DeMorgan Theorem

Design of three-way light control

Soln: three-way light control

Truth table

Row	x	y	z	f
0	0	0	0	0
1	0	0	1	1
2	0	1	0	1
3	0	1	1	0
4	1	0	0	1
5	1	0	1	0
6	1	1	0	0
7	1	1	1	1

$f = \sum m (1, 2, 4, 7) \\= \bar{x} \bar{y} z + \bar{x} y \bar{z} + x \bar{y} \bar{z} + xyz$

	$\bar{x}$		$x$
	$\bar{y}$	$y$		$\bar{y}$
$\bar{z}$	0	1	0	1
$z$	1	0	1	0

$f = \sum m (1, 2, 4, 7) \\= \bar{x} \bar{y} z + \bar{x} y \bar{z} + x \bar{y} \bar{z} + xyz$

Design a multiplexer

Soln: multiplexer

Row	s	x_1	x_2	f
0	0	0	0	0
1	0	0	1	0
2	0	1	0	1
3	0	1	1	1
4	1	0	0	0
5	1	0	1	1
6	1	1	0	0
7	1	1	1	1

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

	$\bar{s}$		$s$
	$\bar{x_1}$	$x_1$		$\bar{x_1}$
$\bar{x_2}$	0	1	0	0
$x_2$	0	1	1	1

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

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

$\overline{m}_0 = M_0$

Soln: Prove that for two variable function

$\bar{m}_0 = M_0$

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

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 three variable function

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

Soln: Prove that for three variable function

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

Find the minimum cost expression for

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

Solution

$\begin{align} f &= \sum m(0, 2, 3, 4, 5, 6, 7) \\ &= \overline{\sum m(1)} \\ &= \prod M(1) \\ &= (x + y + \bz) \end{align}$

SOP in terms of NAND gate

POS in terms of NOR gate

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ECE 275: DeMorgan Theorem