For time series, a simple high-pass filter is obtained by subtracting the previous value from each value:

$y(n) = x(n) - x(n-1)$

If I take a multi-period difference:

$y(n) = x(n) - x(n - a)$ where $a > 1$

Is this equivalent to a band-pass filter which attenuates frequencies significantly above and below $1 / a$? If not, how can it be described in terms of high/low/band-pass filters?

  • $\begingroup$ "Explanation" here might help. $\endgroup$ Sep 24, 2022 at 5:09
  • $\begingroup$ @OverLordGoldDragon Thanks for the link. I ran your code and got what appears to be a filter that selects frequencies that are a multiple of 1/a. Is there a name for this type of filter? $\endgroup$
    – HAL
    Sep 24, 2022 at 6:05

1 Answer 1


What you are describing is two cases of a more general form of a Comb Filter (I encourage you to go through the link, but I'll adapt to your particular case here): $$y(n) = x(n) + \alpha x(n-K) $$ with $K$ the delay in samples, and $\alpha$ the scaling factor applied to the delayed signal. In your case, you have $\alpha = -1$, which gives you $$y(n) = x(n) - x(n-K)$$

To figure out what kind of filter this is, let's move to the frequency domain:

  1. Take the $\mathcal{Z}$-transform and derive the transfer function:

\begin{align*} Y(z) &= X(z) - z^{-K}X(z)= X(z)(1-z^{-K})\\\\ \implies H(z) &= \frac{Y(z)}{X(z)} = 1 - z^{-K} \end{align*}

  1. Substitute $z = e^{j\omega}$ to get the frequency response: $$H(\omega) = 1 - e^{-j\omega K}$$ At this point, you can note that for $\omega = 0$, $H(\omega) = 0$ so you can rule out "low-pass".

  2. Let's go further, and compute the magnitude response:

\begin{align*} \vert H(\omega) \vert &= \vert 1 - e^{-j\omega K} \vert \\\\ &= \vert 1 - \cos{(\omega K)} + j\sin{(\omega K)} \vert \\\\ &= \sqrt{1 - 2 \cos{(\omega K)} + \cos^2{(\omega K}) + \sin^2{(\omega K})}\\\\ &= \sqrt{1 - 2 \cos{(\omega K)} + 1}\\\\ &= \sqrt{2 - 2 \cos{(\omega K)}}\\\\ &= \sqrt{2\vphantom{1 - \cos{(\omega K)}}} \cdot \sqrt{1 - \cos{(\omega K)}} \end{align*}

  1. Let's now analyze the magnitude response:
  • $\vert H(\omega) \vert$ is periodic (since you have a cosine term).

  • $\vert H(\omega) \vert = 0$ for $\omega = 2\pi / K$ since $ \cos{(2\pi)} = 1$, but ALSO for $\omega = 4\pi / K$ since $\cos{(4\pi)} = 1$
    As a matter of fact, $\vert H(\omega) \vert = 0$ (the nulls) for every integer $k$ such that: $$\omega = \frac{2\pi k}{K}$$

  • Similarly, the maxima (the peaks) happen when $\cos{(\omega K)} = -1$, at $k$ such that: $$\omega = \frac{\pi k}{K}$$ At these peaks, $\vert H(\omega) \vert = 2$

The resulting magnitude response looks like a comb, hence the term Comb Filter.

Here are a few examples for different values of $K$ on a logarithmic scale, restricting ourselves to the normalized Nyquist frequency $\pi$, and normalizing by the peak value ($\vert H(\omega) \vert = 2$) so that the maxima fall at $0\,\texttt{dB}$:

enter image description here


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