Question

Assume that X consists of 3 bits, x2 x1 x0, and Y consists of 3 bits, y2 y1 y0. Write logic functions that are true if and only if

■$\mathbf{X}\mathbf{<}\mathbf{Y}$, where X and Y are thought of as unsigned binary numbers

■$\mathbf{X}\mathbf{<}\mathbf{Y}$, where X and Y are thought of as signed (two’s complement) numbers

■$\mathbf{X}\mathbf{=}\mathbf{Y}$

Use a hierarchical approach that can be extended to larger numbers of bits. Show how can you extend it to 6-bit comparison.

Short answer

The Logic function that are true if and only if

• $X<Y$, where X and Y are unsigned binary numbers, let the equation be E1.

$\mathrm{E}1=\overline{\mathrm{x}2}.\mathrm{y}2+(\mathrm{x}2\odot \mathrm{y}2).(\overline{\mathrm{x}1}.\mathrm{y}1)+(\mathrm{x}2\odot \mathrm{y}2).(\mathrm{x}1\odot \mathrm{y}1).(\overline{\mathrm{x}0}.\mathrm{y}0)$

• $\mathrm{X}<\mathrm{Y}$ , where X and Y are signed numbers

$\mathrm{E}2=\mathrm{x}2.\overline{\mathrm{y}2}+(\mathrm{x}2\odot \mathrm{y}2).\overline{\mathrm{x}1}.\mathrm{y}1+(\mathrm{x}2\odot \mathrm{y}2).(\mathrm{x}1\odot \mathrm{y}1).\overline{\mathrm{x}0}.\mathrm{y}0$

• $\mathrm{X}=\mathrm{Y}$

$\mathrm{E}3=(\mathrm{x}2\odot \mathrm{y}2).(\mathrm{x}1\odot \mathrm{y}1).(\mathrm{x}0\odot \mathrm{y}0)$

Step-by-step solution

Determine the logical gates and the functions

The logical functions are the combinations of the inputs and outputs of the truth table. The inputs be operated through the logical operators and the outputs are processed. Each Boolean expression can be represented through gates and the combinational circuits. The logical functions or expressions can be written from the conditions of the input and the output provided.

Determine the logic functions

Given, that X consist of three bits x2, x1, x0 and Y consists of three bits y2, y1, y0.

The logic functions should be written in way that are true if and only if

$\mathrm{X}<\mathrm{Y}$, where X and Y are thought of as unsigned binary numbers

$\mathrm{X}<\mathrm{Y}$, where X and Y are thought of as signed (two’s complement) numbers

First, the 1-bit comparison will be made as follows:

The truth tables for these conditions are as follows:

$\mathrm{X}<\mathrm{Y}$

$\mathrm{X}>\mathrm{Y}$

$\mathrm{X}==\mathrm{Y}$

For $\mathrm{X}<\mathrm{Y}$

X	Y	$\mathrm{X}<\mathrm{Y}$
0	0	0
0	1	1
1	0	0
1	1	0

$(\mathrm{X}<\mathrm{Y})=\overline{\mathrm{X}}.\mathrm{Y}$

For $\mathrm{X}>\mathrm{Y}$

X	Y	$\mathrm{X}>\mathrm{Y}$
0	0	0
0	1	0
1	0	1
1	1	0

$(\mathrm{X}>\mathrm{Y})=\mathrm{X}.\overline{\mathrm{Y}}$

For $\mathrm{X}==\mathrm{Y}$

X	Y	$\mathrm{X}==\mathrm{Y}$
0	0	1
0	1	0
1	0	0
1	1	1

$(\mathrm{X}==\mathrm{Y})=\mathrm{X}\odot \mathrm{Y}$

From the above conditions, the logical equations can be derived.

$\mathrm{X}<\mathrm{Y}$, where X and Y are thought of as unsigned binary numbers

For an unsigned binary number, any of the following conditions must be satisfied.

$\begin{array}{l}\mathrm{x}2<\mathrm{y}2\\ \mathrm{x}2=\mathrm{y}2\&\mathrm{x}1<\mathrm{y}1\\ \mathrm{x}2=\mathrm{y}2\&\mathrm{x}1=\mathrm{y}1\&\mathrm{x}0<\mathrm{y}0\end{array}$

Hence, the logical function is:

$\mathrm{E}1=\overline{\mathrm{x}2}.\mathrm{y}2+(\mathrm{x}2\odot \mathrm{y}2).(\overline{\mathrm{x}1}.\mathrm{y}1)+(\mathrm{x}2\odot \mathrm{y}2).(\mathrm{x}1\odot \mathrm{y}1).(\overline{\mathrm{x}0}.\mathrm{y}0)$

The logical network:

By the hierarchical approach, the 6 bit extension is as follows:

$\mathrm{X}<\mathrm{Y}$, where X and Y are thought of as signed (two’s complement) numbers

For signed binary number, any of the following conditions must be satisfied.

$\begin{array}{l}\mathrm{x}2>\mathrm{y}2\left(\mathrm{sign}\mathrm{bit}\right)\\ \mathrm{x}2=\mathrm{y}2\&\mathrm{x}1<\mathrm{y}1\\ \mathrm{x}2=\mathrm{y}2\&\mathrm{x}1=\mathrm{y}1\&\mathrm{x}0<\mathrm{y}0\end{array}$

Hence, the logical function is:

$\mathrm{E}2=\mathrm{x}2.\overline{\mathrm{y}2}+(\mathrm{x}2\odot \mathrm{y}2).\overline{\mathrm{x}1}.\mathrm{y}1+(\mathrm{x}2\odot \mathrm{y}2).(\mathrm{x}1\odot \mathrm{y}1).\overline{\mathrm{x}0}.\mathrm{y}0$0

The logical network:

By the hierarchical approach, the 6 bit extension is as follows:

,$X==Y$

The logical function is,

$\mathrm{E}3=(\mathrm{x}2\odot \mathrm{y}2).(\mathrm{x}1\odot \mathrm{y}1).(\mathrm{x}0\odot \mathrm{y}0)$

The logical network is shown below: