- Asked 11/23
- 2081 Bhadra · 8 marks
- 2081 Baisakh · 8 marks
- 2076 Bhadra · 10 marks
- 2074 Bhadra · 10 marks
- 2073 Magh · 5 marks
- 2072 Magh · 5 marks
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)
Related questions
- Rectangular waveguide TE mode field equations · RF and Microwave Engineering, chapter 4 · asked 11×
- Circular waveguide TM field strength and dominant mode · RF and Microwave Engineering, chapter 4
- Circular waveguide TE mode field equations · RF and Microwave Engineering, chapter 4 · asked 2×
- Rectangular TM equations; TE10 vs TE20 · RF and Microwave Engineering, chapter 4
- TE vs TM mode in circular waveguide · RF and Microwave Engineering, chapter 4
- Circular waveguide TE equations and degenerate modes · RF and Microwave Engineering, chapter 4
- Rectangular waveguide cut-off and phase velocity · RF and Microwave Engineering, chapter 4
- Define TM mode; TM11 cut-off 5×3 cm · Electromagnetism, chapter 6
- Modes in rectangular waveguide · Electromagnetism, chapter 6 · asked 4×