# 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:
$
\begin{tabular}{c|c|c|c|c|}
AB/CD & 00 & 01 & 11 & 10 \\
\hline
00 & 1 & 1 & 0 & 1 \\
\hline
01 & 0 & 0 & 0 & 0 \\
\hline
11 & 1 & 1 & 1 & 1 \\
\hline
10 & 0 & 0 & 1 & 1 \\
\hline
\end{tabular}
$

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.