Aperture Leakage Cutoff
f_c = c / (2 × slot_length)
Quick answer
This calculator solves f_c = c / (2 × slot_length) for Cutoff Frequency. With Slot Length 10 mm, the result is 14.99 GHz. It solves in 2 directions, so you can enter the values you have and get the one you need.
Calculator
Formula
Description
An aperture or slot in a shielded enclosure acts as a waveguide with a cutoff frequency determined by its longest dimension. Below the cutoff frequency, electromagnetic waves are attenuated exponentially and cannot efficiently pass through the opening. Above the cutoff frequency, the slot becomes an effective antenna that radiates or admits energy. This is why long narrow slots (like those along enclosure seams) are more problematic than round holes of the same area. The critical dimension is the longest linear extent of the opening.
Variables
- f_c — Cutoff frequency above which the slot leaks efficiently (Hz)
- slot_length — Longest dimension of the aperture (m)
Practical Notes
A 15 cm slot (typical enclosure seam) has a cutoff of 1 GHz, meaning it radiates efficiently at Wi-Fi frequencies (2.4/5 GHz). To mitigate, use overlapping seams, EMI gaskets, or conductive tape. Multiple small holes have much less leakage than one large hole of the same total area. The rule of thumb is to keep slot lengths below λ/20 of the highest frequency of concern.
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| fc | Cutoff Frequency | MHz | Hz, kHz, MHz, GHz |
| slot | Slot Length | mm | mm, cm, m, km |
Step by step
- 1. What you know: Slot Length 10 mm
- 2. The relation: f_c = c / (2 × slot_length)
- 3. Substitute: slot = 10 mm
- 4. Result: Cutoff Frequency = 14.99 GHz
Worked examples
Minimum range
Given: Slot Length 0.1 mm
Result: Cutoff Frequency = 1499 GHz
Low range
Given: Slot Length 1 mm
Result: Cutoff Frequency = 149.9 GHz
Typical
Given: Slot Length 10 mm
Result: Cutoff Frequency = 14.99 GHz
High range
Given: Slot Length 10 cm
Result: Cutoff Frequency = 1.499 GHz
Maximum range
Given: Slot Length 1 m
Result: Cutoff Frequency = 149.9 MHz
Solving for each variable
Solve for Cutoff Frequency
Slot Length 0.1 mm → 1499 GHz
Solve for Slot Length
Cutoff Frequency 100 kHz → 1.499 km
How Cutoff Frequency changes with Slot Length
| Slot Length (input) | Cutoff Frequency (result) |
|---|---|
| 1 mm | 149.9 GHz |
| 2.5 mm | 59.96 GHz |
| 5 mm | 29.98 GHz |
| 7.5 mm | 19.99 GHz |
| 10 mm | 14.99 GHz |
| 15 mm | 9.993 GHz |
| 2 cm | 7.495 GHz |
| 4 cm | 3.747 GHz |
| 7 cm | 2.141 GHz |
| 10 cm | 1.499 GHz |
The same result in other units
| Unit | Value |
|---|---|
| MHz | 14990 |
| GHz | 14.99 |
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<p><a href="https://vbatpower.com/formulas/aperture-leakage/">Aperture Leakage Cutoff</a> Calculator by vbatpower.com</p>Frequently asked questions
How do you calculate Aperture Leakage Cutoff?
Use f_c = c / (2 × slot_length). With Slot Length 10 mm, Cutoff Frequency works out to 14.99 GHz.
What is the formula for Aperture Leakage Cutoff?
The formula is f_c = c / (2 × slot_length). An aperture or slot in a shielded enclosure acts as a waveguide with a cutoff frequency determined by its longest dimension.
How do you find Slot Length when you know the other values?
Rearrange for Slot Length: with Cutoff Frequency 100 kHz, Slot Length is 1.499 km.
What units does Aperture Leakage Cutoff use?
Cutoff Frequency is shown in MHz, GHz. Values are held in SI base units internally, so you can mix prefixes without converting by hand.
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