What you could do instead is save the last n values and average them. Depending upon the nature of the data that is an easy way to do it that can work well.
skyjumper:
No, that makes no sense as written. Are you sure nothing is omitted? That code uses an uninitialized variable, which can never ever work.
I think he asked if the algorithm is correct, not if the code was properly written. And it is only the initialization of the variable that is missing (which most likely is cleared and set in more than one place in the code).
That formula you have is a sort of averaging filter. Weighed average I think may be a better name for it.
Basically you have this:
smooth = ((smooth * 9) + analogRead(0))/10;
So every new sample has an effect of 10% in your signal. A little Excel sheet can do wonders to show the result of this and different weights applied to this formula.
Can a filter like this be used to smooth the last n samples, so that old data is at some point essentially discarded? Currently I am storing the last 20 samples and averaging them, but I have been looking for a way to achieve this without consuming 80 bytes (20 floats)...
Can a filter like this be used to smooth the last n samples, so that old data is at some point essentially discarded?
I think you are missing the point after many iterations the old data is so diluted that it has no effect, this is not homeopathic computing.
The alternative is to use an array and apply a different weighting to each sample. In this way any required filter response can be created providing you have enough terms. However, the increase in processing time reduces the top frequency of the filter. This sort of programming is known as digital signal processing.
skyjumper:
Can a filter like this be used to smooth the last n samples, so that old data is at some point essentially discarded?
A follow-up to @Grumpy_Mike's answer: Assuming your samples are integers (e.g. from the analog-to-digital converter), using db2db's ? = 0.9, the contribution for the seventh previous value becomes so small that it is not present in a float. In other words, for practical purposes it's zero. ? = 0.9 gives a history of six samples (the contribution of the six is 1 / 1E-06; not much).
Currently I am storing the last 20 samples and averaging them, but I have been looking for a way to achieve this without consuming 80 bytes (20 floats)...
What are you trying to achieve? Is the goal to filter out noise when reading a stable signal (in which case the "windowed average" you're doing now is probably the right choice)? Do you need the average to "respond quickly" to signal changes?
This seems to work, but is very slow to respond. How can I speed up the response?
float smooth ;
void loop()
{
SSerial.print(analogRead(5)); //show raw output of pot value
SSerial.print(" "); // add a space for readibility
smooth = 0.9 * smooth + 0.1 * analogRead(5);
SSerial_println(smooth); //show filtered output of pot value
delay (100);
}
I am getting speed readings several times each second from a transducer. Each of these readings tends to vary a bit from the prior one, and I just want to get a stable reading to use to present to the driver.
Seems like it can be used for minor potentiometer output jitter. If I initialize smooth to the same value of the pot, it starts the calculation there, which is good.
I now see the relationship between the two side of that formula. One side is fast to react, the other is slow but better at smoothing. Like the petrol needle in your car, very good at smoothing, but too slow to be used for updating a value based on a pot turning.
In case anyone wants to try this easily, here's my test code.
If you change the pot value - which has +5/-5 value jitter added - you'll quickly see the data output of 'smooth' stabilize.
float smooth;
int randomized_potval;
void setup()
{
smooth = analogRead(5); // grab sensor value to be initial value in calculation.
Serial.begin(115200);
}
void loop()
{
float potval = analogRead(5); // read pot on pin 5
randomized_potval = random( potval - 5, potval + 5); // add some jitter
smooth = (0.99 * smooth) + (0.01 * randomized_potval); // smooth it out
serial.print(randomized_potval); // show jittered version of analog 5 value
serial.print(" ");
serial.println(int(smooth)); // OUTPUT - we are looking for this smoothed number to be stable.
delay (2);
}