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K map help and minterms

  1. Apr 19, 2015 #1
    1. The original expression is:

    F(A,B,C,D) = Σm(0,1,2,5,8,9,10)

    If I changed the order of ABCD to:

    F(C,B,A,D)

    What would be the minterms now?


    2. Use the truth table for 0 to 15


    3. The attempt at a solution
    I used the truth table in the regular order

    ABCD and for CBAD.


    ABCD | F
    0000 | 1
    0001 | 1
    0010 | 1
    0011 | 0
    0100 | 0
    0101 | 1
    0110 | 0
    0111 | 0
    1000 | 1
    1001 | 1
    1010 | 1
    1011 | 0
    1100 | 0
    1101 | 0
    1110 | 0
    1111 | 0

    So, here I basically thought that if C was 1 in the regular column then in the new truth table I just look for 1 under the C column. Is this right or do I need to redo the columns so that everything under the A column from the regular truth table is shifted over to 2 columns over?

    CBAD | F
    0000 | 1
    0001 | 1
    0010 | 1
    0011 | 1
    0100 | 0
    0101 | 1
    0110 | 0
    0111 | 0
    1000 | 1
    1001 | 0
    1010 | 1
    1011 | 0
    1100 | 0
    1101 | 0
    1110 | 0
    1111 | 0

    I got F(C,B,A,D) = Σm(0,1,2,3,5,8,10)

    is this correct? I thought that if you changed the order than the minterms should change and only one minterm changed.
     
  2. jcsd
  3. Apr 24, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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