This is the format in 5.19 Convertthefollowingbinaryandhexadecimalnumberstooatin

Question

This is the format in 5.19 Convertthefollowingbinaryandhexadecimalnumberstooatingpointformat. Assume a binary format consisting of a sign bit (negative = 1), a base 2, 8-bit, excess-128 exponent, and 23 bits of mantissa, with the implied binary point to the right of the rst bit of the mantissa. t

1. Convert the following to the floating point representation using the format in 5.19 of your text: 12526588.125

2. Convert the following from the floating point representation to a decimal (base 10) representation using the format in 5.19 of your text: 10010110010010100011011010101011

please i need all the solutions.

Explanation / Answer

1.
So we have decimal (base 10)12526588.125

To start with we convert the whole number part to its binary counterpart:
12526588 = 101111110010001111111100

next we convert the fractional part:
0.125 =
0.125 * 2 = 0.25   0
0.25 * 2 = 0.5    0
0.5   * 2 = 1      1
so 0.125 = 0.001

therefore 12526588.125 = 101111110010001111111100.001
= 1.01111110010001111111100001 * 2^(23)10

So
Signbit = 0
exponent = (23 + 128) = (151) = 10010111
mantisa = 01111110010001111111100

So in binary: 0 10010111 01111110010001111111100

2.
from 10010110010010100011011010101011 we get

sign = 1
exponent = 00101100 (44)
mantisa = 10010100011011010101011

So actual exponet = 44 - 128 = -84

the number is:- 1.10010100011011010101011 * 2^-84
= 1 + 0.5 + 2^-4 ... (and so on)
= 1.57979333400726318359375 * 2^-84

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