Lowpass filter formula question.

say you have a sound of 4 samples

0.0 , 0.9 , 0.5 , 0.0

and you apply above formula on it - which calculates the mean of each sample to its neighbour - you will end up with something like this:

0.0 , 0.475 , 0.24125 , 0.120625

the more far from eath other two sucessive values are, the more they get approached to each other.

or in other words: the higher a frequency is, the more it gets suppressed.

an input of

0.3 , 0.303 , 0. 306 , 0.309

will not get changed so dramatically like the first example does.

that is why [slide~] or [rampsmooth~] are already a form of "low pass filter".

-110

Thanks for taking your time to help me. I have two questions. The first one is about the math and the equation itself. My math is awful and when i try to calculate the equation with your samples my output isnt the same as yours. I will try to type out my thought process so you can see where im going wrong.

The equation can be translated as follows: the sample output by the filter, y(n), is defined as the sum of its input sample x(n) multiplied by coefficient a0 and the previous input sample x(n-1) multiplied by the coefficient a1.

lets take your sample list and put it trough the equation.

0.0 , 0.9 , 0.5 , 0.0

0.0 goes in:

0(n) = 1/2 0(n) + 1/2 0(n-1)

0 comes out

0.9 goes in :
0.0 as the previous input sample.

0.45(n) = 1/2 0.9(n) + 1/2 0(n-1) ( 0.9 x 0.5 = 0.45)

0.5 goes in
0.9 as the previous input sample.

0.7(n) = 1/2 0.5(n) + 1/2 0.9(n-1) ( 0.5 x 0.5 = 0.25 ) ( 0.9 x 0.5 = 0.45) 0.25+ 0.45 = 0.7

etc
so my output samples would be:

0 , 0.45 , 0,7

compared to yours which are:

0.0 , 0.475 , 0.24125

so what am i doing wrong here?

my second question is about this thing you said:

or in other words: the higher a frequency is, the more it gets suppressed.
an input of 0.3 , 0.303 , 0. 306 , 0.309

will not get changed so dramatically like the first example doe

Those sample values correspond to the amplitude value of those samples right?

So how do they relate to a higher or lower frequency?

Thanks again for taking your time to help me.

it might be that my values are not correct (i calculated the mean mto both the last and the next sample) - and it is hard to tell in what situation the first value reacts, because sometimes there is a sample number -1 and sometimes not. :)

just look at only the two middle values 0.9 and 0.5. they represent a spectral content which includes more high frequencies than the second example.

yes, sample values represent amplitude. if you are able and willing to learn abstract stuff from books (i am not) you should look up "sampling theorem".

the values in the second example, 0.303, 0.306 ... could be, for example a small part of a 16 Hz sinewave.

as opposed to a sinewave at 5.5 kHz, which would look like that: 0. 0.5 1.0 0.5 0. -0.5 -1. -0.5 0.

now imagine you would have a sound sample which is a mix of both. (a+b)

0.303 , 0.806 , 1.309 , 0.9.01 ...

what would happen if you average all the samples with the next one?

i just tried to make a patch but of course with that formula and scope it can only fail. :)

let me exaggerate the example a bit more:

when you perform that formula to 0.303 , 0.306 , 0.309 you basically only delay the incoming signal for half a sample.

but when you performa that to a sine wave at samplingrate/2, which is -1. 1. -1. 1, you will get 0. 0. 0. 0. -> i.e. silence.

that is maybe not a "correctly explained" but a working example why higher frequencies are affected more by interpolation than lower ones. (sinewaves are all but ideal to test and explain algorithms in real life)

Roman, I don't think it's the "last and the next sample".
It's n and n-1, so to calculate the current output sample value we use an average of the current sample and the previous one.

I forgot to give the full context my bad. Pedro would be correct see the attached image this is where i got the equation from i hope that clears more up.

Would my calculations be correct in this context? i never have gotten algebra in school sadly. I think i might pick up a book on it since i have read people saying linear algebra is very important in dsp.

And about the frequency

the values in the second example, 0.303, 0.306 ... could be, for example a small part of a 16 Hz sinewave.
as opposed to a sinewave at 5.5 kHz, which would look like that: 0. 0.5 1.0 0.5 0. -0.5 -1. -0.5 0.

So the more amplitude values the higher the frequency? Otherwise i find it kind of odd?
dont all Full amplitude sine waves (for example) have the amplitude values 0.303 and 0.306.

Would the bottom part of this text ( see link ) correspond to the more amplitude values the higher the frequency? or has that nothing to do with it.

https://gyazo.com/9acd9202a8f6e5b6a1bd275f3eb47438

Thanks for trying to help.

I think what Roman meant is: the bigger amplitude variation between samples, the higher the frequency. This is due to sampling.
See this image:
https://upload.wikimedia.org/wikipedia/commons/5/50/Signal_Sampling.png

Sampling happens at a costant pace (44100 Hz, for example), so the "grid" in which you measure amplitude is regular. You can deduce that a lower frequency would produce more "similar" adjacent values, while a higher frequency would likely have higher differences between adjacent samples.
Hence, a low-pass filter will smooth higher "jumps" between samples, resulting in a lower frequency wave.

actually it is the same for analog signals. ;)

POAj Nei dont worry about these numbers beein right or wrong. the priciple is (almost) always the same: digital frequency filters perform operations on a bunch of successive samples/time in order to be able to distinguish between different frequency content.

see the helpfile of slide~ for a similar kind of algorithm. it is birectional and supports sub-sample resolution, but basically it is also just a form of interpolation.

y(n) = y(n-1) + ((x(n) - y(n-1))/slide)

if you feed it symetrically with F/samplingrate you can control the filterfrequency in hertz: *)

#N comlet AUDIO;
#P outlet 47 180 15 0;
#N comlet Hz f;
#P inlet 122 48 15 0;
#N comlet AUDIO;
#P inlet 47 48 15 0;
#P window setfont "Sans Serif" 9.;
#P window linecount 1;
#P newex 157 72 62 196617 dspstate~;
#P newex 122 104 62 196617 !/ 44100.;
#P newex 47 134 56 196617 slide~;
#P connect 2 1 1 1;
#P connect 4 0 1 0;
#P connect 1 0 0 2;
#P connect 1 0 0 1;
#P connect 0 0 5 0;
#P connect 3 0 0 0;
#P window clipboard copycount 6;

*) actually F/samplingrate/2 would be correct, but i had reasons to design my object like that