Rectangular Waveguide Calculator

Rectangular Waveguide Calculator Tutorial: Step-by-Step TE_m,n and TM_m,n Analysis of Rectangular Waveguides

1. How the Rectangular Waveguide Calculator Works

Calculator inputs and outputs

Cutoff wavenumber and cutoff frequency

Solving the wave equation inside a perfectly conducting rectangular tube with the walls at x = 0, a and y = 0, b gives sine and cosine standing waves across the cross-section. The boundary conditions fix the cutoff wavenumber of each mode.^[1][2][3]

$$k_c=\sqrt{{\left(\frac{m\pi }{a}\right)}^2+{\left(\frac{n\pi }{b}\right)}^2}$$

k_c = √[(mπ/a)² + (nπ/b)²]

$$f_c=\frac{k_c}{2\pi \sqrt{\mu \epsilon }}=\frac{c}{2\sqrt{{\epsilon }_r{\mu }_r}}\sqrt{{\left(\frac{m}{a}\right)}^2+{\left(\frac{n}{b}\right)}^2}$$

f_c = k_c / (2π√(με)) = [c / (2√(ε_r·μ_r))]·√[(m/a)² + (n/b)²]

Equation 2 has no TE or TM label, so TE_m,n and TM_m,n are degenerate: they always cut off at the same frequency. The index rules differ, though:^[1][5]

• TE_m,n modes (E_z = 0): m = 0, 1, 2… and n = 0, 1, 2…, but not both zero. TE₁₀ is the dominant mode when a > b.
• TM_m,n modes (H_z = 0): m ≥ 1 and n ≥ 1. TM_m0 and TM_0n do not exist because every field component would vanish. The lowest TM mode is TM₁₁.

Propagation constant, guide wavelength and velocities

Above cutoff the mode propagates with a real phase constant β. Below cutoff β becomes imaginary and the mode is evanescent.^[1][4] These relations are identical for TE and TM modes.

$$\beta =\sqrt{k^2-k_c^2}=k\sqrt{1-{\left(\frac{f_c}{f}\right)}^2},k=\frac{2\pi f\sqrt{{\epsilon }_r{\mu }_r}}{c}$$

β = √(k² − k_c²) = k·√(1 − (f_c/f)²), k = 2πf·√(ε_r·μ_r) / c

$${\lambda }_g=\frac{2\pi }{\beta }=\frac{\lambda }{\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}},\lambda =\frac{c}{f\sqrt{{\epsilon }_r{\mu }_r}}$$

λ_g = 2π/β = λ / √(1 − (f_c/f)²), λ = c / (f·√(ε_r·μ_r))

$$v_p=\frac{\omega }{\beta }=\frac{v}{\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}},v_g=\frac{d\omega }{d\beta }=v\sqrt{1-{\left(\frac{f_c}{f}\right)}^2},v=\frac{c}{\sqrt{{\epsilon }_r{\mu }_r}}$$

v_p = ω/β = v / √(1 − (f_c/f)²); v_g = dω/dβ = v·√(1 − (f_c/f)²); v = c/√(ε_r·μ_r); v_p·v_g = v²

Characteristic (wave) impedance: where TE and TM differ

The transverse wave impedance is the first output where TE and TM modes separate. TE impedance is always higher than the material impedance η and rises toward infinity at cutoff; TM impedance is always lower and falls toward zero.^[1][2]

$$Z_{TE}=\frac{k\eta }{\beta }=\frac{\eta }{\sqrt{1-{\left(\frac{f_c}{f}\right)}^2}},\eta ={\eta }_0\sqrt{\frac{{\mu }_r}{{\epsilon }_r}}$$

Z_TE = kη/β = η / √(1 − (f_c/f)²), η = η₀·√(μ_r/ε_r)

$$Z_{TM}=\frac{\beta \eta }{k}=\eta \sqrt{1-{\left(\frac{f_c}{f}\right)}^2},Z_{TE}·Z_{TM}={\eta }^2$$

Z_TM = βη/k = η·√(1 − (f_c/f)²), Z_TE·Z_TM = η²

Attenuation from wall metal and dielectric

Conductor loss is found by the perturbation method: compute the ideal-mode wall currents, then the power they dissipate in the wall's surface resistance.^[1][2][3]

$$R_s=\sqrt{\frac{\omega {\mu }_0}{2\sigma }}=\sqrt{\frac{\pi f{\mu }_0}{\sigma }}$$

R_s = √(ωμ₀ / 2σ) = √(πfμ₀ / σ)

$${\alpha }_c=\frac{R_s}{b\eta \sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}[1+\frac{2b}{a}{\left(\frac{f_c}{f}\right)}^2]\text{Np/m}$$

α_c(TE_m0) = [R_s / (b·η·√(1 − (f_c/f)²))]·[1 + (2b/a)(f_c/f)²] Np/m

$${\alpha }_c=\frac{2R_s}{b\eta \sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}\{(1+\frac{b}{a}){\left(\frac{f_c}{f}\right)}^2+[1-{\left(\frac{f_c}{f}\right)}^2]\frac{\frac{b}{a}(\frac{b}{a}{m}^2+n^2)}{\frac{b^2}{a^2}{m}^2+n^2}\}$$

α_c(TE_mn) = [2R_s / (b·η·√(1 − (f_c/f)²))]·{(1 + b/a)(f_c/f)² + [1 − (f_c/f)²]·[(b/a)((b/a)m² + n²) / ((b²/a²)m² + n²)]}

$${\alpha }_c=\frac{2R_s}{b\eta \sqrt{1-{\left(\frac{f_c}{f}\right)}^2}}\frac{m^2{\left(\frac{b}{a}\right)}^3+n^2}{m^2{\left(\frac{b}{a}\right)}^2+n^2}\text{Np/m}$$

α_c(TM_mn) = [2R_s / (b·η·√(1 − (f_c/f)²))]·[m²(b/a)³ + n²] / [m²(b/a)² + n²] Np/m

$${\alpha }_d=\frac{k^{2}tan\delta }{2\beta }\text{Np/m},\alpha \left(\text{dB/m}\right)=8.686({\alpha }_c+{\alpha }_d)$$

α_d = k²·tan δ / (2β) Np/m; α (dB/m) = 8.686·(α_c + α_d)

Below cutoff: the evanescent mode

$$\alpha =\sqrt{k_c^2-k^2}=\frac{2\pi f_c}{v}\sqrt{1-{\left(\frac{f}{f_c}\right)}^2}\text{Np/m},A\left(\text{dB}\right)=8.686\alpha \ell$$

α = √(k_c² − k²) = (2πf_c/v)·√(1 − (f/f_c)²) Np/m; A(dB) = 8.686·α·ℓ

2. Tutorial 1: Full TE₁₀ Analysis of a WR-90 Waveguide

Problem: An air-filled copper WR-90 waveguide carries a 10 GHz signal. Find the cutoff frequency, wave impedance, guide wavelength, propagation constant, phase and group velocity, and loss of the dominant TE₁₀ mode, and confirm the operating frequency is in the recommended band.

1. Look up the inside dimensions. WR-90 (IEC R100) has inside dimensions of 0.900 in × 0.400 in, so a = 22.86 mm and b = 10.16 mm.^[6] Always use inside dimensions; flange or outside dimensions shift every result.
2. Enter the geometry and mode. In the calculator, enter a = 22.86 mm, b = 10.16 mm, m = 1, n = 0. (The calculator reports TE and TM results together; with n = 0 only the TE₁₀ results describe a physical mode, see Tutorial 3.)
3. Enter the frequency and materials. Enter f = 10 GHz, ε_r = 1, μ_r = 1, tan δ = 0, and copper walls (σ = 5.8 × 10⁷ S/m).^[9]
4. Read the results. The calculator returns the TE₁₀ values below.

   Expected TE₁₀ results: copper WR-90, air-filled, 10 GHz

   Output	Value	Equation
   Cutoff frequency fc	6,557.14 MHz	Eq. 2
   Wave impedance ZTE	498.97 Ω	Eq. 6
   Guide wavelength λg	39.707 mm	Eq. 4
   Propagation constant β	158.24 rad/m	Eq. 3
   Phase velocity vp	3.9707 × 108 m/s (1.324 c)	Eq. 5
   Group velocity vg	2.2635 × 108 m/s (0.755 c)	Eq. 5
   Conductor attenuation αc	0.01248 Np/m = 0.1084 dB/m	Eq. 9
   Dielectric attenuation αd	0 (air)	Eq. 12

5. Verify the cutoff by hand. For TE₁₀ in air, Equation 2 reduces to c/(2a):

   $$f_c=\frac{c}{2a}=\frac{2.99792\times {10}^8}{2\left(0.02286\right)}=\mathrm{6,557.1}\text{MHz}$$

   f_c = c/(2a) = 2.99792×10⁸ / [2(0.02286)] = 6,557.1 MHz

6. Verify λ_g and Z_TE. The common factor is √(1 − (6.5571/10)²) = 0.75501. With λ = 29.979 mm and η₀ = 376.73 Ω:

   $${\lambda }_g=\frac{29.979\text{mm}}{0.75501}=39.707\text{mm},Z_{TE}=\frac{376.73\text{Ω}}{0.75501}=498.97\text{Ω}$$

   λ_g = 29.979 mm / 0.75501 = 39.707 mm; Z_TE = 376.73 Ω / 0.75501 = 498.97 Ω

7. Verify the conductor loss. At 10 GHz, Equation 8 gives R_s = 0.02609 Ω for copper. Substituting in Equation 9 with b/a = 0.4444:

   $${\alpha }_c=\frac{0.02609}{\left(0.01016\right)\left(376.73\right)\left(0.75501\right)}[1+0.8889(0.42996\left)\right]=0.01248\text{Np/m}=0.1084\text{dB/m}$$

   α_c = [0.02609 / ((0.01016)(376.73)(0.75501))]·[1 + 0.8889(0.42996)] = 0.01248 Np/m = 0.1084 dB/m

8. Check the operating band. Rectangular waveguide is normally used from about 1.25 f_c to 1.9 f_c of TE₁₀: high enough to avoid the steep loss and dispersion near cutoff, low enough to stay below TE₂₀ (13,114.3 MHz) and TE₀₁ (14,753.6 MHz).^[1][6] For WR-90 that is 8.2 GHz to 12.4 GHz; 10 GHz sits near the center.

3. Tutorial 2: Compare TE₁₁ and TM₁₁ in the Same Waveguide

Problem: A WR-90 component is driven at 18 GHz, well above its normal band. Analyze the degenerate TE₁₁ and TM₁₁ pair: which properties are shared and which differ?

1. Enter the inputs. a = 22.86 mm, b = 10.16 mm, m = 1, n = 1, f = 18 GHz, ε_r = μ_r = 1, tan δ = 0, copper walls.
2. Read the TE₁₁ and TM₁₁ columns side by side.

   Expected TE₁₁ and TM₁₁ results: copper WR-90, air-filled, 18 GHz

   Output	TE11	TM11	Same or different?
   Cutoff frequency fc	16,145.09 MHz	16,145.09 MHz	Same (Eq. 2)
   Propagation constant β	166.80 rad/m	166.80 rad/m	Same (Eq. 3)
   Guide wavelength λg	37.670 mm	37.670 mm	Same (Eq. 4)
   Phase velocity vp	6.7806 × 108 m/s	6.7806 × 108 m/s	Same (Eq. 5)
   Group velocity vg	1.3255 × 108 m/s	1.3255 × 108 m/s	Same (Eq. 5)
   Wave impedance	852.07 Ω	166.57 Ω	Different (Eq. 6, 7)
   Conductor attenuation αc	0.4552 dB/m	0.3264 dB/m	Different (Eq. 10, 11)

3. Check the impedance pair. At 18 GHz, √(1 − (16.145/18)²) = 0.44213. Equations 6 and 7 give:

   $$Z_{TE}=\frac{376.73}{0.44213}=852.07\text{Ω},Z_{TM}=376.73\left(0.44213\right)=166.57\text{Ω},Z_{TE}{Z}_{TM}={376.73}^2$$

   Z_TE = 376.73 / 0.44213 = 852.07 Ω; Z_TM = 376.73(0.44213) = 166.57 Ω; Z_TE·Z_TM = 376.73²

4. Check the conductor losses. With R_s = 0.03500 Ω at 18 GHz and (f_c/f)² = 0.80452, the common prefactor 2R_s/(bη√…) = 0.04137 Np/m. Equations 10 and 11 then give:

   $${\alpha }_{c,TE11}=0.05241\text{Np/m}=0.4552\text{dB/m},{\alpha }_{c,TM11}=0.03758\text{Np/m}=0.3264\text{dB/m}$$

   α_c(TE11) = 0.05241 Np/m = 0.4552 dB/m; α_c(TM11) = 0.03758 Np/m = 0.3264 dB/m

5. Compare against TE₁₀. Re-run with m = 1, n = 0 at 18 GHz: TE₁₀ has Z_TE = 404.53 Ω and only 0.0954 dB/m. Both higher-order modes lose 3.4 to 4.8 times more power per meter than the dominant mode.

4. Tutorial 3: Find Every Propagating TE and TM Mode at a Given Frequency

Problem: Which modes can propagate in WR-90 at 18 GHz?

1. List candidate index pairs. Start with the lowest-order pairs: (1,0), (2,0), (0,1), (1,1), (3,0), (2,1), (3,1), (4,0), (0,2).
2. Run the calculator once per pair. Keep a, b and f fixed and change only m and n. Record f_c each time.
3. Apply the TE/TM existence rules. A pair with m = 0 or n = 0 is a TE mode only; the TM result for that pair has no physical meaning. A pair with both indices ≥ 1 gives a TE mode and a TM mode.
4. Mark each mode as propagating or cut off.

   WR-90 mode cutoffs (air-filled) and mode status at 18 GHz

   (m, n)	Modes that exist	fc (MHz)	At 18 GHz
   (1, 0)	TE10	6,557.1	Propagates
   (2, 0)	TE20	13,114.3	Propagates
   (0, 1)	TE01	14,753.6	Propagates
   (1, 1)	TE11, TM11	16,145.1	Both propagate
   (3, 0)	TE30	19,671.4	Cut off
   (2, 1)	TE21, TM21	19,739.6	Cut off
   (3, 1)	TE31, TM31	24,589.3	Cut off
   (4, 0)	TE40	26,228.6	Cut off
   (0, 2)	TE02	29,507.1	Cut off

5. Stop when cutoffs exceed the operating frequency. Because f_c grows with both m and n, once every neighboring pair is cut off, no higher pair can propagate.

5. Tutorial 4: Evanescent TE₁₁/TM₁₁ Decay Below Cutoff

Problem: A probe transition in WR-90 operating at 10 GHz excites small TE₁₁ and TM₁₁ fields near the probe. How far away must the next component be for those fields to fall by 60 dB?

1. Enter the inputs. a = 22.86 mm, b = 10.16 mm, m = 1, n = 1, f = 10 GHz, air filling.
2. Confirm the modes are cut off. The calculator returns f_c = 16,145.1 MHz, above 10 GHz, so both TE₁₁ and TM₁₁ are evanescent. The propagation constant is real (pure attenuation) and the wave impedance is reactive.
3. Read or compute the attenuation constant. Equation 13 depends only on k_c, so TE₁₁ and TM₁₁ decay at the same rate:

   $$\alpha =\frac{2\pi (16.145\times {10}^9)}{2.99792\times {10}^8}\sqrt{1-{\left(\frac{10}{16.145}\right)}^2}=265.66\text{Np/m},\ell =\frac{60\text{dB}}{8.686\left(265.66\right)}=26.0\text{mm}$$

   α = [2π(16.145×10⁹) / 2.99792×10⁸]·√(1 − (10/16.145)²) = 265.66 Np/m = 2,307 dB/m; ℓ = 60 / (8.686 × 265.66) = 26.0 mm

4. Check the lower-order evanescent modes too. Repeat for TE₂₀ (m = 2, n = 0) and TE₀₁ (m = 0, n = 1). TE₂₀ has the lowest cutoff above 10 GHz, so it decays the slowest and usually sets the minimum spacing.

6. Tutorial 5: Dielectric-Filled Waveguide with Metal and Dielectric Loss

Problem: The same WR-90 housing is completely filled with PTFE (ε_r = 2.1, tan δ = 0.0003) and operated at 8 GHz. Find the new mode cutoffs, the TE₁₀ properties and the total loss.

1. Enter the loaded-guide inputs. a = 22.86 mm, b = 10.16 mm, m = 1, n = 0, f = 8 GHz, ε_r = 2.1, μ_r = 1, tan δ = 0.0003, copper walls.
2. Read the scaled cutoffs. Every cutoff is divided by √(ε_rμ_r) = 1.4491. TE₁₀ drops to 4,524.9 MHz, TE₂₀ to 9,049.7 MHz, TE₀₁ to 10,180.9 MHz and TE₁₁/TM₁₁ to 11,141.2 MHz. At 8 GHz the guide is still single-mode.
3. Read the TE₁₀ results.

   Expected TE₁₀ results: copper WR-90, PTFE-filled (ε_r = 2.1, tan δ = 0.0003), 8 GHz

   Output	Value
   Cutoff frequency fc	4,524.86 MHz
   Material impedance η	259.97 Ω
   Wave impedance ZTE	315.24 Ω
   Guide wavelength λg	31.357 mm
   Propagation constant β	200.37 rad/m
   Phase / group velocity	2.5086 × 108 / 1.7061 × 108 m/s
   Conductor attenuation αc	0.01376 Np/m = 0.1195 dB/m
   Dielectric attenuation αd	0.04419 Np/m = 0.3839 dB/m
   Total attenuation	0.5034 dB/m

4. Verify the dielectric loss. With k = 2π(8 × 10⁹)√2.1 / c = 242.97 rad/m:

   $${\alpha }_d=\frac{{\left(242.97\right)}^2\left(0.0003\right)}{2\left(200.37\right)}=0.04419\text{Np/m},\alpha =8.686(0.01376+0.04419)=0.5034\text{dB/m}$$

   α_d = (242.97)²(0.0003) / [2(200.37)] = 0.04419 Np/m; α = 8.686(0.01376 + 0.04419) = 0.5034 dB/m

7. Common Input Mistakes

• Swapping a and b. Enter the larger inside dimension as a. Swapping them exchanges TE₁₀ and TE₀₁ and changes every loss value.
• Reading TM results for m = 0 or n = 0. TM_m0 and TM_0n do not exist. Only the TE results are meaningful for those index pairs.
• Using outside or flange dimensions. Cutoff depends only on the inside dimensions.
• Mixing length or frequency units. Confirm the units the calculator expects for a, b and f before reading results.
• Ignoring degenerate modes. TE_m,n and TM_m,n with m, n ≥ 1 always cut off together; check both.
• Operating too close to cutoff. Below about 1.25 f_c, λ_g, Z_TE and conductor loss rise steeply (Equations 4, 6, 9).
• Treating theoretical loss as measured loss. Equations 9–11 assume smooth walls. Surface roughness and plating typically add 10–30% at microwave frequencies.

8. Rectangular Waveguide Equation Summary

Need help with a waveguide transition, filter, coupler or mode-suppression problem? Contact SiberSci, LLC. Related tools: Circular Waveguide Calculator, Rectangular Waveguide TE_m,n Calculator, Rectangular Waveguide TM_m,n Calculator, Skin Depth Calculator and Wavelength Calculator.

9. 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. 3.3, "Rectangular Waveguide" (TE₁₀ conductor attenuation, Eq. 3.96).
2. C. A. Balanis, Advanced Engineering Electromagnetics, 2nd ed. Hoboken, NJ: Wiley, 2012. Ch. 8, "Rectangular 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. 3 (skin effect and surface resistance) and 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," Sec. 2.2, "Rectangular Waveguides."
6. IEC 60153-2, Hollow Metallic Waveguides — Part 2: Relevant Specifications for Ordinary Rectangular Waveguides. Geneva: International Electrotechnical Commission (R100 / WR-90: 22.86 mm × 10.16 mm, 8.2–12.4 GHz).
7. 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
8. National Institute of Standards and Technology, "CODATA Value: characteristic impedance of vacuum," Z₀ ≈ 376.730 Ω. https://physics.nist.gov/cgi-bin/cuu/Value?z0
9. IEC 60028, International Standard of Resistance for Copper. Geneva: International Electrotechnical Commission, 1925 (annealed copper, 100% IACS ≈ 5.8 × 10⁷ S/m at 20 °C).
10. 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, η₀ = 376.730 Ω and copper σ = 5.8 × 10⁷ S/m; theoretical losses assume smooth walls.