How to process 20-bit sensor data

Question

I have an acceleration sensor (ADXL355) that returns 20-bit data in three bytes, formatted as 2s complement :

• acc3 (bits 19-12) in byte register 3 (mapped to bits 7-0)
• acc2 (bits 11-4) in byte register 2 (mapped to bits 7-0)
• acc1 (bits 3-0) in byte register 1 (mapped to bits 7-4 while 3-0 are unused)

the acceleration value is taken as a signed 32 bit integer from the above 3 registers as follows :

int32_t acc32 = ((uint32_t)acc3 << 12) | ((uint16_t)acc2 << 4) | (acc1 >> 4);
if (acc32 & (1UL << 19)) acc32 -= 1UL << 20;

Then, I am doing some statistical analysis on acceleration values to calculate an offset trim as a signed 32 bit integer.

int32 off32;

This offset has to be written back to sensor as a 16 bit 2s complement value as follows :

• offset high byte (bits 15-8)
• offset low byte (bits 7-0)

The datasheet says that "The significance of offset bits 15-0 matches the significance of acceleration bits 19-4".

I found this rather tricky so I am asking how to convert this signed 32 bit offset integer to a 16 bit offset integer that follows the above specification.

I had this idea but don't know whether it is correct :

int16_t off16[i] = (int16_t)(off32[i] / 16);
sendSPI((uint8_t)(off16 >> 8));
sendSPI((uint8_t)off16);

Any comment should be greatly appreciated !

Answer 1

Other than the subscript [i], which looks suspicious, what you did seems perfectly correct to me. Now, if I go nitpicking, I may suggest:

• use a bit shift, as it is way cheaper than a division
• round to nearest, rather than towards zero or minus infinity.

This gives:

int16_t off16 = (off32 + 8) >> 4;

Note that the bit shift behaves like a division that rounds towards −∞ (regular division rounds towards zero). Adding 8 achieves round-to-nearest, with ties rounded up.

• Vassilis Papanikolaou: Thanks a lot for your answer ! Yes, subscript [i] shouldn't be there ! As far as the bitshift is concerned, I wasn't sure whether it works as a division in signed integers, that's why I chose a division instead.