Question

1. Assume that a and b are 3-bit inputs (let a be a2a1a0 and b be b2b1b0).
   (a) Determine the Boolean expression (in terms of a2,a1,a0, b2,b1 and b0) for

   the relational operation a>b
   (b) Assume that b is a constant and that b = "101". Determine the Boolean

   expression again.
   (c) Assume that a is a constant and that a = "101". Determine the Boolean

   expression again.

Answer 1

Answer 2

(a) Determine the Boolean expression \(a > b\)

For two 3-bit binary numbers \(a = a_2a_1a_0\) and \(b = b_2b_1b_0\), to determine \(a>b\), we can compare from the highest bit to the lowest bit.

If \(a_2 = 1\) and \(b_2 = 0\), then no matter what value \(a_1,a_0,b_1,b_0\) takes, \(a>b\), that is, \(a_2\overline{b_2}\).

If \(a_2 = b_2\), we need to compare the next highest bit \(a_1\) and \(b_1\). When \(a_2 = b_2\) and \(a_1 = 1\), \(b_1 = 0\), there is also \(a>b\), that is, \(a_2b_2a_1\overline{b_1}+\overline{a_2}\overline{b_2}a_1\overline{b_1}=(a_2\oplus\overline{b_2})a_1\overline{b_1}\).

If \(a_2 = b_2\) and \(a_1 = b_1\), the lowest bits \(a_0\) and \(b_0\) need to be compared. When \(a_2 = b_2\), \(a_1 = b_1\) and \(a_0 = 1\), \(b_0 = 0\), \(a>b\), that is, \(a_2b_2a_1b_1a_0\overline{b_0}+\overline{a_2}\overline{b_2}a_1b_1a_0\overline{b_0}+a_2b_2\overline{a_1}\overline{b_1}a_0\overline{b_0}+\overline{a_2}\overline{b_2}\overline{a_1}\overline{b_1}a_0\overline{b_0}\)

To sum up, the Boolean expression \(a > b\) =

\[

\begin{align*}

a>b&=a_2\overline{b_2}+(a_2\odot b_2)a_1\overline{b_1}+(a_2\odot b_2)(a_1\odot b_1)a_0\overline{b_0}\\

\end{align*}

\]

Where \(x\odot y=\overline{x\oplus y}=xy + \overline{x}\overline{y}\) is the XOR operation, \(x\oplus y=x\overline{y}+\overline{x}y\) is the XOR operation.

### (b) Assume \(b\) is a constant, and \(b = "101"\)

At this time \(b_2 = 1\), \(b_1 = 0\), \(b_0 = 1\), substitute it into the expression of \(a > b\):

\[

\begin{align*}

a>b&=a_2\overline{1}+(a_2\odot 1)a_1\overline{0}+(a_2\odot 1)(a_1\odot 0)a_0\overline{1}\\

&=0 + a_2a_1+(a_2)(\overline{a_1})a_0\times0\\

&=a_2a_1

\end{align*}

\]

### (c) Assume \(a\) is a constant, and \(a = "101"\)

At this time \(a_2 = 1\), \(a_1 = 0\), \(a_0 = 1\), substitute it into the expression \(a > b\):

\[

\begin{align*}

a>b&=1\times\overline{b_2}+(1\odot b_2)\times0\times\overline{b_1}+(1\odot b_2)(0\odot b_1)\times1\times\overline{b_0}\\

&=\overline{b_2}+ 0+(b_2)(\overline{b_1})\overline{b_0}\\

&=\overline{b_2}+b_2\overline{b_1}\overline{b_0}

\end{align*}

\]

Summary:

- (a) \(a>b=a_2\overline{b_2}+(a_2\odot b_2)a_1\overline{b_1}+(a_2\odot b_2)(a_1\odot b_1)a_0\overline{b_0}\)

- (b) When \(b = 101\), \(a>b = a_2a_1\)

- (c) When \(a = 101\), \(a>b=\overline{b_2}+b_2\overline{b_1}\overline{b_0}\)