Question

Find a minimal sum-of-products expansion, given the \({\bf{K}}\)-map shown with don't care conditions indicated with \({\bf{d's}}\).

Short answer

The minimal sum-of-products expansion is\({\bf{w\bar y + \bar wx}}\).

Step-by-step solution

Step 1:Definition

In some circuits you care only about the output for some combinations of input values, because other combinations of input values are not possible or never occur. This gives us freedom in producing a simple circuit with the desired output because the output values for all those combinations that never occur can be arbitrarily chosen. The values of the function for these combinations are called don't care conditions.

Step 2:Largest block

Given:

The largest block containing \({\bf{wx\bar yz}}\) in the graph is the block \({\bf{w\bar y}}\), since all cells corresponding to \({\bf{w}}\) and \({\bf{\bar y}}\) have a \({\bf{1}}\) or \({\bf{d\;}}\)in it.

The largest block containing \({\bf{w\bar xyz}}\) in the graph is the block (row) \({\bf{\bar wx}}\), since all cells corresponding to \({\bf{\bar w}}\) and \({\bf{x}}\) have a \(1\) or \({\bf{d}}\) in it.

Minimal sum-of-products

All \(1{\bf{'s}}\)are contained in the previous two mentioned blocks.

The minimal sum-of-products expansion is then the sum of these blocks.

Therefore, the minimal sum-of-products expansion \({\bf{ = w\bar y + \bar wx}}\).