Recipe

Classify or build an ideal frequency-domain filter mask

Use when: you're told which frequencies should pass and which should be blocked, and need $H(\omega)$ 's exact low-pass, high-pass, or band-pass form

Goal: the piecewise formula for $H(\omega)$ , or, run the other way, the correct classification of a given $H(\omega)$

Taught in Low-pass, high-pass, and band-pass filters, defined by their frequency response

Quick recipe

Identify the shape of the kept region
Find the cutoff, or the center and half-width
Write $H(\omega)$ 's piecewise formula, remembering both signs

1
Decide whether the kept frequencies sit next to $\omega=0$ and stop at a finite value, start at a finite value and run to infinity, or form a single finite band away from both.

Example

$H(\omega)=1$ for $|\omega|\ge7$ , $0$ otherwise: the kept region is the outer, unbounded part, so this is high-pass.
- kept region touches $\omega=0$ and stops at a finite value — low-pass.
- kept region starts at a finite value and extends to infinity — high-pass.
- kept region is one finite interval, touching neither $0$ nor infinity — band-pass.
Why: the three filter types are defined entirely by this shape, so naming it correctly narrows everything that follows.
2
For a low-pass or high-pass filter, read off the single boundary value. For a band-pass, average the two endpoints for the center and halve their difference for the half-width.

Example

A band that keeps $2\le|\omega|\le6$ has center $\omega_0=\frac{2+6}2=4$ and half-width $\Delta=\frac{6-2}2=2$ .

Why: the shape from Step 1 alone doesn't pin down one specific filter; these numbers make it concrete.
3
Assemble $1$ on the kept region and $0$ elsewhere, stated for $|\omega|$ so both positive and negative frequencies are covered at once.

Example

With center $4$ and half-width $2$ : $H(\omega)=1$ for $2\le|\omega|\le6$ , and $H(\omega)=0$ otherwise.

⚡ Shortcut

Given a low-pass mask $H_{\text{low}}$ at cutoff $\omega_c$ , the high-pass mask at that same cutoff is just $H_{\text{high}}=1-H_{\text{low}}$ , with no new formula to derive. This only works for low-pass and high-pass sharing one cutoff; a band-pass needs two cutoffs, so it has no single-formula complement this way.

Why: the mask must give a value at every $\omega$ , and stating both what's kept and what's blocked is what actually defines it.

Common mistakes

Swapping which side of the cutoff is kept, writing high-pass's formula where low-pass was asked for.
Forgetting the mask must cover negative $\omega$ too, not only the positive frequencies named in the problem.
Reporting a band-pass's raw endpoints, like $[2,6]$ , as if they were the center and half-width instead of computing $4$ and $2$ .
Re-deriving a high-pass mask from scratch instead of using the low-pass complement shortcut when one is already known.