Question

Question 18 Assurning IEEE 754 single-precision floating point number representation, calculate the floating point number the following bit pattern represent. Show your work to get credit. 1100 0000 0011 0000 0000 0000 0000 0000 Ara 3117 TTTT Paragraph X DOO f Mashup BUSS Path: Click Submit to complete this assessment.

Answer 1

The Given iT 100 0000 000 0000 0000 0000 0000 0000 sakits D. The The IEEE 754 has 3 Parts. Sign 2) The Biased exponent 3) The fraction Part. The 32 bits are divided into these 3 Parts. L 32 a3 bits is bits fign Exponent Mantissa C-1bit- C 8 bits The sign bit which occupies I bit. If it is o then the Sign Positive, if it is I then it is negative. The biased exponent part occupies & bits. As the exponent can be both positive and negative and with a capacity of & bits it Ranges from 127 to +127. So, we need to add Ianto actual bias to get the biased exponento

The normalised mantissa (cu fractional Part which occupies a3 bits. It is just binar stepti Sign 100 0000 0011 0000 0000 0000 0000 0000 The Sign bit is 1. so, the number is negative steps: Exponent 1,100 0000. doll 0000 0000 0000 0000 0000 The sign bits are 100000008 We have to Convert this into decimal (128), 11x2=128 bias=127. we fing, exponent = 128-bias 10 (10000000) 2 128-ban exponent al ie=1

step 3: fractional Parti 1 100 0000 0011 0000 0000 0000 0000 0000 This is Fractional Port Since it is floating point, we have to convert it into decimal form by multiply with negative Powers of a. Mondiora = (oxo vixz4(")+Cow) --- 0+ 0.25 +0.125to. = O. 375 ma stepui final part S = Sign = 1 e = exponenti mantissa = 0.375 We have to apply these in the following formula. (-) X (1+m) x2 (-) x (1 + 0:31) x 1' = -1 < (131) x 2 5)1-1.375x2 => -2.75 e