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Question

Match the following :

List – I List – II 
a. Shift Registeri. Frequency Division
b. Multiplexerii. Memory chip select generation
c. Decoderiii. Parallel to serial conversion
d. Counteriv. Many to one switch

Codes :

This question was previously asked in
UGC NET 2014 Paper 1 Question Paper (28-Dec-2014)
The correct answer is

a-iii, b-iv, c-ii, d-i

 Each block's structure dictates its use: a-iii, b-iv, c-ii, d-i — option 2.

BlockStructureApplication
a. Shift registerCascaded flip-flopsiii. Parallel to serial
b. Multiplexer2n inputs, one outputiv. Many to one switch
c. Decodern inputs, 2n outputsii. Memory chip select
d. CounterToggling flip-flopsi. Frequency division

b is the easiest anchor: "many to one switch" is a literal restatement of what a multiplexer is. Its select lines name which of the inputs reaches the single output, and the whole is a Boolean sum of minterms:

\(Y=\overline{S_{1}}\,\overline{S_{0}}I_{0}+\overline{S_{1}}S_{0}I_{1}+S_{1}\overline{S_{0}}I_{2}+S_{1}S_{0}I_{3}\)

c — why the decoder generates chip selects. It is the multiplexer's dual: n inputs, and exactly one of \(2^{n}\) outputs active for each input code. Feed it the high-order address bits and each output enables one memory device, so a 74138 splits an address space into eight blocks. Since memory CS pins are active low, decoders are built with NAND gates and give active-low outputs — the two conventions were designed to match.

a — parallel to serial. Load all n flip-flops of a shift register at once, then clock n times, and the word leaves the last stage one bit per clock. This is what a UART transmitter does, and it is the general method of converting a wide internal bus to a narrow external link.

d — frequency division. A flip-flop wired to toggle needs two input edges for one output cycle, so each stage halves the frequency and n stages divide by \(2^{n}\). "Counting" and "dividing" are the same operation viewed differently — the count is the state, the division is the rate at which the top bit changes.

The multiplexer and shift register are close cousins, which is why a is worth checking against b: a MUX driven by a counter also converts parallel to serial, but it is combinational and needs the counter to sequence it, whereas a shift register does the job by itself with a single clock.

Hence, the correct code is a-iii, b-iv, c-ii, d-i.

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Similar Questions

  1. Assertion (A) : A demultiplexer can be used as a decoder.

    Reason (R) : A demultiplexer is built by using AND gates only.

  2. Match the following :

     List - I List - II
    (a) Minimization(i) Multiplexer
    (b) Successive Approximation(ii) Shift Register
    (c) Data Selector(iii) A/D Convertor
    (d) Sequence Generator(iv) QM Metdod

    Codes :

  3. List – IList – II  
    a. Multiplexeri. Selects one of many inputs to a single output
    b. Flip-Flopii. Combinational circuit
    c. Shift Registeriii. Memory
    d. Parallel Adderiv. Sequential logic

    Codes :

  4. Assertion (A) : A decoder is dissimilar to a demultiplexer.

    Reason (R) : There is no data input in a decoder.

    Select your answer using the codes given below.


Important Questions from Combinational Circuits

  1. Which of the following represents the borrow in a half subtraction?

  2. Carry input is present in which of the following?

  3. The number of control lines for 8-to-1 multiplexer is:

  4. X/Y (where X represents the input code and Y represents the output code) as a qualifying symbol does NOT relate to:

  5. Consider 4-bit gray code representation of numbers. Let h3 h2 h1 h0 be the gray code representation of a number n and g1g2g3g0 be the gray code representation of the number (n + 1) modulo 16. Which one of the following functions is correct?

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