Circular Waveguide Calculator

Circular Waveguide Calculator Tutorial: Step-by-Step Analysis of Circular Waveguide Geometries

1. How the Circular Waveguide Calculator Works

Calculator inputs and outputs

Cutoff frequency equation

Solving the wave equation in a hollow conducting cylinder gives field solutions in Bessel functions. The boundary condition at the wall fixes a cutoff wavenumber k_c = 2χ_nm/D for each mode.^[1][2][3] The calculator converts this to frequency:

The root χ_nm comes from the boundary condition for each mode family:^[1][4]

• TE_nm modes (tangential E = 0 at the wall): χ_nm = p′_nm, the m-th zero of J′_n(x).
• TM_nm modes (axial E_z = 0 at the wall): χ_nm = p_nm, the m-th zero of J_n(x).

TE₀₁ and TM₁₁ share a cutoff because J′₀(x) = −J₁(x), so both modes use the zero 3.83171.^[7] For air-filled guides, Equation 1 reduces to a quick rule of thumb:

2. Tutorial 1: Analyze an Existing Circular Waveguide

Problem: You have a round tube with a 19.05 mm (0.750 in) inside diameter. Which modes propagate at 10 GHz, what is its single-mode band, and what is the guide wavelength?

1. Measure the inside diameter. Measure or look up the inside diameter: D = 19.05 mm. Wall thickness does not affect cutoff, but using the outside diameter will shift every result low.
2. Enter the diameter and material properties. In the calculator, enter D = 19.05 mm, ε_r = 1 and μ_r = 1 for an air-filled guide.
3. Read the ten cutoff frequencies. The calculator returns the values below, in MHz.

   Calculator output: 19.05 mm inside diameter, air-filled

   Mode	fc (MHz)	At 10 GHz
   TE11	9,223.0	Propagates
   TM01	12,046.5	Cut off
   TE21	15,299.5	Cut off
   TE01	19,194.1	Cut off
   TM11	19,194.1	Cut off
   TM21	25,725.8	Cut off
   TE12	26,706.7	Cut off
   TM02	27,651.7	Cut off
   TM31	31,960.1	Cut off
   TE22	33,592.9	Cut off

4. Verify the dominant-mode result by hand. Substitute into Equation 1 with χ₁₁ = 1.84118:
   Equation 3 — TE₁₁ check for D = 19.05 mm $$f_{c,\text{TE11}}=\frac{(2.99792\times {10}^8)\left(1.84118\right)}{\pi \left(0.01905\right)\sqrt{\left(1\right)\left(1\right)}}=9,223.0\text{ MHz}$$ f_c,TE11 = (2.99792×10^8)(1.84118) / [π(0.01905)√(1·1)] = 9,223.0 MHz
   The shortcut in Equation 2 agrees: 175.70 / 19.05 = 9.223 GHz.
5. Identify which modes propagate. A mode propagates when f > f_c.^[1] At 10 GHz only TE₁₁ is above cutoff, so the guide is single-mode. All other modes are evanescent and decay along the guide.
6. Find the single-mode band. The single-mode band runs from the TE₁₁ cutoff to the TM₀₁ cutoff. Their ratio depends only on the Bessel zeros:
   Equation 4 — Single-mode bandwidth ratio [1], [2] $$\frac{f_{c,\text{TM01}}}{f_{c,\text{TE11}}}=\frac{p_{01}}{p_{11}^{\prime }}=\frac{2.40483}{1.84118}=1.306$$ f_c,TM01 / f_c,TE11 = p01 / p′11 = 2.40483 / 1.84118 = 1.306
   For this guide the single-mode band is 9.223 GHz to 12.046 GHz. A common design practice is to stay above about 1.15 f_c,TE11, where attenuation and dispersion rise steeply, and below about 0.95 f_c,TM01. That gives a practical band of 10.6 GHz to 11.4 GHz.
7. Compute the guide wavelength at the operating frequency. Use the TE₁₁ cutoff from the calculator in the standard propagation relations:^[1][4]
   Equation 5 — Guide wavelength [1, Sec. 3.1], [4] $${\lambda }_g=\frac{\lambda }{\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}},\lambda =\frac{c}{f\sqrt{{\epsilon }_r{\mu }_r}}$$ λ_g = λ / √(1 − (f_c/f)²), λ = c / (f·√(ε_r·μ_r))
   Equation 6 — Phase constant, phase velocity and group velocity [1], [3] $$\beta =\frac{2\pi }{{\lambda }_g}{v}_p=\frac{v}{\sqrt{1-{(f_c/f)}^2}}{v}_g=v\sqrt{1-{(f_c/f)}^2}$$ β = 2π/λ_g; v_p = v/√(1 − (f_c/f)²); v_g = v·√(1 − (f_c/f)²); v = c/√(ε_r·μ_r)
   At 10 GHz: √(1 − (9.223/10)²) = 0.3865, so λ_g = 29.98 mm / 0.3865 = 77.57 mm, β = 81.0 rad/m, v_p = 2.59c and v_g = 0.387c. A quarter guide wavelength, used for transformers and short-circuit spacing, is 19.39 mm.

Result: The 19.05 mm guide is single-mode at 10 GHz, but 10 GHz is only 8.4% above TE₁₁ cutoff, below the practical band. Expect high dispersion and conductor loss. Tutorial 2 shows how to pick a better diameter.

3. Tutorial 2: Choose a Circular Waveguide Diameter for a Target Frequency

Problem: Pick an air-filled diameter that places 10 GHz near the middle of the practical single-mode band (1.15 f_c,TE11 to 0.95 f_c,TM01).

1. Set the target TE₁₁ cutoff. Center the operating frequency geometrically between the band edges: f = √(1.15 × 0.95 × 1.306) f_c,TE11 = 1.1946 f_c,TE11. For f = 10 GHz, the target is f_c,TE11 = 8.371 GHz.
2. Solve Equation 1 for the diameter.
   Equation 7 — Diameter for a chosen TE₁₁ cutoff (Eq. 1 rearranged) $$D=\frac{c{p}_{11}^{\prime }}{\pi f_{c,\text{TE11}}\sqrt{{\epsilon }_r{\mu }_r}}=\frac{175.70\text{ GHz·mm}}{8.371\text{ GHz}}=20.99\text{ mm}$$ D = c·p′11 / (π·f_c,TE11·√(ε_r·μ_r)) = 175.70 / 8.371 = 20.99 mm (air-filled)
3. Round to an available size and enter it in the calculator. Enter D = 21.0 mm, ε_r = 1, μ_r = 1. The calculator returns TE₁₁ = 8,366.6 MHz, TM₀₁ = 10,927.9 MHz, TE₂₁ = 13,878.9 MHz and TE₀₁/TM₁₁ = 17,411.8 MHz.
4. Confirm the margins. 10 GHz is 1.195 times the TE₁₁ cutoff and 0.915 times the TM₀₁ cutoff, both inside the practical limits. The practical band for this guide is 9.62 GHz to 10.38 GHz.
5. Recompute guide wavelength. With Equation 5, √(1 − (8.367/10)²) = 0.5477, so λ_g = 54.73 mm, much less dispersive than the 77.57 mm of the 19.05 mm guide.

Design limit: The practical single-mode window is only about 8% wide (0.95 × 1.306 / 1.15 = 1.079). Systems that need more bandwidth usually choose rectangular waveguide, which has roughly a 2:1 single-mode range.^[1] Compare with the Rectangular Waveguide Calculator.

4. Tutorial 3: Analyze a Dielectric-Filled Circular Waveguide

Problem: The 19.05 mm tube from Tutorial 1 is filled with PTFE (ε_r ≈ 2.1). Find the new cutoffs and the guide wavelength at 8 GHz.

1. Enter the loaded-guide inputs. Enter D = 19.05 mm, ε_r = 2.1, μ_r = 1.
2. Read the scaled cutoffs. Every cutoff is divided by √(ε_rμ_r) = 1.449:
   Equation 8 — Effect of the filling material (from Eq. 1) $$f_c^{\text{filled}}=\frac{f_c^{\text{air}}}{\sqrt{{\epsilon }_r{\mu }_r}}=\frac{9,223.0\text{ MHz}}{1.449}=6,364.5\text{ MHz}$$ f_c(filled) = f_c(air) / √(ε_r·μ_r) = 9,223.0 / 1.449 = 6,364.5 MHz

   19.05 mm guide: air-filled vs. PTFE-filled (first five modes)

   Mode	Air fc (MHz)	PTFE fc (MHz)
   TE11	9,223.0	6,364.5
   TM01	12,046.5	8,312.8
   TE21	15,299.5	10,557.7
   TE01	19,194.1	13,245.2
   TM11	19,194.1	13,245.2

3. Check the operating frequency. At 8 GHz only TE₁₁ propagates (6.364 GHz < 8 GHz < 8.313 GHz). The single-mode ratio is still 1.306, because dielectric loading scales all cutoffs equally.
4. Compute the guide wavelength in the dielectric. In Equation 5, use the wavelength in the filling: λ = c/(f√ε_r) = 25.86 mm. With √(1 − (6.364/8)²) = 0.6059, λ_g = 42.68 mm.

Dielectric loading lets a guide be about 31% smaller in diameter for the same cutoff (1 − 1/1.449), at the cost of added dielectric loss.^[2]

5. Tutorial 4: Design a Waveguide-Below-Cutoff Vent

Problem: A 19.05 mm round vent must give at least 100 dB of shielding at 2.45 GHz. How long must it be?

1. Confirm the operating frequency is below cutoff. From Tutorial 1, the lowest cutoff is TE₁₁ at 9,223.0 MHz. 2.45 GHz is 27% of cutoff, so every mode is evanescent.
2. Calculate the attenuation constant of the TE₁₁ mode. Below cutoff, the propagation constant is real and the field decays exponentially:^[1][9]
   Equation 9 — Attenuation below cutoff [1], [9] $$\alpha =\sqrt{k_c^2-k^2}=\frac{2\pi f_c}{v}\sqrt{1-{\left(\frac{f}{f_c}\right)}^2}\text{ Np/m}A\left(\text{dB}\right)=8.686\alpha \ell$$ α = √(k_c² − k²) = (2πf_c/v)·√(1 − (f/f_c)²) Np/m; A(dB) = 8.686·α·ℓ
   With f_c = 9.223 GHz and f = 2.45 GHz: α = 186.4 Np/m = 1,619 dB/m, or 30.8 dB per diameter of length.
3. Solve for the required length. ℓ = 100 dB / (8.686 × 186.4 Np/m) = 61.8 mm, or 3.24 diameters. Add margin for manufacturing tolerances and for any conductor (cable, rod) passing through the vent, which turns it into a coaxial line with no cutoff.^[9]

The often-quoted rule of about 32 dB per diameter (8.686 × 2 × 1.84118 = 31.98 dB) is the limit as f → 0. Use the calculator's TE₁₁ cutoff in Equation 9 for the exact value at your frequency.

6. Common Input Mistakes

• Using the outside diameter. Cutoff depends only on the inside diameter. On thin-wall tubing the error is small; on heavy-wall tubing it can be several percent.
• Misreading units. The calculator reports cutoffs in MHz. Divide by 1,000 for GHz.
• Using the dielectric constant of a partial filling. Equations 1 and 8 assume the guide is completely filled. A partially filled guide or a dielectric rod needs a full-wave solution.
• Ignoring degenerate modes. TE₀₁ and TM₁₁ always cut off together. If you use TE₀₁ for low-loss transmission, plan for mode filtering of TM₁₁.^[2]
• Operating too close to cutoff. Just above TE₁₁ cutoff, λ_g and conductor loss grow without bound. Keep at least 15% margin.

7. Circular Waveguide Equation Summary

Need help with a circular waveguide feed, rotary joint, mode converter or shielding design? Contact SiberSci, LLC. Related tools: Rectangular Waveguide Calculator, Wavelength Calculator and Skin Depth Calculator.

8. References

1. D. M. Pozar, Microwave Engineering, 4th ed. Hoboken, NJ: Wiley, 2012. Sec. 3.1, "General Solutions for TEM, TE, and TM Waves," and Sec. 3.4, "Circular Waveguide."
2. C. A. Balanis, Advanced Engineering Electromagnetics, 2nd ed. Hoboken, NJ: Wiley, 2012. Ch. 9, "Circular Cross-Section Waveguides and Cavities."
3. R. E. Collin, Foundations for Microwave Engineering, 2nd ed. New York: McGraw-Hill, 1992 (reissued IEEE Press/Wiley, 2001). Ch. 3, "Transmission Lines and Waveguides."
4. S. Ramo, J. R. Whinnery, and T. Van Duzer, Fields and Waves in Communication Electronics, 3rd ed. New York: Wiley, 1994. Ch. 8, "Waveguides with Cylindrical Conducting Boundaries."
5. N. Marcuvitz, Waveguide Handbook, MIT Radiation Laboratory Series, vol. 10. New York: McGraw-Hill, 1951; reprinted London: IEE/Peter Peregrinus, 1986. Ch. 2, "Transmission-Line Modes."
6. M. Abramowitz and I. A. Stegun, Eds., Handbook of Mathematical Functions, NBS Applied Mathematics Series 55. Washington, DC: U.S. Government Printing Office, 1964. Table 9.5, "Zeros and Associated Values of Bessel Functions and Their Derivatives."
7. F. W. J. Olver et al., Eds., NIST Digital Library of Mathematical Functions, §10.21, "Zeros." https://dlmf.nist.gov/10.21
8. National Institute of Standards and Technology, "CODATA Value: speed of light in vacuum," c = 299,792,458 m/s (exact). https://physics.nist.gov/cgi-bin/cuu/Value?c
9. H. W. Ott, Electromagnetic Compatibility Engineering. Hoboken, NJ: Wiley, 2009. Ch. 6, "Shielding" (waveguide below cutoff).

Tutorial by Dr. Scott Best, SiberSci, LLC. Values computed with c = 299,792,458 m/s and Bessel zeros to five decimal places.