This transfer is possible in two ways: direct transfer and using the decimal system.
first, let\'s make a direct transfer.
let\'s do a direct translation from hexadecimal to binary like this:
C43.9D16 = C 4 3. 9 D = C(=1100) 4(=0100) 3(=0011). 9(=1001) D(=1101) = 110001000011.100111012
answer: C43.9D16 = 110001000011.100111012
now let\'s make the transfer using the decimal system.
let\'s translate to decimal like this:
12∙162+4∙161+3∙160+9∙16-1+13∙16-2 = 12∙256+4∙16+3∙1+9∙0.0625+13∙0.00390625 = 3072+64+3+0.5625+0.05078125 = 3139.6132812510
got It: C43.9D16 =3139.6132812510
Translate the number 3139.6132812510 в binary like this:
the Integer part of the number is divided by the base of the new number system:
3139 | 2 | | | | | | | | | | | |
-3138 | 1569 | 2 | | | | | | | | | | |
1 | -1568 | 784 | 2 | | | | | | | | | |
| 1 | -784 | 392 | 2 | | | | | | | | |
| | 0 | -392 | 196 | 2 | | | | | | | |
| | | 0 | -196 | 98 | 2 | | | | | | |
| | | | 0 | -98 | 49 | 2 | | | | | |
| | | | | 0 | -48 | 24 | 2 | | | | |
| | | | | | 1 | -24 | 12 | 2 | | | |
| | | | | | | 0 | -12 | 6 | 2 | | |
| | | | | | | | 0 | -6 | 3 | 2 | |
| | | | | | | | | 0 | -2 | 1 | |
| | | | | | | | | | 1 | | |
|
the Fractional part of the number is multiplied by the base of the new number system:
|
0. | 61328125*2 |
1 | .22656*2 |
0 | .45313*2 |
0 | .90625*2 |
1 | .8125*2 |
1 | .625*2 |
1 | .25*2 |
0 | .5*2 |
1 | .0*2 |
the result of the conversion was:
3139.6132812510 = 110001000011.100111012
answer: C43.9D16 = 110001000011.100111012