Math problem

[Christopher4187]

I need to understand this code and convert it to PBP code. I *think* it's for another microcontroller.


v[0]=0.1*(((d[0]&#0111)-(d[0]&#1000))*256+d[1])

I don't understand what the & and # signs are really doing. I also don't understand why they have 7 in the left and 8 in the right. Why not just just use 1? I'm sure there's a reason for it but I don't understand why.

[keithdoxey]

Just a guess that the & is a Logical AND and that the # is used as an indicator for binary so the the variable d[0] is being masked to get bits 0-2 in the first case and bit three in the second case. eg (d[0] AND 7) -(d[0] AND 8)

Of course I could be way off base LOL

[Christopher4187]

Just a guess that the & is a Logical AND and that the # is used as an indicator for binary so the the variable d[0] is being masked to get bits 0-2 in the first case and bit three in the second case. eg (d[0] AND 7) -(d[0] AND 8)

Of course I could be way off base LOLUsing the binary numbers as a mask is a good guess. I'm not a math expert. When the term logical is used, what exactly does it mean?

[keithdoxey]

It means the mask is applied to the variable being masked and bacause it is an AND, only the bits that are a 1 in both number appear in the result eg assume the number 26 for d[0]

00011010 = 26 = d[o]
00000111 = 7 = 1st mask

00000010 = 2 = 1st result

00011010 = 26 = d[o]
00001000 = 8 = 2nd mask

00001000 = 2 = 2nd result

Hope that makes sense

[Christopher4187]

I see why they use that formula. What's an easy way (fast/less code) to convert this number into usable information.

The first byte will go from 0 to 7 but also from 15 to 8. When it is 0-7, the number is negative....0, -1, -2, -3..... When it is 15 to 8 (stepping down), the number is positive. In this case 15=1, 14=2, 13=3.....

I'm still not sure I understand the original equation correctly but when I was driving I got something like this:

D0=15 and D1=243. I know this is charging current and D0=1, which is a rollover from D1. In other words, once D1 increases from 255 to 256, D1 goes to 0 and D0 increases by 1. In this case, it went from 15 to 0. More current stepped down from 15 to 14. How would I apply this to that formula:

v[0]=0.1*(((d[0]&#0111)-(d[0]&#1000))*256+d[1])

v[0]=.1*(((7)-(8))*256+243

v[0]=.1*(-1)*256+243

v[0]=-.1*256+243

v[0]= -25.6 + 243

v[0] = 217.4

This number doesn't seem correct from a functional perspective but also when plugging in 14 in D0, I would expect the number to increase, not decrease. Is my math above correct?

[Jerson]

Hi Christopher

Your equation simply does this calculation

v[0] = 0.1 * number

where
Number is the value in d[0], d[1] MSB, LSB
Number ranges from -7 to +7 where bit 3 (weight 8) is the sign bit.

How does it work? Let's break it up

v[0]=0.1*(((d[0]&#0111)-(d[0]&#1000))*256+d[1])

Keep aside the 0.1 for the moment

Starting with ( ( (d[0] & #111) ....
you keep only bits 0,1,2 of the MSB (max value of #111 is 7)
......-(d[0]&#1000))
subtract 0 for d[0].3 = 0 or 8 if d[0].3 is 1
d[0].3 is 1 when the number is negative.

( ( (value)-(weight of sign bit) ) * 256 ) ...

Now, multiply this number by 256 or in binary 100hex. Simply means the author wants to move the MSB number by 8 bits left.
Equivalent code would be "MSB number left shifted by 8 bits"

(...... + d[1] )
Now, add in the LSB of the number to get the whole 12bit (4 from msb, 8 from lsb) value

Now you have the entire number with its sign taken care of in some temporary variable generated by the compiler.

v[0] = 0.1 * temporary number

This is the part PBP will not be able to handle for you since it does not accept fractions.

What you can do to overcome this is

v[0] = temp number / 10 which is the same as 0.1 * temp number


Cheers
Jerson