boolean expressions

Question

Construct a circuit for the boolean expression

I am having issues on how to start this. Not sure if we do a graph that we put the 1 or 0 depending on the not x or not y

Answer 1

$$\begin{align} \overline {x\bar y + \overline{\bar x y}}& = \overline{ x\bar y} \cdot \bar xy\tag{DeMorgan's Law} \\ \\ &=(\bar x + y)\cdot \bar x y \tag{DeMorgan's Law}\\ \\ & = \bar x \bar x y+( y\bar x y)\tag{Distributive Law}\\ \\ &= \bar x y + \bar xy \tag {absorption} \\ \\ & = \bar x y\tag{absorption}\\ \\ \end{align}$$

Hence you need only a NOT gate, and an AND gate.

See Wolfram Alpha for the associated circuit:

Answer 2

Assuming $+$ denotes XOR then: $\overline{x\overline{y} + \overline{\overline{x}y}} = 1 + x\overline{y} + \overline{\overline{x}y} = 1 + x\overline{y} + 1 + \overline{x}y = x\overline{y} + \overline{x}y$

Now what gates do we have at our disposal?

Answer 3

While you could just mock up a representation of the circuit then simplify, you could also solve and simplify the equation given.

Starting with $\overline{x\overline{y}+\overline{\overline{x}y}}$, apply DeMorgan's Theorum

$\overline{(x\overline{y}) }\cdot\overline{\overline{(\overline{x}y)}}$

Removing the double negation on the second half, and applying DeMorgan's Theorum to the first half:

$ (\overline{x} + \overline{\overline{y}})\cdot (\overline{x}y) $

Removing the double negation:

$(\overline{x} + y) \cdot (\overline{x} \cdot y)$

Distributing the AND along the OR we get:

$\overline{x}\cdot\overline{x} \cdot y + y\cdot y \cdot \overline{x}$

Removing duplicates in the ANDs, we get:

$\overline{x} \cdot y + \overline{x} \cdot y$

Which simplifies to:

$\overline{x}\cdot y$

Which is just an AND gate, with the X input inverted

Answer 4

If you have to stick to the formula as is, here is the circuit generated by WolframAlpha:

One thing you can do is to just have one $x$ and one $y$ on thew left, but to split them as required by the diagram.