Essentials of Computer Organization and Architecture
Essentials of Computer Organization and Architecture
5th Edition
ISBN: 9781284123043
Author: Linda
Publisher: Jones & Bartlett Learning
Question
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Chapter 2, Problem 27E
Program Plan Intro

Types of representation for signed number:

  • Signed magnitude representation
  • One’s complement
  • Two’s complement
  • Excess-M representation

Signed magnitude representation:

In signed magnitude representation, an extra bit is used to represent the sign of the number. This extra bit is called as sign bit.

  • The Most Significant Bit (MSB) in the binary number is referred as sign bit.
  • In the signed magnitude representation, if the MSB is 0, then it is treated as positive sign and if the MSB is 1, then it is treated as negative sign.

Example:

Consider the decimal number 36 that can be represented as 00100100 and the decimal number -36 can be represented as 10100100.

One’s complement:

The binary numbers can be represented using one’s complement. Here, the value of every binary digit is complemented that is if the value is 1, it becomes 0 and if the value is 0 it becomes 1.

Example:

Consider the binary number 1101; the one’s complement of the given number is 0010.

Two’s complement:

Two’s complement is another way of representing the binary numbers. To implement the two’s complement to the number, the given binary number should be one’s complemented and then add 1 to the result obtained after one’s complement.

Example:

Consider the binary number 1011. First implement the one’s complement to the given binary number. The number becomes 0100. Then add 1 to the resultant obtained after the one’s complement. The result becomes 0101.

Excess-M Representation:

To get the excess-M representation of a given number,

  • Add the given number with the M.
  • Convert the obtained value to the binary representation.

Example:

To change to excess-7 representation, add the number 77 with the 7.

77+7=84

Binary value of 84 is 1010100.

Hence, the result of excess-7 representation for 77 is 1010100.

Expert Solution & Answer
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Explanation of Solution

Representing the binary pattern using various formats in the table:

Unsigned Integer4-bit binary valueSigned MagnitudeOne’s complementTwo’s complement

Excess-7

000000007
100011118
200102229
3001133310
4010044411
5010155512
6011066613
7011177714
81000-0-7-815
91001-1-6-716
101010-2-5-617
111011-3-4-518
121100-4-3-419
131101-5-2-320
141110-6-1-221
151111-7-0-122

Explanation:

First column represent the unsigned integer, and the corresponding 4-bit binary value is represented in the second column.

While representing the singed magnitude, if the Most Significant Bit is “+”, the given number is positive number and if the Most Significant Bit is “-”, the given number is negative number. For example, 1001  the MSB bit is “1” and so the number is negative and the decimal equivalent for “001” is “1”.

While representing one’s complement, in binary number 1001 the MSB bit is “1” and so the number is negative and the one’s complement is applied for the number 1001 and the number becomes 0110. So, the decimal equivalent for “0110” is 6”.

While representing two’s complement, in binary number 1001 the MSB bit is “1” and so the number is negative. First one’s complement is applied for the number 1001, the number becomes 0110 and by adding “1” in the least significant bit, the equivalent two’s complement is attained. So, the number becomes 0111. So, the decimal equivalent for 0111 is “7”.

While representing excess-7, the given unsigned integer is added with the decimal number 7. Here, “1” is represented as “7”, “2” as “8”, and so on.

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Chapter 2 Solutions

Essentials of Computer Organization and Architecture

Ch. 2 - Prob. 7RETCCh. 2 - Prob. 8RETCCh. 2 - Prob. 9RETCCh. 2 - Prob. 10RETCCh. 2 - Prob. 11RETCCh. 2 - Prob. 12RETCCh. 2 - Prob. 13RETCCh. 2 - Prob. 14RETCCh. 2 - Prob. 15RETCCh. 2 - Prob. 16RETCCh. 2 - Prob. 17RETCCh. 2 - Prob. 18RETCCh. 2 - Prob. 19RETCCh. 2 - Prob. 20RETCCh. 2 - Prob. 21RETCCh. 2 - Prob. 22RETCCh. 2 - Prob. 23RETCCh. 2 - Prob. 24RETCCh. 2 - Prob. 25RETCCh. 2 - Prob. 26RETCCh. 2 - Prob. 27RETCCh. 2 - Prob. 28RETCCh. 2 - Prob. 29RETCCh. 2 - Prob. 30RETCCh. 2 - Prob. 31RETCCh. 2 - Prob. 32RETCCh. 2 - Prob. 33RETCCh. 2 - Prob. 34RETCCh. 2 - Prob. 1ECh. 2 - Prob. 2ECh. 2 - Prob. 3ECh. 2 - Prob. 4ECh. 2 - Prob. 5ECh. 2 - Prob. 6ECh. 2 - Prob. 7ECh. 2 - Prob. 8ECh. 2 - Prob. 9ECh. 2 - Prob. 10ECh. 2 - Prob. 11ECh. 2 - Prob. 12ECh. 2 - Prob. 13ECh. 2 - Prob. 14ECh. 2 - Prob. 15ECh. 2 - Prob. 16ECh. 2 - Prob. 17ECh. 2 - Prob. 18ECh. 2 - Prob. 19ECh. 2 - Prob. 20ECh. 2 - Prob. 21ECh. 2 - Prob. 22ECh. 2 - Prob. 23ECh. 2 - Prob. 24ECh. 2 - Prob. 25ECh. 2 - Prob. 26ECh. 2 - Prob. 27ECh. 2 - Prob. 29ECh. 2 - Prob. 30ECh. 2 - Prob. 31ECh. 2 - Prob. 32ECh. 2 - Prob. 33ECh. 2 - Prob. 34ECh. 2 - Prob. 35ECh. 2 - Prob. 36ECh. 2 - Prob. 37ECh. 2 - Prob. 38ECh. 2 - Prob. 39ECh. 2 - Prob. 40ECh. 2 - Prob. 41ECh. 2 - Prob. 42ECh. 2 - Prob. 43ECh. 2 - Prob. 44ECh. 2 - Prob. 45ECh. 2 - Prob. 46ECh. 2 - Prob. 47ECh. 2 - Prob. 48ECh. 2 - Prob. 49ECh. 2 - Prob. 50ECh. 2 - Prob. 51ECh. 2 - Prob. 52ECh. 2 - Prob. 53ECh. 2 - Prob. 54ECh. 2 - Prob. 55ECh. 2 - Prob. 56ECh. 2 - Prob. 57ECh. 2 - Prob. 58ECh. 2 - Prob. 59ECh. 2 - Prob. 60ECh. 2 - Prob. 61ECh. 2 - Prob. 62ECh. 2 - Prob. 63ECh. 2 - Prob. 64ECh. 2 - Prob. 65ECh. 2 - Prob. 66ECh. 2 - Prob. 67ECh. 2 - Prob. 68ECh. 2 - Prob. 69ECh. 2 - Prob. 70ECh. 2 - Prob. 71ECh. 2 - Prob. 72ECh. 2 - Prob. 73ECh. 2 - Prob. 74ECh. 2 - Prob. 75ECh. 2 - Prob. 76ECh. 2 - Prob. 77ECh. 2 - Prob. 78ECh. 2 - Prob. 79ECh. 2 - Prob. 80ECh. 2 - Prob. 81ECh. 2 - Prob. 82E
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