In the previous project, you created a low-pass filter, and the first project, you created an averaging filter. We will now introduce a simple recursive averaging low-pass filter.

Given a value $0 < \alpha < 1$, the output of a signal is given by

$y[n] = \alpha x[n] + (1 - \alpha)y[n - 1]$

meaning that the next output is created by taking a weighted average of of the current value and the previous output.

Enter a value of alpha: 0.5
Enter a file name: data.txt

For example, if the input file contains the following ten numbers:

0.396
0.193
0.0224
0.800
0.428
0.843
0.412
0.996
0.386
0.695

your output will be:

0.198
0.1955
0.10895
0.454475
0.4412375
0.64211875
0.527059375
0.7615296875
0.57376484375
0.634382421875

If your input file is a constant value, this filter does not change that value—it allows a constant function to go through unchanged.

On the other hand, if you give it an alternating input:

 1
-1
 1
-1
 1
-1
 1
-1
 1
-1
 1
-1
 1
-1

you will notice something odd about the output:

 0.5
-0.25
 0.375
-0.3125
 0.34375
-0.328125
 0.3359375
-0.33203125
 0.333984375
-0.3330078125
 0.33349609375
-0.333251953125
 0.3333740234375
-0.33331298828125

You will note that it allows the high frequency to go through, but it attenuates that signal by a factor of $\frac{1}{3}$.

Exercise: How is the alternating signal affected if the value of $\alpha$ is closer to $1$, and how is the signal affected if the value of $\alpha$ is closer to $0$?

Thus, this filter is another example of a low-pass filter.

Infinite impulse response filters

This filter is said to have a infinite impulse response, meaning that if you give it a single "1", the effect on the output never ends; it may be diminished, but it never really disappears.

0
1
0
0
0
0
0
0
0
0

the output will be:

0
0.5
0.25
0.125
0.0625
0.03125
0.015625
0.0078125
0.00390625
0.001953125

References

For further reading on this filter, please see this excellent web site:

zipcpu.com