Switch circuit, finding truth table, SOP and POS

Discussion in 'Homework Help' started by ramil13, Nov 20, 2013.

  1. ramil13

    Thread Starter New Member

    Nov 14, 2013
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    0
    1. A switching circuit has four inputs as shown. A and B represent the first(MSB) and second(LSB) bits of a binary number N1, respectively. C and D represent the first(MSB) and second(LSB) bits of a binary number N2, respectively. The output is to be 1 only if the product N1 x N2 is less than or equal to 2.
    (a) Write the truth table for the system.
    (b) Write the canonical SOP and POS expressions for F.

    a)
    A B C D F Minterm Maxterm
    0 0 0 0 1 A’B’C’D’ A+B+C+D
    0 0 0 1 1 A’B’C’D A+B+C’+D
    0 0 1 0 1 A’B’CD’ A+B+C’+D
    0 0 1 1 1 A’B’CD A+B+C’+D’
    0 1 0 0 1 A’BC’D’ A+B’+C+D
    0 1 0 1 1 A’BC’D A+B’+C+D’
    0 1 1 0 1 A’BCD’ A+B’+C’+D
    0 1 1 1 0 A’BCD A+B’+C’+D’
    1 0 0 0 1 AB’C’D’ A’+B+C+D
    1 0 0 1 1 AB’C’D A’+B+C+D’
    1 0 1 0 0 A’B’CD’ A’+B+C’+D
    1 0 1 1 0 AB’CD A’+B+C’+D’
    1 1 0 0 1 ABC’D’ A’+B’C+D
    1 1 0 1 0 ABC’D A’+B’+C+D’
    1 1 1 0 0 ABCD’ A’+B’+C’+D
    1 1 1 1 0 ABCD A'+B’+C’+D’

    b)
    canonical POS: Z = (A+B’+C’+D’)( A’+B+C’+D)( A’+B+C’+D’)( A’+B’+C+D’)
    ( A’+B’+C’+D)( A'+B’+C’+D’)

    canonical SOP: Z = A’B’C’D’ + A’B’C’D + A’B’CD’ + A’B’CD + A’BC’D’ + A’BC’D + A’BCD’ + AB’C’D’ + AB’C’D + ABC’D’


    This is a homework question I'm working on, I'm not sure if my truth table is correct based on what the question is asking. But if my truth table is correct then I know for sure my POS and SOP is correct. Does the truth table look like it makes sense?
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,720
    4,788
    Well, what are all the possible combinations of {0,1,2,3} that yield values equal to or greater than 2? Are those properly identified as False in your truth table? Are all others identified as True?
     
  3. ramil13

    Thread Starter New Member

    Nov 14, 2013
    9
    0
    The truth table follows this binary order:
    Since A is MSB if it is 1 then it will be 2(Decimal value), and since B is LSB if it is 1 it is 1(Decimal value).
    Since C is MSB if it is 1 then it will be 2(Decimal value), and since D is LSB if it is 1 it is 1(Decimal value).


    I believe it is correct

    For example if it A = 0, B = 0, C = 1, D = 1
    that means 00(decimal = 0) x 11(decimal = 3) = 0 x 3 = 0(Product = 0)
    => F = 1
     
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