0 votes 0 votes closed with the note: resolved Can someone please help in highlighted part. Thanks Digital Logic floating-point-representation ieee-representation + – tusharp asked Dec 3, 2018 reshown Jan 5, 2019 by tusharp tusharp 829 views comment Share Follow See all 12 Comments See all 12 12 Comments reply srestha commented Dec 3, 2018 reply Follow Share what is problem there? 0 votes 0 votes tusharp commented Dec 3, 2018 reply Follow Share Not getting 2^149 and 2^1074. 0 votes 0 votes Raj Singh 1 commented Dec 16, 2018 reply Follow Share Did you get this? I am also stuck on this. :( 0 votes 0 votes tusharp commented Dec 24, 2018 reply Follow Share Yes. U still need explanation or got it resolved? 0 votes 0 votes Raj Singh 1 commented Dec 24, 2018 i edited by Raj Singh 1 Dec 24, 2018 reply Follow Share Not yet understood :( How those $2^{-126}$ and $2^{-149}$ came? 0 votes 0 votes tusharp commented Dec 25, 2018 reply Follow Share smallest fractional form in 32 bit will be 0.00000.....1(23rd bit). Weight of this bit is 2^-23. Just used fractional form for denormalized case. You will get 2^-149 1 votes 1 votes Raj Singh 1 commented Jan 3, 2019 reply Follow Share I got smallest fractional form in 32 bit will be 0.00000.....1(23rd bit). Weight of this bit is 2^-23. But. I did not get Just used fractional form for denormalized case. You will get 2^-149 I am interpreting it as you added -126 to -23 to get -149. But what is this -126? To get unbiased exponent in normalized numbers, we do $e-b$, where $e$ is biased exponent and $b$ is bias. $b$ in IEEE 754 is 127. So it will be $e-127$ for normalized numbers. I know exponent in denormalized numbers is zero. So if we are to calculate unbiased exponent of denormalized numbers in similar way that in normalized numbers, it would have been $0-127=-127$, not $-126$. I just checked this link. In the table of summary section, it says value of positive denormalized real is $0.f\times 2^{(-b+1)}$, which will correctly give $2^{-149}$ . But, I didnt get from where that $-b\color{red}{+1}$ came. Especially that $+1$. Also isnt denormalized numbers unbiased? Or they indeed are biased, just that their exponent is 0, more precisely a biased zero. 0 votes 0 votes Raj Singh 1 commented Jan 4, 2019 i edited by Raj Singh 1 Jan 6, 2019 reply Follow Share bro super thanks!!!! Maaan you seems to be enlighted one!!!! Did you just come up with that reasoning? Or you found it somewhere? Anyways thanks. 0 votes 0 votes tusharp commented Jan 4, 2019 reply Follow Share http://steve.hollasch.net/cgindex/coding/ieeefloat.html read this properly. 0 votes 0 votes Raj Singh 1 commented Jan 4, 2019 reply Follow Share That link only I put in earlier comment and asked doubt on it: Just tell from where $-b\color{red}{+1}$ came on that page in summary table at the end 0 votes 0 votes tusharp commented Jan 6, 2019 reply Follow Share Denormalized numbers are used to fill the gap between 0 and smallest positive number in normalized form. what is smallest positive number in normalized form? S=0 M =0 and E =1 so it will become 1.0 * 2 ^ -126 Now if number is smaller than this, it will be of form 0.M * 2 ^ -126 Hence when E=0 to make it -126 than +1 is added. i.e E- bias +1 = 0-127+1= -126. 2 votes 2 votes Raj Singh 1 commented Jan 7, 2019 reply Follow Share super thanks 0 votes 0 votes Please log in or register to add a comment.