Simply an equation

Discussion in 'General Electronics Chat' started by Kayne, Mar 22, 2009.

  1. Kayne

    Thread Starter Active Member

    Mar 19, 2009
    105
    0
    Hi just would like some clarification on whether or not either way i have tried to simplfy this equation is correct

    ! = NOT (Lines above letters)
    & = AND


    Method 1


    !A&!B&!C + !B&C&!D + A&C + A&B

    !A&!B(C+!C) +!B&!D + A&C + A&B

    !A&!B + !B&!D + A&C + A&B

    Is this as far as I can go or can

    !A&!B + A&B Cancel out also leaving

    !B&!D + A&C




    Or is this correct way



    Method 2



    !A&!B&!C + !B&C&!D + A&C + A&B

    !B((!A+A)(C+!C) +!B&C&!D + A&B


    !B + C&!D + A


    if someone can clear this up that would be helpful



    Thanks
     
  2. dsp_redux

    Active Member

    Apr 11, 2009
    182
    5
    If I'm right, the equation you'd like to solve is: \overline{ABC} + \overline{B}C \overline{D} + AC + AB. Correct me if I'm wrong but from a Karnaugh table you get:
    <br />
\begin{tabular}{c|c|c|c|c|}<br />
AB/CD & 00 & 01 & 11 & 10 \\<br />
\hline<br />
00 & 1 & 1 & 0 & 1 \\<br />
\hline<br />
01 & 0 & 0 & 0 & 0 \\<br />
\hline<br />
11 & 1 & 1 & 1 & 1 \\<br />
\hline<br />
10 & 0 & 0 & 1 & 1 \\<br />
\hline<br />
\end{tabular}<br />
    That should give the equation (using the summation principle): S = AB + AC + \overline{ABC} + \overline{AB}C \overline{D}. This is about the simplest you can get from that method... and looks almost identical to the starting equation.

    For your method 1. You can't cancel everthing like you said. For exemple when  ABCD = 0001 you should have S=1. In \overline{BD} + AC = 0 when ABCD=0001. You second method is also false. Try it with  ABCD = 0011. Finally try method 1... with ABCD=1000. You get S=1 but should get S=0. So I think none of your simplifications are good.
     
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