Floating-point notation (contd):

The exponent of numbers represented in floating-point notation is represented in excess notation. You may quickly determine the excess notation representation of any number, for any word size, by applying the following procedure. Note that the procedure defined below is for a general word size of k bits where k can be any integer. Note also that below, the phrase "excess binary" refers to excess notation and "binary" to the usual binary notation as discussed in week # 1.

1. Binary to decimal:
   • Convert the "excess binary" bit pattern as if it was a "binary" pattern.
   • Subtract (2^k-1)₁₀ from the resulting decimal number.
2. Decimal to binary:
   • Add (2^k-1)₁₀ to the decimal number.
   • Convert the resulting decimal number using "binary" notation.

Note:

Hence for k=3 subtract or add (2^3-1)₁₀= (2²)₁₀=4₁₀ and for k=4 subtract or add (2^4-1)₁₀= (2³)₁₀=8₁₀ and so on for different values of k.

To understand this procedure, remember that in "excess binary" notation 0₁₀ is represented by the bit pattern for the decimal number (2^k-1)₁₀. That is, for a 3 bit system, as in fig 1.23 of the text (some copies of the text have a printing error), 0₁₀ is represented by the bit pattern 100 which, if interpreted using "binary" notation, is (2^3-1)₁₀= (2²)₁₀=4₁₀. Similarly, for a 4 bit system, as in fig 1.22 of the text, 0₁₀ is represented by the bit pattern 1000 which, if interpreted using "binary" notation is (2^4-1)₁₀= (2³)₁₀=8₁₀. It is because of this why the author of your text refers to these systems as excess four and excess eight notation respectively.

e.g.

We will use simple examples to illustrate. That is, we will use words of 3 and 4 bits (k=3 and k=4). Notice that this procedure really only becomes useful for values of k greater than 4.

1. Let k=3. Convert the "excess binary" patterns 111, 101, 000, 011 to decimal. Use fig 1.23 to verify.
   • Note that 111 interpreted in "binary" is 7₁₀. Since we subtract 4₁₀ then the the answer is 7₁₀ - 4₁₀ = 3₁₀.
   • Note that 101 interpreted in "binary" is 5₁₀. Since we subtract 4₁₀ then the the answer is 5₁₀ - 4₁₀ = 1₁₀.
   • Note that 000 interpreted in "binary" is 0₁₀. Since we subtract 4₁₀ then the the answer is 0₁₀ - 4₁₀ = -4₁₀.
   • Note that 011 interpreted in "binary" is 3₁₀. Since we subtract 4₁₀ then the the answer is 3₁₀ - 4₁₀ = -1₁₀.
2. Let k=4. Convert the "excess binary" patterns 1110, 0010 to decimal. Use fig 1.22 to verify.
   • Note that 1110 interpreted in "binary" is 14₁₀. Since we subtract 8₁₀ then the the answer is 14₁₀ - 8₁₀ = 6₁₀.
   • Note that 0010 interpreted in "binary" is 2₁₀. Since we subtract 8₁₀ then the the answer is 2₁₀ - 8₁₀ = -6₁₀.
3. Let k=3. Convert the decimal numbers -3₁₀, 1₁₀ to "excess binary" notation. Use fig 1.23 to verify.
   • We first add 4₁₀ to -3₁₀. Hence -3₁₀ + 4₁₀ = 1₁₀. The required pattern is 001.
   • We first add 4₁₀ to 1₁₀. Hence 1₁₀ + 4₁₀ = 5₁₀. The required pattern is 101.
4. Let k=4. Convert the decimal numbers -3₁₀, -7₁₀ to "excess binary" notation. Use fig 1.22 to verify.
   • We first add 8₁₀ to -3₁₀. Hence -3₁₀ + 8₁₀ = 5₁₀. The required pattern is 0101.
   • We first add 8₁₀ to -7₁₀. Hence -7₁₀ + 8₁₀ = 1₁₀. The required pattern is 0001.

Representing text:

See page 36 of the text.

Representing images:

See page 42 of the text.

Communication errors:

We will only cover Parity Bit schemes. See page 66 of the text.