Shift Registers MCQs: 10 Solved Questions on Timing, Modes and LFSRs

Shift-register definitions can feel easy until one question mixes a timing formula with serial movement, or an LFSR asks for the state several clocks later. Circuit recognition and loading modes lead into numerical timing and feedback-state tracing. Attempt each item before reading its explanation, and write the state after every clock in feedback-register questions. More than 10 Shift Registers practice questions are available on KnowledgeGate. For adjacent practice on the storage elements behind registers, use Sequential Circuits MCQs: 11 Solved Flip-Flops for flip-flop and counter behaviour; use the register questions below for loading modes, timing and feedback states.

Related reading: Shift register types and Flip-flops.

1. Shift Registers MCQs on circuit type, loading and data movement

Stored state, loading, and data movement are key shift-register concepts.

Q1. A register is a sequential circuit

PYQ: TPSC System Analyst 2026

Which of the following is a SEQUENTIAL circuit ?

• A. PARITY

• B. ADDER

• C. REGISTER

• D. MULTIPLEXER

Correct answer: C. REGISTER

A register contains memory elements, so its output depends on stored state and current input. The other three circuits normally depend only on current inputs. Memory is the deciding test.

Open Q1 in the Shift Registers module.

Q2. Identify the PIPO diagram

PYQ: DSSSB TGT 2023

The given diagram expresses ______.

• A. Parallel-In Parallel-Out Shift Register

• B. Serial-In Parallel-Out Shift Register

• C. Serial-In Serial-Out Shift Register

• D. Parallel-In Serial-Out Shift Register

Correct answer: A. Parallel-In Parallel-Out Shift Register

Several input lines feed the stages, and every stage output is exposed, so bits enter and leave together. This is parallel-in, parallel-out. SIPO has serial entry, while PISO has serial exit.

Open Q2 in the Shift Registers module.

Q3. Synchronous parallel loading takes one pulse

PYQ: KVS 2018

To load a byte of data parallelly into a shift register with a synchronous load, there must be _____.

• A. one clock pulse

• B. one clock pulse for each 1 in the data

• C. eight clock pulses

• D. one clock pulse for each 0 in the data

Correct answer: A. one clock pulse

A synchronous parallel load captures all eight bits on the same active edge. One pulse therefore loads the byte. Eight pulses describe serial loading, with one bit entering per pulse.

Open Q3 in the Shift Registers module.

Q4. Reading several bytes through a shift register

PYQ: BPSC NB 2024

In order to read multiple bytes of a row at the same time, we make use of ______.

• A. Memory extension

• B. Latch

• C. Shift register

• D. More than one of the above

• E. None of the above

Correct answer: C. Shift register

A shift register can accept a stream over successive clocks and expose its stored bits together, or perform the reverse conversion. A latch stores a value without controlled shifting. Memory extension changes capacity, not transfer mode.

Open Q4 in the Shift Registers module.

2. Shift register timing MCQs with complete clock calculations

Use T = 1/f. Under the SISO loading-plus-reading convention, total clocks are 2N - 1.

Q5. Delay through an 8-bit register at 400 MHz

PYQ: ISRO 2023

The time delay obtained through an 8-bit serial register with 400 MHz clock is:

• A. 20 ns

• B. 2.5 µs

• C. 20 µs

• D. 2.5 µs

Correct answer: A. 20 ns

T = 1/f = 1/(400 × 10^6) s = 2.5 × 10^-9 s = 2.5 ns

Delay = 8T = 8 × 2.5 ns = 20 ns

Here 2.5 ns is one clock period, not the delay through eight stages. The duplicated 2.5 µs choices appear twice in the printed options.

Open Q5 in the Shift Registers module.

Q6. Loading and reading 32 bits in SISO mode

Consider a 32- bit shift register which uses a clock of 1 GHz. If register is operated in SISO mode, find total time required to perform loading and reading?

• A. 1 ns

• B. 31 ns

• C. 32 ns

• D. 63 ns

Correct answer: D. 63 ns

T = 1/(1 × 10^9) s = 1 × 10^-9 s = 1 ns

Loading takes 32 pulses. Once the first output bit appears, the remaining 31 bits need 31 more pulses, so total time = (32 + 31) × 1 ns = (2 × 32 - 1) × 1 ns = 63 ns.

Practise Q6 from the module hub.

3. LFSR MCQs solved by tracing every register state

Write every post-clock LFSR state, follow the shown shift direction, and apply XOR before insertion. Review Sequential Circuits: Flip-Flops and Counters if the flip-flop transitions need a refresh.

Q7. Six input bits through a four-bit LFSR

PYQ: GATE Information Technology 2007

What is the final value stored in the linear feedback shift register if the input is 101101?

• A. 0110

• B. 1011

• C. 1101

• D. 1111

Correct answer: A. 0110

Begin at 0000, shift left, and insert input XOR old rightmost bit on the right. Inputs 1 0 1 1 0 1 give 0001 → 0011 → 0110 → 1101 → 1011 → 0110. The final content is 0110; 1101 and 1011 are intermediate states.

Open Q7 in the Shift Registers module.

Q8. Run an XOR-feedback register four clocks backward

Consider the following circuit with present state Q0=Q1=0 and Q2=Q3=1.

What was the state of the register before 4 clock cycles?

• A. Q0 = 0, Q1 = 1, Q2 = 0, Q3 = 0

• B. Q0 = 1, Q1 = 0, Q2 = 1, Q3 = 0

• C. Q0 = 1, Q1 = 0, Q2 = 0, Q3 = 0

• D. None

Correct answer: A. Q0 = 0, Q1 = 1, Q2 = 0, Q3 = 0

Read states in Q0Q1Q2Q3 order and invert the shown XOR-feedback relation. From 0011, step backward through 0110 → 1101 → 1010 → 0100. Thus Q0=0, Q1=1, Q2=0, Q3=0; forward replay gives 0100 → 1010 → 1101 → 0110 → 0011.

Practise Q8 in the Shift Registers section.

4. Feedback shift register MCQs with nine-clock and three-clock traces

These longer traces are safest when every intermediate state is visible.

Q9. Three-bit XOR register after nine input bits

PYQ: GATE 2006

Consider the circuit in the diagram. The ⊕ operator represents Ex-OR. The D flipflops are initialized to zeroes (cleared).

The following data: 100110000 is supplied to the “data” terminal in nine clock cycles. After that the values of q2q1q0 are:

• A. 000

• B. 001

• C. 010

• D. 101

Correct answer: C. 010

Starting at q2q1q0 = 000, apply q2(new)=q1(old), q1(new)=q0(old) and q0(new)=data XOR q2(old). Inputs 1 0 0 1 1 0 0 0 0 produce 001 → 010 → 100 → 000 → 001 → 010 → 100 → 001 → 010. The ninth-clock state is therefore 010.

Open Q9 in the Shift Registers module.

Q10. Five-bit reducer after three clocks

Consider 5-bit left shift reducer and D flip flop shown in figure below:

The bits present in queue after 3 clocks:

• A. 01010

• B. 01110

• C. 01011

• D. 10110

Correct answer: C. 01011

Follow the shift and recirculating D-input connection exactly as drawn. Starting from the figure's 01101, the clock-by-clock states are 01101 → 11010 → 10101 → 01011. After three clocks the register contains 01011; choosing 10110 means the direction or bit order was reversed.

Practise Q10 in the Shift Registers section.

5. Shift register mistakes that turn correct arithmetic into a wrong option

For clock-by-clock practice, use Sync Counter Analysis MCQs.

6. Common shift-register question types

Shift-register questions commonly test circuit-class identification, clock-pulse counting, frequency-to-time conversion, and XOR-feedback state tracing. A PYQ label identifies the cited exam and year; it does not establish topic frequency or weightage. Use the same three-step order each time: mark shift direction and state order, write one update rule, then complete the full timing or state table on paper before checking the options.

7. Shift registers MCQs: the short version and next step

Memory distinguishes a register from combinational logic. Parallel load takes one active edge, timing begins with T = 1/f, and feedback questions require a written state after every clock. For a structured route from the basics through Digital Electronics and the wider syllabus, see GATE Guidance by Sanchit Sir. To compare the broader preparation catalogue, use GATE CS Exam Preparation.