Chapter 1.7 Storing Fractions. Excess Notation, continued… In this notation, "m" indicates the total number of bits. For us (working with 8 bits), it

Chapter 1.7 Storing Fractions

Excess Notation, continued…

In this notation, "m" indicates the total number of bits. For us (working with 8 bits), it would be excess 2^7. To represent a number (positive or negative) in excess 2^7, begin by taking the number in regular binary representation. Then add 2^7 (=128) to that number. For example, 7 would be 128 + 7=135, or 2^7+2^2+2^1+2^0, and, in binary,10000111. We would represent -7 as 128-7=121, and, in binary, 01111001.

Note: Unless you know which representation has been used,

you cannot figure out the value of a number. A number in excess 2^(m-1) is the same as that number

in two's complement with the leftmost bit flipped.

http://www.math.grin.edu/~rebelsky/Courses/152/97F/Readings/student-binary.html#excess

Chapter 1.7

Storing Fractions

Chapter 1.7 Storing Fractions

In contrast to the storage of integers, the storage of a value with a fractional part requires that we store not only the pattern of 0s and 1s representing its binary representation, but also the position of the radix point.

A popular way of doing this is based on scientific notation and is called floating-point notation.

Floating-Point Notation

Let us explain floating-point notation with an example using only one byte of storage.

Although machines normally use much longer patterns, this eight-bit format is representative of actual systems and serves to demonstrate the important concepts without the clutter of long bit patterns.


We first designate the high-order bit of the byte as the sign bit. 0 – nonnegative 1 – negative


Next, we divide the remaining seven bits of the byte into two groups, or fields Exponent field Mantissa field


Let us designate the three bits following the sign bit as the exponent field and the remaining four bits as the mantissa field.


We can explain the meaning of the fields by considering the following example: Suppose a byte contains the bit pattern

01101011


Analyzing this pattern with the preceding format, we see that the sign bit is 0, the exponent is 110, and the mantissa is 1011.


To decode the byte, we first extract the mantissa and place a radix point on its left sideleft side, obtaining .1011


Next, we extract the contents of the exponent field and interpret it as an integer stored using the three-bit excess method.


Thus the pattern in the exponent field in our example represents a positive 2 110 -> 2 excess notation. Remember what the others are?

100 -> 0 etc


The exponent field is +2 This tells us to move the radix in our

solution to the RIGHT by two bits. A positive exponent means to move the

radix to the right A negative exponent means to move the

radix to the left


So we have: .1011 10.11

What is it represent in decimal? 2 ¾ Remember fractions in binary?


Review of fractions in binary:


10.11 The part at the left to the radix point is

10

The part at the right to the radix point is 11

10 -> 2 11 -> 1* ½ + 1* ¼ -> ¾ So the number altogether is 2 ¾


Don’t forget the SIGN BIT In this example, it is 01101011 So, the sign bit is 0 It is nonnegative So, we can conclude that the byte

01101011 represents 2 ¾


1 byte -> 8 bits This need to be memorized


Let’s do another example 10111100

First of all, we extract it Sign bit 1 Exponent 011 Mantissa 1100


Mantissa .1100 Exponent 011 which means -1 move the radix to the left .01100 which represent 3/8 SIGN BIT 1 which means negative


The pattern 10111100 represents – 3/8

Some more examples


To store a value using floating point notation.

We reverse the process.

For example, to encode 1(1/8)


First we express it in binary notation and obtain: 1.001

Because it is positive, so the sign bit is 0 Now we can copy the bit pattern into the

mantissa field from left to right, starting with the left most 1 in the binary representation.

_ _ _ _ 1001


_ _ _ _ 1001 Also we have the sign bit is 0 since it’s

positive number, so we have 0 _ _ _ 1001 Since it is 1.001, .1001->1.001, the radix

point moves to the right one bit. So, we need the 1 at the exponent field

Which is 101


So, we filled out all the fields at this point Number 1 1/8 is 0 101 1001


Note, there is a subtle point you may have missed when filling in the mantissa field. The rule is to copy the bit pattern appearing in the binary representation from left to right, STARTING WITH THE LEFTMOST 1

Example, 3/8 is .011 in binary, in mantissa it will be 1100

Instead of 0110


This is because we fill in the mantissa field starting with the leftmost 1 that appears in the binary representation.

This rule eliminates the possibility of multiple representations for the same value.

This representation is said to be in normalized form.


If we don’t normalize them, for the same number, there can be more than one representation.

For example: 3/8 It can be 000111100 Or 000000110 Etc.


Truncation Error or Round-off Error


Homework #6

Page 60Question 1 a,e 2 b,e 3Due next Thursday.Remember to write your name and student

ID on your homework.

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Chapter 1.7 Storing Fractions. Excess Notation, continued… In this notation, "m" indicates the total number of bits. For us (working with 8 bits), it