The formula for finding the Exponent is: 2^(k-1) - 1. Since you have 8 exponent bits, replace "k" with 8 in that formula. You get 2^7, which is 128. Subtract 1 from that, and you get 127 as your final answer. :)
We have 8 bits to represent the exponent. We can represent 2^8 = 256 numbers with 8 bits. The bias is considered to be half of it minus 1 so that we can represent both positive and negative exponents.
People with PhDs in computers get swirled into talking about irrelevant and overly complex things very easily. Its hard for them to stick to the point.
Professors get so much better at what they do that they forget what it's like to learn it for the first time or what it's like to not knowing these things, it becomes very obvious to them.
You explained how to do this in less than 10 minutes and were very clear. It took my professor almost 45 minutes and no one understood what was happening. Thank you so much!
I felt like tearing my hair out when this was being taught in my computer architecture and organization classes. Its actually unreal how well you taught this topic as opposed to working professions in some universities. My sincerest thank you
5 years later , who would've thought someone from the other side of the world would find this extremely helpful and saving . thank you so much for doing this
yes yes ofcourse studenst are stuck in university professors lecture which cointain 45 min and no one can understand any of this. Allah kry hamry teachers ko samj ajy ky student ky mind ko kessy prhaty hn we have not good enough teacher but good enogh enough degree personn p.h.d , masters etc
You made such a difficult question so easy! I am linking this video to my professor so people who have difficulty with this format can get help as well! :)
Best explanation ever! I watched several videos and followed many tutorials, and I did not learn how to convert to binary32. Now, Abishalini's excellent method of explaining this makes it very clear. Thank you Abishalini!
I don't actually hesitate to subscribe to such a serious and rich content channel. This lady is very powerful. Am doing computer science at Copperbelt University of Zambia, you're very helpful, keep producing contents more ❤❤
I have to say, I have rarely seen such a splendid, easy to understand explanation on how to convert decimals. Good job on you for making this vid; there's a reason it has so many views.
My man, this is genius - been around PLCs for going on 13 years and never have I fully understood how 32 bit floating point decimal registers work haha. Well done! Wish I could give you double thumbs up :)
for some coding projects it's good to know for example if you have limited space on any device, it's good to know that floating point in this notation only requires 4 bytes (32 bits)
In embedded systems programming, a programmer is often writing code which operates at a low level by dealing with bytes and bits. Suppose you are writing code for a product which employs a microcontroller and you are using some communication protocol, e.g., SPI or I2C, so the microcontroller can communicate with an external device, e.g., a shift register. Suppose we are transmitting the value of a floating point number, e.g., you are programming in C and the data type of the value is double, which is represented in IEEE 754 notation. During testing you discover that the shift register is receiving a value which is different than the value stored in a variable in the microcontroller, i.e., you have a bug in the transmission code. So, you hook up your oscilloscope so you can view the signal being transmitted. The signal should go high when transmitting a 1 bit and it goes low when transmitting a 0 bit. You need to determine where a wrong bit is being transmitted because that will help you locate the bug. Therefore, you need to know the IEEE 754 binary floating point representation of the double value being transmitted so you will know when a 1 or 0 bit should be sent.
Thank you very much. Hard to believe students are becoming better at teaching other student than professors right now. For those wondering about the exponent bits. IEEE exponent - 127 = actual exponent
I have one question for you.I will appreciate it if you will answer it.In this IEEE standard there are two terms called Min exponent and Max exponent.In single precision(32 binary format) values of these terms are -126 and 127 respectively.So we get exponent bias by Max exponent.Could you please tell me what is the use of Min exponent and how is it determined?
The IEEE standard has 4 different concepts to handle. 1. Non infinite normalized numbers 2. Non infinite denormalized numbers 3. positive or negative infinity. 4. Not a number. Used for error indications that propagate so it doesn't have to be checked for every mathematical operation. Notice that the range -126 t0 127 represents a total of 254 distinct values (don't forget the 0). But 8 bits can represent 256 distinct values. The 2 extra values are used to handle infinities and denormalized numbers. Number format s eeeeeeee mmmmmmmmmmmmmmmmmmmmmmm S = s E = eeeeeeee M = mmmmmmmmmmmmmmmmmmmmmmm if E is 00000000, the number is denormalized, value is (-1)^S * 0.M * 2^(-126) 00000001 - 11111110, number is normalized, value is (-1)^S * 1.M * 2^(E-127) 11111111, number is either infinity (M=0), or a NaN (Not a Number) if M 0. Notice that for the denormal numbers, that the most significate digit of the mantissa is 0, while for the normalized numbers, the most significant digit is 1. The most significate digit IS NOT EXPLICITLY STORED. That lets the IEEE standard squeeze in an extra digit of precision. So binary32 is consider to have 24 binary digits of precision, while only storing 23 of those digits. Same for binary64. Considered to have 53 digits of precision, while storing only 52. By the IEEE standard, Min will always have a magnitude one less than Max. So for binary32, we have 8 bits for the exponent. Since we have to account for positive and negative exponents, we have Max = 2^(8-1)-1 = 2^7-1 = 128-1 = 127 Min = -(Max-1) = -(127-1) = -126 For binary64, we use 11 bits for the exponent, so Max = 2^(11-1)-1 = 2^10 - 1 = 1024 -1 = 1023 Min = -(Max-1) = -(1023-1) = -1022 Doing it that way allows for the all zeros or all ones value of the exponent to handle denormal numbers as well as the special value infinity and the Not a Number error indications.
@@TheAntonioclewis I'm willing to help you, but don't require any money. Perhaps this book drive.google.com/file/d/1-nWgEU-RcliZGmYIEQmoy43UocDUhHUQ/view?usp=sharing will help a bit.
I appreciate this video a lot. My professor has a very thick accent, and while I know he really tries his hardest to help us learn the material, sometimes it can be difficult to parse what he's saying. This explained it very well, thank you!
Wow! Thank you so much. The explanation was very clear. One of the best technical videos I've seen on RU-vid. You have a real talent for explaining. Please keep helping us : )
Very nice! Clearly explained and sets the stage for discussing roundoff error (reverse the process and compare the result with the original) as well as for showing oddball bit patterns indicating floating point exceptions, denormal numbers, etc.
Stumbled upon this video out of curiosity but the explanation is so good that I watched the full video, and I'm probably never ever going to need this but the tutorial was so good that I got a fairly good understanding of IEEE 754 anyhow 😅
