As I know, the shape of a low pass filter in time and frequency are as follow:

enter image description here

  • But how averaging work like a normal function in time domain?
  • Is there any intuition in it?
  • $\begingroup$ Can you explain using DFT as to why an averaging filter is low pass? $\endgroup$
    – Anonymous
    Oct 27, 2021 at 5:15

3 Answers 3


First intuition: If you symmetrize the top plot across the $y$-axis, you will see a bell-shaped function. For real filters, one usually only plots the right-most part, since the Fourier magnitude is symmetric. The most notable example of a bell-shaped function is the Gaussian, and the Fourier transform of a Gaussian is a Gaussian too. So half a Gaussian on the left of the Fourier domain is like full Gaussian in time.

Second intuition: perform a moving average of two-sample width on a signal. Do that again, and again, again (like in a forest). Those combined averages tend to a Gaussian.

Finally, a caveat: the curvature of your top plot is only one possibility for a low-pass filter. Moving average filters (with $M$ length) are brick-like, and their Fourier domain counterparts do not look like Gaussians.

Moving average filters


Consider a sequence of numbers

x = [1, 100, 1, 100, 1, 100]

If we send this sequence through an averaging transformation

y(i) = 0.6*x(i) + 0.4*x(i-1)

we get

y = [1, 60.4, 40.6, 60.4, 40.6, 60.4]

Original signal x had a lot of sudden changes 1 -> 100, 100 -> 1 where as signal y has smaller changes 60 -> 40, 40 -> 60

When we say that a signal has high frequency components we mean that the values change rapidly with time. So x had rapid changes in amplitude, while y does not have that much of rapid changes in values. This is the intuition behind why averaging is equivalent to low-pass filtering (disallowing high frequencies).


Another way to look at it

x = [1, 100, 1, 100, 1, 100]

could be written as sum of two sequences, x1 and x2

Low(zero?) frequency component

x1 = [50, 50, 50, 50, 50, 50]

High frequency component

x2 = [-49, 50, -49, 50, -49, 50]

After averaging transformation, we get

y = [1, 60.4, 40.6, 60.4, 40.6, 60.4]

Which has low frequency component of

y1 = [50, 50, 50, 50, 50, 50]

and high frequency component

y2 = [-49, 10.4, -9.4, 10.4, -9.4, 10.4]

As you can see that y2 has much smaller absolute amplitudes than x2, meaning that we have diminished the high frequency components in x by doing averaging.


Well, as far as intuitive thinking goes, I was thinking something along these terms: if you have two numbers, $a\neq b$, and you do an average between them, the average will always come between those numbers, which means that its value will always be lower than the highest (but also higher than the lowest). So the average will never converge upwards of the highest numbers, even if you have $999$ terms of $1e6$ and $1$ term of $1e-6$. The average will be less than the highest numbers, so the high numbers will get diminished, while the lower numbers increased. Just another way of looking at things.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.