It only takes a minute to sign up. It can therefore be considered the right definition. The reason is that there are two conventions for ADCs depending on the code transition levels used. The transfer functions of both types are shown below, the dotted lines indicate the range for mid-tread type and the dashed lines are for the mid-riser.
So one LSB is removed from the upper end. What this means is that the voltage you are measuring could be anywhere within that range, so you must decide how you want to represent it - for example you could floor and pick the bottom of the range 1ceil and pick the top of the range 2or pick the mid value 3.
That would give calculations of:. If you refer to page of the datasheetthe equation relating voltage to ADC code is given. Rearranging that into your form, it is:. Sign up to join this community. The best answers are voted up and rise to the top. Home Questions Tags Users Unanswered. Asked 3 years, 7 months ago. Active 1 year, 7 months ago.
Viewed 5k times. Full scale input maplestory kinesis leveling guide Has this anything to do with the location of the first code transition?
Can someone clarify this problem for me? Olin Lathrop k 33 33 gold badges silver badges bronze badges. Don Don 1 1 gold badge 2 2 silver badges 8 8 bronze badges. Active Oldest Votes. Mario Mario 7, 1 1 gold badge 11 11 silver badges 17 17 bronze badges. Tom Carpenter Tom Carpenter Olin Lathrop Olin Lathrop k 33 33 gold badges silver badges bronze badges. Sign up or log in Sign up using Google. Sign up using Facebook.
Sign up using Email and Password. Post as a guest Name. Email Required, but never shown. The Overflow Blog. Featured on Meta. Feedback on Q2 Community Roadmap. Linked 2.We start counting from the Zero position. Zero has a value, the starting value. So from 0 to 10 there are 11 numbers. We start from the LSB powered by the position.
This is a decimal number because it has 10 digits : 0 1 2 3 4 5 6 7 8 9 In binary we have 2 digits: 0 and 1. The first position is 0, the second position is 1, the third position is 2 etc and the base power is number 2 the number of digits.
For example: this can be written omitting the 00 no value at the end of the number. There is no point to take care of them. So, the number is The pattern here is that the next number is double the previous. There are many questions based on sub-netting here. So you must be able to know all these powers of 2.
Before the Cisco exams, be sure to take down all the numbers just in case you need them later — once you enter the class, you write down anything you want. First off, ask yourself: how many bits do I need?
The number is The power indicates the number of bits you need; yes 9 bits from 0 to 8. What this means for us? It means that if I have 4 digits 4 bits that can take 2 values 0 or 1 then the combinations are Using 4 bits I can have 9 combinations. Using 9 bits, I can have combinations. So, to convert to binary I need 9 bit. Every bit has a value. Write the value of its position for 9 bits:.
Pay attention that 9 bits are from 0 to 8. This is only due to the calculation of the combination number of bits. This is how we convert it.Many times this leads to confusion, as a few messages I received from my readers show.
What is an LSB? Both converters are used at the boundaries between the analog and digital realms, making it possible for the analog circuits to talk to the digital ones and backwards.
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The manufacturer writes in the data sheet all the information one needs about that particular component. The ADC needs a voltage reference to convert an analog signal into a digital word. Depending on the number of bits it has, the ADC divides the voltage reference in small levels called counts. For example, if this is an 8-bit ADC, the counts will look like those in Figure 1. Figure 1. The ADC outputs a digital word that shows how many counts are in its input voltage level.
As the ADC counts the input level, it never reaches the voltage reference. Its full scale FS is calculated with the following formula:. Figure 2. As you can see, and ADC can never reach its Vref but, as the number of bits is higher, it gets very close to its reference voltage. The same can be said about a DAC. Moreover, from equation 1we can write the mathematical relation between Vref and LSB as follows:.
This is the trouble, as the LSB has two definitions, equations 1 and 7.Endianness: Bit and Byte Ordering
Both of them are valid, and some authors are ambiguous or confused about them. I have seen articles in which Vref is considered the component full-scale, which is the premise that generates subsequent wrong definitions. Thank you for this post. I needed to convert lsb to fs and searched on google.
Boom, the second article on the first page. Thank you for doing this. Great website also. Naveen and Chris. Thank you for your comments. I am glad you liked this article and the website in general.
I have been looking for clear explanation of this for some time. Use equation 7 in this article to determine the least significant bit. At minimum, you need to know the number of bits the ADC is made for. Usually, the full scale is not 5V. The voltage reference, Vref, is.What is Bitrate?
Understanding the most and least significant bit. In computer science and information theory, a bit is the smallest possible meaningful piece of information. It is most often expressed as a digit of the binary numeral system: either 0 or 1.
A string of 8 bits is called a byte. If we take, for example, the binary number in decimaland send it as a string of data of a network, we can send it in two ways: starting from left to right, or starting right to left.
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In this case, we referred to the first, or left-most bit as the Most Significant Bit msb for short. The msb is the bit in a binary sequence that carries the greatest numerical value. For simpler reference, if we take a look at the equivalent decimal number,the most significant digit is the leading 2. Compared to the other two digits, the leading 2 determines the greatest part of the number's numerical value, as it signifies the hundreds in the number.
Analogous to this, the leading 1 in our binary number is it's most significant bit. The msb can also be used to denote the sign bit of a binary number in it's one's or two's complement notation, with 1 meaning it's a negative number, and 0 meaning it's a positive number. One's compliment of a binary number is obtained by simply inverting all of it digits. In order to calculate a binary number's two's complement, we first determine it's one's complement by swapping it's digits like beforeand then adding 1 to it.
In basic calculus, this two's complement will behave like the negative number of the starting bit sequence, and it enables the use of adders digital circuits used for addition for subtraction. For example: we wish to subtract from First, we expand the number so it has the same amount of digits as the minuend.
Software Engineering Stack Exchange is a question and answer site for professionals, academics, and students working within the systems development life cycle. It only takes a minute to sign up. I have a hard time understanding the use of MSB Most significant bit vs. LSB Least significant bit and why they're used at all. It is my understanding that it has to do with Endianness, but I cannot get my head around it.
Without a specification telling you which side the most significant and least significant decimals were, you wouldn't know whether. Note for additional curiosity that "twenty-one" is MSD whereas "nineteen" is LSD, so it's certainly not that universally obvious even within the same community which one is the right one.
Again, you have to pick some order, and you have to communicate which order you picked, and there is no obviously "right" order. I would argue though, that the US date format is obviously wrong :-D. When a byte is serialized into a stream of bits and transmitted serially, it becomes important to know whether it's transmitted LSbit-first or MSbit-first.
The transmitter can send the bits in either order, and the contract whether it's LSbit-first or MSbit-first is established in the spec or datasheet. For example, the receiver receives this:. If the transmitter was sending LSbit-first, then the value is 0x If the transmitter was sending MSbit-first, then the value is 0x The receive has to know which one it is.
Bit order and byte order are two separate things, but they may affect your preferences. Say you have a little-endian machine, and you want to transmit a bit number. Also transmitting the bits LSB-first means overall bit order for the entire int with lsb numbered 0 is:. Looking at this signal on an oscilloscope just got more difficult, and designing shift registers in hardware just got a little more tricky. The reverse occurs for big-endian machines.
So endianness doesn't dictate bit order, but it sure does make one bit order easier to work with. As far as preferring one endianness over another, it's more than picking one at random. Big endian is easier to understand conceptually, but little endian means you don't have to offset the address in order to treat a byte as a word.
For example, during addition, the carries flow from the addition of two LSB's toward the next higher bit. The LSB itself receives no carry because it starts the addition; whereas all the other bits get a carry from the next least significant bit position. Overflow is when a carry of value 1 happens from the MSB, because there are no more bits in the byte or word size left to carry over to. The MSB is also considered the sign bit for signed data types: if the MSB is 1 the value is negative, if 0 the value is positive or zero.
Endianess is a matter of the storage or transmission order for bits within a byte and also more significantly as it turns out for the order of bytes within a word or long word. These differences can be seen when data is moved from one system to another. Fortunately bit order within a byte is standardized on hard drives and networks, so we don't have to worry about bit endianess.
However, byte order within a word is different for some processors than others, and bytes become swapped when changing architectures; as this causes problems it requires mitigation e.
Little endian stores least significant bytes first followed by more significant bytes in higher storage or packet address order, whereas big endian is the opposite.Remember Me? So how can I do this in minimum?
Because my micro is 1k Flash please give best a solution. If it has the following commands: - shift left - shift right - conditional jump if negative then the best approach would be to make continious left shift of initial operand with conditional jump if the result of shifted operand was negative. After 8 cycles of this algorithm you will get the final result.
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For most uC this algorithm could be implemented in less then 10 instructions. Part and Inventory Search. Welcome to EDABoard. Design Resources. New Posts. Chronology for understanding computer architecture 1. Schottky diode on ADS 2.
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The dark mode beta is finally here. Change your preferences any time. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. All bits must be reversed; that is, this is not endianness-swapping.
NOTE : All algorithms below are in C, but should be portable to your language of choice just don't look at me when they're not as fast :.
Low Memory bit intbit machine from here :. From the famous Bit Twiddling Hacks page :. You can extend this idea to bit int s, or trade off memory for speed assuming your L1 Data Cache is large enoughand reverse 16 bits at a time with a 64K-entry lookup table. If you want to do this on a bit intjust reverse the bits in each byte, and reverse the order of the bytes.
That is:. I benchmarked the two most promising solutions, the lookup table, and bitwise-AND the first one. I used gcc 4.
I tried both approaches at several different optimizations, ran 3 trials at each level, and each trial reversed million random unsigned ints. For the lookup table option, I tried both schemes options 1 and 2 given on the bitwise hacks page. Results are shown below. Use the lookup table, with option 1 byte addressing is unsurprisingly slow if you're concerned about performance. If you need to squeeze every last byte of memory out of your system and you might, if you care about the performance of bit reversalthe optimized versions of the bitwise-AND approach aren't too shabby either.
Yes, I know the benchmark code is a complete hack. Suggestions on how to improve it are more than welcome. Things I know about:. The assembly code is shown below for the former case, reversing bits for two bit int types at a time :.
This thread caught my attention since it deals with a simple problem that requires a lot of work CPU cycles even for a modern CPU. I had to flip millions of bytes.
However I know all my target systems are modern Intel-based so let's start optimizing to the extreme!!! So I used Matt J's lookup code as the base. I then went ahead and tried to see if Intel's ISPC compiler could vectorise the arithmetics in the reverse. I'm not going to bore you with my findings here since I tried a lot to help the compiler find stuff, anyhow I ended up with performance of around 0. It's a great reduction but for my application that's still way way too slow.
The code takes 32 bytes then masks out the nibbles. The high nibble gets shifted right by 4. I could use a single lookup table but then I would have to shift left before ORing the nibbles together again. There are even faster ways of flipping the bits.