I was trying to solve the following question:

  1. Calculate the DFT of the given filter impulse response $h(n,m)$.
  2. Based on the result, determine if the given filter is a high-pass or a low-pass filter.

In the solution they proved that: $$ H(k,l)=\sum_{m=-1}^{1}\sum_{n=-1}^{1}h(m,n)e^{-i2\pi\left(\frac{m}{3}k+\frac{n}{3}l\right)}=\frac{1}{9}\left(2\cos\left(\frac{2}{3}\pi k\right)+1\right)\left(2\cos\left(\frac{2}{3}\pi l\right)+1\right) $$ Then for the second part, they said that because $H(0,0)=1$ and $H(\pi,\pi)=0$ we can figure it's a lowpass filter. I'm trying to understand why. I could not find such "definition" online. Mathematically speaking, given filter $h(n,m)$ how can I know if it's a lowpass or highpass filter?

  • 1
    $\begingroup$ Highpass means "passes high frequencies, blocks low ones". Lowpass means "passes low frequencies, blocks high ones". Does it make sense now? $\endgroup$
    – TimWescott
    Feb 9, 2022 at 23:51

2 Answers 2


This is more a quick examination test or a rule of thumb than an actual proof that a filter has a "low-pass" or "high-pass" behavior.

For a low-pass filter, it is expected that:

  • a flat or constant signal will be untouched (or keep constant amplitude), hence the sum of filter coefficients should be one (or close to)
  • a maximally-varying zero-mean signal of alternating $-1$ or $1$ (like a checkerboard pattern) should vanish, hence the sum of filter coefficients should be zero (or close to).

This can be understood as follows: for a flat data of value $C$, the average over $K$ values will be $\overline{C}=\frac{1}{K}\sum_{k\in K} S_k =C$ (untouched) while $\overline{D}=\frac{1}{K}\sum_{k\in K} (-1)^k S_k $ is either $0$ or $\pm C/K$, which vanishes as the length or support $K$ increases (the larger the differentiating filter, the larger the attenuation.

And you get the converse statement of high-pass filters. But I think this does not suffice. A globally decreasing behavior in the frequency domain is a plus.

The answer given is just the translation of the above hints in the Fourier domain, because the sum of coefficients is $H(0,0)$ and the same multiplied by alternating signs is $H(\pi,\pi)$.

To start with, a couple of more details is given here: Filter coefficients to know high pass and low pass filter


I think the detail to understand is that the frequencies are given in normalized radian frequencies in units of radians/sample. We can convert frequencies given in Hz or radians/sec to normalized frequency by dividing the frequency by the sampling rate.

Normalized frequency in cycles/samples: $f/f_s$

Normalized frequency in randians/sample: $\omega/f_s$

Where $f$ is the frequency of interest in Hz and $\omega$ is the same frequency of interest in radians/sec.

Thus we have a frequency range from DC to half the sampling rate (Nyquist) as either extending from $0$ to $0.5$ for cycles/sample units, or $0$ to $\pi$ for radians/sample units.

The OP has a 2-dimensional filter (image filter I assume?) where $H[0,0]$ represents the frequency response for an input frequency at DC (f=0), and had a result as "1" meaning it would pass low frequencies, and then $H[\pi,\pi]=0$ meaning it would not pass the Nyquist frequency at $f_s/2$, and thus represents a low pass filter.


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