Skip to main content

Rectangular waveguide TM mode field equations

Short answer

For TMmn, Hz=0 and Ez=E0 sin(mπx/a)sin(nπy/b)e^-jβz; transverse fields follow from Maxwell's equations. Cutoff fc=(c/2)√((m/a)^2+(n/b)^2), guide wavelength λg=λ0/√(1-(fc/f)^2), wave impedance Z_TM=η√(1-(fc/f)^2). Lowest TM mode is TM11.

Explain the field and characteristic equations of a rectangular waveguide working in TM mode.

A rectangular waveguide is a hollow metal pipe of inner width a (x-direction) and height b (y-direction), a > b, with the wave travelling along z. In a TM (transverse magnetic) mode, Hz = 0 and Ez ≠ 0.

   y
   ^
 b |+-----------------+
   ||                 |   wave travels in +z
   ||    air (μ0, ε0)  |   (into the page)
   |+-----------------+
   0------------------> x
                      a

Wave equation and solution

For fields varying as e^(j(ωt − βz)), Ez satisfies

 ∂²Ez/∂x² + ∂²Ez/∂y² + kc² Ez = 0
 kc² = k² - β²,   k = ω√(με)

By separation of variables, Ez = X(x)Y(y) with X = A1 cos kx·x + A2 sin kx·x, Y = B1 cos ky·y + B2 sin ky·y.

Boundary conditions: tangential E is zero on the walls, so Ez = 0 at x = 0, a and y = 0, b. This gives A1 = B1 = 0, sin kx·a = 0 and sin ky·b = 0:

 kx = mπ/a,  ky = nπ/b,  m, n = 1, 2, 3 ...
 kc² = (mπ/a)² + (nπ/b)²

Field equations (TMmn)

 Ez = E0 sin(mπx/a) sin(nπy/b) e^(-jβz)
 Ex = -(jβ/kc²)(mπ/a) E0 cos(mπx/a) sin(nπy/b) e^(-jβz)
 Ey = -(jβ/kc²)(nπ/b) E0 sin(mπx/a) cos(nπy/b) e^(-jβz)
 Hx =  (jωε/kc²)(nπ/b) E0 sin(mπx/a) cos(nπy/b) e^(-jβz)
 Hy = -(jωε/kc²)(mπ/a) E0 cos(mπx/a) sin(nπy/b) e^(-jβz)
 Hz = 0

Since Ez contains sin(mπx/a)·sin(nπy/b), m = 0 or n = 0 gives zero field. So TM00, TM10, TM01 do not exist; the lowest TM mode is TM11.

Characteristic equations

  • Propagation constant: β = √(ω²με − kc²); the mode propagates only when ω²με > kc².
  • Cut-off frequency:
 fc = (1/(2√(με))) √((m/a)² + (n/b)²)
    = (c/2) √((m/a)² + (n/b)²)   (air)
  • Cut-off wavelength: λc = 2/√((m/a)² + (n/b)²)
  • Guide wavelength: λg = λ0/√(1 − (fc/f)²) (longer than free-space λ0)
  • Phase velocity: vp = ω/β = c/√(1 − (fc/f)²) > c
  • Group velocity: vg = c·√(1 − (fc/f)²) < c, with vp·vg = c²
  • Wave impedance:
 Z_TM = Ex/Hy = β/(ωε) = η·√(1 - (fc/f)²)
 (η = 377 Ω in air)

Z_TM is always less than η and becomes zero at cut-off.

Example

For an X-band guide (a = 2.286 cm, b = 1.016 cm), the TM11 cut-off is fc = (3×10⁸/2)√((1/0.02286)² + (1/0.01016)²) ≈ 16.2 GHz, much higher than the TE10 cut-off (6.56 GHz). This is why TM modes are not used as the operating mode in standard guides.

✓ Answer checked by Claude Opus 5.5 (high effort)

All RF and Microwave Engineering chapter 4 questions →