Practice Question

AI Summary

An AI-generated summary of this video lecture.

This GATE-2005 lecture segment teaches how to encode a decimal floating-point number into a custom 16-bit format. The slide defines the layout as bit 15 for the sign, bits 14-8 for an excess-64 exponent, and bits 7-0 for a pure-fraction sign-magnitude mantissa. The problem asks for the hexadecimal representation of 0.239 × 2^13 without normalization or rounding off, with options (A) 0D24, (B) 0D4D, (C) 4D0D, and (D) 4D3D. The instructor writes the value beneath the question, converts the mantissa to binary as .0011110101110, sets the sign bit to 0 for a positive number, and computes the biased exponent as B.E. = T.E + Bias = 13 + 64 = 77. The bits are then grouped into nibbles and converted to hexadecimal, with option (D) 4D3D underlined as the final answer.

Chapters

1. 0:00 – 2:00 00:00-02:00

   The opening slide presents the GATE-2005 floating-point question and format diagram. On-screen labels identify bit 15 as the "Sign bit", bits 14-8 as the "Excess - 64 Exponent", and bits 7-0 as the "Mantissa". The question asks for the hexadecimal representation of "0.239 x 2^13" without normalization or rounding off, listing options (A) 0D24, (B) 0D4D, (C) 4D0D, and (D) 4D3D. The instructor handwrites "0.239 x 2^13" in blue ink beneath the question, emphasizing the given value before conversion.

2. 2:00 – 4:44 02:00-04:44

   The solution proceeds by converting the mantissa 0.239 into binary, shown as ".0011110101110", and setting the sign bit to 0 because the number is positive. The biased exponent is calculated on screen as "B.E = T.E + Bias", then "= 13 + 64", giving "= 77". The resulting sign, exponent, and mantissa bits are placed into the format diagram, grouped into 4-bit nibbles, and converted to hexadecimal. The final answer "4D3D" is written at the top right, and option (D) 4D3D is underlined as correct.

The central concept is encoding a decimal number into a non-standard 16-bit floating-point format. The key method has three steps: first, convert the mantissa fraction to binary without normalization; second, compute the biased exponent by adding the excess-64 bias to the true exponent (13 + 64 = 77); third, concatenate the sign bit, biased exponent bits, and mantissa bits, then group into nibbles for hexadecimal conversion. The underlined phrases "without normalization" and "rounding off" are important constraints, meaning the mantissa is used as-is rather than shifted to a normalized form. The final answer 4D3D follows from the sign bit 0, exponent 77 in binary, and the truncated mantissa bits.