Calculator
Rectangular uses (m,n) indices. Circular uses common Bessel roots.
Calculation History
Downloads export this table. Stored only for your current session.
| # | Timestamp | Type | Mode | Geometry | εr | μr | fc (GHz) | λc (m) |
|---|---|---|---|---|---|---|---|---|
| No rows yet. Run a calculation to populate history. | ||||||||
Example Data Table
Typical rectangular waveguides (air-filled). Dominant mode is usually TE10.
| Waveguide | a (mm) | b (mm) | TE10 Cutoff (GHz) |
|---|---|---|---|
| WR-90 | 22.86 | 10.16 | 6.55714 |
| WR-75 | 19.05 | 9.525 | 7.868568 |
| WR-62 | 15.80 | 7.90 | 9.487103 |
| WR-42 | 10.668 | 4.318 | 14.051015 |
| WR-28 | 7.112 | 3.556 | 21.076523 |
Formula Used
How to Use This Calculator
- Select Waveguide Type (Rectangular or Circular).
- Pick TE or TM; rectangular TM needs m>0 and n>0.
- Enter dimensions and choose units (mm, cm, m, inches).
- Set εr and μr for your filling material.
- Click Calculate Cutoff to show results above the form.
- Use Download CSV or Download PDF to export history.
Cutoff Fundamentals
Waveguides support discrete field patterns that propagate only above a cutoff frequency. Below cutoff, power decays exponentially and behaves like an evanescent field. For any filling medium, the relevant wave speed is v = c/√(εrμr), so raising εr lowers cutoff and increases wavelength. Engineers often start with the dominant mode to maximize bandwidth and reduce loss.
Rectangular Mode Planning
For a rectangular guide, fc = (v/2)·√((m/a)²+(n/b)²). The TE10 mode is usually dominant because it has the smallest cutoff when a>b. Using WR‑90 (a=22.86 mm, air filled), TE10 cutoff is about 6.56 GHz, placing the practical operating band comfortably above that point. Higher m or n raises cutoff and narrows usable bandwidth.
Circular Mode Planning
Circular guides use Bessel roots: fc = (v·X)/(2πr). The most common dominant choice is TE11 with X≈1.841. TM01 uses X≈2.405 and typically cuts off higher for the same radius. Because circular symmetry can simplify rotation, designers still check mode separation carefully; two nearby cutoffs may lead to mode conversion if the launch is imperfect.
Material Filling Effects
Dielectric loading shifts cutoffs and impedance. If εr=2.2 and μr≈1, the medium speed drops by √2.2, so every cutoff frequency drops by the same factor. This is useful for compact components but increases dielectric loss and can reduce power handling. When comparing designs, keep geometry constant and vary εr to see how quickly the cutoff margin changes.
Interpreting the Plot
The interactive Plotly chart visualizes how cutoff changes with a key dimension. Rectangular plots sweep the broad wall a, while circular plots sweep radius r, holding other inputs fixed. Expect a near 1/a or 1/r trend, so doubling size roughly halves cutoff. The highlighted marker shows your current point, helping you choose manufacturing tolerances with a clear safety margin.
Documentation and Reporting
Session history records each run with timestamp, geometry, material, and computed fc and λc. CSV export supports quick filtering, while the PDF report is convenient for design reviews and lab notebooks. For best traceability, record the selected mode family and confirm that the intended operating frequency stays well above cutoff, typically by at least 20–30% for stable single‑mode behavior. Include units for every dimension and note any assumptions about conductivity, temperature, or finish during testing.
FAQs
1) Which mode should I start with?
Start with the dominant mode (rectangular TE10 or circular TE11) because it has the lowest cutoff and typically offers the widest single‑mode bandwidth. After that, check the next higher cutoff to ensure adequate separation for your operating band.
2) Why are rectangular TM modes restricted?
Rectangular TM modes require both indices non‑zero because the longitudinal electric field must satisfy boundary conditions on all conducting walls. If m or n is zero, the required field components collapse and the TM solution becomes invalid for that geometry.
3) How do εr and μr change cutoff?
Cutoff frequency scales with wave speed in the filling medium: v = c/√(εrμr). Increasing εr or μr reduces v, which lowers cutoff proportionally. Geometry and mode indices set the shape term; material scales the final value.
4) Does the calculator include losses or dispersion?
No. It computes ideal cutoff frequency and cutoff wavelength for the selected mode and medium. Conductor roughness, dielectric loss tangent, bends, flanges, and frequency‑dependent dispersion affect real systems and should be evaluated with datasheets or full‑wave simulation.
5) Can I solve for a or r from a desired cutoff?
Yes by rearranging. For rectangular TE10, a ≈ v/(2fc). For circular TE11, r ≈ (v·X)/(2πfc) using X≈1.841. Use this tool by iterating dimensions until the plotted marker reaches your target.
6) What does cutoff wavelength tell me?
Cutoff wavelength λc is v/fc. Near cutoff, guide wavelength becomes large and the mode is sensitive to tolerances. Keeping operating frequency well above cutoff reduces dispersion and makes phase and impedance behavior more stable.