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Chapter 12: Q4E (page 827)

In Exercises 1–5 find the output of the given circuit.

Short Answer

Expert verified

The output of the circuit is\(\overline {{\bf{xyz}}} {\bf{(x + y + z)}}\).

Step by step solution

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01

Defining of gates

There are three types of gates.

It is also called NOT gate.

02

Find the output.

For the result we can check by the diagram.

Therefore, the output is \(\overline {{\bf{xyz}}} {\bf{(x + y + z)}}\).

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Most popular questions from this chapter

\({\bf{a)}}\)Draw a \(K{\bf{ - }}\)map for a function in three variables. Put a \(1\) in the cell that represents \(\bar xy\bar z\).

\({\bf{b)}}\)Which minterms are represented by cells adjacent to this cell\(?\)
  1. How many cells does a \(K{\bf{ - }}\)map in six variables have\(?\)
  2. How many cells are adjacent to a given cell in a \(K{\bf{ - }}\)map in six variables\(?\)

How many different Boolean functions \({\bf{F(x,y,z)}}\) are there such that \({\bf{F(\bar x,y,z) = F(x,\bar y,z) = F(x,y,\bar z)}}\) for all values of the Boolean variables \({\bf{x,y}}\), and \({\bf{z}}\)\({\bf{?}}\)

Show that you obtain De Morgan's laws for propositions (in Table \(6\) in Section \(1.3\)) when you transform De Morgan's laws for Boolean algebra in Table \(6\) into logical equivalences.

For each of these equalities either prove it is an identity or find a set of values of the variables for which it does not hold.

\(\begin{array}{l}a)x|(y\mid z){\bf{ = }}(x\mid y)|z\\b)x \downarrow (y \downarrow z){\bf{ = }}(x \downarrow y) \downarrow (x \downarrow z)\\c)x \downarrow (y\mid z){\bf{ = }}(x \downarrow y)\mid (x \downarrow z)\end{array}\)

Define the Boolean operator \( \odot \) as follows: \(1 \odot 1{\bf{ = }}1,1 \odot 0{\bf{ = }}0,0 \odot 1{\bf{ = }}0\), and \(0 \odot 0{\bf{ = }}1\).
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