Logic Circuit help

[Harald Kapp]

I have no idea how to test my circuit to see if it matches the truth table.

Start by writing down the truth table for your circuit's function Y=A*B + NOT C.
Then check against the truth table Laplace has given you.

And is it the And gate?

Right. That's the gate that goes to 0 when the control input is 0.

Now go on to the gate or gates that create the control signal which is 0 when A=B=C=0 ´, and only for this case.
Hint: This gate will have 1 for any other combination of A, B and C, so whenever one or more of A, B or C are 1, the output (the control signal) will be 1, too.
Oh my good, I've almost given away the solution with this hint,.

 

[Madz2007]

A*B + NOT C

Doesnt work, I tested it for both the outputs and its not that,

 

[Harald Kapp]

No, it doesn't work.
You need a function of A, B and C that gives 0 when every input is 0
Start wth two inputs only Y=A ??? B where ??? is the function you're looking for (e.g. OR, EXOR, AND, NAND, NOR ...) Which one will give 0 for A=0 and B=0?
Use the result of this function and the input C to arrive at the final control signal.

 

[Madz2007]

Or and and will give 00

 

[Madz2007]

Im one hour away of submitting this and i still have no idea... i think ill fail this time but if you can carry on helping it will be much appreiciated

 

[Laplace]

I also got this, this MAY NOT be correct, please tell me if it is or not.

It is not correct.

Specifically, the Boolean functions do not match the gate types. (e.g., "+" is not AND)

 

[Harald Kapp]

Or and and will give 00

OR will, and it wil also fulfill the other condition:

Now go on to the gate or gates that create the control signal which is 0 when A=B=C=0 ´, and only for this case.

as OR will give 1 for any of A, B or C being 1.

You now have CONTROL = A OR B OR C.
You also know that RED = NOT GREEN Except for A=B=C=0 here RED = 0, too instead of 1.
How will you use this CONTROL signal to make RED=0 ?

 

[Harald Kapp]

Boolean functions do not match the gate types

This escaped my attention, I'll be forever struggling with the American way of drawing logic gates .
I only loked at the equation, which is unsuitable nevertheless.