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Below is a question in my homework:

\$F(P,Q,R,S,T)=(P+Q)S+(R+T)\bar S\$ using one or more 2x2 AOI." />


Here's my attempt for (a):

Step 1: simplify boolean expression.

\$F(P,Q,R,S,T)\$

\$=(P+Q)S+(R+T)\bar S\$

\$=PS+QS+R\bar S+\bar ST\$

Step 2: expand boolean expression so that it fits into the AOI gate logic:

\$F=PS+QS+R\bar S+\bar ST\$

\$=\overline{\overline{(PS+QS)+(R\bar S+\bar ST)}}\$

\$=\overline{(\overline{PS+QS})(\overline{R\bar S+\bar ST})}\$


I want to know:

  1. Do I continue manipulating the boolean expression, or can I just implement the result from Step 2 with AOI gates?

  2. Is this double-negate manipulating method suitable for implementing boolean functions with AOI gates (or maybe even all gates in general)?

Any guidance is appreciated!

asked Oct 12, 2019 at 6:58
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1 Answer 1

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The result of Step 2 has obvious two 2X2 AOI. One for the left hand side let us call it X and one for right hand side let's call it Y.

\begin{equation} X = \overline{PS + QS} \; (one AOI22)\\ Y = \overline{R\overline{S} + \overline{S}T} \; (one AOI22) \end{equation}

Also you can implement the inverter using AOI22 as below: \begin{equation} \overline{(S.1 + 0.0)} = \overline{S} \; (one AOI22) \end{equation}

What you need is \begin{equation} \overline{X} + \overline{Y} \end{equation}

which you can perform by one 2x2 AOI gate as:

\begin{equation} \overline{(X.Y + 0.0)} = \overline{X} + \overline{Y} \; (one AOI22) \end{equation}

This makes 4 AOI22 for the result of Step 2.

answered Oct 12, 2019 at 20:05
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  • \$\begingroup\$ Thanks for the help!! I forgot that an AOI can be used as an inverter as well, which was what got me stuck. \$\endgroup\$ Commented Oct 13, 2019 at 14:45

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