This is a hw question. However, here is what I have thought about.(adsbygoogle = window.adsbygoogle || []).push({});

Here is my formula for the output filtered signal:

Yk = a*Uk + (1-a)Yk-1

Where "a" is coeffiecient that "weights" the current value of the unfiltered signal. And Yk-1 is the previous output signal. For some reason, low frequencies signals can pass through while high ones get attenuated. My job is to explain why.

Now this is for a lab where we had a random noise generator that gets added to a sine wave. The ONLY thing I can guess is that the high frequencies are so random that they get averaged out by the equation. But if the noise generator gave off low freq. while my original signal was high, then the Butterworth would be highpass.

I've been all over the web to figure this out. If anyone can point me in the right direction, that would be great.

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# Why is Digital Butterworth lowpass?

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