Question

Implement the four functions described in Exercise B.11 using a PLA

Short answer

X contains only one 0:

$\mathrm{X}=~\mathrm{x}2.\mathrm{x}1.\mathrm{x}0+\mathrm{x}2.~\mathrm{x}1.\mathrm{x}0+\mathrm{x}2.\mathrm{x}1.~\mathrm{x}0$

X contains an even number of zero

$\mathrm{X}=~\mathrm{x}2.~\mathrm{x}1.\mathrm{x}0+~\mathrm{x}2.\mathrm{x}1.~\mathrm{x}0+\mathrm{x}2.~\mathrm{x}1.~\mathrm{x}0$

X interpreted as an unsigned binary number less than 4

$Y=~x2$

X interpreted as a negative number

$\mathrm{Y}=\mathrm{x}2$

Step-by-step solution

Determine the logical gates

The PLA is the programmable logical array that is used to illustrate the logical circuits, The basic logic gates are AND, OR, and NOT. AND gate will take two inputs and produce output as 1 only if both the inputs are 1. OR gate will take two inputs and produce output 1 if any of the inputs is 1. NOT gate will take only one input and gives the opposite of the input as the output. 1.

Determine the logic functions

Given that the X consists of 3 bits x0, x1, and x2. The logic functions for which that are true if and only if

• X contains only one 0
• X contains an even number of 0s
• X when interpreted as an unsigned binary number is less than 4
• X when interpreted as a signed (two’s complement) number is negative.

For X contains only one zero, only one of the three bits should be zero. The logical function that satisfies this condition is as follows:

$\mathrm{X}=~\mathrm{x}2.\mathrm{x}1.\mathrm{x}0+\mathrm{x}2.~\mathrm{x}1.\mathrm{x}0+\mathrm{x}2.\mathrm{x}1.~\mathrm{x}0$

The logical circuit is as follows:

For X contains an even number of 0s, two bits of the three bits should be zero. The logical function that satisfies this condition is as follows:

$\mathrm{X}=~\mathrm{x}2.~\mathrm{x}1.\mathrm{x}0+~\mathrm{x}2.\mathrm{x}1.~\mathrm{x}0+\mathrm{x}2.~\mathrm{x}1.~\mathrm{x}0$

The logical circuit is as follows:

For X when interpreted as an unsigned binary number is less than 4 is as follows:

$Y=~x2$

The logical circuit is as follows:

ForX when interpreted as a signed (two’s complement) number is negative. The values will be complements and the logical function is as

$\mathrm{Y}=\mathrm{x}2$

The logical circuit is as follows: