Stamp II math note #5

Statistics and digital data filtering

(c) 1998 EME Systems, Berkeley CA U.S.A.
Tracy Allen
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Contents (updated 1/8/2002)



The following routine implements a digital first order low pass filter. It filters an input data stream, so that noise or fluctuations are not so pronounced in the output data stream. There is a time constant. If a step change in the input value occurs, the time constant is the number of iterations it takes for the response to slew 2/3 of the way from the initial to the final value. A larger value of the time constant means more filtering, but slower response to changes.

Suppose the reading to be filtered comes, as it often does, from an analog to digital converter. Suppose further that the readings are in the range of 0 to 4095. Here is a filter with a time constant of 16. It implements the following numerical filter:

Y0 = X0*16                initial accumulator =16* initial datum
Yn = Yn-1 * 15/16 + Xn iteration adds new value to accumulator
Zn = Yn/16 filter output ' Filter with time constant=16 iterations
X var word ' current value, filter input
Y var word ' smoothed value, accumulator
Z var word ' filter output
gosub getX ' get current value, not shown
Y=X*16 'initialize, note multipication *16
gosub getX ' get current value, not shown
Y = Y ** 61441 + X ' smoothing function
Z = Y/16
debug dec X, tab,dec Z, CR ' show values
goto loop

Notes and explanation:

Here is another method:, which actually has better numerical convergence properties at the expense of little more calculation. This algorithm operates on the difference between the new input value ( times 16) and the output value. At each iteration, it acts to remove 1/4 of the difference. However, at the end when the values are close together it converges linearly at 1 unit per iteration and ends up at exactly zero difference. Say the input jumps from 1000 to 1002. Then the filter output stays at 1000 for 16 iterations, then at 1001 for 16 iterations while the filter "fractional" value progresses on step at a time from 16000 to 16032. So if the input value bobbles around from 1000 to 1001, the filter output will hover close to 1001. If the input makes a much larger jump, the convergence to the new value is initially much faster, making 1/4 of the difference in each step. (This is similar to the low pass filter)

' Filter with final time constant=16 iterations (linear)
X var word ' current value, filter input
Y var word ' smoothed value, accumulator
Z var word ' deviation of current value from the smoothed one
' also the filter output
S var bit ' sign of the difference

gosub getX ' get current value, not shown
Y=X*16 'initialize, note multipication *8
gosub getX ' get current value, not shown
debug dec X,tab ' show it
Z = X*16-Y ' difference from smoothed value
sign= Z.bit15 ' sign of the difference
Z=abs Z ' magnitude of the difference
Z = (Z+3)/4 ' 1/4 of the difference, or 1, or 0
Y= Y + (-S^Z+S) ' update smoothed value
' term in () is conditional subtraction
debug dec Y, tab,dec Y*/160, CR ' show values
goto loop

The */160 in the debug shows the decimal value in tenths, so you can visualize the fractional part tracking slowly behind the input value, filtering out the rapid changes, and catching up if the input value stays steady at one value.

Here is a filter that works for larger numbers, such as barometer data that has to be filtered for altimetry. This filter uses a double precision accumulator, it too converges well without numerical problems. This can be used for small numbers too.

example with time constant kR=8:
Y0 = X0*65536 initial accumulator =216* initial datum
Yn = Yn-1 * 7/8 + Xn*65536/8 iteration adds new value to accumulator
Zn = Yn/65536 filter output ' Double precision filter for input numbers up to 65535
kR con 3 ' the filter time constant is 23=8, done a shift right kR
kL con 16-kR ' for the computation shift lefts
X var word ' current value, filter input
Y0 var word ' smoothed value, accumulator low word
Y1 var word ' smoothed value, accumulator high word
Z var word ' scratch variable for calculations
' also final filter output
gosub getX ' get current value, not shown
Y1=X 'initialize, note effective multipication *65536
gosub getX ' get current value, not shown
Z=(Y0>>kR)|(Y1<<kL) ' form low word of accumulator, divided by (2^kR).
Y0=Y0-Z ' subtract it from low word of accumulator
' next do the same for high word, minus borrow from low word
Y1=Y1-(Y1>>kR)+(Y0 max -Z +Z max 1 -1)
Z=X<<kL ' form input data times divided by (2^kR), part to low word
Y0= Y0 + Z ' add it to low word of the accumulator
Y1= Y1+(X>>kR)+(Y0 max Z -Z max 1) ' do the same for the high word, plus carry
Z = Y1 + (X0 max 1) ' filter output, high word of accumulator with adjustment.
debug dec X, tab, dec Z, CR ' show the raw value and the filtered value
goto loop

Statistics: mean, min, max, and S.D.


This is an example of how to deal with the large numbers that arise from maintaining sums of values that can quickly exceed the 16 bit limit of the Stamp word. It potentially takes only 16 samples from a 12-bit A/D converter to reach that limit. But there is no problem if we can sum the values to a double precision total. On the Stamp, this takes routines for addition, and also division in order to divide the sum of the values by the number of samples to get the mean.

The sum of the squares of the data values is even more demanding, since it can exceed 16 bits in one sample. For example 1000*1000=1,000,000, which is well above the 65535 limit of 16 bit math. But it is well within the 4,294,901,760 limit of 32 bit double precision math. The sum of the squares is essential for the calculation of the standard deviation and the variance. Once having the sum, it is necessary to subtract from that the square of the sum and take its square root, so that means double precision routines for subtraction, multiplication and the square root.

In this example I arranged to use a sample size of 64, so that all the di visions can be done as shifts. (Note, I implemented this routine before I understood how to do the general double precision math on the BASIC Stamp. I would do it a little differently next time around.) I cheated a little on the standard deviation, and used a division by 64 where it should really be a division by 63. The samples in this example are the result of a 12-bit A/D conversion. Each sample is less than 2^12. The sum of the 64 samples will be less than 2^18. The sum of the squares will be less than 2^30. A special trick has to be to calculate the square of the sum, which can be as large as 2^36.

The original application for this calculation was for an optical backscatter instrument, measuring the turbidity (cloudiness) of water. Currents and turbulance in the water kicked up silt off the bottom, and the fluctuations in turbidity on the scale of seconds were an important component of the research.

In this note, additions are done with a 15-bit representation, that is, numbers are adjusted so that the most significant bit of the word variable represents the carry out of an addition or subtraction. Multiplications are done in a 16-bit representation, so that the * and ** operators will work as intended. The representation of the numbers will have to change from 15 to 16 bit and vice versa as called for. And divisions are done using shifts operations.


sum0    var     word          ' sum of the samples, least sig. word
sum1 var word ' most significant word
sumc var sum0.bit15 ' internal carry for sum
ssq0 var word ' sum of squares, least sig. word
ssq1 var word ' most significant word.
ssqc var ssq0.bit15 ' internal carry for squares
omax var word ' maximum sample value
omin var word ' minimum sample value
ADres var word ' A/D converter result, 12 bit
wordj var word ' temporary variable
NN var byte ' loop counter and remporary variable

' 64 samples are taken and summary stats accumulated
omin=$ffff ' initialize accumulations

serout 16,b96,[CR," sampling.. "]
for NN=1 to 64 ' 64 samples
pause 75 ' for timed sampling.
' each time around the loop eats up 25 milliseconds
' pause 75 for 0.1 second, pause 975 for 1 second etc.
gosub ADread ' get sample
omin = ADres max omin ' minimum
omax = ADres min omax ' maximum
sum0=sum0+ADres ' sum of the values
sum1=sum1+sumc ' carry to high word
sumc=0 ' zero the carry bit (15 bit representation)
wordj=ADres*ADres ' calculate sample^2 (low 16 bits)
ssq0=wordj & $7fff + ssq0
' sum of square, low word (15 bit representation)
wordj=ADres**ADres << 1 + wordj.bit15
' calculate sample^2 (high 16 bits of r15 representation)
' note that the lsb comes from the * calculation
ssq1=wordj+ssq1+ssqc ' accumulate high word, include carry from low sum
ssqc=0 ' zero the carry bit (15 bit representation)

' show minimum and maximum
serout 16,b96,["min=",dec omin," max=",dec omax]
' calculate mean; r15->word (4096 max)
ADres=sum0>>6+(sum1<<9) ' divide by 64, by shift and OR
' the result is in 16-bit representation
serout 16,b96,[" avg=",dec ADres]
' calculate Standard Deviation
' the notation r15 means 15-bit representation
' the notation r16 means 16-bit representation
' there will be a lot of switching back and forth
' compute square of the sums
' first divide by 8 and save remainder
' this is a trick--the square of the sums can be as large as 2^36
NN=sum0//8 ' save remainder
sum0=sum0>>3+(sum1<<12) ' divide by 8; r15->r16 word
wordj=sum0 ' save a copy
sum1=sum0**sum0 ' square of sum/8
sum1=sum1<<1+sumc ' convert r16->r15
wordj=wordj/4*NN ' compute remainder times sum
sum0=sum0+wordj ' add remainder
' subract 64*mean^2 from sum of squares r15->r15
' divide by 64 (should really divide by 63 here, oh well)
ssq1=ssq0>>6|(ssq1<<9) ' r15--> r16 word
' take square root
' & interpolate; r16 word->r16 word
ssq0=sqr ssq1 ' first pass, result is byte
' see separate note on interpolation of sqr
serout 16,b96,[" SD=",dec ssq1]

median filter, also maximum, minimum and mode


The "median" (the middle value in a list of numbers when they are sorted in ascending order) is a robust statistic. It is often less sensitive to outliers than is the average value of the numbers. It is often used as the very first stage of digital signal processing. A small fifo buffer holds raw data samples. At each step, the output to subsequent stages of the filter is the median value of the numbers currently in the buffer. At each step, the oldest value in the filter is replaced with the newest value.

Another statistic is the "mode", the most common value in a list of numbers. This is an iffy statistic in a small sample. It can happen that more than one number appears the same number of times, or that no number appears more than once.

The most familiar statistic is the average , the sum of the numbers divided by the length of the buffer. The average tends to be strongly affected by outlying values that might come from noise or instrument errors, so it is not a good choice for the initial stages of a digital filter. However, the output of a median filter based on 5 to 15 samples maybe fed to a second level of filtering that does in fact compute an average over a longer fixed time interval, or a moving (windowed) average.

I will build up to the median filter by developing a couple of subroutines.

Here's a subroutine that counts occurances of a particular value. For example, there might be 3 instances of the particular value y=29 in a list of numbers. This routine "marks" the values in the list as they are counted, and in the later routines, this will allow us to avoid counting the same list element more than once. The variables in this subroutine are used in all the following routines in this section.

  m   con   7          ' number of samples in the list
m1 con m-1 ' one less than that (for index)
x var word(m) ' an m-byte list of samples
y var word ' value to match for count
i var nib ' index
n var nib ' count of occurances (cumulative)
fx var byte ' 8 bits for marking
f var fx.bit0 ' for bit addressing of marks
cnty: ' enter with y=value to count
for i=0 to m1 ' scan the list
if f(i)=0 then cnty1 ' already counted
if x(i)-y max 1 then cnty1 ' not =y
n=n+1 ' count it
f(i)=0 ' mark it as counted
return ' exit with n=# of occurances


The above example works on a list of 7 word-size entries. It should be obvious what changes need to be made for different sized lists or for different sized variables. The byte variable, fx has to be at least as large as the number of elements in the list. It is the bits of that byte that are used to "mark" the list elements as "counted".

If you use external memory, or even the scratchpad memory in a BS2sx or BS2e, you could modify the routine for a larger list. fx has to be expanded to match, and the variables i and n (here nibs) too have to be sized to count the elements. As a digital filter, the list is usually going to be 5 to 15 samples only.

Note that the function (x(i)-y max 1) in the above is equal to zero only if x(i)=y. This is a "trick" function that allows the algorithm to avoid using an IF-THEN statment.

Here are routines that locate the maximum or minimum value in the list of numbers. This only searches the elements in the list that have their flag bit, f(i) =1 (uncounted). This allows the median routine to first find the highest value, then the second highest, and so on.

  ' finds the maximum value in a list of m samples
' skips values that have been counted as marked.
' same variables as cnty above
for i=0 to m1

y=x(i)*f(i) min y ' maximum among the uncounted
' or use y=x(i)&-f(i) min y
return ' maximum in list is y

' finds the minimum value in a list of m samples
' same variables as cnty above
y=-1 ' maximum value, $f, $ff or $ffff
for i=0 to m1
y=x(i)|~-f(i) max y
return ' minimum in list is y

In the above, it may seem strange to see the min operator used to compute the maximum, and the max operator to compute the minimum, but that is the way it works in the BASIC Stamp. The logic expressions assure that only the flagged list elements contribute. This could be done with IF-THEN, but again it takes longer and uses more code on the BS2.

Here at last is a median routine. It works by finding the maximum value in the list, counting how many instances there are of that value, then the second largest value, how many of those, and so on. It continues until the total number counted exceeds 1/2 of the values in the list. That value is the median. We assume that there are an odd number of samples in the list, so that there really is an unambiguous middle element.

  ' finds the median value in a list of m samples
' same variables & constants as cnty above, plus..
m2 con m/2+1 ' middle sample in the list
' e.g. m=7 in list, m2=4th in the middle
n2 var nib ' to hold the cumulative count
fx=-1 ' mark entire array as "uncounted"
n=0 ' none yet counted
gosub findmax ' maximum value in list, as yet uncounted
gosub cnty ' number of instances of that value
if n<m2 then findmedian1 ' not done yet
return ' median=y

Now suppose that the elements in the list come from an analog to digital converter or from some other means of reading data. The list will be filled dynamically, with each new entry replacing the oldest in the list, and the median value being output to the rest of the process. The median filter is an effective first stage for digital signal processing. This is a simple example of a filtering routine.

  ' median filter
j var nib ' pointer to the oldest data in the table
' (this is different from variable "i"!)
' do not use j for anything else in the program
for i=0 to m1 ' fill the table with readings
gosub getdata ' not shown, returns a reading, y
x(i)=y ' fill the table with readings
gosub findmedian ' now find the median
debug ?y ' show the median
' might do some additional calculations here
' or feed the data to some other process
gosub getdata ' not shown, returns a reading, y
x(j)=y ' replace the oldest reading with y
j=j+1//m ' point to the new oldest reading, circular buffer
pause 1000 ' for demo loop
goto loop

Here is a mode routine. It counts the number of occurances of each value in the list, and returns the value and the frequency of the most common one. The mode is not a very strong statistic, unless there are a relatively large number of samples distributed over a relatively small number of catagories. There may be other uses for this routine, other than statistical.

  k   var   nib    ' frequency of most common value
z var word
j var nib ' index for mode scan
fx=-1 ' mark all as uncounted, $ffff
k=0 ' frequency
j=ncd fx ' find an uncounted value
if j=0 then mode2 ' done
y=x(j-1) ' value to count
n=0 ' start at zero
gosub cnty
if n<=k then mode1 ' try again if count is low
k=n ' this count is higher
z=y ' this value is most frequent (so far)
goto mode1 ' do until no values left to count
return ' mode=z frequency=k

If there is more than one value that qualifies, the mode routine returns the first one it encounters. The variable k comes back with the count of the most common value. If k is small, that means that the "peak" is not very strong. If k=1, then no value appeared more than once. You have to decide what to do about those situations.

And finally, here is a routine to plot a frequency distribution, using ascii graphics. This of course is more interesting if there is a strong node, that is, if there are a relatively large number of samples distributed over a relatively small number of catagories, so that you can see a definite peak with tails. Samples can be combined into fewer catagories, so that the frequency in each catagory becomes significant.

  ' plots a frequency distribution of values in the list
miny var word
maxy var word
fx=-1 ' mark all as uncounted
gosub findmin
gosub findmax
for y=miny to maxy
n=0 ' zero count to start
gosub cnty
debug dec5 y," ",rep "*"\n ' number of stars=# of instances

' example
'00779 **
'00780 *** <-- median and mode
'00781 * <-- average
'00789 * <-- outlier

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