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Answer to GPK Question ________________________________ Consider a 16 bit register to store floating point numbers. The mantissa is an implicitly normalized signed fraction number. Exponent is represented in excess-64 form. What is the 16-bit value for -27.625 in this register? The given register has 16 bits. 1 bit is always used for sign (S). Now we have 15 bits left. E is represented in excess-64 form. i.e. Bias = 64 2^k-1 = 64 2^k-1 = 2^6 i.e. k-1 = 6 k = 6 + 1 = 7 Here k = no. of bits used for storing the exponent = 7 Out of the total 16 bits we use 1 for Sign 7 for Exponent and thr left (16 - (7+8)), i.e. 8 for Mantissa. The given no. is + 27.625 [Sign bit is 0] => 27 => 11011 => .625 => .101 => 27.625 => 11011.101 Now we normalize it implicitly (there must be a bit 1 before the binary point)... 11011.101 1.1011101 * 2^4 [Notice that we have our e = 4 we have our Mantissa = 1011101] Lets get our biased exponent [E]... E = e + Bias E = 4 + 64 = 68 Binary form of E = 1000100 Now we have S = 1 E = 1000100 M = 1011101 Since we have 8 bits for M we need to pad it by adding a spare 0 just after the LSB Padded M = 10111010 At last our final bit sequence for representing -25.625 in conventional floating point representation => S = 1 E = 1000100 M = 10111010 1100010010111010 or 0xC4BA {In Hex}
Awesome teaching style sir....!! 👌👌👌👌👌.... I feel very lucky that I am studying from you sir ... thank you so much sir for this quality content for free and for helping student like us....😇😇😇😇
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@@MrSagarBiswas yeah i see...the minimum positive value that sir wrote comes out to be 0.1x(2^-32) but why do we say it as minimum value even though we CAN in fact represent the value 0.000000001×(2^-32) in that register?
GPK (HomeWork) 1:14:41 Answer => [ 1 | 1000100 | 10111010 ] => C4BA => 0xC4BA Max number can be stored = +511*2^(55) Min number can be stored = -511*2^(55) Min +ve number can be stored = +2^(-64) Max -ve number can be stored = -2^(-64)
@MrSagarBiswas Thank you for this comment. It inspired me to answer all corollary questions and that helped me in my understanding of the taught concept.
I got min +ve +(256*2^(-72)) In the formula I took (-1)^0 * 1.00000000 * 2^(0-64) As it is implicit normalization I took 1.M Could you help me out with this
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And of GPKBias is 64, K=7, e=6 ,E=70, since no is -ve s=1. ...Answer calculated is 1 (s bit) 1001010(Exponent 7 bits). 11011101(Mantissa 8bits) Hex equivalent is- 0xCADD
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GPK ans is 0xC4BA || exponent is 4+64 and mantissa is 1011101
Mantissa is going to be of 8 bits. 10111010 is the correct mantissa.
0xC6BA is answer
Ans is 0xC5DD mentisaa 11011101
@@Nikhilmod440 bhai gate 2025 ki tyari kr rha hai? Break mtt lena bss agar kr rhe ho toh...mera hogya clear gate 2024
@@dashingcricket accha bhai
Aap kis year se taiyari kar rhe hai??
Amazing teaching
U explained everything about floating point representation
You have explained this in easiest way possible thank you for this amazing lecture 😊
Suno vo k-1 krne par value jyada kaise ho gyi 2 ki power 5-1 =6 kaise hua please btado
maybe we do not watch it live but we will definitely see this on our own time...keep up
Amazing
Start -> 07:38
thnks brother
Answer to GPK Question
________________________________
Consider a 16 bit register to store floating point numbers. The mantissa is an implicitly normalized signed fraction number. Exponent is represented in excess-64 form. What is the 16-bit value for -27.625 in this register?
The given register has 16 bits.
1 bit is always used for sign (S).
Now we have 15 bits left.
E is represented in excess-64 form.
i.e. Bias = 64
2^k-1 = 64
2^k-1 = 2^6
i.e. k-1 = 6
k = 6 + 1 = 7
Here k = no. of bits used for storing the exponent = 7
Out of the total 16 bits we use
1 for Sign
7 for Exponent and
thr left (16 - (7+8)), i.e. 8 for Mantissa.
The given no. is + 27.625 [Sign bit is 0]
=> 27 => 11011
=> .625 => .101
=> 27.625 => 11011.101
Now we normalize it implicitly (there must be a bit 1 before the binary point)...
11011.101
1.1011101 * 2^4
[Notice that
we have our e = 4
we have our Mantissa = 1011101]
Lets get our biased exponent [E]...
E = e + Bias
E = 4 + 64 = 68
Binary form of E = 1000100
Now we have
S = 1
E = 1000100
M = 1011101
Since we have 8 bits for M we need to pad it by adding a spare 0 just after the LSB
Padded M = 10111010
At last our final bit sequence for representing -25.625 in conventional floating point representation =>
S = 1
E = 1000100
M = 10111010
1100010010111010
or
0xC4BA {In Hex}
thenks
your teaching is just amazing sir
7:42
Awesome teaching style sir....!! 👌👌👌👌👌.... I feel very lucky that I am studying from you sir ... thank you so much sir for this quality content for free and for helping student like us....😇😇😇😇
Suno vo k-1 krne par value jyada kaise ho gyi 2 ki power 5-1 =6 kaise hua please btado
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Sir i m literally enjoying your lecture
Sir, your teaching is awesome and it can be fully understandable
1:07:41 - why did he take mantissa 100 000 000 instead of 000 000 001?
Because mantissa is numbers after the point. So, 0.1 => 0.100000000 and here 100000000 is mantissa.
@@MrSagarBiswas yeah i see...the minimum positive value that sir wrote comes out to be 0.1x(2^-32)
but why do we say it as minimum value even though we CAN in fact represent the value 0.000000001×(2^-32) in that register?
please anyone tell why we write 2 raised to power 64-32, 32 is bias value but How E value is
64 here please tell
How to find exponent and matissa and bias value if only size of a register is given?
Amazing lecture🙏🙏
Video starts at 7:41
Mantissa is normalised "Signed" fraction number. Since mantissa just refers to number after the point, What does signed mean here ?
GPK Ans is 0xC4BA || E=68 and M=10111010
Implicit hai bhai
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ans of GPK - (C4BA) base 16
u have to do implicit normalization bro
Your teaching is awesome
GPK (HomeWork) 1:14:41 Answer => [ 1 | 1000100 | 10111010 ] => C4BA => 0xC4BA
Max number can be stored = +511*2^(55)
Min number can be stored = -511*2^(55)
Min +ve number can be stored = +2^(-64)
Max -ve number can be stored = -2^(-64)
bro how will u store 7 bits in a 6 bit sub register ??
in ur exponent part it should be of 6 bits right?? but urs is of 7 bits
@@SayanjitMukherjee we can't store 7 bits in 6 bit register
@MrSagarBiswas Thank you for this comment. It inspired me to answer all corollary questions and that helped me in my understanding of the taught concept.
I got min +ve +(256*2^(-72))
In the formula I took
(-1)^0 * 1.00000000 * 2^(0-64)
As it is implicit normalization
I took 1.M
Could you help me out with this
@@tarun_gudipalli According to the values placed by you in the formula , the minimum positive value should turn out to be + (2^(-64)).
Sab samajh ane laga . Love you sir
your are the best sir
for coa
35:33 kaise sir multiply karne par aayega
Can someone please provide the links to the prerequisite classes Sir is talking about?
type coa pre requisite on youtube you will get it youtube doesn't allow to comments links
@@ritvikreddy3959 thanks
Class start 7:42
target 2024
What is the formula that used to find the range of biased ?
-2^n to (2^n )-1 as one bit used for sign , n=4 so -16 to 15
Pdf kaha milega Koi bata do please
Awasome...Thanks alottt sir!
GPK Answer will be (C5DD)16 right??
yeaa
But I am getting C4BA
i m getting C8DD
@@sushmitaghosh6917 Iam also getting OXC4BA
Amazing lecture
GPK ans =: (C4BA) base 16
C5DD Hoga na
@@agnidebmukherjee9756
Nahi C4BA hi aayega
@@harygaming.8805 pls Bhai explain karo na kaise..
@@harygaming.8805 bhai mera toh C4B2 aya hai.. MANTISAA MERA 10110010 aya
@@replyingtomemeansyourstupi8996 bro check onces again na
Bias=7, S=1, E= 1000100, M=10111010
0.625 ka binary .101 check karioo to
link to the prerequisite class anyone?
he looks as well as behave like Sandeep Maheshwari ❤️🥺 love u sir
Totally Interesting Sir!
Can you solve this question 11*11 multiplication booth algorithm ?
Sir You Are GREAT 🙌🔥
Awesome lecture sir 🔥🔥😌
Thankyou for make these amazing tutorials on a complex subject these all videos all so helpful for us
I'm persuing diploma in I.T. but I use to see your video sir. Excellent way of explaination 💯👍
GPK ANS: 0x265D ?? Please confirm
Great way of explanation
❤️❤️
this was tough ngl ,appreacite your hardwork
p+2
Lecture 2 completed. Amazing section
Sir can u get more question on the concept of denormaliseing floating point value
Sir I am preparing for nimcet 2023 can I use this uncademy app to get note of computer awareness topic
Is in 37 is 100101. In binary formate
Sir please tell me that bias is 15 for 5 bit exponent or not?
Hn bhai mujhe bhi doubt h ye btao n kya hoga iska meine pta kia to 15 hi aarha h
51:45
Great teacher💖💖💖💖💖💖
Why E is taken 63 when solving the max value representation on that 16 bit register
Sir ye slides bhi provide kr dijiye please.
Started at 7:20
Homework Ans:- C4B8
is the full course hindi. I know only english can i learn from this?
no
Thanks for video
thanks sir 💌
Excellent sir
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Sir can i get more questions on explicit and implicit normalisation with solutions so I can check my answer also
59:17 😂😂😂😂
1100010010111010
Sign bit:- 1
Exponent bit:- 1000100
Mantissa bit:- 10111010
hexadecimal form:- 0xC4BA
Nicely explain ❤️🙂❤️
Thanks sir. Nice video
Mantissa 9-bit kaise laye sir
Great ...
start at 7:42
Thank you so much sir
Great sir 👍
Sir can I get more question on this topic to solve
15.62 ka normalized form?...
Thank you sir !
readyyyy!!!!
ausome sir jii
E is 37 and 37 in binary os is 100101 but you have written 100111 which equals to 39
Arre behan poora lecture dekh lo correct kr di thi sir ne
Tqsm sir
63-32 kaise aaya sir 1:04:18
I have also same doubt
Thank you sir
Sir, 2^9=512?
Yes
Ans is 0.x (C4BA) s =1 and M = 10111010 and E=68!
sorry sir but there is a mistake 37binray will be 100101
How is 9 times 1 equal to 511
And of GPKBias is 64, K=7, e=6 ,E=70, since no is -ve s=1. ...Answer calculated is 1 (s bit) 1001010(Exponent 7 bits). 11011101(Mantissa 8bits)
Hex equivalent is- 0xCADD
e=5
1:10:19
Starting time 7.42
Ans is 0xC4BA
❤
❤️❤️❤️
Nice
homework question answer bta do?
Ur looking like Sandeep Maheshwari 🤭
bhul gaya hi sir dobara playlist dekh rha 2x me 😂
the answer is:1|1000101|11011101
Correct
Answer of GPK Question is--> 1000101110111010 💯
vro tu bahut achha padhata hai, lekin unacademy nhi lunga hue hue
1:00:00 tumhara kuch ni ho skta/..... hahahahahahhahah
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