Digital circuits miscellaneous

Digital circuits miscellaneous

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Digital Circuits

Digital circuits miscellaneous
  1. A 3 line to 8 line decoder, with active low outputs, is used t o implement a 3- variable Boolean function as shown in the figure given below.

    The simplified form of Boolean function F (A, B, C) implemented in ‘Product of Sum’ form will be








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    Let us consider moment active high input F = ∑(1, 3, 5, 6)
    K-map

    ∴ product of sum
    F = (X + Z) (Y + Z) (X + Y + Z) ...(1) for active high output
    But given, signal is active low output. So each variable in equation (1) will be complemented
    ∴ F = (X + Z) (X + Y + Z) (Y + Z) ...(2) for active low output

    Correct Option: B

    Let us consider moment active high input F = ∑(1, 3, 5, 6)
    K-map

    ∴ product of sum
    F = (X + Z) (Y + Z) (X + Y + Z) ...(1) for active high output
    But given, signal is active low output. So each variable in equation (1) will be complemented
    ∴ F = (X + Z) (X + Y + Z) (Y + Z) ...(2) for active low output

  1. The octal equivalent of the HEX number AB.CD is








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    First converting HEX to binary
    A = 1010
    B = 1011
    C = 1100
    D = 1101
    (AB.CD)16 = (10 10 10 11 1100 1101)2
    Grouping for octal conversion =

    = (253.632)8

    Correct Option: B

    First converting HEX to binary
    A = 1010
    B = 1011
    C = 1100
    D = 1101
    (AB.CD)16 = (10 10 10 11 1100 1101)2
    Grouping for octal conversion =

    = (253.632)8

  1. A, B, C and D are input bits, and Y is the out put bit in the XOR gate circuit of the figure given below. Which of the following statements about the sum S of A, B, C, D and Y is correct?








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    Y = A ⊕ B ⊕ C ⊕ D
    = ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD
    S = Y + A + B + C + D
    Drawing K-map for Y,

    When sum of A, B, C, D is odd, Y = 1
    and, S = 0, CY = 0 or 1
    when sum of A, B, C, D is even, Y = 0
    and, S = 0, CY = 0 or 1
    Hence, S is always either zero or even.

    Correct Option: D

    Y = A ⊕ B ⊕ C ⊕ D
    = ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD
    S = Y + A + B + C + D
    Drawing K-map for Y,

    When sum of A, B, C, D is odd, Y = 1
    and, S = 0, CY = 0 or 1
    when sum of A, B, C, D is even, Y = 0
    and, S = 0, CY = 0 or 1
    Hence, S is always either zero or even.

  1. A 4 × 1 MUX is used to implement a 3– input Boolean function as shown in figure given below. The Boolean function F (A, B, C) implemented is








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    The truth-table for given MUX is given as,

    Y = ABC + ABC + BC
    Y = ABC + ABC + ABC + ABC
    Again drawing K-map for Y having three variables

    As shown in k -map, the boolean function F(A, B, C) implemented as,
    F(A, B, C) = ∑(1, 2, 4, 6)

    Correct Option: A

    The truth-table for given MUX is given as,

    Y = ABC + ABC + BC
    Y = ABC + ABC + ABC + ABC
    Again drawing K-map for Y having three variables

    As shown in k -map, the boolean function F(A, B, C) implemented as,
    F(A, B, C) = ∑(1, 2, 4, 6)

  1. The TTL circuit shown in the figure given below is fed with the waveform X (also shown). All gates have equal propagation delay of 10 ns. The output Y of the circuit is








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    At point 1, γ1 = X.X = X delayed 10 n sec.
    γ2 = X. X = X= x delayed 20 n sec.
    Due to 10n sec propagation delay, it will take 10 n sec at ON and 10 n sec at OFF

    2 i.e. output is high only when one input high and other is low.

    Correct Option: A


    At point 1, γ1 = X.X = X delayed 10 n sec.
    γ2 = X. X = X= x delayed 20 n sec.
    Due to 10n sec propagation delay, it will take 10 n sec at ON and 10 n sec at OFF

    2 i.e. output is high only when one input high and other is low.