topic: normalised floating point

Question 1

using normalised floating point binary representation using 8 bits for the matisssa and 4 for the exponent, represent the denary value -6.5

(i wrote the full detailed steps on the answer side)

Answer 1

1. write num in 2s complement - by writing it normally then flipping all the binary nums except the first one in the 1/2 column (0110.1 -> 1001.1)
2. move the point so its between the 01 (1.0011)
3. the exponent is how many places the point has moved, since it moves 3 to the left the exponent is positive
4. write 3 using 2s complement (0011)
5. put it together, making sure the mantissa has 8 bits by fiilling up the right side with 0s and write the exponent next to mantissa (10011000 0011)

Question 2

Convert the following number from denary to normalised floating point binary (answer to be written as x.xxxxxxx xxxx e.g. 0.1000000 0000) -13.25

Answer 2

1.0010110 0100

Question 3

Convert the following numbers from denary to normalised floating point binary (8 bits for the mantissa, 4 bits for exponent)
13.25

Answer 3

0.1101010 0100

Question 4

The part of the floating point number that represents the significant digits of that number.

Answer 4

mantissa

Question 5

The power to which the number in the mantissa is to be raised.

Answer 5

exponent

Question 6

Convert the following numbers from floating point binary to denary.
1.0101101 0101

Answer 6

-20.75

Question 7

why are floating point numbers normalised (2)

Answer 7

1. there will be a unique representation for a number, the format will ensure the number is represented with the greatest possible accuracy/ precision
2. and multiplication is performed more accurately

Question 8

explain trade off between mantissa and exponent

Answer 8

the choice will affect size/ range and accuracy
more bits for mantissa means a greater accuracy
however more bits for exponent means greater range and size in the number

Question 9

Show the denary number −5.25 in floating point binary form representing the mantissa and exponent in two’s complement, using as few bits as possible. Show your working.

Answer 9

In fixed point is 1010.11
Mantissa becomes 1.01011
Exponent of 3 / 11
Giving answer of 101011 011

Question 10

convert -13.25 from denary to normalised floating point binary

Answer 10

1.0010110 0100