Using the formula given at the start of the question, I had to convert the numbers in part (a) to floating point notation.

My answers for part (a) were: (i) 1111.2^11 + 111 (ii) 1111.2^11 + 000 (iii) 0000.2^00 + 000 (iv) 1100.2^00 + 000 (v) 1010.2^11 + 101

But now i have no clue on what to write for 2(b), because I didn't really use a specific formula for finding out the answers for part (a). Any help would be greatly


1 Answer 1


To get the exponent you need to get the index of the highest selected bit of the top 3:

This is the truth table with $B_6$ $B_5$ $B_4$ -> exponent:

1** -> 11
01* -> 10
001 -> 01
000 -> 00

For the mantissa you shift the bits to the right $E$ bits. The underflow lands in T after a $4 - E$ shift to the right.

Or in other words if $E = 3$ then $M$ will take $B_6$ $B_5$ $B_4$ $B_3$

if $E = 2$ then $M$ will take $B_5$ $B_4$ $B_3$ $B_2$

and so one

$T$ will take whatever bits M didn't take.

  • $\begingroup$ Sorry, could you reclarify the the mantissa bit again? I dont understand what youre saying. $\endgroup$
    – James
    Sep 3, 2015 at 12:58
  • $\begingroup$ @james in psuedo code M = B >> E and the under flow from that is T shifted by some bits. $\endgroup$ Sep 3, 2015 at 13:01
  • $\begingroup$ we havent learnt about pseudo code or underflow haha. Could you possibly give an example of what you mean? Thanks $\endgroup$
    – James
    Sep 3, 2015 at 13:06
  • $\begingroup$ Could you use one of the examples already given please $\endgroup$
    – James
    Sep 3, 2015 at 13:20

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