## Basic gates

AND gate
$$L_{\text{AND}}(x_1, x_2) = x_1 \cdot x_2$$
OR gate
$$L_{\text{OR}}(x_1, x_2) = x_1 + x_2$$

NOT gate
$$L_{\text{NOT}}(x_1) = \bar{x}_1$$
Source:Verilog 3rd, Brown and Vranesic 2014

## Boolean algebra

#### Single variable Theorems

• $$x \cdot 0 = ?$$

$$x \cdot 0 = 0$$

• $$x + 1 = ?$$

$$x + 1 = 1$$

• $$x \cdot 1 = ?$$

$$x \cdot 1 = x$$

• $$x + 0 = ?$$

$$x + 0 = x$$

• $$x \cdot x = ?$$

$$x \cdot x = x$$

• $$x + x = ?$$

$$x + x = x$$

• $$x \cdot \bar{x} = ?$$

$$x \cdot \bar{x} = 0$$

• $$x + \bar{x} = ?$$

$$x + \bar{x} = 1$$

• $$\bar{\bar{x}} = x$$

## Sample homework problem

Use algebraic manipulation to find the minimum product of sums expression for the function

$f = x_1 \bx_3 \bx_4 + x_2 \bx_3 \bx_4 + x_1 \bx_2 \bx_3$
$f = x_1 \bx_3 \bx_4 + x_2 \bx_3 \bx_4 + x_1 \bx_2 \bx_3$

## Soln

\begin{align} f &= x_1 \bx_3 \bx_4 + x_2 \bx_3 \bx_4 + x_1 \bx_2 \bx_3 & \\ &= x_1 (x_2 + \bx_2) \bx_3 \bx_4 + x_2 \bx_3 \bx_4 + x_1 \bx_2 \bx_3 & \because 1 = x_2 + \bx_2 \end{align}
$$\bar{x}_1$$ $$x_1$$
$$\bar{x}_2$$ $$x_2$$ $$\bar{x}_2$$
$$\bar{x}_3$$$$\bar{x}_4$$ 0 1 1 1
$$x_4$$ 0 0 0 1
$$x_3$$ 0 0 0 0
$$\bar{x}_4$$ 0 0 0 0
\begin{align} &= x_1 x_2 \bx_3 \bx_4 + x_1 \bx_2 \bx_3 \bx_4 + x_2 \bx_3 \bx_4 + x_1 \bx_2 \bx_3 & \text{ by distributive prop. } \\ &= (x_1 + 1) x_2 \bx_3 \bx_4 + x_1 \bx_2 \bx_3 (x_4 + 1) & \text{ by distributive prop. } \\ &= x_2 \bx_3 \bx_4 + x_1 \bx_2 \bx_3 & \text{ by absorption } \end{align} This is the simplest sum-of-product form.

## Sample homework problems

Use algebraic manipulation to prove that

$x + yz = (x+y)(x+z)$

## Soln

To prove: $$x + yz = (x+y)(x+z)$$
\begin{align} \text{RHS} &= (x+y)(x+z) \\ &= x(x+z)+ y(x+z) & \text{ by distributive prop} \\ &= x\cdot x + x\cdot z + y \cdot x + y \cdot z & \text{ by distributive prop } \\ &= x + xz + yx + yz & \because x \cdot x = x \\ &= x(1 + z + y) + yz & \text{by commutative prop} \\ &= x + yz = \text{LHS} & \text{ by absorption } \end{align}

## Soln: three-way light control

#### Truth table

Rowxyz f
0 0000
1 0011
2 0101
3 0110
4 1001
5 1010
6 1100
7 1111
$f = \sum m (1, 2, 4, 7) \\= \bar{x} \bar{y} z + \bar{x} y \bar{z} + x \bar{y} \bar{z} + xyz$
$$\bar{x}$$ $$x$$
$$\bar{y}$$ $$y$$ $$\bar{y}$$
$$\bar{z}$$ 0 1 0 1
$$z$$ 1 0 1 0

## Soln: multiplexer

Rowsx_1x_2 f
0 0000
1 0010
2 0101
3 0111
4 1000
5 1011
6 1100
7 1111
$$f = \sum m(2, 3, 5, 7)$$
$$\bar{s}$$ $$s$$
$$\bar{x_1}$$ $$x_1$$ $$\bar{x_1}$$
$$\bar{x_2}$$ 0 1 0 0
$$x_2$$ 0 1 1 1

$$f = sx_1 + \bar{s} x_2$$

## Sample homework problem

Determine whether or not the following expressions is valid

$\bx_1 x_3 + x_1 x_2 \bx_3 + \bx_1 x_2 + x_1 \bx_2 = \bx_2 x_3 + x_1 \bx_3 + x_2 \bx_3 + \bx_1 x_2 x_3$

## Prove that

If $$x + y = 1$$ and $$xy = 0$$ then $$\bx = y$$

## Soln: Prove that

If $$x + y = 1$$ and $$xy = 0$$ then $$\bx = y$$
\begin{align} &x + y = 1 \\ &\implies x \bx + y \bx = \bx & \text{ multiply both sides by } \bx \\ &\implies y \bx = \bx & \because x \bx = 0 \\ &\implies y \bx + y x = \bx + y x & \text{ add } y x \text{ to both sides } \\ &\implies y (\bx + x) = \bx & \text{by dist. prop. and } xy = y x = 0 \\ &\implies y = \bx & \end{align}

## Prove that for two variable function

$$\overline{\sum m(0, 2)} = \sum m(1, 3)$$

## Review: Minterms

Row numABCDf
000000
100011
200100
300110
401000
501011
601100
701110
810000
910010
1010100
1110110
1211000
1311011
1411100
1511110

#### Minterms

List of rows where function is 1

$$f = m_1 + m_5 + m_{13}$$

$$f = \sum m(1, 5, 13)$$

## Review: Sum of products

#### Minterms

$$f = \sum m(1, 5, 13)$$

 $$m_0$$ $$\triangleq \bar{A} \bar{B} \bar{C} \bar{D}$$ $$m_1$$ $$\triangleq \bar{A} \bar{B} \bar{C} D$$ $$m_2$$ $$\triangleq \bar{A} \bar{B} C \bar{D}$$ $$m_3$$ $$\triangleq \bar{A} \bar{B} C D$$ $$m_4$$ $$\triangleq \bar{A} \bar{B} \bar{C} \bar{D}$$ $$m_5$$ $$\triangleq \bar{A} \bar{B} \bar{C} D$$ $$m_6$$ $$\triangleq \bar{A} \bar{B} C \bar{D}$$ $$m_7$$ $$\triangleq \bar{A} \bar{B} C D$$ $$m_8$$ $$\triangleq A \bar{B} \bar{C} \bar{D}$$ $$m_9$$ $$\triangleq A \bar{B} \bar{C} D$$ $$m_{10}$$ $$\triangleq A \bar{B} C \bar{D}$$ $$m_{11}$$ $$\triangleq A \bar{B} C D$$ $$m_{12}$$ $$\triangleq A B \bar{C} \bar{D}$$ $$m_{13}$$ $$\triangleq A B \bar{C} D$$ $$m_{14}$$ $$\triangleq A B C \bar{D}$$ $$m_{15}$$ $$\triangleq A B C D$$

#### Sum of products

$$f = \bar{A} \bar{B} \bar{C} D + \bar{A} \bar{B} \bar{C} D + A B \bar{C} D$$

## Soln: Prove that for two variable function

$$\overline{\sum m(0, 2)} = \sum m(1, 3)$$
We know that $$\sum m(0, 2) + \sum m(1, 3) = 1$$
\begin{align} \left(\sum m(0, 2) \right)& \cdot \left(\sum m(1, 3) \right)\\ &= (\bx_1 \bx_2 + x_1 \bx_2)(\bx_1 x_2 + x_1 x_2) \\ &= \bx_1 \bx_2 \bx_1 x_2 + \bx_1 \bx_2 x_1 x_2 + x_1 \bx_2 \bx_1 x_2 + x_1 \bx_2 x_1 x_2 \\ &= 0 \end{align}

## Prove that for three variable function

$$\sum m(0, 2, 3, 7) = \prod M(1, 4, 5, 6)$$

## Soln: Prove that for three variable function

\begin{align} \sum m(0, 2, 3, 7) &= m_0 + m_2 + m_3 + m_7 \\ &= \overline{ m_1 + m_4 + m_5 + m_6} \\ &= \bar{m}_1 \bar{m}_4 \bar{m}_5 \bar{m}_6 \\ &= M_1 M_4 M_5 M_6 \\ &=\prod M(1, 4, 5, 6) \end{align}

## Design an expression for

$$f = \sum m(0, 2, 3, 4, 5, 6, 7)$$

## Thanks, Questions, Feedback?

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