Lecture 19

Lecture 19 - ECE52 Spring 11 Lecture 19 1 In binary IEEE...

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1 ECE52 Spring 11 Lecture 19 2/25/11
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2 In binary: IEEE Standard Floating Point Format Sign 32 bits 23 bits of mantissa excess-127 exponent 8-bit 52 bits of mantissa 11-bit excess-1023 exponent 64 bits Sign S M S M (a) Single precision (c) Double precision E + E 0 denotes 1 denotes
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3 Examples what about 0? Problem is with this scheme there is no representation of 0 – the bit pattern all 0 is 1x2 -127 , so an exponent field of 00000000 is a special case indicating the number is exactly 0. Exponent field 11111111 – the largest possible exponent – will also be special – it means . & 0 ... 0010111000 . 1 2 2 7 10 1 011 . 1 0 ... 01100 . 0 1 2 2 000 ... 0010111000 10000110 0 2 0010111 . 1 10010111 151 75 . 2 2 375 . 1 2 011 . 1 000 0000000000 0110000000 10000000 0 7 127 134 1 127 128 = + = + = + - - × = = = × = × + =
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4 Range of Floating point numbers 1.111…111x2 127 10 38 1.000…000x2 -126 10 -38 So range of roughly ±10 ±38 What about 1/3? In base 10, exact decimal representation is 0.3333333…… - i.e. infinite repeating decimal!
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5 Repeating fractions… no exact representation! in binary, So best IEEE approximation is 1.0101…011x2 -2 (why 011 at end of 24-bit string?) NB 2/5 has a finite representation in base 10 but not in base 2 – why? Introduces notion of precision ) ! PowerPoint in do to hard too - board (on ... 01010101 . 0 ... 00000000 . 1 11 3 1 =
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6 What about precision? Base 10 example: in an 8-digit system with a 2-digit exponent, how do you count one new person in the U.S.?
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Lecture 19 - ECE52 Spring 11 Lecture 19 1 In binary IEEE...

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