Boolean Expressions

A general realization of a Boolean expression is shown in 12.15

Figure 12.15: Realization of a Boolean Expression: shown as a black box

Example:
In a car, we have the following components:

A

Day-night sensor: Day-1, Night-0

B

Lamps on: On-1, Off-0

C

Ignition on: On-1, Off-0

D

Warning light for lamps-on

In this case, the truth table for the logic D would be

A	B	C	D
0	0	0	0
0	0	1	1
0	1	0	1
0	1	1	0
1	0	0	0
1	0	1	0
1	1	0	1
1	1	1	1

Therefore, $D=\overline{A}\overline{B}C+\overline{A}B\overline{C}+A\overline{B}\overline{C... ...e{C}+ABC=\overline{A}\overline{B}C+\overline{A}B\overline{C}+AB\overline{C}+ABC$, which can be written as $\sum\;1,2,6,7$ in the sum of product form. We arrive at this by looking at the combinations when the outout is one.

We can alternatively, express this in the product of sums form by looking at the combinatins when the outout is low as $D=(A+B+C)\centerdot (A+\overline{B}+\overline{C})\centerdot (\overline{A}+B+C)\centerdot (\overline{A}+B+\overline{C})=\Pi 0,4,5,6$

Using SOP and POS, it can be implemented as follows:

Next, we will try to reduce the number of gates by combining terms suitably.

$$D = \overline{A}\overline{B}C+\overline{A}B\overline{C}+AB\overline{C}+ABC$$

$$= \overline{A}\overline{B}C+\overline{A}B\overline{C}+AB$$

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

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

$$= AB+B\overline{C}+\overline{A}\overline{B}C$$

We can get the above by clubbing the $1'$s in the k-map shown.

Now, if we club the zeroes together in the k-map,

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

Check that we get the same expression by simplifying the product of sums expression (by using (X+Y)(X+Z)=X+YZ)

Multiplexer

The truth table for the multiplexer is as follows:

$S_1$	$S_2$	$O_0$
0	0	$I_0$
0	1	$I_1$
1	0	$I_2$
1	1	$I_3$

$$C_0 = \overline{S_1}\overline{S_0}$$

$$C_1 = \overline{S_1}S_0$$

$$C_2 = S_1\overline{S_0}$$

$$C_3 = S_1S_0$$