Waveguide Attenuation Calculator
Compute rectangular waveguide TE10 cutoff, single-mode band, dB/m attenuation and transmitted power for any size, metal and frequency. Free, in-browser.
Loss curve across the operating band
- Total α (dB/m)
- Wall αc (dB/m)
- Dielectric / cutoff αd (dB/m)
Tolerance-driven insertion loss distribution
- Paths
Waveguide assessment
The selected WR-90 X-BAND section is operating at 10.000 GHz with a dominant-mode cutoff of 6.555 GHz. This is inside the single-mode window, so the TE10 interpretation is clean.
The wall-loss term follows the NIST TE10 perturbation model with surface resistance Rs = √(ωμ0 / 2σ). The dielectric/cutoff term is computed from the complex propagation constant and then separated into cutoff and dielectric increments for reporting.
Over 1.00 m the magnitude transmission is 98.75% and the corresponding insertion loss is 0.1094 dB.
Wall conductivity is temperature-adjusted with a linear coefficient. Higher wall temperature lowers conductivity and increases wall loss.
Equation list
Output summary
A clear path from inputs to results.
Find out how much signal a rectangular metal waveguide loses and whether your frequency actually propagates. Enter the guide size, wall metal, fill dielectric, frequency, and length, and the tool returns the TE10 cutoff frequency, the single-mode band, the attenuation in dB per metre, and the fraction of power that reaches the far end.
How to use
- Choose a standard waveguide (for example WR90) or enter the broad wall a and narrow wall b dimensions directly.
- Select the wall conductor (its conductivity and temperature coefficient) and set the operating temperature; the tool adjusts conductivity for temperature before computing wall loss.
- Pick the fill dielectric (air by default) to set the relative permittivity and loss tangent.
- Enter the operating frequency and the guide length.
- Read the TE10 cutoff frequency and the single-mode band, then the surface resistance, the attenuation in dB/m, the total insertion loss, the transmission coefficient S21, and the percentage of power transmitted.
The formula
Reference: Pozar, Microwave Engineering, 4th ed., TE10 attenuation of a rectangular waveguide (Ch. 3); conductor loss cross-checked against NIST Technical Note 2109; surface resistance R_s = √(ωμ0/2σ).
Worked example
WR90 copper guide, air fill, 10 GHz over 1 m
- Dimensions a = 22.86 mm, b = 10.16 mm (WR90); copper walls (σ20 = 5.8×10⁷ S/m, temperature coefficient 0.00393 /°C) at 25 °C; air fill (εr ≈ 1.00059); frequency 10 GHz; length 1 m.
- 1
Temperature first lowers the conductivity to σ = 5.8×10⁷ / (1 + 0.00393 × (25 − 20)) = 5.6882×10⁷ S/m. The dominant-mode cutoff is f_c = c / (2a√εr) = 6.5552 GHz, so at 10 GHz the guide is single-moded (the next mode, TE20, cuts on at 13.110 GHz). The surface resistance is R_s = √(ωμ0 / 2σ) = 0.026345 Ω. Feeding R_s, the free-space wavenumber k0, the cutoff wavenumber kc = π/a, and the propagation constant β into the TE10 conductor-loss formula gives α = 0.012592 Np/m, which is 8.686 × α = 0.10940 dB/m. Over 1 m that is 0.10940 dB of insertion loss, so S21 = 10^(−0.10940/20) = 0.98748 and the transmitted power is 10^(−0.10940/10) = 0.97512.
TE10 cutoff 6.5552 GHz (single-mode band 6.56–13.11 GHz); attenuation 0.1094 dB/m; insertion loss 0.1094 dB over 1 m; S21 = 98.75 %; transmitted power 97.51 %; guide wavelength 39.69 mm.
Interpreting the result
A frequency propagates only above the TE10 cutoff; below it the mode is evanescent and the wave decays without carrying power. The clean single-mode band runs from the TE10 cutoff up to the TE20 cutoff (twice the TE10 value for an air guide), and you normally operate in the middle of that band. The attenuation shown is the ohmic wall loss of the dominant mode; it rises near cutoff, dips to a broad minimum, and climbs again at high frequency. Multiply the dB/m figure by your run length to get total loss, and remember that a few tenths of a dB per metre is typical for a well-made copper guide.
Assumptions & limitations
- Only the dominant TE10 mode is modeled; the result is physically meaningful inside the single-mode band and the tool reports the TE20/TE01 cutoffs that bound it.
- Walls are treated as a good conductor with the classical skin-effect surface resistance R_s = √(ωμ0/2σ); surface roughness, plating, and anomalous skin effect are not included and will add loss in practice.
- Conductivity is scaled linearly with temperature using the entered coefficient; it is accurate near room temperature but not at cryogenic or very high temperatures.
- Dielectric loss is added from the fill's loss tangent; for air it is negligible, so the quoted loss is essentially all conductor loss.
- The guide is assumed uniform, straight, and perfectly matched; bends, flanges, joints, and reflections from mismatch are not accounted for.
Frequently asked questions
What is the cutoff frequency and why does it matter?
The cutoff frequency is the lowest frequency the waveguide will carry in a given mode. For the dominant TE10 mode it is c / (2a√εr), set almost entirely by the broad wall a. Below cutoff the mode cannot propagate — it decays exponentially — so you must operate above it. For WR90 in air the TE10 cutoff is 6.5552 GHz.
Why does raising the temperature increase the loss?
Metal conductivity falls as temperature rises, following σ = σ20 / (1 + α(T − 20)). Lower conductivity raises the surface resistance R_s = √(ωμ0/2σ), and attenuation is directly proportional to R_s. In the worked example, 25 °C instead of 20 °C drops copper's conductivity from 5.80×10⁷ to 5.69×10⁷ S/m and slightly raises the dB/m figure.
Does this include surface roughness and plating losses?
No. The model uses the ideal smooth-wall surface resistance for the chosen bulk conductor. Real guides have surface roughness and may be silver- or gold-plated, which changes the effective loss. Treat the result as the theoretical best case; measured loss is usually somewhat higher.
What is the single-mode bandwidth?
It is the frequency range where only the TE10 mode propagates, running from the TE10 cutoff up to the next higher cutoff (TE20, at twice the TE10 frequency for an air-filled guide). Operating inside this band avoids multi-mode interference. For WR90 that band is roughly 6.56 to 13.11 GHz, and 10 GHz sits comfortably within it.
How do I turn dB/m into the power that actually arrives?
Multiply the per-metre attenuation by your length to get total loss in dB, then convert: transmitted power fraction = 10^(−loss_dB/10) and voltage transmission S21 = 10^(−loss_dB/20). At 0.1094 dB over 1 m that is 97.51 % of the power and 98.75 % of the field amplitude reaching the far end.
Methodology last reviewed 2026-10-06. Results are informational and not financial or engineering advice — see our disclaimer.