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Chapter 12: Problem 4

Use a K-map to find a minimal expansion as a Boolean sum of Boolean products of each of these functions of the Boolean variables \(x\) and \(y .\) a) \(\overline{x} y+\overline{x} \overline{y}\) b) \(x y+x \overline{y}\) c) \(x y+x \overline{y}+\overline{x} y+\overline{x} \overline{y}\)

Short Answer

Expert verified
a) \(\bar{x}\) b) \x\ c) 1

Step by step solution

01

Express the Functions in Canonical Form

Rewrite each function to show all the minterms explicitly. For example: (a) \(\bar{x}y + \bar{x}\bar{y}\)(b) \(xy + x\bar{y}\)(c) \(xy + x\bar{y} + \bar{x}y + \bar{x}\bar{y}\)
02

Create Karnaugh Maps

Draw a 2x2 K-map for each function with variables x and y. The K-map will have 4 cells corresponding to the minterms. Label the rows and columns with x values (0,1) and y values (0,1).
03

Mark the Minterms in the K-map

For each function, fill in the cells of the K-map with 1s and 0s based on the minterms:(a) \(\bar{x}y\) and \(\bar{x}\bar{y}\) yield a K-map where cells (0,1) and (0,0) are 1.(b) \(xy\) and \(x\bar{y}\) yield a K-map where cells (1,1) and (1,0) are 1.(c) All combinations yield a K-map where all cells are 1.
04

Identify Groups

Group the 1s in each K-map into the largest possible power of 2 groups. This minimizes the Boolean expressions:(a) Group the two 1s in the row \(\bar{x}\) together.(b) Group the two 1s in the column \x\ together.(c) The entire K-map is a group of 4 since all cells are 1.
05

Write the Minimal Expressions

Convert each group to its Boolean product term:(a) The group \(\bar{x}\) corresponds to \(\bar{x}\).(b) The group \x\ corresponds to \x\.(c) The entire K-map corresponds to 1 (since it contains all minterms of x and y).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boolean algebra
Boolean algebra is a branch of mathematics dealing with variables that have only two distinct values: true (1) or false (0). In this context, we're dealing with logical operations such as AND, OR, and NOT.
  • The AND operation (denoted by juxtaposition, e.g., \(xy\)) is true if both operands are true.
  • The OR operation (denoted by +, e.g., \(x + y\)) is true if at least one of the operands is true.
  • The NOT operation (denoted by a bar or tilde, e.g., \(\bar{x}/\text{~}x\)) inverts the value of the variable.
Understanding these operations is crucial for working with Karnaugh maps because they simplify how Boolean expressions are represented and manipulated.
Minterms
Minterms are the standard form of expressing every possible combination of variables in a Boolean function. Each minterm represents a distinct combination of variables set to true or false. In a function with two variables, \(x\) and \(y\), there are four possible minterms:
  • \(\bar{x}\bar{y}\)
  • \(\bar{x}y\)
  • \(x\bar{y}\)
  • \(xy\)
These combinations cover every scenario the variables can represent. The expression of functions in terms of minterms helps in plotting on Karnaugh maps and aids in the simplification process.
Minimal Boolean expression
Simplifying a Boolean expression to its minimal form means reducing it to the smallest number of terms without changing its functionality. This is done by combining terms using logical equivalences based on Boolean algebra rules. For example:
  • The expression \(xy + x\bar{y}\) can be simplified by grouping, resulting in \x\.
  • Similarly, \(\bar{x}y + \bar{x}\bar{y}\) minimizes to \(\bar{x}\).
The minimal Boolean expression is easier to understand and implement, especially in digital circuits.
K-map grouping
Karnaugh Map (K-map) grouping is a visual method used for simplifying Boolean expressions. Each cell in a K-map corresponds to a minterm of the variables. Here鈥檚 the process:
  • Draw the K-map grid based on the number of variables (for two variables, a 2x2 grid).
  • Label rows and columns with variable combinations (e.g., for \(x\) and \(y\), rows and columns represent \(0\) or \(1\).
  • Mark the cells with \1\ or \0\ based on the minterms provided in the Boolean function.
  • Group the adjacent 1s in powers of 2 (1, 2, 4, ...) as large as possible.
These groups help in reducing the Boolean expression to fewer terms, ultimately leading to the minimal Boolean expression.
Canonical form
The canonical form of a Boolean expression involves writing it as a sum of minterms (Sum of Products, SOP) or as a product of maxterms (Product of Sums, POS). For example:
  • The expression \(xy + x\bar{y}\) is already in SOP.
  • Similarly, \(\bar{x}y + \bar{x}\bar{y}\) is in SOP.
The canonical form makes it straightforward to map the function onto a K-map and aids in the simplification process. It's a standardized way to represent all possible scenarios covered by the Boolean variables.

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Most popular questions from this chapter

A multiplexer is a switching circuit that produces as output one of a set of input bits based on the value of control bits. Construct a multiplexer using AND gates, OR gates, and inverters that has as input the four bits \(x_{0}, x_{1}, x_{2},\) and \(x_{3}\) and the two control bits \(c_{0}\) and \(c_{1} .\) Set up the circuit so that \(x_{i}\) is the output, where \(i\) is the value of the two-bit integer \(\left(c_{1} c_{0}\right)_{2} .\)

Find the sum-of-products expansion of the Boolean function \(F\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)\) that has the value 1 if and only if three or more of the variables \(x_{1}, x_{2}, x_{3}, x_{4},\) and \(x_{5}\) have the value \(1 .\)

Draw the 3 -cube \(Q_{3}\) and label each vertex with the minterm in the Boolean variables \(x, y,\) and \(z\) associated with the bit string represented by this vertex. For each literal in these variables indicate the 2 -cube \(Q_{2}\) that is a subgraph of \(Q_{3}\) and represents this literal.

In Exercises \(35-42,\) use the laws in Definition 1 to show that the stated properties hold in every Boolean algebra. Prove that in a Boolean algebra, the law of the double complement holds; that is, \(\overline{\overline{x}}=x\) for every element \(x .\)

a) Show that \((\overline{1} \cdot \overline{0})+(1 \cdot \overline{0})=1\) b) Translate the equation in part (a) into a propositional equivalence by changing each 0 into an \(\mathbf{F}\) , each 1 into a \(\mathbf{T}\) , each Boolean sum into a disjunction, each Boolean product into a conjunction, each complementation into a negation, and the equals sign into a propositional equivalence sign.
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