# Computer Organisation Part 3

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Information about Computer Organisation Part 3
Technology

Published on July 29, 2009

Author: kalakar_2k

Source: slideshare.net

## Description

Computer Organisation by Mukesh Upadhyay from Lachoo Memorial College Jodhpur

01001001000001001001001000100010010000100101111010101010100101010100101001010100100100000100100100100010001001000010010111101010101010010101010010100101001001010010010000010010010010001000100100001001011110101010101001010101001010010101001001000001001001001000100010010000100101111010101010100101010100101001010100100100000100100100100010001001000010010111101010101010010101010010100101010010010000010100100100000100100100100010001001000010010111101010101010010101010010100101010010010000010010010010001000100100001001011110101010101001010101001010010100100100100010001001000010010111101010101010010101010010100101010010010000010010010010001000100100001001011110101010101001010101001010010101001001000001001001001000100010010000100101111010101010100101010100101001010010000010010010010001000100100011001011110101010101001010101001010010110010010000010010010010001000100100001001011110101010101001010101001010010101001001000001001001001000100010010000100101111010101010100101010100101001010100100100000100100 Number system Simplicity of complexity Krishna Kumar Bohra (KKB), MCA LMCST www.selectall.wordpress.com

NUMBERS FIXED POINT FLOATING POINT REAL NUMBERS FLOATING POINT Number means, Value assigned to a particular symbol. 35 Tens Unit 3 X 10 + 5 X 1 23.45 Whole Part Fractional Part Krishna Kumar Bohra (KKB), MCA LMCST www.selectall.wordpress.com

23.4500 X 100 234.500 X 10-1 2345.00 X 10-2 2.34500 X 101 .234500 X 102 Representation of Floating Point Number : +/- m X b+/-e exponent mantissa base . power . power ( Therefore, it is floating point ) Krishna Kumar Bohra (KKB), MCA LMCST www.selectall.wordpress.com

Arithmetic Operation : Addition and Subtraction i.) make e1 = e2 ii.) Add/Subtract m1 and m2 Example : N1 N2 29.32 2.48 29.32 X 100 2.48 X 100 2.932 X 101 .248 X 101 2.932 X 101 .248 X 101 -------------------------- 3.580 X 101 -------------------------- Multiplication i.) Add e1 = e2 ii.) Multiply m1 and m2 Division i.) Subtraction e1 = e2 ii.) Divide m1 and m2 Krishna Kumar Bohra (KKB), MCA LMCST www.selectall.wordpress.com

Normalization of Numbers : After decimal there is always non zero value than representation is called NORMALIZED But, ZERO can not be normalized, because zero can not contain any non zero value therefore it is so. For –ve 0 For +ve __ . __ __ __ __ __ __ __ __ __ __ __ m Exp sign sign Krishna Kumar Bohra (KKB), MCA LMCST www.selectall.wordpress.com

Errors are of 2 types : 1.) Truncation Error .0003899  .00038 here, 99 X 10-7 2.) Round off Error .0003899  .00039 here, 99 X 10-7 In both the above cases there is intolerable. Therefore we go for After decimal there is always non zero value than representation is called NORMALIZED Krishna Kumar Bohra (KKB), MCA LMCST www.selectall.wordpress.com

Examples : .0003899  .3899000 X 10-3  0 3 8 9 9 0 1 0 3 + . 3 8 9 9 0 - 1003 235.8  .2358 X 103  0 2 3 5 8 0 0 0 4 1 9 9 9 9 9 0 0 2 + . 3 8 9 9 0 + 1003 99.999  .99999 X 102  - . 9 9 9 9 9 + 1002 Krishna Kumar Bohra (KKB), MCA LMCST www.selectall.wordpress.com

Examples : -.00836  .836 X 10-2  1 8 3 6 0 0 1 0 2 - . 3 8 9 9 0 - 1002 -00235.7  .23587 X 105  1 2 3 5 8 7 0 0 5 1 9 9 9 9 9 1 0 2 - . 2 3 5 8 7 + 1005 -.00999  .99999 X 10 -2  - . 9 9 9 9 9 - 10 -02 Krishna Kumar Bohra (KKB), MCA LMCST www.selectall.wordpress.com

+/- m X b +/-e 23 8 1 IEEE 754 Standard : Double Single for exponent biased value 127 (default number) EXPONENT MANTISSA 32 32 bits because it is decided in standard in 1EEE 754 floating point standard Krishna Kumar Bohra (KKB), MCA LMCST www.selectall.wordpress.com

Examples (11)10 (1)10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 Krishna Kumar Bohra (KKB), MCA LMCST www.selectall.wordpress.com

Ccontd…

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