Converting 2 8 bit integers into a 16 bit?

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Mustafa Abu-Mallouh el 23 de Dic. de 2018
Comentada: Mike Faulkner el 1 de Jun. de 2023
I am pulling data from an MPU6050 accelerometer using an Arduino uno and the outputs from the device are 2 8 bit integers. Here is my code for pulling the data from the accelerometer (example for the X-acceleration):
sensitivity = 16384;
g_level = 2;
X = swapbytes( [x2 x1] );
X_accel(i) = double( typecast(X,'uint16') ) / sensitivity - g_level;
The variable x1 contains bits 15 to 8 and x2 contains bits 7:0. The conversion I am using puts the code on the proper ±2 g scale I was expecting to receive but the actual values make no sense.
Attached is a plot of the acceleration values I am getting; the y-axes of the subplots should be in units of G and the accelerometer was fixed (except for the pulse at around index 65-70) and oriented with the positive z-axis set vertical (gravity in the negative z direction).
Am I converting the 8 bit integers into the 16 bit integer correctly?
Thank you!
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Mike Faulkner el 1 de Jun. de 2023
Try this: Take the Byte with the MSBs shift it to the left by 8 bit places (x by 256) and then add the Byte with the LSBs
X= X1*256 + X2 (if bits 15:8 are the MSBs)
This worked for me ... good luck

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Guillaume el 23 de Dic. de 2018
swapbytes on an array of uint8 is never going to swap anything. If you want to change the order of the two bytes, then simply concatenate them the other way round:
X = [x1, x2]; %instead of [x2, x1]
Your conversion from X to 16-bit integer is correct, so the above is the only thing I can see going wrong. If x1 is indeed bit 15:8 then it should be [x1, x2].
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