This question deals with floating-point arithmetic. Assume that for our floating

Question

This question deals with floating-point arithmetic. Assume that for our floating- point representation here, we have a ceg3156-precision as shown in figure Sign Exponent Mantissa 12 Figure 1: Ceg3156-precision floating-point number representation Here are a couple of notes to keep in mind while working with ceg3156- precision Sign 1-bit wide, and used to denote the sign of the number (0 signifies and 1 signifies -) Exponent 5-bit signed exponent in excess-15 representation; Mantissa 6-bit fractional component.

Explanation / Answer

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The given number is : -1023.6666

Our number is negative therefore sign bit = 1.

Now converting number to binary

1023 = 01111111111

.666=.1010101001111110111110011101101100100010111

Then 1023.666 = 1111111111.101010100111111…..

Now shifting decimal to first 1 bit

=1.111111111101010100111111… * 2^9.

Now for 5 bit exponent, the value of exponent will 16 + 9 = 25.

25 = 110011

And taking first six digits of mantissa ( number after decimal point) as we are given 6 bit representation.

The representation is

1 11001 111111.

The reprepresentat is not precise as the number if bits are low to represent this big number.

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