GATE CS 2005 Q85-a — Number Representations (Fixed and Floating Point)

The 16-bit floating-point format consists of two parts:

• Exponent (8 bits): Represented in excess-64 (biased) format.
• Mantissa (8 bits): Represented as a pure fraction in sign-magnitude form (1 sign bit + 7 magnitude bits).

Step 1: Convert the Exponent

The given exponent is $13$. With a bias of $64$, the biased exponent is: $$\text{Biased Exponent} = 13 + 64 = 77$$ Converting $77$ to 8-bit binary: $$77_{10} = 0100\,1101_2 = \text{4D}_{16}$$

Step 2: Convert the Mantissa

The mantissa is $+0.239$. Converting the fractional part to binary: $$\begin{align*} 0.239 \times 2 &= 0.478 \implies 0 \\ 0.478 \times 2 &= 0.956 \implies 0 \\ 0.956 \times 2 &= 1.912 \implies 1 \\ 0.912 \times 2 &= 1.824 \implies 1 \\ 0.824 \times 2 &= 1.648 \implies 1 \\ 0.648 \times 2 &= 1.296 \implies 1 \\ 0.296 \times 2 &= 0.592 \implies 0 \\ 0.592 \times 2 &= 1.184 \implies 1 \end{align*}$$ Thus, $0.239 \approx 0.00111101_2$.

Since the number is positive, the sign bit is $0$. The 8-bit sign-magnitude representation is: $$\text{Mantissa} = 0\,0111101_2 = 0011\,1101_2 = \text{3D}_{16}$$

Step 3: Combine Exponent and Mantissa

Combining the two parts gives: $$\text{4D 3D}_{16}$$