Derive Boolean Expression for 3-Input, 2-Output Truth Table

corty_d

Joined Feb 13, 2010
2
Hi guys, first post on here. Im doing a HNC and currently doing the Digital Techniques module. Im looking for a little confirmation rather than something answering so here goes:

ABC X Y
000 0 0
001 1 1
010 0 0
011 0 0
100 0 1
101 1 1
110 1 0
111 1 0

a) Find the Boolean expression for the truth table shown and then implement each output seperately using the minimum number of two input Nand gates.

b) Show how you could combine the two circuits to make a three input two output circuit with two input Nand gates using less gates than the original two circuits.

Ok... The main bit im stuck on is creating a Boolean expression for the full table.

For X output - X = AB + B'C (Simplified)
For Y output - Y = AB' + B'C (Simplified)

To simplify these I have used the Idempotent law as I need to build these using only two input Nand gates.

For the full table I think the simplified expression would be:

B'C + AC' + AB

Could anyone confirm these answers for me? Also do you think for b) it just means design a circuit for the full expression?

Cheers,
Simon.

LoganFife

Joined Feb 7, 2010
13
your answers for X and Y are correct, but you need to make them NAND only,do you know De Morgans Theory?

because X and Y are independant outputs, you can't simply combine thier logic expressions.

part a is asking for two seperate NAND circuits, one for X and one for Y

Part b is asking you to combine those circuits, reducing the overall number of NAND gates required.

corty_d

Joined Feb 13, 2010
2
I know demorgans theory but how does that relate to my output for X and Y? Im pretty sure these answers are ok to make Nand gate circuits for.

The main but thats puzzling me is combing the circuits to make a 2 output circuit... could you explain the process using Boolean for the second circuit?

Regards,
Simon.

LoganFife

Joined Feb 7, 2010
13
any expressions can be made using only NAND gates, i was only suggesting that you alter the equations to better reflect the circuit.
your first equation looks something like:
X = AB + B'C ---> $$X = \bar{ ( \bar{AB} )( \bar{ \bar{B}C})}$$

the combining of the circuit is simply identifying the common components to both individual circuits. I don't think you are being asked to manipulate Boolean Algebra to do this, merely recognise that for example; C must be inverted for each individual circuit, but when you combine the circuits, you only need to do this once.

this shows up in the Eqns where two expressions have identical components.