Traffic Light 1
_ _ _ _ _ _ _ _ _
R = X·Y·Z + X·Y·Z + X·Y·Z + X·Y·Z + X·Y·Z + X·Y·Z
This is the Boolean equation for when the red light is on there are 6 states and a load of different gates which will cost far to much money and time so we can simplify this so we can use the least amount of gates possible to keep the cost down but not the time so if I create the Boolean equation for when the red light is off and the put a NOT gate after it, it will work out exactly the same way but saves a whole lot of time.
I will need to apply De Morgan’s Theorem to following equation.
So now,
_ _ _
R1 = X·Y·Z + X·Y·Z
_ _ _
R1 = X·Y·Z · X·Y·Z
= _ = = _ _
R1 = X+Y+Z · X+Y+Z
_ _ _
R1 = X+Y+Z · X+Y+Z
The De Morgan’s Theorem is now finished but this Boolean Equation can no be simplified.
So now this is Boolean Equation is much more easier to simplify and will take half the time the previous equation would of took.
_ _ _
R1 = X+Y+Z · X+Y+Z
_
R1 = X+Y
This now is the smallest equation I could possibly get, I can not simplify this no more and the reason why we no longer have ‘Z’ in the equation is because ‘Z’ and ‘Z’ cancel each other out of the equation. I will now simplify the rest of the equations.
Traffic light 1
The following equations will be simplified so then I can create the circuit using the least amount of gates.
_ _ _
R1 = X+Y+Z · X+Y+Z
_
R1 = X+Y
_ _ _
A1 = X·Y·Z + X·Y·Z
_
A1 = X·Z
_ _
G1 = X·Y·Z (This equation cannot be simplified anymore.)
Traffic light 2
I have only applied De Morgan’s Theorem to the red light on traffic light 1, I will now need to do the same for the red light on traffic light 2.
De Morgan’s Theorem
_
R2 = X·Y·Z + X·Y·Z
_
R2 = X·Y·Z · X·Y·Z
_ _ = _ _ _
R2 = X+Y+Z · X+Y+Z
_ _ _ _ _
R2 = X+Y+Z · X+Y+Z
I will now need to simplify this equation.
So now,
_ _ _ _ _
R2 = X+Y+Z · X+Y+Z
_ _
R2 = X + Y
_
A2 = X·Y·Z + X·Y·Z
A2 = X·Z
_
G2 = X·Y·Z
Now I have all the equation I need to create my circuit.
Key:
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? =
· =
+ =
Circuit
