Complément à 2 sur 12 bits/ 2's complement on 12 bits

[hamzdiou]

Bonjour à tous,

Après avoir essayé différentes méthodes trouvées dans le forum sans succès, je viens poster ma question ici à qui pourrait m'aider 🙂

Voici le contexte :

- Je reçois des trames d'un micro contrôleur (sur bus série) de ce type là : S_90FE7000903D_P

- S_ correspond au header et _P au footer

- Une fois la data extraite j'ai : 90FE pour X, 7000 pour Y et 903D pour Z (X,Y,Z axes de l'accéléromètre en question)

- La datasheet de l'accéléro montre que le traitement qui doit être fait est le suivant :

                      90FE -> FE9 -> puis complément à 2 pour avoir en décimal : -16

Aujourd'hui je reste bloqué à l'étape du complément à 2 sur 12 bits car sur 16 je sais faire mais sur 12 je n'ai plus d'idée.

Vous trouverez le VI ne question en copie.

Merci d'avance pour votre aide

[nathand · Accepted solution]

I believe there's an error in your math: FE9 should be -23, not -16. I would do it as shown in this snippet:

Don't worry about removing the unnecessary byte; it's the low-order byte and will get shifted away in the x2^n node. What this does is gets 4 characters (16 bits), converts from text to a number, swaps the bytes, then shifts right by 4 bits while preserving the sign.

[hamzdiou]

Thank you nathand !!!

Your code works well but can you detail me a little more how does it work ?
I find quite hard to understand from hexa to decimal directly, but when I try to understand how does it work in binary, I fail ...

You have 90FE so in binary : 10010000 11111110 so after the swaping : 11111110 10010000 and then with the shifting 11111110 1001 (so that's great) but when the 2's complement has been done ?

Thanks for your help

[nathand · Accepted solution]

What mathematical operation do you think you need to do in order to take the 2's complement?

The first bit of a 2's complement value indicates the sign (although it is not enough to say it's a sign bit). To make a 12-bit value into a 16-bit value, you need to sign-extend, that is, put the first bit from the 12-bit value in all the upper bits of the 16-bit value.

After the byte swap, you have your desired value in the upper 12 bits of the 16-bit value; that is, you have your number, with the correct sign, multiplied by 2^4. So, to get the actual number, you need to multiply by 2^-4. This shifts the value right by 4 bits, and importantly, unlike a simple bit shift, it also preserves the sign.

[Bob_Schor]

I went looking all over for that Power of 2 function -- I'd never before noticed it on the Numeric Palette ...

Bob (always learning something) Schor

[hamzdiou]

Thank you very much Nathand, it's clear for me now !