Converting two decimal numbers into binary, concatenating the resulting 8 bit integers to get 16 bit signed integer and converting to signed float (IMU)

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Swapnil Sayan Saha
Swapnil Sayan Saha 2019-11-23
评论： Swapnil Sayan Saha 2019-11-23
So I have this 6DOF IMU that spits out data packets as follows:
sensorData = [85 13 109 85 12 147 12 0 156 0 71 255 229 10 144 252];
The acceleration in x axis for example, would be found by converting 12 and 147 to 8 bit binary numbers, concatenating them to get signed 16 bit binary number and converting them to signed floating point number.
I am not sure how to properly write the code to do the job. Here's my code for example, which doesn't work as expected:
sensorData = read(characteristic(b, "FF06", "FF08"));
z(i,1) = (typecast(uint16(bin2dec(strcat(dec2bin(sensorData(5),8), dec2bin(sensorData(6),8)))),'int16'))/SFacc; %SFacc = 8192 LSB/ms^-2
Am I doing something wrong? The values I receive are not what I am expecting.

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Walter Roberson
Walter Roberson 2019-11-23
double( swapbytes(typecast(uint8(sensorData), 'int16')) ) / SFacc
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Walter Roberson
Walter Roberson 2019-11-23
Yes, it takes groups of two uint8 to form the int16.
All supported versions of MATLAB for a number of years have been on Intel architecture, which is "little endian". For little endian, the byte pair A, B when bundled together into a single 16 bit entity, are mathematically A + 256*B not 256*A + B . The swapbytes() reorders the byte pairs within the 16 bit numbers.

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