19bit Int?

Matthias von Zorn · Oct 15, 2009

Hey All.
Im using a positrack system from pepperl+fuchs and the location info coming to me from the header is a 19 bit int. Im on a SLC505 and im a tad confused on how im going to get this into an integer. Could a float be used in this case? Can you do a mask move on a float? It appears that floats cant be used in a MVM. And no double ints. What to do?

Matt

OkiePC · Oct 15, 2009

Is there a sign bit involved? I think you can MUL the most significant 4 bits by 32768 with a float destination, and then add the lower 15 bits to the result...but the sign bit (if any) may add some extra steps.

Matthias von Zorn · Oct 16, 2009

No, there is no sign bit, its unsigned. Its divided into a 16bit word and a 3 bit word.

So, first I would convert the 16 bit unsigned to a float by moving the value of the first 15 bits into it, then adding 32768 to it if the 16th bit is on. Then, examine the state of the next three bits (bits 17,18, and 19) (which are bits 1,2, and 3 on the next word) and add either 65536, 131072, or 262144 of each bit.

Damn, wish I just had DINTs. This is the last slc project for me, its all compact logix from here out. Thanks

matt

Matthias von Zorn · Oct 16, 2009

Heres what I have. Looks ok, but i end up off 1 when im over 15 bit unsigned in value. Hmm. That should be ok I think.

OkiePC · Oct 16, 2009

Here's what I had in mind

Try it like this...see if that helps with accuracy...

OkiePC · Oct 16, 2009

And the rest of it...

Two posts so they're visible on screen:
EDIT: Not sure I understood the original bit pattern, so that may need "adjustment"...

Matthias von Zorn · Oct 16, 2009

Very similar, but you take the 16th bit and mask it out so u can MUL the first 15, where I converted the entire 16 to float.
Then do the same to the new 4 bit high word.

I would bet your version scans faster.

TConnolly · Oct 16, 2009

This should work too. ANDing the high order bits with 7 might be superfluous if there is not possibility N18:0 could have any more than the lowest three bits set and could be omitted.

TConnolly · Oct 16, 2009

Taking a second look at the CPT instructions above I see it could be simplified.

CPT F8:0 ( N18:0 * 65536.0 ) + ( N18:1 AND 32767 )

XIC N18:1/15 ADD F8:0 32768.0 F8:0

19bit Int?

Matthias von Zorn

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