For Digital Electronics I have been trying for hours to figure out a way to simplify this circuit: X=(STxV)+(PS+PT+PV) I have tested it in MultiSim and it works perfect. The scenario is that you have to make a voting machine so that the President's (P) vote counts as 2, and the Vice President's (V), the Secretary (S) and the Treasure (T) counts as 1. 3 Votes are needed to pass and make X = 1 The fun part is that it needs to be as efficient as possible by using only AND, OR, and NOT gates for a total of less than 8. I have tried all sorts of wacky things and have not yet figured this out. Any help or tips would be greatly appreciated.. I am using MultiSim 10.1
X is the boolean value. Here is a truth table of the values that need to work. P V S T 0 1 1 1 = 1 1 1 1 1 = 1 1 1 0 0 = 1 1 0 1 0 = 1 1 0 0 1 = 1 1 1 1 0 = 1 1 0 1 1 = 1 Diagram of what I have now is attached.
According to your truth table, no matter what the inputs are, the output is a 1. That can't be right. Are you guys studying Karnaugh maps?
Not all possible input's were put in the table, just the ones where the output equals one. I think he forgot P V S T 1 1 0 1 = 1
Given you original equation: X=(STxV)+(PS+PT+PV) assuming it is X=((ST)V) + (PS+PT+PV) Is there anything common to the terms in the second half of your equation? If the President votes, what else needs to happen...
"The fun part is that it needs to be as efficient as possible by using only AND, OR, and NOT gates for a total of less than 8." Check out the 74HC21 dual 4-input AND gate and the 74HC4075 triple 3-input OR gate. (Along with an inverter.) You should be able to solve X = (P' * V * S * T) + (P * (V+S+T)) readily, using only 4 gates. --Rich P.S. -- If V, S, & T are TRUE, then the state of P is immaterial. The solution becomes: X = (V*S*T) + (P * (V+S+T))