Rectangular PEC waveguide

Take a straight waveguide with 0<x<a, 0<y<b, a>b, and uniform +z propagation. Its side walls are perfect electric conductors; the filling is a homogeneous, isotropic, linear, lossless, nondispersive medium with ε>0 and μ>0. A real physical guide has finite losses, but the ideal eigenproblem must be solved first.

The time dependence is Re{F̲(x,y) exp(+jωt−jβz)}. Inside the guide, Maxwell gives ∇×E̲=−jωμH̲ and ∇×H̲=+jωεE̲. The PEC side wall imposes n×E̲=0; β denotes longitudinal propagation, and k_c the transverse eigenvalue. They are not interchangeable.

Write the longitudinal generating field as ψ(x,y)=X(x)Y(y). Division by XY away from its zeros gives X''/X+Y''/Y=−k_c². Since one summand depends only on x and the other only on y, each must be constant. Therefore X''+k_x²X=0 and Y''+k_y²Y=0, with k_c²=k_x²+k_y².

Before applying any boundary condition, both factors are arbitrary linear combinations of sine and cosine. It is the PEC boundary condition—not an arbitrary choice of symmetry—that selects the allowed basis. The identical transverse eigenvalue formula for TE and TM does not mean that their index sets are identical.

For a TE-to-z mode E_z=0 and H_z generates the other components. The PEC walls require ∂H_z/∂n=0, a Neumann condition. At x=0,a and y=0,b this selects cos(mπx/a)cos(nπy/b). Indices m,n may each vanish, but not simultaneously.

The pair (0,0) would make H_z spatially constant and all reconstructed transverse fields vanish for positive k_c; it is not a physical hollow-guide TE00 traveling mode. At fixed indices, the longitudinal dependence exp(−jβz) multiplies the transverse eigenfunction.

For a TM-to-z mode H_z=0 and E_z is nonzero. Because E_z is tangential to every side wall, the PEC condition requires E_z=0 along x=0,a and y=0,b. Separation therefore selects sine factors in both transverse directions.

If m=0 or n=0, one sine factor vanishes everywhere. Hence the TM family requires m,n≥1. In particular TM10 and TM01 do not exist for the ideal closed rectangular cross-section, while TM11 does.

The sine and cosine solutions have the same numerical eigenvalue k_c,mn=π√[(m/a)²+(n/b)²]. Setting β=0 gives f_c,mn=(v/2)√[(m/a)²+(n/b)²] with v=1/√(με). For a>b, the lowest allowed cutoff belongs to TE10, f_c,10=v/(2a).

The next propagating possibility is the lower of TE20 and TE01: f_c,20=v/a and f_c,01=v/(2b). Therefore strictly one propagating mode exists only for f_c,10<f<min(f_c,20,f_c,01). At a=2b the two upper-bound competitors are degenerate. Exactly at their cutoff they are threshold modes, not additional traveling modes with positive real β.

TE11 and TM11 have the same transverse cutoff value but different longitudinal fields and PEC conditions; frequency degeneracy never implies identical electromagnetic fields.

Take H_z=H₀ cos(πx/a)exp(−jβz), E_z=0 and k_c=π/a. Using the chapter 02 transverse-field reconstruction with exp(+jωt−jβz), E_y=−j(ωμ/k_c)H₀ sin(k_cx)exp(−jβz) and H_x=+j(β/k_c)H₀ sin(k_cx)exp(−jβz). The remaining TE10 components E_x, E_z, H_y vanish.

At x=0 and x=a the tangential field E_y vanishes. At the horizontal walls y=0,b, E_y is normal and may be nonzero; the tangential electric components there vanish. A sign audit gives H_x=−(β/(ωμ))E_y and consequently positive +z average power for the forward wave. The magnetic H_z does not need to vanish at the walls.

For f>f_c,10, β=(2πf/v)√[1−(f_c,10/f)²]. The longitudinal guide wavelength is λ_g=2π/β and increases without bound at cutoff. The TE field-wave impedance is Z_TE=ωμ/β=η/√[1−(f_c,10/f)²], with η=√(μ/ε). Neither quantity is determined solely by the waveguide outer dimensions when the filling material changes.

Using peak phasors, P_z=(1/2)Re∫(E×H*)·ẑ dA. For the TE10 amplitudes above and real β, the result is P_z=(ab/4)(ωμβ/k_c²)|H₀|²>0. Equivalently P_z=ab|E_y,max|²/(4Z_TE). This Z_TE is a transverse field ratio, not a calibrated 50-ohm VNA port impedance.

Below cutoff use β=−jα_z, where α_z=√[k_c²−ω²με]>0. Then exp(−jβz)=exp(−α_z z) decays with distance; this geometric evanescence is distinct from conductor and dielectric losses.

Choose a=22.86 mm, b=10.16 mm, vacuum filling, and f=10 GHz. The eigenvalue calculation gives f_c,10≈6.557 GHz, f_c,20≈13.114 GHz, and f_c,01≈14.754 GHz. Since 6.557<10<13.114 GHz, only TE10 propagates in this ideal guide.

At 10 GHz the free-space wavelength is λ₀≈29.98 mm, the guide wavelength is λ_g≈39.7 mm, and the TE field-wave impedance is near 499 Ω. These results must not be confused with a coaxial fixture's 50 Ω reference. Finite wall conductivity gives attenuation and slightly alters phase; abrupt transitions can excite higher evanescent modes even inside the nominal single-mode band.

All numerical results assume perfect walls and an empty guide. Dielectric loading replaces c by 1/√(με) in the nondispersive approximation, while inhomogeneous or anisotropic loading requires a new vector eigenproblem.

1. Starting from the transverse Helmholtz equation, derive the separated equations for X and Y and explain why k_c²=k_x²+k_y².
2. Use the PEC sidewall condition to derive the permitted TM indices and explain why TM10 is impossible.
3. Derive the Neumann conditions for TE modes and explain why TE00 is not a propagating hollow-guide mode.
4. Derive f_c,mn for homogeneous filling, then calculate f_c,10 for a=22.86 mm in vacuum.
5. For a=22.86 mm and b=10.16 mm, determine the next cutoff after TE10 and the strict single-mode band.
6. Find the aspect ratio at which TE20 and TE01 are degenerate. Explain why TE11/TM11 field patterns are still different.
7. Reconstruct H_x and E_y from H_z for the forward TE10 field, and check the sign in the phasor curl equations.
8. Derive the TE10 carried power from the Poynting integral for peak phasors and H_z amplitude H₀.
9. At 10 GHz in the stated guide, calculate β, λ_g and Z_TE. Why is this not the VNA reference impedance?
10. At 5 GHz, decide whether TE10 propagates, determine the sign of Im β for forward decay, and state the physical assumptions.

The next chapter will close a guide with conducting end walls, creating a three-dimensional cavity eigenproblem. A resonance frequency is not by itself a complete mode: the vector fields and all six wall conditions remain essential.