Thursday 28 January 2016

Revision

Binary is base 2 number system to represent numbers. Right to left each digit represents


Binary can represent negative numbers there are two way to do this:
Sign and magnitude:
  • Simplest way
  • The most significant bit represents the sign (whether it is positive or negative)
  • 1 means it is negative
  • 0 means it is positive
  • However the MSB means that the largest number for 8 bits that can be represented as 127, as the 8th bit is used for the positive or negative sign.
  • Only represent a 7 bit number
  • It is harder to do calculations as some bits are numbers and some are for signs.
  • 0 is represented twice
  • 127 to -127 range

Represent -4 in 8 bits:
4=00000100
-4= 10000100



Two's Complement:
If in the exam it says it represented in Two's Complement the whole binary number will be wrong.
  • Overcomes the shortcomings of S/M as the whole number range can be represented
  • Write out as a positive number then starting from the right hand side invert bits.
  • The most significant bit becomes a negative number
Represent -90 in binary using 8bits:
90=01011010
-90=10100110

0+0 = 0
1+0 = 1
1+1 = 10
1+1+1= 11

Hexadecimal:
  • 0 to 15 using 0-9 and A to F
  • Represented in 4 bits per hex character
Floating point Binary
  • We use this to represent really small or really big numbers accurately using 8 bits
92=
0.92 x10^2
0.92 is the mantissa this represents significant digits of that number
10 is the base
2 is the exponent (How far the mantissa is to be raised)

01101 011
Where 01101 is the mantissa and 011 is the exponent

Steps to take:
1. Convert the exponent into denary
2. The mantissa started as 0.1101. Move the decimal point 3 places to get 0110.1
3. Write 0110.1 as denary 6.5 (4+2+0.5)

If there is more than one leading 0 this will be classed as an un-normalised number. The decimal goes after the first 0
Shift by the exponent of 3
0110.1
Left bits go by 1,2,4
Right bits go by 1/2, 1/4, 1/8, 1/6


There is a trade off between accuracy and range depending on how many bits are used for the exponent and mantissa.
More bits for the mantissa = more accuracy
More bits for the exponent = more range

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