3 Bit Decoder and Truth Tables

[Capt.Jamesbloke]

So far I have managed to get up to the # bit decoder challenge without the use of truth tables as I could do the logic in my head. I'm pretty sure this is the breaking point whereas I have spent hours trying to work this one out but to no avail. I am not after the answer as I would like to work this out. I'm pretty sure I'm lacking method here. So I have a few Questions!

(1) Is a truth table necessary and the right thing, or do I need to try harder?
(2) How do I write a truth table for this?
(3) Could someone suggest online resources for learning to write truth tables?
(4) What are some suggested methods in formulating a circuit, etc.

Any help would be greatly appreciated!

[UnsignedRobin]

The truth table is shown at the bottom of the screen. You only look at each condition where the output should be green. So a 3 bit decoder has 8 outputs. The first one is on, when all of the inputs are off.

You can deduce the circuit that you need for the first output in a quite literal way (expressing 'off' as 'NOT on'):

"Output1 is on when Input1 is NOT on, AND Input2 is not on, AND input3 is NOT on".

Solution: You to need a NOT for each input and plug that into a 3-way AND.

Lets do an example for wire number 7. That wire should be on when the inputs form a binary 6, which is 110. So you do the same thing again: "The output should be on, when Input1 is NOT on AND Input2 is on, AND Input3 is on" So another 3-way AND in with a NOT for Input1 and normal wires for Input 2 and 3.

That should be enough to solve the 3-Bit Decoder. But in later levels, you can expand that literal approach when ORs come into play. You just build the curcuit as you would describe it in words (using NOT, AND, OR) and focussing only on the "on's".

[Capt.Jamesbloke]

Originally posted by Emeser:

The truth table is shown at the bottom of the screen. You look at each condition where the output should be green. So a 3 bit decoder has 8 outputs. The first one is on, when all of the inputs are off.

You can deduce the circuit that you need for the first output in a quite literal way (expressing 'off' as 'NOT on'):

"Output1 is on when Input1 is NOT on, AND Input2 is not on, AND input3 is NOT on".

Solution: You to need a NOT for each input and plug that into a 3-way AND.

Lets do an example for wire number 5. That wire should be on when the inputs form a binary 6, which is 110. So you do the same thing again: "The output should be on, when Input1 is NOT on AND Input2 is on, AND Input3 is on" So another 3-way in with a NOT for Input1 and normal wires for Input 2 and 3.

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Okey, I was way off track! I'll have a crack at it and thank you oh so much!!!

[UnsignedRobin]

Got a few minor mistakes in my answer that i corrected, but i think the general idea was clear :)