how does the number of mantissa bits are related to significant decimal digits after decimal point ? and also how are we loosing the precision even though number of bits are 23 ? do you mean precision of the 32 bit floating point is less than 32 fixed point ?
What is we won't add biased and instead of that we convert the exponent of a number in its binary form and then represent it in the exponent bit of single/ double precision from right to left like 100 if we'll represnt this in exponent part then it will be 0000 0100? Will this be correct if we won't find it's biased ?
It’s a rule To calculate the exponent For single : add 127 to the numbers of bits in mantisa after removing one bit for sign For double: add 1023 in the same way Notice that number of bits in double must be 11 while single is 8
Floating point representation 11001000110100011110100000000000, a. - 0.1101000111101 x 2^9 b.- 0.1101000111101 x 2^ 8 c.- 0.1101000111101 x2^7 d.- 0.1101000111101 x 2^6 Can you figure and explain how to reduce the exponent . THANK YU VERY MUCH
After binary point, the weight reduces by factor of 2. So first 1 will have weight 1/2. Then next digit will have weight 1/4, then 1/8 and then 1/16. So here if you add the weights where there is 1, the sum would be 0.5 + 0. 125=0.625. I hope, it will clear your doubt.
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ngl you just taught me my profs whole weeks lecture, in 22 mins. Genuinely appreciate it!
Actually, real explanation and summarizing. 👍
Example 1 really helped me out, thank you so much
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before commenting like this compare your concentration levels while listening to UA-cam videos and listening to college lecture
That is true
@@anithathota2310 nope, she's 100% right
Amazing explanation every time 🎉
Very nice as usual 🎉🎉
so good explaination, If only clg prof taught like this !
how does the number of mantissa bits are related to significant decimal digits after decimal point ? and also how are we loosing the precision even though number of bits are 23 ? do you mean precision of the 32 bit floating point is less than 32 fixed point ?
Please do a video on multiple sequence detecting fsm in one Sequence
Man u are great, thank god for you and godbless you
What is we won't add biased and instead of that we convert the exponent of a number in its binary form and then represent it in the exponent bit of single/ double precision from right to left like 100 if we'll represnt this in exponent part then it will be 0000 0100? Will this be correct if we won't find it's biased ?
Very Nice Continue 👍👍👍👌👌👌
Nice explanation bruh 🎉
Hello man, welcome to you.
Woooow amazing really thank u are u also expline Mips datapath R, I, J type
Please someone tell me why he add(+) 127 why he subtract ?
It’s a rule To calculate the exponent
For single : add 127 to the numbers of bits in mantisa after removing one bit for sign
For double: add 1023 in the same way
Notice that number of bits in double must be 11 while single is 8
11 and 8 bits .. I mean for exponent
To make it clear
Floating point representation 11001000110100011110100000000000,
a. - 0.1101000111101 x 2^9
b.- 0.1101000111101 x 2^ 8
c.- 0.1101000111101 x2^7
d.- 0.1101000111101 x 2^6
Can you figure and explain how to reduce the exponent . THANK YU VERY MUCH
Thanks!
The Scientific Notations
How .75
How 1010 is .625
😂
After binary point, the weight reduces by factor of 2. So first 1 will have weight 1/2. Then next digit will have weight 1/4, then 1/8 and then 1/16. So here if you add the weights where there is 1, the sum would be 0.5 + 0. 125=0.625.
I hope, it will clear your doubt.
♥️♥️❤️🙏