Essentials of Computer Organization and Architecture
Essentials of Computer Organization and Architecture
4th Edition
ISBN: 9781284074482
Author: Linda Null, Julia Lobur
Publisher: Jones & Bartlett Learning
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Chapter 3, Problem 51E
Program Plan Intro

 JK flip flop:

  • JK stands for “Jack Kilby” who invented the integrated circuit. It is similar to SR flip flop but when the “J” and “K” inputs are both 0, there is no modification in the state.
  • When the “J” and “K” inputs are both 1 at the clock edge then the output will toggle from one state to another state.

 Circuit diagram:

Essentials of Computer Organization and Architecture, Chapter 3, Problem 51E , additional homework tip  1

 Truth table:

Essentials of Computer Organization and Architecture, Chapter 3, Problem 51E , additional homework tip  2

 Explanation:

In order to avoid undefined as an output value, the user must eliminate the condition at the very source by ensuring that both inputs never have the value “1”. This is done by creating an extension to the circuit where inputs to the SR are filtered to remove the possibility that both values are “1”. For this reason the JK flip flop often to prefer the SR flip flop.

D flip flop:

One type of a flip flop which truly reflects how the computer storage works is the D flip flop, which is a sequential circuit. It stores one bit of information. The value stored in D flip flop is same as the input.

 Circuit diagram:

Essentials of Computer Organization and Architecture, Chapter 3, Problem 51E , additional homework tip  3

 Truth table:

 

DOutput (Q(t + 1))
00
11

 XOR gate:

 The “X” in the XOR gate means “exclusive”. The result of the XOR gate will be 1, if any one of the input is 1 and 0 if both the inputs are 1 or 0.

Essentials of Computer Organization and Architecture, Chapter 3, Problem 51E , additional homework tip  4

 The truth table for XOR gate is given below:

 

ABC = A XOR B
000
011
101
110

 Here, the Boolean expression of “A XOR B” gives 0 if both the inputs are 0 or 1 and 1 if any one of the input is 1.

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Chapter 3 Solutions

Essentials of Computer Organization and Architecture

Ch. 3.3A - Prob. 11ECh. 3.3A - Prob. 12ECh. 3 - Prob. 1RETCCh. 3 - Prob. 2RETCCh. 3 - Prob. 3RETCCh. 3 - Prob. 4RETCCh. 3 - Prob. 5RETCCh. 3 - Prob. 6RETCCh. 3 - Prob. 7RETCCh. 3 - Prob. 8RETCCh. 3 - Prob. 9RETCCh. 3 - Prob. 10RETCCh. 3 - Prob. 11RETCCh. 3 - Prob. 12RETCCh. 3 - Prob. 13RETCCh. 3 - Prob. 14RETCCh. 3 - Prob. 15RETCCh. 3 - Prob. 16RETCCh. 3 - Prob. 17RETCCh. 3 - Prob. 18RETCCh. 3 - Prob. 19RETCCh. 3 - Prob. 20RETCCh. 3 - Prob. 21RETCCh. 3 - Prob. 22RETCCh. 3 - Prob. 23RETCCh. 3 - Prob. 24RETCCh. 3 - Prob. 25RETCCh. 3 - Prob. 26RETCCh. 3 - Prob. 1ECh. 3 - Prob. 2ECh. 3 - Prob. 3ECh. 3 - Prob. 4ECh. 3 - Prob. 5ECh. 3 - Prob. 6ECh. 3 - Prob. 7ECh. 3 - Prob. 8ECh. 3 - Prob. 9ECh. 3 - Prob. 10ECh. 3 - Prob. 11ECh. 3 - Prob. 12ECh. 3 - Prob. 13ECh. 3 - Prob. 14ECh. 3 - Prob. 15ECh. 3 - Prob. 16ECh. 3 - Prob. 17ECh. 3 - Prob. 18ECh. 3 - Prob. 19ECh. 3 - Prob. 20ECh. 3 - Prob. 21ECh. 3 - Prob. 22ECh. 3 - Prob. 23ECh. 3 - Prob. 24ECh. 3 - Prob. 25ECh. 3 - Prob. 26ECh. 3 - Prob. 27ECh. 3 - Prob. 28ECh. 3 - Prob. 29ECh. 3 - Prob. 30ECh. 3 - Prob. 31ECh. 3 - Prob. 32ECh. 3 - Prob. 33ECh. 3 - Prob. 34ECh. 3 - Prob. 35ECh. 3 - Prob. 36ECh. 3 - Prob. 37ECh. 3 - Prob. 38ECh. 3 - Prob. 39ECh. 3 - Prob. 40ECh. 3 - Prob. 41ECh. 3 - Prob. 42ECh. 3 - Prob. 43ECh. 3 - Prob. 44ECh. 3 - Prob. 45ECh. 3 - Prob. 46ECh. 3 - Prob. 47ECh. 3 - Prob. 48ECh. 3 - Prob. 49ECh. 3 - Prob. 50ECh. 3 - Prob. 51ECh. 3 - Prob. 52ECh. 3 - Prob. 53ECh. 3 - Prob. 54ECh. 3 - Prob. 55ECh. 3 - Prob. 56ECh. 3 - Prob. 58ECh. 3 - Prob. 59ECh. 3 - Prob. 60ECh. 3 - Prob. 61ECh. 3 - Prob. 62ECh. 3 - Prob. 63ECh. 3 - Prob. 64ECh. 3 - Prob. 65ECh. 3 - Prob. 66ECh. 3 - Prob. 67ECh. 3 - Prob. 68E
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