Digital electronics miscellaneous

Digital electronics miscellaneous

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Digital Electronics

Digital electronics miscellaneous
  1. The octal equivalent of the HEX number AB. CD is—








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    NA

    Correct Option: B

    NA

  1. The addition of two binary variables A and B results into a SUM and a CARRY output. Consider the following expressions for the SUM and CARRY outputs—








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    SUM = A ⊕ B = A B + BA
    CARRY = AB

    Correct Option: B

    SUM = A ⊕ B = A B + BA
    CARRY = AB

  1. In the figure, as long as X1 = 1 and X2 = 1, the output Q remains—










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    As long as X1 = 1 and X2 = 1
    Since the output is feedback to input NAND gate. It means output is toggle between 0 → 1 → 0 → 1…… makes the output Q unstable.


    Correct Option: D

    As long as X1 = 1 and X2 = 1
    Since the output is feedback to input NAND gate. It means output is toggle between 0 → 1 → 0 → 1…… makes the output Q unstable.


  1. A digital circuit which compares two numbers A3 A2A1 A0 and B3 B2 B1 B0 is shown below. To get the output
    X = 0, choose one pair of correct input numbers—










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    To get the output zero i.e., X = 0.
    Applying hit and trial method in order to choose the pair of number.
    (A) For input 1 0 1 0, 1 0 1 0 output
    Y1 = 0, Y2 = 0, Y3 = 0, Y4 = 0
    Now, X = Y1 Y2 Y3 Y4 + Y1 Y2 Y3 Y4
    or X = 0. 0. 0. 0 + 0 0 0 0 = 1 (Not required)
    (B) For input 0 1 0 1, 0 1 0 1 output
    Y1 = 0, Y2 = 0, Y3 = 0, Y4 = 0
    Now, X = Y1 Y2 Y3 Y4 + Y1 Y2 Y3 Y4
    or X = 0. 0. 0. 0 + 0 0 0 0 = 1 (Not required)
    (C) For input 0 0 1 0, 0 0 1 0 output
    Y1 = 0, Y2 = 0, Y3 = 0, Y4 = 0
    Now, X = Y1 Y2 Y3 Y4 + Y1 Y2 Y3 Y4
    or X = 0. 0. 0. 0 + 0 0 0 0 = 1 (Not required)
    (D) For input 1 0 1 0, 1 0 1 1 output
    Y1 = 0, Y2 = 0, Y3 = 0, Y4 = 0
    Now, X = Y1 Y2 Y3 Y4 + Y1 Y2 Y3 Y4
    or X = 0. 0. 0. 0 + 0 0 0 1 = 0 + 0 = 0
    This is required condition. Hence alternative (D) is the correct choice.


    Correct Option: D

    To get the output zero i.e., X = 0.
    Applying hit and trial method in order to choose the pair of number.
    (A) For input 1 0 1 0, 1 0 1 0 output
    Y1 = 0, Y2 = 0, Y3 = 0, Y4 = 0
    Now, X = Y1 Y2 Y3 Y4 + Y1 Y2 Y3 Y4
    or X = 0. 0. 0. 0 + 0 0 0 0 = 1 (Not required)
    (B) For input 0 1 0 1, 0 1 0 1 output
    Y1 = 0, Y2 = 0, Y3 = 0, Y4 = 0
    Now, X = Y1 Y2 Y3 Y4 + Y1 Y2 Y3 Y4
    or X = 0. 0. 0. 0 + 0 0 0 0 = 1 (Not required)
    (C) For input 0 0 1 0, 0 0 1 0 output
    Y1 = 0, Y2 = 0, Y3 = 0, Y4 = 0
    Now, X = Y1 Y2 Y3 Y4 + Y1 Y2 Y3 Y4
    or X = 0. 0. 0. 0 + 0 0 0 0 = 1 (Not required)
    (D) For input 1 0 1 0, 1 0 1 1 output
    Y1 = 0, Y2 = 0, Y3 = 0, Y4 = 0
    Now, X = Y1 Y2 Y3 Y4 + Y1 Y2 Y3 Y4
    or X = 0. 0. 0. 0 + 0 0 0 1 = 0 + 0 = 0
    This is required condition. Hence alternative (D) is the correct choice.


  1. The simplified form of the Boolean expression Y = ( ABC + D) ( AD + B C) can be written as—








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    Given expression
    Y = (ABC + D) (AD + B C)
    or Y = ABCAD + DA D + ABC B C + D B C
    or Y = ABCD + A D + DB C
    Now, further it can be easily solved by using K-map.
    Now, from K-map, we get
    Y = AD + B CD.


    Correct Option: A

    Given expression
    Y = (ABC + D) (AD + B C)
    or Y = ABCAD + DA D + ABC B C + D B C
    or Y = ABCD + A D + DB C
    Now, further it can be easily solved by using K-map.
    Now, from K-map, we get
    Y = AD + B CD.