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Circular Waveguide Calculator — Cutoff Frequency, TE11 & TM01 Modes

Cutoff frequency of any TE or TM mode of a circular waveguide, the single-mode band between TE₁₁ and TM₀₁, and above cutoff the guide wavelength, phase constant, phase and group velocities, wave impedance and the conductor and dielectric losses; below cutoff, the evanescent attenuation.

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Formula

fc=c χnm2πaεr,β=k2−(χnma)2,ZTE=kηβ,ZTM=βηk,αd=k2tan⁡δ2βf_c = \frac{c\,\chi_{nm}}{2\pi a\sqrt{\varepsilon_r}},\quad \beta = \sqrt{k^2 - \left(\frac{\chi_{nm}}{a}\right)^2},\quad Z_{TE} = \frac{k\eta}{\beta},\quad Z_{TM} = \frac{\beta\eta}{k},\quad \alpha_d = \frac{k^2\tan\delta}{2\beta}

Reference: Pozar, Microwave Engineering, 4th ed. (2012), §3.4, Tables 3.3–3.5, Eqs. (3.127)–(3.140); NIST DLMF §10.21

a— Inner radius (m)
\chi_{nm}— Bessel root: p′ₙₘ, zero of J′ₙ (TE); pₙₘ, zero of Jₙ (TM)
k— Wavenumber in the filling, 2πf√εr/c (rad/m)
\beta— Phase constant (rad/m)
\eta— Wave impedance of the filling, μ₀c/√εr (Ω)
R_s— Wall surface resistance, √(ωμ₀/2σ) (Ω)

How It Works

A circular waveguide is a hollow metal tube that carries microwaves above a cutoff frequency set by its radius. Its fields form modes, transverse electric (TE) or transverse magnetic (TM), each shaped by a Bessel function, and each mode propagates only above its own cutoff:

fc=c χnm2πaεrf_c = \frac{c\,\chi_{nm}}{2\pi a\sqrt{\varepsilon_r}}

Here aa is the inner radius, εr\varepsilon_r the relative permittivity of the filling, and χnm\chi_{nm} a zero of a Bessel function: for TE modes the mm-th zero pnm′p'_{nm} of Jn′J'_n, for TM modes the mm-th zero pnmp_{nm} of JnJ_n (Pozar, Microwave Engineering, 4th ed., section 3.4). The index nn counts the field's variations around the circumference and mm those across the radius. The calculator holds the roots to ten decimals, as tabulated in the NIST Digital Library of Mathematical Functions, section 10.21.

The single-mode band

The dominant mode is TE₁₁, with p11′=1.8412p'_{11} = 1.8412, and the next is TM₀₁, with p01=2.4048p_{01} = 2.4048. A guide therefore carries a single mode only between their two cutoffs, a band of 2.4048/1.8412 = 1.306 to 1 whatever the radius or filling.

Above cutoff

With k=2πfεr/ck = 2\pi f\sqrt{\varepsilon_r}/c, the phase constant is β=k2−(χnm/a)2\beta = \sqrt{k^2 - (\chi_{nm}/a)^2}, the guide wavelength 2π/β2\pi/\beta, the phase velocity ω/β\omega/\beta, and the group velocity c2/(εrvp)c^2/(\varepsilon_r v_p). The wave impedance is kη/βk\eta/\beta for TE modes and βη/k\beta\eta/k for TM modes, with η=μ0c/εr\eta = \mu_0 c/\sqrt{\varepsilon_r}.

The dielectric attenuation is αd=k2tan⁡δ/(2β)\alpha_d = k^2\tan\delta/(2\beta). The wall loss, with Rs=ωμ0/2σR_s = \sqrt{\omega\mu_0/2\sigma} and x=fc/fx = f_c/f, is

αc=Rsaη1−x2(x2+n2pnm′2−n2)\alpha_c = \frac{R_s}{a\eta\sqrt{1-x^2}}\left(x^2 + \frac{n^2}{p'^2_{nm} - n^2}\right)

for TE modes, which is Pozar's equation 3.133 for TE₁₁, and Rs/(aη1−x2)R_s/(a\eta\sqrt{1-x^2}) for TM modes. With n=0n = 0 the TE term in brackets is x2x^2 alone, so the wall loss of TE₀₁ keeps falling as the frequency rises.

Below cutoff

Below its cutoff a mode does not propagate. Its fields decay as e−αze^{-\alpha z} with α=(χnm/a)2−k2\alpha = \sqrt{(\chi_{nm}/a)^2 - k^2}, which the calculator reports in dB/m. A short length of guide below cutoff is how a vent blocks low frequencies.

Validity

The guide is a perfect circular cylinder filled with one homogeneous material, and the mode indices run from n = 0 to 5 and m = 1 to 5. The loss formulas come from the perturbation method: they assume a good conductor, with a skin depth far smaller than the radius, and a small loss tangent, and they lose accuracy very close to cutoff, where β tends to zero. A loss tangent above 0.1 or a skin depth above 1% of the radius is shown with a warning.

Worked Example

Problem: An air-filled copper circular guide with a 10 mm inner radius carries a 10 GHz signal. Is it single-mode, what are its guide wavelength and loss, and how strongly is the next mode cut off?

Step 1 - Cutoffs: f_c(TE₁₁) = 1.8412 × 299792458 / (2π × 0.010) = 8.785 GHz f_c(TM₀₁) = 2.4048 × 299792458 / (2π × 0.010) = 11.474 GHz 10 GHz lies in the single-mode band, 8.785 to 11.474 GHz.

Step 2 - Phase constant of TE₁₁: k = 2π × 10e9 / 299792458 = 209.58 rad/m k_c = 1.8412 / 0.010 = 184.12 rad/m β = √(209.58² − 184.12²) = 100.13 rad/m

Step 3 - Guide wavelength: λ_g = 2π / 100.13 = 62.75 mm, against 29.98 mm in free space

Step 4 - Velocities: v_p = 2.093 c, v_g = 0.478 c

Step 5 - Wave impedance: Z_TE = 209.58 × 376.73 / 100.13 = 788.5 Ω

Step 6 - Wall loss, copper at 58.13 MS/m: α_c = 0.1497 dB/m, so 0.15 dB over 1 m

Step 7 - TM₀₁ at 10 GHz: below its 11.474 GHz cutoff, it decays at 1024.3 dB/m, about 10 dB in the first 10 mm.

At 30 GHz the same guide is overmoded, but its TE₀₁ mode loses only 0.0487 dB/m, less than TE₁₁'s 0.0549 dB/m at the same frequency.

Practical Tips

  • ✓Keep the operating frequency well inside the single-mode band: the loss rises steeply near the TE₁₁ cutoff, and above TM₀₁ a second mode can propagate.
  • ✓Filling the guide with a dielectric lowers every cutoff by √εr, which shrinks the guide for a given band at the cost of dielectric loss.
  • ✓For long low-loss runs at millimetre-wave frequencies, an oversized guide carrying TE₀₁ has a wall loss that falls with frequency, but it needs mode filters to keep other modes out.
  • ✓A guide below cutoff makes a simple high-pass filter or a shielded vent: its attenuation is the evanescent figure times its length.

Common Mistakes

  • ✗Putting the diameter into the cutoff formula. It takes the inner radius; with the diameter every cutoff comes out half as high.
  • ✗Assuming the single-mode band is as wide as a rectangular guide's. In a circular guide it runs only from the TE₁₁ to the TM₀₁ cutoff, a ratio of 1.306 to 1.
  • ✗Forgetting that TE₁₁ has no fixed polarisation in a round guide. Slight ellipticity or a bend can rotate it or split it into two polarisations.
  • ✗Trusting the attenuation formulas right at cutoff. They grow without bound as β tends to zero, so very close to cutoff the loss must be computed by an exact method.
  • ✗Using the zero of J′₀ at x = 0 for a TE₀ₘ mode. The TE₀ₘ modes use the non-zero roots, so TE₀₁ uses 3.8317, the same root as TM₁₁, and the two modes share a cutoff.

Frequently Asked Questions

TE₁₁. It has the lowest cutoff, set by the first zero of J′₁, 1.8412: the cutoff frequency is 1.8412 c / (2π a √εr). The next mode is TM₀₁, at 2.4048 c / (2π a √εr).
Multiply the speed of light by the Bessel root of the mode and divide by 2π times the inner radius times the square root of the filling's permittivity. For an air-filled guide of 10 mm radius, TE₁₁ cuts off at 8.785 GHz and TM₀₁ at 11.474 GHz.
Its wall currents run only around the circumference, and its loss keeps falling as the frequency rises above cutoff. TE₁₁, the TM modes and every other mode with circumferential variation see their loss rise again at high frequency. That made TE₀₁ the mode of choice for long low-loss millimetre-wave runs.
In a waveguide the wave travels as a pattern bouncing between the walls, so the phase along the axis advances more slowly than in free space. The phase velocity exceeds the speed of light in the filling while the energy, at the group velocity, travels more slowly, and the product of the two is c²/εr.
It does not propagate. Its fields decay exponentially along the guide at the evanescent attenuation, which grows as the frequency falls further below cutoff. Short lengths of guide below cutoff are used as high-pass filters and as vents that block low-frequency interference.

Methodology & References

References

  • Microwave Engineering, 4th ed. — David M. Pozar (2012), §3.4, pp. 121–130 — circular waveguide TE and TM modes, Tables 3.3–3.5, Eq. (3.133) and Example 3.2; Appendices F and G for conductivities and loss tangents
  • NIST Digital Library of Mathematical Functions — F. W. J. Olver et al., eds., Release 1.2.8 (2026-09-15), §10.21 — zeros of Bessel functions and their derivatives link

Reproduces Pozar's Example 3.2 (12.19 and 15.92 GHz; at 14 GHz β = 208.0 m⁻¹ and 2.07 dB/m) within the 0.07% his c = 3×10⁸ m/s explains; the TM₀₁/TE₁₁ ratio is 1.3061 for any radius or filling, and the attenuation formulas agree with a numerical integration of the wall loss.

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