Normalisation

Question 1

What is meant by a real number?

Answer 1

A number with a fractional part

Question 2

What are the two different ways we can write down a real number in denary?

Answer 2

Simple representation or Scientific Notation

Question 3

Which method is best with dealing in very large/small numbers?

Answer 3

Scientific Notation

Question 4

What is another name for scientific notation?

Answer 4

Exponential Notation

Question 5

Binary code has to store real numbers in computer systems. What are the two possible methods for this?

Answer 5

Fixed Point Representations

Floating Point Representation

Question 6

State what is meant by fixed point representation?

Answer 6

An overall number of bits is chosen with a defined number of bits for the whole number part and the remainder for the fractional part.

Question 7

State what is meant by floating-point representation?

Answer 7

A defined number of bits are used for what is called the mantissa.
The remaining bits are used for the exponent.
The R is the base (2).

+M x RE

Question 8

How does fixed point store a number using fixed point representation?

Answer 8

Most Significant Bit = Sign
5 bits = whole number
2 bits = fraction

Question 9

How does floating-point representation store a number?

Answer 9

select appropriate number of bits for mantissa and exponent in 2s complement

Question 10

How is the mantissa stored?

Answer 10

Fixed point value

Question 11

How is the exponent stored?

Answer 11

Signed integer

Question 12

How many bits does an exponent have?

Answer 12

8-bit (signed) / 127 bits

Question 13

How many bits does the mantissa have?

Answer 13

23-bit fraction

Question 14

What are the steps for converting a denary number using single precision? i.e. 51.7

Answer 14

Take the 51 and divide it by 2. Ensure you are collecting all the remainders.
51/2 = 25 r 1
25/2 = 12 r 1
12/2 = 6 r 0
6/2 = 3 r 0
3/2 = 1 r 1
1 / 2 = 0 r 1
= 110011 (write the binary backwards)
2. Take the .7 and x2
7x2 = 14          1
4x2 = 08          0
Done as we have reached 0

110011.1 is our mantissa
3.Move decimal point to furthest 1…
1.100111 = 5dcp
4. add the dcp to 127 = 132
5. Turn 132 into binary =
10000100 = exponent
6. (sign)01000100.1001110000000000000000000000 (32 BIT) is our answer as we combined the exponent and the mantissa

Question 15

How does the mantissa and the exponent affect precision?

Answer 15

The greater the mantissa bits = the larger the number you can store, but the least accurate it would be.
The greater the exponent number of bits = the smaller the number range you can store, but the more accurate the number would be (more numbers after the DCP)

Question 16

Why use normalisation?

Answer 16

You can have the same value represented in more than one way. This is not good.

Question 17

What are the rules of normalisation?

Answer 17

The binary point always starts with the 1st two digits
Positive numbers = 01
Negative numbers = 10
The mantissa is the number.
The exponent is the position of the binary point.
Positive exponent move right
Negative exponent move left

Question 18

How do we convert a number from denary to floating point using normalisation i.e. 3.5

Answer 18

Convert 3 into binary including flag.
Turn the .5 into a fraction.
011.1
Move the decimal point past the furthest 1.
0.111 (2DCP)
Write out the new number
01110 (mantissa)
010 (exponent) = 2

Question 19

What issues can come from normalisation? such as using 6.75 for an 8-bit number (5 for mantissa) and (3 for exponent)

Answer 19

The overall number will be 6.5
01101 (mantissa)
011 (exponent)
due to the number being too big for the bits available thus reducing accuracy