Rectangular Waveguide TE_m,n Calculator

Rectangular Waveguide TE_m,n Calculator Tutorial: Step-by-Step Analysis of Waveguide Wavelength Problems

1. How the Rectangular Waveguide TE_m,n Calculator Works

Calculator inputs and outputs

Cutoff frequency and cutoff wavelength

Solving the wave equation inside a rectangular conducting pipe, with the tangential electric field forced to zero on all four walls, gives standing-wave patterns across the guide described by the cutoff wavenumber k_c = √((mπ/a)² + (nπ/b)²).^[1][2][3] The calculator converts this to the cutoff frequency:

Every wavelength problem in a waveguide is easier when the cutoff is expressed as a wavelength. The cutoff wavelength depends only on the geometry and the mode, not on the filling material:^[1][4]

Guide wavelength

Above cutoff the mode travels down the guide as a pair of plane waves bouncing between the walls, so the distance between equal-phase points along the axis, the guide wavelength, is always longer than the wavelength in the unbounded filling material.^[1][4] The calculator uses:

Because f_c/f = λ/λ_c, Equation 4 can be rearranged into a frequency-free form that links the three wavelengths. This form is the fastest way to work wavelength problems backward, for example from a measured guide wavelength to frequency:^[2][5]

Propagation constant, impedance and velocities

Attenuation

Wall losses are found with the perturbation method: the lossless fields are used to compute the surface current, and the power lost in the surface resistance R_s sets α_c.^[1][2] For the dominant TE₁₀ mode the result reduces to the familiar form below; the calculator applies Collin's general TE_mn expression, which gives the same value for TE₁₀.

2. Tutorial 1: Find the Guide Wavelength of WR-90 at 10 GHz

Problem: An air-filled, annealed-copper WR-90 waveguide carries the TE₁₀ mode at 10 GHz. Find the cutoff frequency, the guide wavelength, the half- and quarter-guide wavelengths, and check the results by hand.

1. Look up the inside dimensions. WR-90 has inside dimensions a = 0.900 in (22.86 mm) and b = 0.400 in (10.16 mm), with a recommended TE₁₀ band of 8.2 to 12.4 GHz.^[8]
2. Select the units and filling material. Select Inches for the waveguide dimensions. Enter ε_r = 1.0, permittivity loss tangent = 0.0, μ_r = 1.0 and permeability loss tangent = 0.0 for air.
3. Enter the dimensions and mode. Enter a = 0.9, b = 0.4, m = 1 and n = 0. The calculator returns a cutoff frequency of 6,557.14 MHz.
4. Verify the cutoff by hand. For TE₁₀, Equations 1 and 2 reduce to f_c = c/(2a) and λ_c = 2a = 45.72 mm:
   Equation 11 — TE₁₀ cutoff check for WR-90 $$f_{c,\text{TE10}}=\frac{2.99792458\times {10}^8}{2\left(0.02286\right)}=6,557.14\text{ MHz}$$ f_c,TE10 = c/(2a) = 2.99792458×10^8 / [2(0.02286)] = 6,557.14 MHz
5. Enter the operating frequency and wall metal. Enter the frequency in hertz: 10E9 (or 10,000,000,000). It must be greater than f_c. Select Annealed Copper (5.8E7 S/m).
6. Read the wavelength results. The calculator returns the values below.

   Calculator output: air-filled WR-90, TE₁₀, 10 GHz, annealed copper

   Output	Calculator value	Rounded
   Cutoff frequency	6,557.14 MHz	6.557 GHz
   Wave impedance ZTE	498.97 Ω	499 Ω
   Guide wavelength λg	0.039707 m	39.71 mm
   Half guide wavelength λg/2	0.019854 m	19.85 mm
   Quarter guide wavelength λg/4	0.0099268 m	9.93 mm
   Propagation constant β	158.24 rad/m	158.2 rad/m
   Phase velocity vp	3.9707×108 m/s	1.324c
   Group velocity vg	2.2635×108 m/s	0.755c
   Conductor attenuation αc	0.1084 dB/m	0.108 dB/m
   Dielectric attenuation αd	0 dB/m	0

7. Verify the guide wavelength by hand. The free-space wavelength is λ = c/f = 29.979 mm and f_c/f = 0.65571. Substitute into Equation 4:
   Equation 12 — Guide wavelength check for WR-90 at 10 GHz $${\lambda }_g=\frac{29.979\text{ mm}}{\sqrt{1-{\left(0.65571\right)}^2}}=\frac{29.979\text{ mm}}{0.75501}=39.707\text{ mm}$$ λ_g = 29.979 mm / √(1 − 0.65571²) = 29.979 / 0.75501 = 39.707 mm
   The impedance, velocities and β follow from the same square-root factor 0.75501: Z_TE = 376.73/0.75501 = 498.97 Ω, v_p = c/0.75501 = 1.3245c, v_g = 0.75501c, and β = 2π/0.039707 = 158.24 rad/m.
8. Check the wavelength identity. Equation 5 must balance: 1/λ_g² + 1/λ_c² = 1/39.707² + 1/45.72² = 1.1126×10⁻³ mm⁻² = 1/29.979². It does, which confirms the inputs were entered correctly.

Result: At 10 GHz the TE₁₀ guide wavelength in WR-90 is 39.71 mm, 32.4% longer than the 29.98 mm free-space wavelength. Use λ_g, not λ, for every length along a waveguide: stub lengths, iris spacings, transformer sections and slot spacings.^[1]

3. Tutorial 2: Guide-Wavelength Dispersion, Slot Spacing and Electrical Length

Problem: A broadside slotted-waveguide array in WR-90 is designed at 10 GHz. How does the guide wavelength change across the X-band, what slot spacing is required, and what is the electrical length and delay of a 1 m feed run?

1. Keep the geometry and change only the frequency. Leave a = 0.9 in, b = 0.4 in, m = 1, n = 0, air and annealed copper from Tutorial 1. Enter each frequency in turn: 8.2E9, 9E9, 10E9, 11E9 and 12.4E9.
2. Record the wavelength outputs.

   WR-90 TE₁₀ guide wavelength across the recommended band (calculator output)

   f (GHz)	λ (mm)	√(1−(fc/f)²)	λg (mm)	λg/2 (mm)	λg/4 (mm)	αc (dB/m)
   8.2	36.56	0.6005	60.89	30.44	15.22	0.140
   9.0	33.31	0.6850	48.63	24.32	12.16	0.121
   10.0	29.98	0.7550	39.71	19.85	9.93	0.108
   11.0	27.25	0.8029	33.94	16.97	8.49	0.102
   12.4	24.18	0.8487	28.49	14.24	7.12	0.097

   Over the band the free-space wavelength changes by a factor of 1.51, but the guide wavelength changes by 2.14. This extra change is waveguide dispersion, and it grows rapidly toward cutoff.^[1][4]
3. Set the slot spacing from the half guide wavelength. In a resonant (broadside) slot array, adjacent slots are spaced λ_g/2 apart and placed on alternating sides of the broad-wall centerline so they radiate in phase.^[7] Read λ_g/2 directly from the calculator:
   Equation 13 — Broadside slot spacing at the design frequency [7] $$d=\frac{{\lambda }_g}{2}=\frac{39.707\text{ mm}}{2}=19.85\text{ mm}$$ d = λ_g/2 = 39.707 mm / 2 = 19.85 mm (WR-90, TE10, 10 GHz)
   At 9.5 GHz λ_g/2 is 21.81 mm and at 10.5 GHz it is 18.28 mm. Because the machined spacing stays at 19.85 mm, the slots fall out of phase away from 10 GHz, which is why resonant slot arrays are narrowband.
4. Find the electrical length of a 1 m run. Use β from the calculator. The phase shift is βL = 158.24 rad = 9,066.4°, which is 25.18 guide wavelengths, or 66.4° after removing whole cycles. The same run measured in free-space wavelengths would be only 33.36 wavelengths long because λ_g > λ.
5. Find the group delay and loss. Signals travel at the group velocity, so the delay is τ_g = L/v_g = 1 m / 2.2635×10⁸ m/s = 4.418 ns, compared with 3.336 ns in free space. The loss is 0.108 dB for annealed copper.

4. Tutorial 3: Find the Frequency from a Measured Guide Wavelength

Problem: A slotted WR-90 line terminated in a short shows standing-wave minima spaced 20.00 mm apart. What is the source frequency?

1. Convert the minima spacing to a guide wavelength. Adjacent minima of a standing wave are λ_g/2 apart, so λ_g = 2 × 20.00 mm = 40.00 mm.^[6]
2. Get the cutoff wavelength from the calculator. With a = 0.9 in, b = 0.4 in, m = 1, n = 0, the calculator returns f_c = 6,557.14 MHz, so λ_c = c/f_c = 45.72 mm (equal to 2a).
3. Solve Equation 5 for frequency.
   Equation 14 — Frequency from a measured guide wavelength (Eq. 5 applied) $$f=(2.99792\times {10}^8)\sqrt{\frac{1}{{\left(0.04000\right)}^2}+\frac{1}{{\left(0.04572\right)}^2}}=9.958\text{ GHz}$$ f = (2.99792×10^8)·√(1/0.04000² + 1/0.04572²) = 9.958 GHz
4. Confirm with the calculator. Enter f = 9.958E9. The calculator returns a guide wavelength of about 0.0400 m and a half-wavelength of about 0.0200 m, matching the measured minima spacing.

Why this matters: A 1% error in the measured spacing moves the computed frequency by only about 0.6% here, but the sensitivity grows sharply near cutoff. Slotted-line frequency measurements are most accurate in the upper half of the waveguide band.^[6]

5. Tutorial 4: Guide Wavelength in a Dielectric-Filled Waveguide

Problem: The WR-90 guide from Tutorial 1 is completely filled with PTFE (ε_r ≈ 2.1, tan δ ≈ 0.0004 near 10 GHz).^[1] Find the new cutoff, guide wavelength and loss at 10 GHz.

1. Enter the dielectric properties. Keep the dimensions and mode. Enter ε_r = 2.1, permittivity loss tangent = 0.0004, μ_r = 1.0 and permeability loss tangent = 0.0.
2. Read the scaled cutoff. The cutoff drops by √2.1 = 1.449 to 4,524.86 MHz. The cutoff wavelength in the dielectric is still 2a = 45.72 mm; only the frequency at which the wavelength in the material equals 2a has moved.
3. Enter 10E9 and read the results.

   WR-90 TE₁₀ at 10 GHz: air-filled vs. PTFE-filled (calculator output, annealed copper)

   Output	Air	PTFE
   Cutoff frequency	6,557.14 MHz	4,524.86 MHz
   Wavelength in filling λ	29.98 mm	20.69 mm
   Guide wavelength λg	39.71 mm	23.20 mm
   Quarter guide wavelength λg/4	9.93 mm	5.80 mm
   Wave impedance ZTE	498.97 Ω	291.52 Ω
   Propagation constant β	158.24 rad/m	270.85 rad/m
   Phase velocity vp	1.324c	0.774c
   Group velocity vg	0.755c	0.615c
   Dielectric attenuation αd	0 dB/m	0.592 dB/m

4. Verify the guide wavelength. λ = c/(f√ε_r) = 20.688 mm and f_c/f = 0.45249, so √(1 − 0.45249²) = 0.89177 and λ_g = 20.688/0.89177 = 23.198 mm. Check with Equation 5: 1/23.198² + 1/45.72² = 1/20.688².
5. Verify the dielectric loss. With k = 2πf√ε_r/c = 303.72 rad/m and β = 270.85 rad/m, Equation 10 gives α_d = (303.72²)(0.0004)/(2 × 270.85) = 0.0681 Np/m = 0.592 dB/m.

Result: Filling the guide shortens the guide wavelength from 39.71 mm to 23.20 mm, a 42% reduction, which makes dielectric-loaded slot arrays, phase shifters and filters more compact. The cost is dielectric loss, which for PTFE is several times the copper wall loss at this frequency.^[1][3]

6. Tutorial 5: Guide Wavelengths of Higher-Order TE_m,n Modes

Problem: A WR-90 component is driven at 15 GHz, above the recommended band. Which TE modes can propagate, and what is the guide wavelength of each?

1. Find the cutoff of each candidate mode. Keep the air-filled WR-90 inputs and change only m and n, reading the cutoff each time.

   WR-90 TE_mn cutoff frequencies (calculator output, air-filled)

   Mode	m, n	fc (MHz)	λc (mm)	At 15 GHz
   TE10	1, 0	6,557.14	45.72	Propagates
   TE20	2, 0	13,114.28	22.86	Propagates
   TE01	0, 1	14,753.57	20.32	Propagates
   TE11	1, 1	16,145.09	18.57	Cut off
   TE30	3, 0	19,671.42	15.24	Cut off
   TE21	2, 1	19,739.61	15.19	Cut off

   The TM₁₁ mode shares the TE₁₁ cutoff; use the Rectangular Waveguide TM_m,n Calculator to analyze TM modes.
2. Enter 15E9 for each propagating mode.

   WR-90 at 15 GHz: wavelength results for each propagating TE mode (annealed copper)

   Mode	λg (mm)	β (rad/m)	ZTE (Ω)	vp	αc (dB/m)
   TE10	22.22	282.75	418.9	1.112c	0.094
   TE20	41.17	152.60	776.1	2.060c	0.251
   TE01	110.71	56.75	2,086.9	5.540c	0.956

   The free-space wavelength is the same 19.99 mm for all three modes, but each mode has its own guide wavelength. The mode closest to its cutoff (TE₀₁) has the longest λ_g, the highest impedance and the highest loss.
3. Find the beat length between two modes. When a discontinuity excites both TE₁₀ and TE₂₀, their fields drift in and out of phase along the guide. The distance for one full cycle of that interference is:
   Equation 15 — Two-mode beat length [2, Ch. 3] $$L_B=\frac{2\pi }{{\beta }_{10}-{\beta }_{20}}=\frac{2\pi }{282.75-152.60}=48.28\text{ mm}$$ L_B = 2π / (β10 − β20) = 2π / (282.75 − 152.60) = 48.28 mm
   Multimode beating produces field patterns, power readings and component responses that vary with position, which is why waveguides are normally operated where only TE₁₀ propagates.

7. Common Input Mistakes

• Entering the frequency in GHz. The operating frequency is entered in hertz. Type 10E9 for 10 GHz, not 10.
• Mixing up the output units. The cutoff frequency is reported in MHz, but guide wavelengths are in meters. Multiply by 1,000 for millimeters.
• Choosing a frequency below cutoff. The frequency-dependent outputs are valid only for f > f_c. Below cutoff the mode is evanescent: β is imaginary and there is no guide wavelength.^[1]
• Swapping a and b. a is the broad wall. If a and b are swapped, TE₁₀ and TE₀₁ exchange cutoffs and the dominant-mode results are wrong.
• Using outside dimensions. Use the inside dimensions; WR-90 tubing measures 1.000 in × 0.500 in outside but 0.900 in × 0.400 in inside.^[8]
• Entering m = n = 0. TE₀₀ does not exist. At least one index must be 1 or higher.
• Using λ instead of λ_g for lengths. Quarter-wave transformers, short-circuit stubs and slot spacings inside a waveguide are set by the guide wavelength, which is always longer than the free-space wavelength.
• Using a partial dielectric fill. The calculator assumes the guide is completely filled. Partially filled guides, such as a dielectric slab, need a full-wave solution.

8. Rectangular Waveguide TE_m,n Equation Summary

Need help with a waveguide feed, slot array, filter, transition or wavelength measurement? Contact SiberSci, LLC. Related tools: Rectangular Waveguide TM_m,n Calculator, Rectangular Waveguide Calculator, Circular Waveguide Calculator, Wavelength Calculator and Skin Depth 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," Sec. 3.3, "Rectangular Waveguide," and the appendix of dielectric constants and loss tangents.
2. 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."
3. C. A. Balanis, Advanced Engineering Electromagnetics, 2nd ed. Hoboken, NJ: Wiley, 2012. Ch. 8, "Rectangular Cross-Section Waveguides and Cavities."
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. E. L. Ginzton, Microwave Measurements. New York: McGraw-Hill, 1957. Slotted-line standing-wave and wavelength measurements.
7. R. S. Elliott, Antenna Theory and Design, rev. ed. Hoboken, NJ: Wiley-IEEE Press, 2003. Slot arrays in rectangular waveguide.
8. Electronic Industries Association, EIA Standard RS-261-B, Rectangular Waveguides (WR3 to WR2300). Washington, DC: EIA, 1979. WR-90: 0.900 in × 0.400 in, 8.2–12.4 GHz.
9. 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

Tutorial by Dr. Scott Best, SiberSci, LLC. Values computed with the calculator's own equations, c = 299,792,458 m/s, ε₀ = 8.8541853×10⁻¹² F/m and μ₀ = 4π×10⁻⁷ H/m.