Converting a number to 16-bit Floating Point Format

Question

I want to convert the number -29.375 to IEEE 745 16-bit floating point format. Here is my solution:

The format of the floating point number is: 1 sign bit unbiased exponent in 4 bits plus a sign bit 10 bits for the mantissa plus the explicit 1

First, I realize that 29.375 = 29 + 3/8.
Then realize that:
29 = 16 +13 = 16 + 8 + 5 = 16 + 8 + 4 + 1 1 1101.011 = 1.1101 0110 * 2^3

This gives us:
mantissa is: 1101 011000
sign: 1
exp: 00011

Hence in 16 bits, we have: 1000111101011000

Do I have this right? I was hoping that there would be a calculator on the web that I could use but I could not find one. They were all at least 32-bits.

Answer 1

First of all, the IEEE 745 16-bit format has:

• 1 bit for the sign.
• 5 bits (not 4) for the biased exponent. The bias is $15$.
• 10 bits for the mantissa. The leading $1$ is implicit (not explicit).

The calculation is incorrect since $(29.375)_2 = (11101.011)_2 = (1.1101011)_2 \cdot 2^4$ (and not $2^3$).

Then:

• The sign bit is $1$ because the number is negative.
• The unbiased exponent is $(4)_{10} = (100)_{2}$ and therefore the biased exponent is $(19)_{10} = (10011)_2$.
• The mantissa, without the implicit leading $1$ and padded to 10 bits with trailing zeros, is $(1101011000)_2$.

Then the final $16$-bit number is: $11001101011000$.