summation question

cplatt

I do not understand how to use numbers over a Word's 16-bit size limit. I want to poll a sensor every 10th of a second, get a number like 3,000, and average the readings over lets say 5 minutes. The total of the averages exceeds a Word's size very quickly. Can someone help or give me a good link??? Thank you!!

Phil Pilgrim (PhiPi)

You can create a 32-bit sum, as follows:

1. Add your reading to the lower 16-bit word. If the result is lower than it was before the add, that means there was a carry out of the high bit, so increment the high-order 16-bit word by one.
2. Do this 16384 times (a power of two, or about 5.46 minutes worth of data.)

Your average will be the lower 14 bits of the high word, concatenated with the upper 2 bits of the low word. You can shift the high word left by two bits and the low word right by 14 bits, then bitwise "or" the two together to get the average in one 16-bit word.

-Phil

kwinn

I take it that you are not using a prop but some uC that is limited to 16 bit math. 50 samples per second times 300 seconds is 15,000 samples. At readings around 3000 that would come to a sum of about 45,000,000. I suspect the readings are probably 12 bits so the worst case would be 61,425,000.

Simplest would be to use two words for double precision math. When word one has a carry increment word 2 by 1.

It would also be simpler if you use a power of two for the number of samples to average so you can use a shift instead of a divide to do the averaging.

What uC and language are you using?

Phil Pilgrim (PhiPi)

I assumed he was using a BASIC Stamp, due to the way he capitalized "Word", but I could be wrong.

-Phil

kwinn

Phil Pilgrim (PhiPi) wrote: »

I assumed he was using a BASIC Stamp, due to the way he capitalized "Word", but I could be wrong.

-Phil

That was my guess as well. Looks like our responses crossed in the ether.

cplatt

Thank you kind Sirs. Yes I am using a Basic Stamp...if you could slightly dumb down your answer it would help me, I am a beginner, or if you could send me a link to a primer I would so appreciate it!!!

cplatt

Here is my best try so far...do not understand the or command yet

FOR x= 1 TO 16384
sensor=3000
IF lowword < sensor + lowword THEN highword = highword + 1
ELSE lowword=lowword + sensor
ENDIF
PAUSE 100
DEBUG, highword << 2 (OR???) lowword >> 14
NEXT x
END

kwinn

Try this:
sensor=3000
FOR x= 1 TO 16384
IF sensor + lowword < lowword THEN highword = highword + 1
ENDIF
lowword=lowword + sensor
PAUSE 100
NEXT x
DEBUG, highword = highword << 2 + lowword >> 14
END

Phil Pilgrim (PhiPi)

Here's some code that summarizes the principle:

' {$STAMP BS2}
' {$PBASIC 2.5}

i         VAR Word
high_sum  VAR Word
low_sum   VAR Word
sensor    VAR Word
average   VAR Word

high_sum = 0
low_sum = 0
FOR i = 1 TO 16384
  GOSUB read_sensor
  IF (low_sum + sensor < low_sum) THEN high_sum = high_sum + 1
  low_sum = low_sum + sensor
NEXT

average = (HIGH_sum << 2) | (low_sum >> 14)
DEBUG DEC average

read_sensor:
  'Read sensor value.
  RETURN

BTW, guys:



-Phil

kwinn

Shouldn't "IF (low_sum < low_sum + sensor) THEN high_sum = high_sum + 1" be "IF ( low_sum + sensor < low_sum) THEN high_sum = high_sum + 1"

Tracy Allen

cplatt,

There is something of a tutorial on double precision math with the Stamp at http://emesystems.com/BS2math6.htm.

It shows methods for using non-power-of-two divisors like 3000 (5 minutes worth of samples at 0.1 sec spacing).

Phil Pilgrim (PhiPi)

kwinn,

Yes, of course you're right. 'Fixed.

Tracey,

Thanks for chiming in. It's always a relief to have your input on BASIC Stamp math questions!

-Phil

Tracy Allen

I usually have the following subroutine for double precision averaging in my Stamp programs. The input at each step is from, say, a 12 bit ADC. The output is the current average value, times 10. I usually do it times 10 or more in order to take advantage of interpolation, if possible. The final step is division of the double precision accumulator by the number of samples. This uses the scratchpad RAM in the BS2pe, which is very convenient in an otherwise memory starved system.

Average4:
' -- entry:
'    wx =  current sample
'    wj = current number of samples
'   wpe1=address of 2 contiguous words (32 bits) allocated in scram for accumulator
' -- exit:
'    wx = current average value of wx * 10
'  -- uses for computation:
'     wy, wz
  wx=wx*10        ' average computed on wx*10 for possible interpolation
  GET wpe1,WORD wy   ' low word of accumulation
  wy=wy+wx
  PUT wpe1,WORD wy
  GET wpe1+2, WORD wz    ' high word of accumulation
  IF wy<wx THEN
    wz=wz+1
    PUT wpe1+2,WORD wz
  ENDIF
 ' compute double precision division, wz:wy / wj       
  wx=65535//wj*wz//wj+wz+(wy//wj)
  wx=65535/wj*wz+(65535//wj*wz/wj)+(wy/wj)+(wx/wj)
  RETURN

cplatt

You guys are fantastic!!! Thank you so much...I think I got it!!!