Double to 48-bit fixed-point number

Projects Programming
1

Here's what I need to do:

Bucket = 0.1 ∗ ((3600 ∗ 1767) / (667 ∗ 50)) = 19.07406

The value to enter in the bucket register should be set as:

Bucket Register = 19.07406 ∗ 2^24
  
The value to set the bucket register is therefore:

High word = 0x000013; Low word = 0x12F5CA

And here's what I have:

double bucketRegister = 19.07406;
int bSum = bucketRegister * pow(2, 24);

char temp[6];
temp[0] = bSum & 0xFF;
temp[1] = (bSum >> 8) & 0xFF;
temp[2] = (bSum >> 16) & 0xFF;
temp[3] = (bSum >> 24) & 0xFF;
temp[4] = (bSum >> 32) & 0xFF;
temp[5] = (bSum >> 40) & 0xFF;

uint32_t lowWord = (temp[2] << 16) + (temp[1] << 8) + temp[0];
uint32_t highWord = (temp[5] << 16) + (temp[4] << 8) + temp[3];

And I get vvvvery close to what the answer should be; 0x000013 and 0x12F598.

I'm losing precision with the bSum variable which has a decimal point after the calculation, but I can't change bSum to a double/float as the compiler throws an error about the >> operator. I'm guessing that I'm following the datasheets workings too literally and there is a way to bypass the multiplication by 224?

What can I do?

Backstory: I need to write two 24 bit words to a MAX78615 which are combined to form a 48-bit fixed-point number with 24 bits for the integer part and 24 bits for the fractional part. Here is the part in the datasheet (bottom of page 31)

2

Most Arduino's have no double type, double is a synonym for float, and int is 16 bit.

Which board are you compiling for?

3

Ah sorry, its a Teensy 3.1, but it uses the Arduino IDE and is mostly compatible.

Looks like the extra capability of the Teensy has saved me thus far.

If its easier, assume I have a normal ATMEGA328 instead :slight_smile:

4

TenderLoins:
Bucket Register = 19.07406 ∗ 2^24

The value to set the bucket register is therefore:

High word = 0x000013; Low word = 0x12F5CA
...
And I get vvvvery close to what the answer should be; 0x000013 and 0x12F598.

So why do you think that 19.07406 ∗ 2^24 is 0x1312F5CA and not 0x1312F598?

0x1312F598 is correct (integer rounding omitted).

5

Looks like the datasheet is wrong!

Pete

6

jurs:
So why do you think that 19.07406 ∗ 2^24 is 0x1312F5CA and not 0x1312F598?

0x1312F598 is correct (integer rounding omitted).

headslap well now you mention it, I just did the maths, and it checks out... I was foolishly following the datasheet and realised that I didn't have the same value using my code so figured something was wrong.

Thanks for pointing out that I shouldn't blindly accept what a datasheet says, especially a Maxim one... :slight_smile:

7

el_supremo:
Looks like the datasheet is wrong!

It's easy to see what they have done.
They did not round the float to = 19.07406, but they used the full result 19.074062968515742128935532233883

19.074062968515742128935532233883 * 2^24 is indeed 0x1312F5CA, as it is written in the datasheet.

So the problem is rounding 19.074062968515742128935532233883 down to 19.07406, which causes the discrepancy.