stages, respectively. Step-by-Step Solution: Per stage capacity = 1 bit.Total capacity for N stages = N * 1 bit.Example: 8 stages → 8 bits (1 byte). Verification / Alternative check: Datasheets label Q0..Q7 for an 8-stage register, explicitly indicating 8 single-bit outputs. Why Other Options Are Wrong: Two bits / four bits / eight bits: these would require 2, 4, or 8 flip-flops per stage, which is not how a stage is defined. Common Pitfalls: Confusing total register width with per-stage capacity. Final Answer: One bit

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