filter slope of multipass moving average

[Slev1n]

Hey guys,

here is a picture of the frequency response of a multipass moving average filter. The number of stages tells you how often the data is passed through the filter.

As I am interested in the different slope behaviours of the different stages I wanted to ask you how to determine the slope.

Do I take the envelope to calculate the roll off (=slope) of the filter?

Actually it should be 1 pass (= 1 stage) is 20dB/Dekade, 2 passes = 40dB/dekade and so on...

I hope you can help me.

Kind regards

Slev1n

[Bob_Schor]

I think you need an upper level course on Signal Theory.

BS

[Slev1n]

I am not sure how I should interpret your answer 🙂

so there is no simple answer to this problem? Anyone with experience, who can tell me how to read out the roll off?

greats Slev1n

[jamiva · Accepted solution]

You should set the plots to semi-log: vertical axis in dB, and horizontal (frequency) axis in log scale. The slope should be the envelope of that curve.

For example, a 20dB/decade falloff from 0.003Hz 0.009Hz is 9.5dB (20*log10(.009/003)). Your 1 stage filter (white line) is about -13.5dB at 0.003Hz and about -23dB at 0.009Hz, a difference of -9.5dB.

[Slev1n]

Hey,

thanks for your help. But I do not understand why 0,003Hz to 0,009Hz is a decade?

But I converted my x-axis to log and although I do not have a real envelope, but it is much easier to read out the roll off.

So if you take a line/function which connects all the maxima you would have the envelope and therefor the slope I guess.

Any comments regarding my theory?

The approach approves at least the 20dB and 40dB per decade.

kind regards

Slev1n

[dan_u]

Nobody said .003 to .009 Hz is a decade, jamiva stated that if you have 20dB/decade you should have around 9.5dB in the range .003 to .009 Hz, which corresponds to your data.

But you can see the 20dB/decade easily in the log/log plot.

[Slev1n]

You are right. I just misunderstood him/her 🙂

[Slev1n]

So, the first problem is solved.

Now I wanted to check the frequency at -3dB (cut off frequency).

Can someone of you tell me, why the freuqencies are not the same at 0,5|H(w)| (left) and -3dB (right)?

I hope you guys can help me again 🙂

kind regards Slev1n

[dan_u]

Why do you think it's not the same? -3dB corresponds to a gain of 0.708, so it looks right to me.

[Slev1n]

Yes, you are right again. I just calculated 20*log(0,7) = -3dB

But I had in mind that at -3dB attenuation, the amplitude of the signal is the half of the original signal, which is right if you calculate 10*log(0,5).

As the signal I am filtering is a voltage signal I use the 20*log(x) for a conversion from Volt to dB. If I would have a power signal/spectrum I would use 10*log(x) for a conversion from Watt into dB.

Can you confirm my conclusion?