# In time series analysis, is taking a multi-period difference equivalent to a band-pass filter?

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?

• "Explanation" here might help. Commented Sep 24, 2022 at 5:09
• @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?
– HAL
Commented Sep 24, 2022 at 6:05

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 $$0\,\texttt{dB}$$: