Difficulty with i2c Read Register. Not enough information in the doc

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Robert 2017-1-24

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评论： Walter Roberson 2020-4-7

采纳的回答： Walter Roberson

So i am working with an I2C device and reading in 8 bit register data

The problem is the following. The doc does not allow for data to be read in a specific format.

The register is currently housing data in a 2,s complement format

When i read from the register its giving decimal answers.

This is a problem because i have no idea what conversion its doing in the background. It sure doesn't know that what its reading is 2s complement so I'm assuming the decimal number I'm getting back is wrong also.

How can i just read the register in pure binary?

There is nothing in the doc

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Answer 1

Walter Roberson 2017-1-24

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If you are using arduino and read() of an i2cdev object, then the transfer always binary. As is always the case, by default MATLAB displays binary numbers in decimal . You can use dec2bin() to display it as binary.

I suspect that want you want is to pass 'int8' to the read() routine.

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Robert 2017-1-24

在 MATLAB Online 中打开

Hello Walter

no I'm using the readRegister function to read the registers of a device that is connected to an arduino. The arduino is just acting as a data aquisition model for me and or a communication buffer to the hardware.

The data that is in the register is 2s complement So when the function converts it to decimal its incorrectly converting it because it thinks its an unsigned 8 bit binary number

How do i get it back into the proper format

Right now I'm doing the following and its not working

Im taking the decimal number it gives me and using

dec2bin(,8) so i get an 8 bit number back

Then I'm taking that binary number and sticking it in a 2s complement function i found online.. However its giving me the wrong answer

I don't know if its a bogus function i found or what? Here is the code for the function i found

function i = twoscomp2int(b)
% TWOSCOMP2INT(B) Convert a twos-complement binary number, B,
% represented as a row vector of zeros and ones, to an integer. Twoscomp2int
% operates on the rows of a matrix.
[m, n] = size(b);
i = -b(:,1).*2.^(n-1) + b(:,2:end)*transpose(2.^(n-2:-1:0));

Walter Roberson 2017-1-24

在 MATLAB Online 中打开

There is no electrical difference at all between a byte used to store -87 (decimal, 2's complement) or 169 (decimal, unsigned) or 10101001 (binary) or '©' (character). The difference is only in interpretation. readRegister is returning what really is in the register. The question is how you interpret it. If you want the signed byte interpretation then add the 'int8' option to readRegister. If you want to look at the binary that is stored in the result, then typecast() the result to uint8 and dec2bin()

vals = readRegister(c, 20, 'int8');   %register 20 of device c, returned as signed byte
valu = typecast(vals, 'uint8');  %does not change value, just interpretation
val_as_bitstring = dec2bin(valu, 8)   %build one of the binary representations of that, this time as a string of '0' and '1'
val_as_bits = val_as_bitstring - '0'  %build a different binary representation, this time as a vector of 0 and 1

What you would get in val_as_bitstring and val_as_bits are not the value itself: they would be ways of displaying the values in different forms.

Think of the number 10 (decimal). One of the ways of representing it is '10', the pair of characters. One of the ways of representing it is by writing the numeral 1 and then the numeral 0, which is the base 10 representation. The base 9 representation of the same number could be written as 11; the base 8 representation as 12; base 7 as 13; base 6 as 14; base 5 as 20; base 4 as 22; base 3 as 101; base 2 as 1010; base 1 as 1111111111 . Those are all representations that are useful for different purposes, but none of them is the number; the number is itself, a pure idea.

You might find it convenient, for checking purposes, to

dec2bin(typecast(vals, 'uint8'))

I promise you that this will show you a correct binary representation of the signed 8-bit byte. For example,

vals = int8(-87)
dec2bin(typecast(vals, 'uint8'))

This will return '10101001' . You can cross-check that this is correct by performing 2's complement: negate each bit individually, and add 1 to the last bit, giving '01010111' . Now bin2dec('01010111') and you will get 87, so '10101001' really is a 2's complement representation for the negative of 87. The binary stream 10101001 is how -87 is stored in a signed byte. That also happens to be the way that 169 is stored in an unsigned byte. This is not a contradiction or an error: this is a matter of how the actual binary is to be interpreted.

Fawwaz Hosein 2020-4-7

Thank you Walter for such a lengthy response. However, consider the case when the register returns the 2's complement value as '00110000'. Now, reading this value as a signed value is correct. However even changing the type to unsigned then converting using dec2bin could still end up in error. Since with 2's comp we essentially are looking for the MSB bit (i.e. the sign bit). Referring to the previous 8 bit number, we would be unsure if the MSB is a 0 or a 1. Therefore performing dec2bin would automatically make the assumption that the MSB is 1 which would be wrong.

Walter Roberson 2020-4-7

http://www.ti.com/lit/an/slva704/slva704.pdf

Understanding the I2C Bus

Data is transferred Most Significant Bit (MSB) first.

As the bit order for I2C is fixed, there is no possibility that the bit order for any given device should be the other way around. This is not a negotiable property of the device.

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