Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

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 \(\overline {{\bf{xyz}}} {\bf{(x + y + z)}}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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.

Use a \({\bf{K}}\)-map to find a minimal expansion as a Boolean sum of Boolean products of each of these functions in the variables \({\bf{w, x, y}}\) and \({\bf{z}}\).

\(\begin{array}{l}{\bf{a) wxyz + wx\bar yz + wx\bar y\bar z + w\bar xy\bar z + w\bar x\bar yz}}\\{\bf{b) wxy\bar z + wx\bar yz + w\bar xyz + \bar wx\bar yz + \bar w\bar xy\bar z + \bar w\bar x\bar yz}}\\{\bf{c) wxyz + wxy\bar z + wx\bar yz + w\bar x\bar yz + w\bar x\bar y\bar z + \bar wx\bar yz + \bar w\bar xy\bar z + \bar w\bar x\bar yz}}\\{\bf{d) wxyz + wxy\bar z + wx\bar yz + w\bar xyz + w\bar xy\bar z + \bar wxyz + \bar w\bar xyz + \bar w\bar xy\bar z + \bar w\bar x\bar yz}}\end{array}\)

Is it always true that \((x \odot y) \odot z{\bf{ = }}x \odot (y \odot z)\)\(?\)

Show that \({\bf{x}} \odot {\bf{y = xy + \bar x\bar y}}\).

Find a Boolean product of the Boolean variables x, y,and z, or their complements, that has the value 1 if and only if

a)x=y=0, z=1

b)x=0, y=1, z=0

c)x=0, y=z=1

d)x=y=z=0
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free