Chapter 2.4, Problem 10QE

Explanation of Solution

Determine the addition of two binary numbers if the pattern represents values stored in two’s complement notation:

• In two’s complement system, the leftmost bit of pattern represents the sign of the value. Negative values are indicated by the binary number 1, while positive values are indicated by the binary number 0.
• Similar to the binary addition, the addition in two’s complement system is done by adding both the numbers, but the carry is truncated in this case so that the number of bits in all the patterns including the answer is same.
• The addition of 01101010 and 11001100 is:

  $\begin{array}{l}01101010\\ \underset{\_}{11001100}\\ 100110110\end{array}$

• Since, the number of bits in answer is 9 and number of bits in input values is 8, so the carry part is truncated from the sum and the result is reduced to 00110110.

Therefore, the sum of 01101010 and 11001100 is “00110110”.

Determine the addition of two binary numbers if the pattern represents values stored in floating-point notation:

• Floating-point notation is the method in which numbers are represented as decimal fractions and exponents.
• In floating point notation, the number consists of only 8 bits. First bit from the left is assigned as sign bit, next three bits are assigned as exponents and the last four bits are assigned as mantissa. Negative values are indicated by the binary number 1, while positive values are indicated by the binary number 0.
• The procedure to perform addition in floating-point notation is:
  • Convert both the binary numbers in floating point numbers.
  • Add both the numbers.
  • Convert the result into binary notation.
• Decode the bit pattern 01101010 in floating point notation. The exponent field is 110 and it represents 2 according to the excess notation system for 3 bits.
  • Extract the mantissa to obtain .1010
  • Move the radix 2 bit to the right, since the exponent represents +2. So, the value is 10.10.
  • Convert 10.10 to base 10 representations.
    • First consider the integer part 10,

      $\begin{array}{c}1\times 2^1+0\times 2^0=2+0\\ =2\end{array}$

    • Now consider the fractional part .10,

      $\begin{array}{c}1\times \frac{1}{2}+0\times \frac{1}{2^2}=\frac{1}{2}+0\\ =\frac{1}{2}\end{array}$

  • Combine integer and fractional part results to obtain the base 10 value of 10.10. The value is $2\frac{1}{2}$.
• Decode the bit pattern 11001100 in floating point notation. The exponent field is 100 and it represents 0 according to the excess notation system for 3 bits.
  • Extract the mantissa to obtain .1100
  • The radix will not move in any direction since the exponent represents 0. So, the value is .1100.
  • Convert .1100 to base 10 representations.

    $\begin{array}{c}1\times \frac{1}{2}+1\times \frac{1}{2^2}+0\times \frac{1}{2^3}+0\times \frac{1}{2^4}=\frac{1}{2}+\frac{1}{4}+0+0\\ =\frac{3}{4}\end{array}$

  • The base 10 value is $-\frac{3}{4}$