LabVIEW
Writing data of HEX into EEPROM adresses with particular storage format
Solved!
@shaik89 wrote:
Hello All,
Can anyone help me how to write floating point number with specific storage format and output as HEX i need, i will be writing the data 16 bit data
For example, I have a floating point number -177.609375 and storage format is s8.7 then i suppose to get HEX A732 .
Regards,
Nafees
First off I'm not exactly sure what your "specific storage format" is. Never heard of s8.7, but I looked at your code and I think I understand what the data format is.
Speaking about looking at your code, it is VERY obtuse. You have a main VI that calls 4 subVIs. You have multiple For Loops, multiple case structures. You convert back and forth from U16 to floating point to String. Your algorithm can be greatly simplified if you work directly with the bits. I got it done in a simple block diagram. I'm sure my algorithm can be further optimized.
You only have 1 exemplar for your format: 0xA732 => -177.609375. This block diagram works for this one conversion but I think it will work on others. Try it out and VERIFY it works correctly for other hex numbers BEFORE you use it.
Now on to your next problem: going from Floating Point to Hex. This one was a little trickier. To work properly, you MUST truncate the U16 number (Round Towards -Infinity) to avoid any carry over of fractional part to the whole number part. However, I'm assuming you would like to "Round to Nearest" value. So to get around the truncate requirement, I added an offset of 1/2 of the smallest fractional step size less a small epsilon before the Round Towards -Infinity.
This is the inverse function of the above algorithm. This will work ONLY if my assumptions of the above Hex_2_Float algorithm are correct.
If you can't read 2017 version, I attached a 2014 VI version.
OK that spec made a lot of sense. I also over complicated the algorithm. It is really simple!!! BUT you need to know what format spec you are dealing with ahead of time, hence the "Format Spec" Enum Control. Here is what I have so far for the HEX to Float Conversion. I need to sign off. Will look at the Inverse Float to Hex later.
A little nugget. You can get rid of the reciprocal function and change the input to the Compound Arithmetic function to Invert. Both results are equivalent. Just looks cleaner.
@shaik89 wrote:
Hello jamiva,
the above logic does't work for hex FDCA and format specifier 0.16 * e power-2 i suppose to get value as 0.247840881348 valid response according to the document i attached but am getting value -0.00215911865234375 this is invalid response(in your logic i here changed f as 16 and e as -2) , when you see my main vi you can give the HEX FDCA format as f 16, e is -2 then you will get the value 0.247840881348.
Regards,
Nafees
The png file you attached clipped the spec for the 0.16x2^-2. So I didn't include it in the previous VI.
First:
I don't think you fully understand the syntax of the format specifier. "s" means "signed" so you would cast it to a signed integer (I8 or I16 etc). Without the "s" it implies an unsigned integer so you need to cast it to U8 or U16 et.). For the above format, you need a U16.
Second:
I had a bug in the "e" and "f" constants. They are Unsigned constants, so you can't change it to a value of -2. I updated the VI to have them define as I32 instead. Added a case to the ENUM & fixed the bug. It works fine now. (ver 2014 VI is attached)
As far as the "Inverse Function" goes, it is VERY simple algebra.
If:
Float = Ib x 2^e / 2^f
then the inverse is (solve for Ib):
Ib = Float x 2^f / 2^e
I will try and find time today to help you out. Try coding it up yourself. Shouldn't be too hard.
