Worksheet

A7.3 Fixed point and floating point

Data representation · A level · OCR H446 1.4.1, AQA 7517 4.5.4.4, Eduqas A500QS 2.3 · about 35 min

What this lesson is about

Binary fractions, mantissa and exponent, normalisation, and adding floating point numbers.

• Binary fractions
• Fixed point
• Floating point: mantissa and exponent
• Normalisation
• Adding and subtracting floating point
• The robot's view

Questions 6 marks in all

1. [1 mark]An unsigned fixed point byte has 4 bits before the binary point and 4 after. What is the value of 10110110?

2. [1 mark]A floating point number has an 8-bit two's complement mantissa 01010000 and a 4-bit two's complement exponent 0011. What is its denary value?

3. [1 mark]A floating point number has an 8-bit two's complement mantissa 10100000 and a 4-bit two's complement exponent 0001. What is its denary value?

4. [1 mark]Which of these 8-bit two's complement mantissas are normalised?

   Tick every answer that is true.

   1. A01101000
   2. B00110000
   3. C10110000
   4. D11010000

5. [1 mark]Normalise the floating point number with 8-bit mantissa 00011000 and 4-bit exponent 0011. Give the mantissa, a space, then the exponent.

6. [1 mark]Why are floating point numbers normalised?

   Tick every answer that is true.

   1. AIt gives the greatest precision for the number of mantissa bits
   2. BEach value has only one representation
   3. CIt increases the range of the exponent
   4. DIt stops rounding errors completely

The task: decode a float

Write two functions. - float_value(mantissa, exponent): mantissa is a string of 8 characters 0 or 1, a two's complement fixed point number with the binary point after the first bit (place values -1, ½, ¼ down to 1/128). exponent is a string of 4 characters 0 or 1, a two's complement integer from -8 to 7. It returns the value, mantissa × 2 to the power of exponent, as a float. - is_normalised(mantissa): returns True when the first two bits of the mantissa are different, otherwise False. The loop prints each line as 01101000 0011 = 6.5 normalised or 00011000 0101 = 6.0 not normalised, with the value exactly as Python prints the float.

# the two lines every program starts with: the commands, then the robot
from bugbot import *
connect()

def float_value(mantissa, exponent):
    return 0.0

def is_normalised(mantissa):
    return False

for mantissa, exponent in [("01101000", "0011"), ("10110000", "0010"), ("00011000", "0101"), ("11100000", "1111")]:
    word = "normalised" if is_normalised(mantissa) else "not normalised"
    print(mantissa, exponent, "=", float_value(mantissa, exponent), word)

Plan your program here, then type it in and press Run.

Do it on the robot
www.bugbotlab.com/learn/a7-3-fixed-and-floating-point/
The simulator checks it and tells you when it passes. Nothing to install, no account.

Challenges

1. Write normalise(mantissa, exponent) that returns the normalised pair of strings.
2. Store -0.375 in normalised form with an 8-bit mantissa and 4-bit exponent.
3. What is the largest positive value this 12-bit format can hold, and what is the smallest positive normalised value?