On the naming of Mintems

The minterms and maxterms derive their names from the fact that they are respectively the minimum and maximum areas which may be defined on a Venn diagram .

NBS Technical Note. (1959)

$$m_0 = \bar{x}_1 \bar{x}_2$$
$$m_1 = \bar{x}_1 x_2$$
$$m_2 = x_1 \bar{x}_2$$
$$m_3 = x_1 x_2$$

Truth table to K-map

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

K-map

$$\bar{A}$$ $$A$$
$$\bar{B}$$ $$B$$ $$\bar{B}$$
$$\bar{C}$$$$\bar{D}$$ $$m_0$$ $$m_4$$ $$m_{12}$$ $$m_8$$
$$D$$ $$m_1$$ $$m_5$$ $$m_{13}$$ $$m_9$$
$$C$$ $$m_3$$ $$m_7$$ $$m_{15}$$ $$m_{11}$$
$$\bar{D}$$ $$m_2$$ $$m_6$$ $$m_{14}$$ $$m_{10}$$
$$\bar{A}$$ $$A$$
$$\bar{B}$$ $$B$$ $$\bar{B}$$
$$\bar{C}$$$$\bar{D}$$ 0 0 0 0
$$D$$ 1 1 1 0
$$C$$ 0 0 0 0
$$\bar{D}$$ 0 0 0 0

K-map to expression

K-map

$$\bar{A}$$ $$A$$
$$\bar{B}$$ $$B$$ $$\bar{B}$$
$$\bar{C}$$$$\bar{D}$$ 0 0 0 0
$$D$$ 1 1 1 0
$$C$$ 0 0 0 0
$$\bar{D}$$ 0 0 0 0

Expression

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

Expression to truth table

Expression

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

Truth table

Row numABCD $$\bar{A}\bar{C}D$$ $$B\bar{C}D$$ f
0 0000000
1 0001101
2 0010000
3 0011000
4 0100000
5 0101111
6 0110000
7 0111000
8 1000000
9 1001000
101010000
111011000
121100000
131101011
141110000
151111000

Truth table to timing diagram

Row num0123456789101112131415
A0000000011111111
B0000111100001111
C0011001100110011
D0101010101010101
f0100010000000100

Boolean expression to Verilog

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

Verilog


module examplefunction1(A, B, C, D, f);
input A, B, C, D;
output f;
assign f = (~(A&(~B) | C))&D;
endmodule


Functionally equiv networks

$$f_1 = \overline{(A\bar{B} + C)}D$$
$$f_2 = (\bar{A} + B)\bar{C}D$$

Boolean algebra

Axioms

• $$0 \cdot 0 = 0$$
• $$1 + 1 = 1$$
• $$1 \cdot 1 = 1$$
• $$0 + 0 = 0$$
• $$0 \cdot 1 = 1 \cdot 0 = 0$$
• $$\bar{0} = 1$$
• $$\bar{1} = 0$$

Single variable Theorems

• $$x \cdot 0 = 0$$
• $$x + 1 = 1$$
• $$x \cdot 1 = x$$
• $$x + 0 = x$$
• $$x \cdot x = x$$
• $$x + x = x$$
• $$x \cdot \bar{x} = 0$$
$$x + \bar{x} = 1$$
• $$\bar{\bar{x}} = x$$

Duality: Swap $$+$$ with $$\cdot$$ and $$0$$ with $$1$$ to get another law

Two and three variable properties

 $$x \cdot y = y \cdot x$$ Commutative $$x + y = y + x$$ $$x \cdot (y \cdot z) = (x \cdot y) \cdot z$$ Associative $$x + (y + z)= (x + y) + z$$ $$x \cdot (y + z) = x\cdot y + x \cdot z$$ Distributive $$x + y\cdot z = (x+y) \cdot (x+z)$$

DeMorgan's theorem

• $$\overline{x \cdot y} = \bar{x} + \bar{y}$$
• $$\overline{x + y} = \bar{x} \cdot \bar{y}$$

DeMorgan's theorem

• $$\overline{x \cdot y} = \bar{x} + \bar{y}$$
• $$\overline{x + y} = \bar{x} \cdot \bar{y}$$

Prove that for two variable function

$$\bar{m}_0 = M_0$$

Soln: Prove that for two variable function

$$\bar{m}_0 = M_0$$ \begin{align} m_0 &= \bar{x}_1 \bar{x}_2 & \text{ by definition of } m_0 \\ \implies \bar{m}_0 &= \overline{\bar{x}_1 \bar{x}_2} & \\ &= \bar{\bar{x}}_1 + \bar{\bar{x}}_2 & \text{ by DeMorgan's theorem } \\ &= x_1 + x_2 = M_0 & \text{ by definition of } M_0 \end{align}

Inequality

• Define $$x \le y$$ as $$x = 0$$ whenever $$y = 0$$
• If $$x \le y$$ and $$y \le x$$, then $$x = y$$
• If $$x \le y$$, then $$x \cdot y = x$$ and $$x + y = y$$
• $$x \cdot z \le z$$ and $$x \cdot z \le x$$

Two and three variable properties

 $$x + x \cdot y = x$$ Absorption $$x \cdot (x + y) = x$$ $$x \cdot y + x \cdot \bar{y} = x$$ Combining $$(x + y)\cdot (x + \bar{y}) = x$$

Two and three variable properties

 $$x + \bar{x}y = x + y$$ Concensus $$x \cdot (\bar{x} + y) = x \cdot y$$ $$x \cdot y + y \cdot z + \bar{x} \cdot z = x \cdot y + \bar{x} \cdot z$$ $$(x + y) \cdot ( y + z ) \cdot (\bar{x} + z) = (x + y) \cdot (\bar{x} + z)$$

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

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

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}

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

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

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