COMPUTER SCIENCE ILLUMIN.-TEXT
COMPUTER SCIENCE ILLUMIN.-TEXT
7th Edition
ISBN: 9781284156010
Author: Dale
Publisher: Jones & Barlett
bartleby

Videos

Question
Book Icon
Chapter 4, Problem 60E
Program Plan Intro

Circuit:

  • The circuit is known as the combination of gates that is used to achieve a difficult logical operation.
  • It has two general categories; they are:
    • Combinational circuit
    • Sequential circuit

Expert Solution & Answer
Check Mark

Explanation of Solution

Given circuit diagram:

COMPUTER SCIENCE ILLUMIN.-TEXT, Chapter 4, Problem 60E

Behavior of the circuit:

  • From the circuit diagram:
    • First, the input A is passed to NOT gate to perform the inverse of the A and produces the output as A¯.
    • Next, the input A and input B is passed in the AND gate to perform the product of A and B, to produce the output as AB.
    • Finally, the output of NOT gate and the output of AND gate is passed as the input of XOR gate.
      • Note: when both the inputs are the same, then the output of XOR gate is 0. Otherwise, the output of XOR gate is 1.
      • That is, “A¯” and “AB” are passed as the input for XOR gate and produce as the output A¯(AB).

Truth table for the given circuit diagram:

Step 1:

  • The inputs are A and B for the above circuit diagram:
ABA¯ABA¯(AB)
00   
01   
10   
11   

Step 2:

  • When the inputs are A as 0 and B as 0:
ABA¯ABA¯(AB)
00101
01   
10   
11   
  • First, the input A as 0 is passed to NOT gate to perform the inverse of the A and produces the output as 10=1 .
  • Next, the input A as 0 and input B as 0 is passed to the AND gate to perform the product of A and B and produce the output as  s.
  • Finally, the output of NOT gate and the output of AND gate is passed as the input of XOR gate.
    • Note: when both the inputs are the same, then the output of XOR gate is 0. Otherwise, the output of XOR gate is 1.
    • That is, “1” and “0” are passed as inputs to XOR gate and produces output as  10=1.

Step 3:

  • When the inputs are A as 0 and B as 1:
ABA¯ABA¯(AB)
00101
01101
10   
11   
  • First, the input A as 0 is passed to NOT gate to perform the inverse of the A and produces the output as 0¯=1.
  • Next, the input A as 0 and input B as 1 is passed to the AND gate to perform the product of the A and B and produce the output as  01= .
  • Finally, the output of NOT gate and the output of AND gate is passed as the input of XOR gate.
    • Note: when both the inputs are the same, then the output of XOR gate is 0. Otherwise, the output of XOR gate is 1.
    • That is, “1” and “0” are passed as input for XOR gate and produce output as 10=1.

Step 4:

  • When the inputs are A as 1 and B as 0:
ABA¯ABA¯(AB)
00101
01101
10000
11   
  • First, the input A as 1 is passed to NOT gate to perform the inverse of A and produces the output as  1 ¯=0 .
  • Next, the input A as 1 and input B as 0 is passed in the AND gate to perform the product of the A and B and produce the output as 10=0.
  • Finally, the output of NOT gate and the output of AND gate is passed as the input to XOR gate.
    • Note: when both the inputs of XOR gate are the same, the output of XOR gate is 0. Otherwise, the output of XOR gate is 1.
    • That is, “0” and “0” are passed as input for XOR gate and produce as the output 00=0.

Step 5:

  • When the inputs are A as 1 and B as 1:
ABA¯ABA¯(AB)
00101
01101
10000
11011
  • First, the input A as 1 is passed to NOT gate to perform the inverse of the A and produces the output as  1 ¯=0.
  • Next, the input A and input B as 1 is passed in the AND gate to perform the product of the A and B and produce the output as 11=1.
  • Finally, the output of NOT gate and the output of AND gate is passed as the input of XOR gate.
    • Note: when both the inputs of XOR gate are the same, the output of XOR gate is 0. Otherwise, the output of XOR gate is 1.
    • That is, “0” and “1” are passed as input for XOR gate and produce as the output 01=1.

Final truth table:

Therefore, truth table of the given circuit diagram is:

ABA¯ABA¯(AB)
00101
01101
10000
11011

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
The images I have uploaded are the part 1 to 4 and questions below are continue on the questions uploaded 5. C++ Class Template with Method Stubs #pragma once #include <iostream> #include <string> #include <stdexcept> #include <vector>   template <typename T> class HashTable { private:     struct Entry {         std::string key;         T value;         bool isOccupied;         bool isDeleted;         Entry() : key(""), value(), isOccupied(false), isDeleted(false) {}     };       Entry* table;     size_t capacity;     size_t size;     double loadFactorThreshold;           size_t customHash(const std::string& key) const {         size_t hash = 5381;         for (char c : key) {             hash = ((hash  <<  5) + hash)  +  c;         }         return hash;     }       size_t probe(const std::string& key, bool forInsert = false) const;     void resize();   public:     // Constructor     HashTable(size_t initialCapacity = 101);         // Big…
this project is NOT for graded(marks) purposes, please help me with the introduction. give me answers for the project. i will include an image explaining everything about the project.
Java Graphics (Bonus In this lab, we'll be practicing what we learned about GUIs, and Mouse events. You will need to implement the following: A GUI with a drawing panel. We can click in this panel, and you will capture those clicks as a Point (see java.awt.Point) in a PointCollection class (you need to build this). The points need to be represented by circles. Below the drawing panel, you will need 5 buttons: O о о ○ An input button to register your mouse to the drawing panel. A show button to paint the points in your collection on the drawing panel. A button to shift all the points to the left by 50 pixels. The x position of the points is not allowed to go below zero. Another button to shift all the points to the right 50 pixels. " The x position of the points cannot go further than the You can implement this GUI in any way you choose. I suggest using the BorderLayout for a panel containing the buttons, and a GridLayout to hold the drawing panel and button panels. Regardless of how…
Knowledge Booster
Background pattern image
Computer Science
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, computer-science and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Database System Concepts
Computer Science
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:McGraw-Hill Education
Text book image
Starting Out with Python (4th Edition)
Computer Science
ISBN:9780134444321
Author:Tony Gaddis
Publisher:PEARSON
Text book image
Digital Fundamentals (11th Edition)
Computer Science
ISBN:9780132737968
Author:Thomas L. Floyd
Publisher:PEARSON
Text book image
C How to Program (8th Edition)
Computer Science
ISBN:9780133976892
Author:Paul J. Deitel, Harvey Deitel
Publisher:PEARSON
Text book image
Database Systems: Design, Implementation, & Manag...
Computer Science
ISBN:9781337627900
Author:Carlos Coronel, Steven Morris
Publisher:Cengage Learning
Text book image
Programmable Logic Controllers
Computer Science
ISBN:9780073373843
Author:Frank D. Petruzella
Publisher:McGraw-Hill Education
Boolean Algebra - Digital Logic and Logic Families - Industrial Electronics; Author: Ekeeda;https://www.youtube.com/watch?v=u7XnJos-_Hs;License: Standard YouTube License, CC-BY
Boolean Algebra 1 – The Laws of Boolean Algebra; Author: Computer Science;https://www.youtube.com/watch?v=EPJf4owqwdA;License: Standard Youtube License