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

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

I hope the answer was helpful :)

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