Question

2. Convert 0.0003 to single precision floating point and write the final answer to hexadecimal.

Answer 1

Solni- First we Convert 0.0003 into binary 0.0003 X2 = 0.00061 Q 0.0006 X2 = 0.0012 0 0.0012 X 2 = 0.0024 10 0.0024 X2 = 0.0048 0.0048 X 2 0.0096 0 0.3072 ܥܐ 0.0096 X 2 0.0192 0.0192 X 2 -0.038410 0.038.4 x 2 0.0768 0 0.0768X2 = 0.1536 o 0.1536 X 2 o 0.307 2 X 2 0.6144 0 0.61 44X2 = 1.2288 0.2288X2 3 0.4576 0.4576X2 = 0.9562 o 0.91532 X2 1,8304 0.830 4X2 1.6608 0.6608 X 2 =1.3216 1 053216 X 2= 0.6432 o 0.6432 X 2 1.2864 0.2864 x 2 : 0.5728 1.0 0.9152 1 न 1

0,5728X2=1,1456 0.1456 X 2 = 0.2912 -0.2912X2 = 0.5824 x 2 0.5824 1:1648 D. 1648X 2 = 0.3296 O - OO 2 ܥ 0 ) 0.3296 X2 0,6592 0,6592 X 2 1.3184 0.3184 X 2 = 0.6368 0.6368 X 2 = 1.2736 0.2736X2=0.5472 0,547282 1.0944 0.0944X2 = 0.1888 0.1888X2 2 0. 3776 0.3776X2 0.7552 o O - O Binary: 00000000000100111010100100 lo 1 0 1 0 0 0 - Writing in scientitic notifi notation 0001.00111010100100 TOOTO00 ООООО X look

IEE 754 floating point format X XXXXXXXXXXXX---XX I sign bit exp bits 8 23 fraction bits Mantissa bias-power 127-12 z IIS (115) = (0.1110011), is zero since 0.0003 sign bit is positive Rest 23 bits are copied from bits after decimal in scientific notation Thus required floating point representation in si 32 bits is ☺ 01110.011 00111010100100101010018 = 0011 1001 100111010100 1001 3 Loom Loomino oceno 100 010 0010 5 2 Hexadecimal representation of single precision floating point 3 99 d 4952 z 399 D 4952