16 bit 2's complement to 12 bit 2's complement conversion

Hans Wedemeyer · Joined: 15 Sep 2003 Posts: 226

16 bit 2's complement to 12 bit 2's complement conversion
Sound easy and perhaps I doing somethign wrong.

0111111111111111 = 32767
0000000000000001 = 1
0
1111111111111111 = -1
1000000000000000 = -32767

If the value is shifted right four places to make it 12 bit, it looks like the sign bit is preserved and the conversion is complete.

The original values from 1 through 15 will be lost. this is fine and acceptable.

0000011111111111 = 2047
0000000000000001 = 1
0
0000111111111111 = -1
0000100000000000 = -2047

I think it looks correct but an old program that displays 12 bit 2's Complement data from an ADC chokes on this converted data !

What did I do wrong here... !

PCM programmer · Joined: 06 Sep 2003 Posts: 21708

I'm not sure I understand you problem.
I would assume that you're getting a 12-bit signed value from
your ADC and you want to display it. So it seems to me that
you actually want to sign-extend it to a 16-bit value, and then
give that to printf() to display it. Is that correct ?
Printf won't work with 12-bit signed values. It has to be to
be 8, 16 or 32 bits.

Guest · Posted: Tue Jun 07, 2005 1:15 am

What kind of shifting? There are two types: arithmetic and logic shift. Logic shift is the "unsigned" version: simply shift left or right.
Arithmetic shift preserves the sign bit: it copies the previous MSB to the new MSB.
So after the 4th shift you have to check bit11 of your number. If it is 1 then set the 4 MSBs to 1:
if (bit_test(mynumber,11)) mynumber |= 0xF000;

Hans Wedemeyer · Joined: 15 Sep 2003 Posts: 226

Hans Wedemeyer · Joined: 15 Sep 2003 Posts: 226

sseidman · Joined: 14 Mar 2005 Posts: 159

bfemmel · Joined: 18 Jul 2004 Posts: 40 Location: San Carlos, CA.

So if I understand the problem correctly, your 12th bit will be the sign bit to emulate the output from a signed 12 bit DAC. If you use a shift you will lose the data in the lower bits so that leaves setting the bits yourself. You will need to test the sign bit and either set or clear the 12th bit. I am not sure what needs to be done with bits 13-16, that depends on your application but try something like this.